Synthesis Lectures on

Wave Phenomena in the Physical Sciences

Sanichiro Yoshida

Fundamentals
of Acoustic
Waves and
Applications
Synthesis Lectures on Wave Phenomena in
the Physical Sciences

Series Editor
Sanichiro Yoshida, Department of Chemistry and Physics, Southeastern Louisiana
University, Hammond, LA, USA
The aim of this series is to discuss the science of various waves. An emphasis is laid on
grasping the big picture of each subject without dealing formalism, and yet understanding
the practical aspects of the subject. To this end, mathematical formulations are simplified
as much as possible and applications to cutting edge research are included.
Sanichiro Yoshida

Fundamentals of Acoustic
Waves and Applications
Sanichiro Yoshida
Department of Chemistry and Physics
Southeastern Louisiana University
Hammond, LA, USA

ISSN 2690-2346 ISSN 2690-2354 (electronic)

Synthesis Lectures on Wave Phenomena in the Physical Sciences
ISBN 978-3-031-48199-4 ISBN 978-3-031-48200-7 (eBook)
https://doi.org/10.1007/978-3-031-48200-7

© The Editor(s) (if applicable) and The Author(s), under exclusive license to Springer Nature
Switzerland AG 2024

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This book is dedicated to the memory of my late
father who introduced me to the joy of engineering,
and my mother, Sonoko Yoshida, and my wife, Yuko
Yoshida, for filling my life with love, joy, and
happiness.
Preface

This book discusses the fundamentals of acoustic wave phenomena. I use the word “dis-
cuss” because I intend to describe each topic through discussions from various angles
rather than to “provide” the information about them. I believe we can deepen our under-
standing of natural sciences by discussing subjects. I hope the readers apply the concept
gained from this book to other fields of science and engineering. The target audience is
undergraduate students majoring in physics and engineering.
I wrote this book as part of Synthesis Lectures on Wave Phenomena in the Physi-
cal Sciences, which aims to describe various waves in a short book. It is impossible to
cover many topics with this short-book concept. I selected a few and dug them deeper,
exploring their physical meanings. Mathematical operations play significant roles in this
process. Often we can sense physical phenomena intuitively by digesting the associated
mathematics. I tried my best to break down mathematical operations into small pieces.
This sometimes makes equations long. Please bear with me while I explain a long equation
step by step.
Acoustic technologies have numerous applications in a variety of engineering and sci-
entific fields. Recent advancements in hardware and software technologies facilitate the
use of devices and analyses of the outcomes. On the other hand, this convenience tends
to hide the operation principles of hardware and software modules. This tendency some-
times leads to the inefficient use of a hardware device and the misinterpretation of data.
A proper understanding of the science behind the scene is essential. Upon writing this
book, I always bore this factor in mind.
The first two chapters discuss the basics of waves, where Chap. 1 focuses on the
mathematical aspect of it and Chap. 2 on the physical. Chapters 3 and 4 discuss acoustic
waves propagating in air and solids, with audible frequency in air and ultrasounds in solids
in mind. Chapter 5 describes the operation principles of typical acoustic transducers.
Finally, I would like to express appreciation to people. I owe much to my high school
and college teachers because my understanding originates from what they taught me
decades ago. I am grateful to my parents for allowing me to receive such an excellent

vii
viii Preface

education. I thank my wife, Yuko Yoshida, for her continuous support. Often I spent much
time on weekends for this book project.

Hammond, LA, USA Sanichiro Yoshida

August 2023
Contents

1 General Discussions of Waves . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1

1.1 Wave as a Movement of an Oscillatory Pattern . . . . . . . . . . . . . . . . . . . . . . . 1
1.1.1 Terms to Represent a Wave . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 2
1.1.2 Longitudinal and Transverse Waves . . . . . . . . . . . . . . . . . . . . . . . . . . 4
1.2 Mathematics of Waves . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 6
1.2.1 Wave Function as a Periodic Function of Time and Space . . . . . . . 6
1.2.2 Wave Equations . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 9
1.2.3 Wave Solutions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 12
References . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 21
2 Wave Dynamics . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 23
2.1 Oscillations and Waves . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 23
2.1.1 Harmonic Oscillations . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 24
2.1.2 Equation of Motion of Point Mass-Spring System . . . . . . . . . . . . . . 26
2.1.3 Oscillation to Waves . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 29
2.1.4 Wave Equations . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 29
2.2 Acoustic Wave Equations and Solutions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 34
2.2.1 Acoustic Wave in Air . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 35
2.2.2 Longitudinal and Transverse Vibrations in Solid . . . . . . . . . . . . . . . 39
2.2.3 Decaying Acoustic Wave Equations and Solutions . . . . . . . . . . . . . 43
2.2.4 Nonlinear Wave Equations . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 49
2.3 Wave as a Flow of Energy . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 50
2.3.1 Acoustic Energy . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 50
2.3.2 Intensity of Acoustic Wave . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 56
2.4 Acoustic Wave Reflection and Transmission . . . . . . . . . . . . . . . . . . . . . . . . . 58
2.4.1 Laws of Reflection and Refraction . . . . . . . . . . . . . . . . . . . . . . . . . . . 58
2.4.2 Behavior of Acoustic Waves Near Boundary . . . . . . . . . . . . . . . . . . 60
References . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 63

ix
x Contents

3 Propagation of Acoustic Waves in Air . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 65

3.1 Dynamics of Air Particles and Acoustic Wave Equation in Air . . . . . . . . . 65
3.1.1 Particle Motions in Longitudinal Waves . . . . . . . . . . . . . . . . . . . . . . 65
3.1.2 Wave-Generating Mechanism in Air . . . . . . . . . . . . . . . . . . . . . . . . . . 66
3.1.3 Decaying Pressure Waves in Air . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 71
3.2 Sound We Hear . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 72
3.2.1 Our Ear . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 72
3.2.2 Sounds of Speech and Music . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 74
3.3 Transmission of Audible Sound Waves . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 78
References . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 84
4 Propagation of Acoustic Waves in Solids . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 87
4.1 Acoustic Wave as Solutions to the Equation of Motion . . . . . . . . . . . . . . . . 87
4.1.1 Equation of Motion . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 87
4.2 Propagation of Acoustic Waves in Solid Under Various Conditions . . . . . 91
4.2.1 Plane Waves in an Elastic Medium . . . . . . . . . . . . . . . . . . . . . . . . . . . 91
4.2.2 Plane Waves in a Medium with Free Surface . . . . . . . . . . . . . . . . . . 96
4.2.3 Surface Waves . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 108
4.2.4 Waves Propagating in Plates . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 111
References . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 120
5 Electrical-Mechanical Transduction . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 123
5.1 General Argument . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 123
5.2 Reciprocal Transduction . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 124
5.2.1 Electrostatic Transducer . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 125
5.2.2 Electrical Governing Equation of Electrostatic Transducer . . . . . . 126
5.2.3 Mechanical Governing Equation of Electrostatic Transducer . . . . . 127
5.2.4 Physics Behind Electrostatic Transducer’s Operation . . . . . . . . . . . 128
5.3 Anti-reciprocal Transducers . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 132
5.3.1 Moving-Coil Transducer . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 132
References . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 139

Appendix A: Direction Cosines . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 141

Appendix B: Gradient Operation . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 143
Appendix C: Orthogonality of Sine and Cosine Functions . . . . . . . . . . . . . . . . . . . 149
Appendix D: Notes on Standing Waves . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 151
Appendix E: Generalized Hooke’s Law and Equation of Motion
for Isotropic Media . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 155
Index . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 167
General Discussions of Waves
1

This introductory chapter discusses wave characteristics and related terminologies in general.
We first view waves as a motion of a spatial pattern (a snapshot) of an entity on the time
axis and as a temporal oscillation on the spatial axis. This spatiotemporal behavior leads
to the concept of phase velocity, which characterizes the medium that the wave propagates
through. In the second half of the chapter, we discuss some mathematical aspects of wave
dynamics. We derive wave equations from the spatiotemporal characteristics and discuss
their solutions in general.

1.1 Wave as a Movement of an Oscillatory Pattern

Consider a sample wave in Fig. 1.1. This figure illustrates a wave pattern spread along the
x-axis at three different times, 0, 1, and 2.µs. The wavy pattern at each time step is a snapshot
of the wave. With the elapse of time (shown by the arrows in Fig. 1.1), the pattern moves to
the right. The little circle put at the leftmost trough indicates that each wave pattern moves
rightward in the positive .x-direction with time. The solid line connecting the little circles
shows that it has an angle to the time axis. Here the dashed line is drawn parallel to the time
axis. The angle made by the solid and dashed lines clarifies the rightward movement of the
circle with the passage of time. We say that the wave has a rightward velocity. This type of
wave velocity is referred to as phase velocity for the reason clarified shortly in this chapter.

© The Author(s), under exclusive license to Springer Nature Switzerland AG 2024 1

S. Yoshida, Fundamentals of Acoustic Waves and Applications, Synthesis Lectures on Wave
Phenomena in the Physical Sciences, https://doi.org/10.1007/978-3-031-48200-7_1
2 1 General Discussions of Waves

Fig. 1.1 Schematic illustration

of a wave as a function of
space. Each wave pattern
illustrates the snapshot taken at
0, 1, and 2 .µs

1.1.1 Terms to Represent a Wave

Wavelength; spatial oscillatory periodicity

Figure 1.2 illustrates this rightward motion of the wave pattern more explicitly. Notice that
the pattern has a spatial periodicity. As the dashed box indicates, the periodicity in this
example is 10 mm in length. The spatial period of a wave is referred to as wavelength.
We can say that this wave has a wavelength of 10 mm. From the analysis of the motion of
the periodicity and how long it takes in time, we can calculate the phase velocity. In this
example, the periodicity moves 0.5 mm in 1.µs, and we find that the phase velocity is 500
m/s (0.5 mm .÷ 1 µs .= 500 m/s).

Period of wave; temporal oscillatory periodicity

Observation of Fig. 1.2 at the same location on the .x-axis reveals the temporal oscillation
of the wave. For instance, see the signal at .x = 3.8 mm indicated by a vertical line in Fig.
1.2. The signal at this location is positive at .t = 0 .µs, approximately 0 at .t = 1 .µs, and
negative at .t = 2 .µs. It is clear that the medium oscillates during this duration of 2 .µs. This
oscillation is called particle oscillation and the corresponding velocity is called the particle
velocity.
Figure 1.3 illustrates the temporal oscillation (the particle oscillation) behavior of the
same wave as Fig. 1.1 in the same form of three-dimensional presentation. Here, the oscil-
lations at three representative locations of .x = 0, 1, and 2 mm are presented. You may
notice the similarity in the oscillatory pattern between Figs. 1.1 and 1.3. In fact, as Fig. 1.4
explicitly shows, the oscillatory patterns as a function of space and time are identical to each
other. The temporal periodicity observed in Fig. 1.4 is referred to as period.
1.1 Wave as a Movement of an Oscillatory Pattern 3

Fig. 1.2 Schematic illustration of a wave as a function of space emphasizing the motion of the spatial
periodicity

Fig. 1.3 Schematic illustration of a wave as a function of time. Each wave pattern illustrates the
particle oscillation at a fixed .x location
4 1 General Discussions of Waves

Fig. 1.4 Schematic illustration of a wave as a function of time. Each wavy pattern is a snapshot of
the wave at the corresponding time

1.1.2 Longitudinal and Transverse Waves

Longitudinal waves

Figures 1.1, 1.2, 1.3 and 1.4 illustrate the oscillatory behavior of a wave as a function of space
and time. However, they do not indicate the direction of particle oscillations. It can be parallel
or perpendicular to the direction of the wave propagation. We call a wave a longitudinal
wave when the particles oscillate parallel to the direction of the wave propagation. A wave
traveling through a sling is an example of a longitudinal wave. A sound wave traveling
through air is also a longitudinal wave. (See Sects. 2.2.1 and 3.1.)
Figure 1.5 illustrates the propagation of a sample longitudinal wave. The regions where
the dots are dense (sparse) represent the portion of the medium where the density is high
(low). The illustrations at each row present the pattern at a given time, where the time elapses
from the top towards the bottom row with an increment of 0.25 .µs. Notice that the illustra-
tions on the left column indicate that the dens-sparse pattern moves to the right, whereas
those on the right column represent that it moves to the left. Since the direction of the wave
motion is parallel to the particle motions, these waves are longitudinal.
1.1 Wave as a Movement of an Oscillatory Pattern 5

Fig. 1.5 Propagation of longitudinal waves. The wave moves parallel to the dens-sparse patterns of
the particles’ motions

Transverse waves

If the particle oscillation is perpendicular to the wave propagation, we call the wave a
transverse wave. A wave formed by fans in a soccer stadium is a transverse wave. Figure
1.6 illustrates the propagation of a transverse wave in the same fashion as Fig. 1.5. The dots
represent particles. The rows represent the progression in time, whereas the left and right
columns illustrate the waves moving rightward and leftward, respectively. In this case, the
oscillatory pattern does not exhibit a change in the density. Instead, it indicates a swinging
motion of the particles in the perpendicular direction to the wave motion. A sound wave can
travel through a solid as a longitudinal or transverse wave. (See Chap. 4.) An ocean wave is
a mixture of longitudinal and transverse waves [1]. The particle oscillation (the motion of
the water molecules) has longitudinal and transverse components to the propagation.
6 1 General Discussions of Waves

Fig.1.6 Propagation of transverse waves. The waves move horizontally whereas the particles oscillate
vertically

1.2 Mathematics of Waves

1.2.1 Wave Function as a Periodic Function of Time and Space

The fact that a wave has an identical pattern in its spatial and temporal characteristics allows
us to put a wave function in the following form.

. f (θ ) = f (ωt ± kx) (1.1)

Here, .t is the time coordinate variable, .x is the spatial coordinate variable, and

.θ ≡ ωt ± kx (1.2)

is called the phase of the wave. For the physical meaning of .ω and .k, see the following
paragraphs.
Since . f (θ ) is a periodic function, we obtain the following relation.

. f (θ + Θ) = f (θ ± N Θ) = f (θ ) (1.3)
1.2 Mathematics of Waves 7

Here .Θ is the period of the function . f (θ ) and . N is an integer. Equation (1.3) literally
indicates the fact that every time the phase changes by .±Θ the function takes the same
value. We can always find the same change in .θ by fixing .t and varying .x or fixing .x and
varying .t. This fact stipulates the above observation that a wave has identical temporal and
spatial patterns.
We can interpret the quantities .ω and .k as the temporal and spatial frequencies of the
wave by making the following argument. Find the total derivative of .θ by differentiating
(1.2).
∂θ ∂θ
.dθ = dt ± d x = ωdt ± kd x (1.4)
∂t ∂x
Consider the spatial periodicity using (1.4) in association with Fig. 1.2. Since the spatial
wave patterns in this figure are the snapshots taken at the respective time, we can put .dt = 0
in (1.4). Under this condition, express (1.4) for one period. We can put the left-hand side as
the phase period .dθ = Θ and the right-hand side as the wavelength .d x = λ. Thus,

. Θ = kλ (1.5)

Since period .Θ is a constant for a given wave, (1.5) indicates that .k is proportional to the
reciprocal of .λ. Here the wavelength .λ represents the periodicity in length (m in the SI
unit). The reciprocal of .λ measures the number of wavelengths in the unit length, i.e., how
frequently the wavelength appears as we move over the unit length on the spatial axis. Thus,
it is called the spatial frequency of the wave. In the case of sinusoidal waves (as we will
discuss in more detail in later sections), the periodicity in phase is.2π . In this case, the spatial
frequency becomes as follows.
2π
.k = (1.6)
λ
We can repeat the same type of argument on (1.4) in association with Fig. 1.4. In this
case, .d x = 0 and we obtain the following equation as the temporal version of (1.5).

. Θ = ωτ (1.7)

Here .τ is the period of the wave.

Similarly to (1.6), we can define the temporal frequency .ω for a sinusoidal wave.
2π
ω=
. (1.8)
τ
Repeating the same argument as for .k, we can say that .ω represents how frequently the
period appears as we move along the time axis.
To distinguish from the terms the spatial and temporal frequency in the unit of 1/m or
1/s, we call .k and .ω the angular spatial frequency and angular temporal frequency. They are
in rad/m and rad/s, respectively.
8 1 General Discussions of Waves

Traveling wave and phase velocity

As indicated by (1.3), the phase determines the value of a wave function. The motion of
a wave is the trajectory of a constant phase value. In other words, the wave velocity we
discussed above is the velocity of a phase value; that is why we call the velocity phase
velocity. A surfer stays at a crest of an ocean wave where the phase makes the vertical
position of the water molecule at the highest. So, we can say that the surfer moves at
the phase velocity of the wave he is on. Here we consider the concept of phase velocity
quantitatively.
When we move along with a constant phase, the phase does not change over time.
Mathematically, we can express this situation by saying .dθ/dt = 0. We can discuss it using
(1.4) by differentiating the phase .θ with respect to time and setting it to zero.
dθ dx
. =ω±k =0 (1.9)
dt dt
From (1.9), we obtain the following equation.
ω dx
. =∓ (1.10)
k dt
Consider the meaning of .d x and .dt that appear on the right-hand side of (1.10). These are
the changes in the coordinate variable .x and .t necessary to keep the phase .θ = ωt ± kx
constant. In other words, if the time elapses by .dt, we need to change the location on the .x
coordinate by .d x to keep the phase unchanged. From this viewpoint, we can interpret that
.d x/dt is the velocity of the constant phase location on the . x-axis. Thus, we can say that the

quantity expressed by the left-hand side of (1.10) represents the phase velocity of the wave
.v p .
ω dx
.v p = =∓ (1.11)
k dt
Note that the negative (positive) sign on the right-hand side of this equation corresponds to
the positive (negative) sign on the right-hand side of the phase expression (1.2). This means
that . f (ωt + kx) corresponds to .ω = −d x/dt, i.e., a negative phase velocity for positive
.d x and .dt, indicating that the wave travels in the negative . x direction, and . f (ωt − kx)

represents a wave traveling in the positive .x direction. In other words, to keep the phase
.θ = ωt − kx constant when the time elapses by .dt (i.e., .t increases), we need to increase . x

by .d x = (ω/k)dt.
A wave expressed in the form of (1.1) is referred to as a traveling wave. As clear from the
above discussion, when the sign between the time and space terms of the phase is negative
(. f (ωt − kx)), the wave travels in the positive .x-direction. If the sign is negative, the wave
travels in the negative .x-direction. According to (1.10) and (1.11), we can distinguish the
direction of the wave motion by the sign of the phase velocity .v p .
1.2 Mathematics of Waves 9

1.2.2 Wave Equations

Consider the property of the wave expression (1.1) further by differentiating the wave func-
tion . f with respect to time and space.

∂2 f
. = ω2 f '' (1.12)
∂t 2
∂2 f
. = k 2 f '' (1.13)
∂x2
Here . f '' denotes the second-order derivative of the function . f (t, x), e.g., if . f = cos(ωt −
kx), . f '' = − cos(ωt − kx). Equating . f '' appearing in (1.12) and (1.13), we obtain the
following wave equation.
∂2 f ( ω )2 ∂ 2 f
. = (1.14)
∂t 2 k ∂x2
Note that .ω/k appearing on the right-hand side is the phase velocity (1.11). Expressing the
unit of the quantity represented by . f with .[ f ], we find that the unit of the left-hand side of
(1.14) is .[ f ]/s.2 and the unit of .∂ 2 f /∂ x 2 is .[ f ]/m.2 , indicating that .(ω/k) is in m/s. Indeed,
it has the dimension of velocity. We can express the wave equation using (1.11) in (1.14).

∂2 f 2∂ f
2
. = v p (1.15)
∂t 2 ∂x2
Extension to three dimensions

It is clear that the differential equation (1.14) yields the wave solution . f (ωt ± kx). We call
this equation a wave equation. As discussed in the preceding section, (1.14) yields a wave
solution that travels along the .x-axis. Since its traveling direction is limited in one axis, this
type of wave is known as a one-dimensional wave.
What if the traveling direction is not in line with the .x-axis, e.g., it has some angle to the
. x-axis? Even if the traveling behavior is physically the same, we cannot express the phase

term in the form of .ωt − kx. With the passage of time, the change in the spatial coordinate
account for the same phase cannot be expressed only by the change in .x. In this situation,
we say that the traveling wave is three-dimensional, and call the spatial frequency .k the
propagation constant. Here, the propagation constant has a directionality, i.e., we need to
move along an axis that has an angle to the .x-axis to define the spatial frequency. In other
words, the propagation constant .k is a vector referred to as the propagation vector.
Figure 1.7 illustrates the situation where the same physical wave travels through space
represented by two different coordinate systems. In Fig. 1.7a, the .x-axis is in line with the
direction of the wave propagation, hence the propagation vector .k has one component. In
Fig. 1.7b the direction of propagation has angles of .θx , .θ y , and .θz with the .x, . y and .z-axis,
respectively. The cosines of these angles are called direction cosines [2]. The propagation
10 1 General Discussions of Waves

(a) (b)

Fig. 1.7 The same wave propagates through space represented by two different coordinate systems.
a The propagation is in line with the .x-axis. b The propagation has angles of .θx , .θ y , and .θz with the
. x, . y and .z-axis. . x̂ is the unit vector along the . x-axis. .lˆ is the unit vector along the wave’s propagation

vector’s .x, . y and .z-components are the vector’s magnitude multiplied by the corresponding
directional cosine, e.g., .k x = k cos θx .
The conversion from a one-dimensional to a three-dimensional wave expression leads to
the following change in the phase term of wave function . f (ωt ± kx) defined in (1.1).

. f (θ ) = f (ωt ± k · r) (1.16)

Here .r is the vector that represents the spatial coordinates; the three-dimensional version of
. x in (1.1).
.r = x î + y ĵ + z k̂ (1.17)
Here .î etc. are the unit vectors for the corresponding directions. The scalar product in the
phase term of (1.16) represents the spatial phase change in the direction of propagation
vector .k, i.e., along the propagation path of the wave.
( )
k · r = k x x + k y y + k z z = k cos θx x + cos θ y y + cos θz z
. (1.18)

Here .k is the magnitude of .k, i.e., .k = |k|.

Defining vector .lˆ as the unit vector in the direction of .k, we can express (2.89) in a
compact form as follows.
( )
k · r = k lˆ · r = k l x x + l y y + l z z
. (1.19)

ˆ
Here .l x etc. are the components of .l.
Comparison of (2.89) and (1.19) reveals that .lˆ has direction cosines as its components.

lˆ = cos θx î + cos θ y ĵ + cos θz k̂

. (1.20)

In Appendix A we prove that .lˆ is a unit vector.

1.2 Mathematics of Waves 11

Using (1.19) or (1.20), we can express wave function (1.16)

. f (ωt ± k lˆ · r) = f (ωt ± k(l x x + l y y + l z z)) = f (ωt ± k(cos θx x + cos θ y y + cos θz z))

(1.21)
The extension to three dimensions requires one more procedural change regarding dif-
ferentiation. Differentiating a function is the process to find the change in the value of
the function associated with an infinitesimal change of independent variables. When the
wave function has one spatial independent variable, like the above case of . f (t, x), the
change in the value of function . f is definitely determined by the function’s dependence
on .x, i.e., . f ' = d f /d x. However, if the function has three independent spatial variables
like . f (t, x, y, z), the spatial derivatives become a vector quantity. We need to replace the
differential operation .d f /d x with the gradient operation .∇ f [3]. In Cartesian coordinates,
it is given as follows.
( )
∂ ∂ ∂ ∂f ∂f ∂f
.∇ f = î + ĵ + k̂ f = î + ĵ + k̂ (1.22)
∂x ∂y ∂z ∂x ∂y ∂z

Appendix B discusses some more description of the gradient operation.

By repeating the gradient operation twice, we can derive the following expression as the
three-dimensional version of the second-order spatial differentiation .∂ 2 /∂ x 2 .
( ) ( )
∂ ∂ ∂ ∂ ∂ ∂
.∇ f = ∇ · ∇ f = î + ĵ + k̂ · î + ĵ + k̂ f
2
∂x ∂y ∂z ∂x ∂y ∂z
∂2 f ∂2 f ∂2 f
= + + (1.23)
∂x2 ∂ y2 ∂z 2

The operator .∇ 2 is called the Laplace operator or Laplacian.

Using (1.21) in (1.23) we find the following expression as the three-dimensional version
of (1.13).
2 ''
.∇ f = k (l x + l y + l x ) f
2 2 2 2
= k 2 f '' (1.24)

Here .lˆ = l x î + l y ĵ + l z k̂ is the unit vector in the direction of the propagation vector, and we
used that .(l x2 + l y2 + l x2 ) = 1 on the right-hand side of (1.24).
From (1.12) and (1.24), we derive the following equation.

∂2 f ( ω )2
. = ∇2 f (1.25)
∂t 2 k
Wave equation (1.25) is the three-dimensional version of wave equation (1.14).
It should be noted that above we derived wave equations based on the temporal and
spatial periodicities. As will be discussed in the following section, this type of wave equation
yields solutions that continue unlimitedly spatially and temporally. It is because the periodic
function we started with has such a property. In reality, waves decay due to various causes
such as viscosity. We will discuss this topic in the next chapter by considering the physics
behind wave dynamics.
12 1 General Discussions of Waves

1.2.3 Wave Solutions

Consider solving the wave equations derived above. Since we derived the wave equations by
considering a periodic function in the form of (1.1), the solution should take this form. As a
simple example, we know that sine and cosine functions are periodic. Let’s take a moment
and confirm that a simple cosine function in the following form satisfies the one-dimensional
wave equation (1.14).
. f (t, x) = A cos(at ± bx) (1.26)
Here . A, .a, and .b are constant.
Differentiating this function with respect to time and space twice, we obtain the following
set of equations.

∂2 f
. = −a 2 cos(at ± bx) (1.27)
∂t 2
∂2 f
. = −b2 cos(at ± bx) (1.28)
∂x2
We find that with the following conditions, (1.26) satisfies wave equation (1.14).

.a=ω (1.29)
. b=k (1.30)

Adding a constant phase term .φ does not affect the temporal or spatial differentiation.
Thus, we find the following function is a solution to the wave equation (1.14).

. f (t, x) = A cos(ωt ± kx + φ) (1.31)

Of course, we can extend solution (1.31) into three dimensions.

. f (t, x) = A cos(ωt ± k · r + φ) (1.32)

The freedom to add a phase .φ indicates that a sine function of the same temporal and spatial
frequency can also be a solution.
Above, we casually used sinusoidal functions as examples. However, this discussion does
not lose generality if we remember Fourier’s theorem that state “a periodic function can be
expanded into a series of cosine and sine functions.” In the next section, we briefly discuss
this topic.

Fourier series

According to Fourier’s theorem [4, 5], the temporal periodicity of a wave function . f (t) can
be expanded into a series of cosine and sine functions as follows.
1.2 Mathematics of Waves 13

a0
. f (t) = + a1 cos(ω0 t) + a2 cos(2ω0 t) + · · · + an cos(nω0 t) (1.33)
2
.+ b1 sin(ω0 t) + b2 sin(2ω0 t) + · · · + bn sin(nω0 t) (1.34)

Here .ω0 is the fundamental frequency, .2ω0 , .3ω0 , .. . . . N ω0 are the second, third, .. . . . Nth
harmonics. For a given function . f (t) we can find the coefficients .a0 .· · · .bn using the
orthogonality of the sine and cosine functions. See Appendix C.
Consider these harmonics in the context of wave solution (1.31). We discussed that a
cosine sine function in the form of (1.31) satisfies the wave equation (1.15). This wave
equation indicates that the secondary temporal differentiation of a wave function is pro-
portional to the secondary spatial differentiation, where the constant of proportionality is
the square of the phase velocity. As will be discussed later, the phase velocity is a material
constant. Each material has its unique value of acoustic phase velocity. We also learned that
phase velocity is the ratio of temporal frequency over spatial frequency (see (1.11)). These
facts lead to the following discussion. When an acoustic wave travels through a material at
the frequency determined by the source, the phase velocity determines the spatial frequency,
hence the wavelength. The combination of .ω and .k in solution (1.31) satisfies this relation,
i.e., its ratio is the material’s unique phase velocity.
The above situation indicates that if an acoustic wave at the fundamental frequency .ω0 in
the Fourier series (1.33) satisfies the wave equation, a harmonic of frequency . N ω0 is also a
solution to the same wave equation where the spatial frequency is .(N ω0 )/v p , i.e., . N times
greater than the fundamental wave. Since the wavelength of the fundamental frequency is
.λ0 = v p /ω0 , the wavelength of this harmonic is . N times shorter. This situation is the same

for any . N .
The above discussion indicates that the same physical system (the acoustic source and
the medium through which the acoustic wave travels) can generate harmonics. Shortly, we
will discuss the phenomenon known as resonance, which is an example where the same
physical system generates harmonics (higher resonator modes).

Standing wave

In Sect. 1.2.1, we discussed that a wave function. f (ωt − kx) travels in the positive.x direction
and . f (ωt + kx) travels in the negative .x direction. Call the former a forward-going wave
and the latter a backward-going wave. Since these two waves have the same phase velocity,
.v p = ω/k, the same physical system can generate both waves at the same time, raising the

following questions: “Can a pair of forward-going and backward-going waves coexist in a

medium? If so, do they add each other, and if that is the case, what is the direction of the
combined wave?” The answer is “Yes, they can coexist and add to each other. The combined
wave can be stationary. This type of stationary wave is called a standing wave.” I say “can be
stationary” because, under some conditions, the combined wave can travel. We will discuss
such conditions in Appendix D. Here, we focus our discussions on standing waves.
14 1 General Discussions of Waves

Before starting the discussions here, I would like to note some physical significance of
standing waves. As will be discussed later, a wave carries energy. Therefore, if a wave stays
at the same location, the energy carried by the wave does not flow and thereby the local
energy grows in time. Indeed this phenomenon occurs under the condition of resonance. We
can construct an instrument to embody such a resonance phenomenon, and the instrument
is called a resonator [6]. Most musical instruments [7] (see Sect. 3.2.2) are resonators [8] to
generate standing waves [9].
Consider the formation of a standing wave by adding two oppositely traveling waves.
Assume the two waves have the same amplitude . A.

. A sin(ωt − kx) + A sin(ωt + kx) = 2 A cos(kx) sin(ωt) (1.35)

Equation (1.35) indicates that when the forward-going and backward-going waves have the
same amplitude, the superposed wave oscillates without traveling in either direction (forward
or backward). The time function part (.sin ωt) of the right-hand side of (1.35) tells us every
time the time .t makes .ωt = N π (. N is an integer), the wave function becomes zero over the
entire.x. This situation is referred to as complete destructive interference. The forward-going
and backward-going waves (call these waves component waves) interfere with each other
so that the superposed wave becomes zero at all .x locations.
When time .t satisfies .ωt = π/2 + N π , on the other hand, .sin ωt takes .±1 and thereby
the component waves interfere with each other constructively, maximizing the amplitude of
the superposed wave. The “2” in front of . A on the right-hand side of (1.35) indicates that
under the condition of complete constructive interference, the resultant amplitude is doubled
as compared with the component waves. Note that destructive and constructive interference
each repeats at a period of .π (an increase of 1 in the integer . N changes the phase by .π ).
This period is half of that of the component waves.
Figure 1.8 is an example of a wave expressed in the form of Eq. (1.35). Here the left
column illustrates oppositely traveling component waves and the right column the standing
wave resulting from the superposition of the component waves. Each row represents time at
every one-eighth of the period .τ . It is seen that the first complete constructive interference
occurs at .t = 2τ/8, i.e., a quarter period after .t = 0 when complete destructive interference
is seen. At .t = 4τ/8, i.e., half a period later the second complete destructive interference
occurs. We can easily imagine that the second complete constructive interference occurs
a quarter period later, i.e., at .t = 4τ/8 + 2τ/8 = 6τ/8. As mentioned in the preceding
paragraph, the completely destructive and constructive interference each repeats every half
period or .π in phase.
The left column illustrates that the component waves are completely out of phase under
complete destructive interference. Under complete constructive interference, they are com-
pletely in phase. The complete destructive and constructive interferences alternate at every
quarter period (.t = 2τ/8), meaning that destructive (constructive) interference occurs in the
middle of consecutive constructive (destructive) interferences.
1.2 Mathematics of Waves 15

Fig. 1.8 An example of a standing wave

In the above sections, we assumed that the amplitude . A is a constant. This assumption
means that the wave has the same strength over the plane perpendicular to the .x-axis (the
axis along which the component waves travel). Since every wave carries energy, this in turn
indicates that the energy is infinite. Such a wave is unrealistic, but we often use this type
of solution as it exhibits important characteristics of waves. We call a wave with a constant
amplitude a plane wave.
The standing wave solution (1.35) has the form of the product of a temporal function and
spatial function. This reminds us of the variable separation method. Let’s solve the wave
equation (1.15) using the variable separation method [10]. Put the solution as the product
of time function .T (t) and space function . X (x) as follows.

. f (t, x) = T (t)X (x) (1.36)

Substitution of (1.36) it into the wave equation (1.15) yields the following set of equations.

∂2 ( )
. f¨ = 2 T (t)X (x) = T̈ X (1.37)
∂t
'' ∂2 ( )
.f = 2 T (t)X (x) = T X '' (1.38)
∂x
16 1 General Discussions of Waves

. T̈ X = v 2p T X '' (1.39)

Divide (1.39) by .T X .
T̈ X ''
. = v 2p (1.40)
T X
Here.T̈ /T is a function of time independent of.x, and.v 2p X /X '' is a function of.x independent
of .t. (1.39) says these two terms are equal to each other. This means that both terms must
be a constant. Let .−c2 represent that constant and put (1.40) in the following form.

T̈
. = −c2 (1.41)
T
''
2 X
.v p = −c2 (1.42)
X
Put .T (t) in the following form.

. T (t) = Tc cos(ct) + Ts sin(ct) (1.43)

Differentiate .T (t) twice with respect to time.

( )
. T̈ = −c2 Tc cos(ct) + T s sin(ct) = −c2 T (t) (1.44)

From (1.44), we can easily find that (1.43) satisfies (1.41).

Similarly, we find the following form of . X (x) satisfies (1.42).
( ) ( )
c c
. X (x) = X c cos x + X s sin x (1.45)
vp vp

Thus, by substituting (1.43) and (1.45) into (1.36), we find that the wave function takes the
following form.
[ ( ) ( )]
[ ] c c
. f (t, x) = Tc cos(ct) + Ts sin(ct) X c cos x + X s sin x (1.46)
vp vp

We now in a position to determine amplitudes .Tc , .Ts , . X c , and . X s , and the constant .c. The
values of these quantities depend on the initial and boundary conditions.

Initial condition

In Fig. 1.8, we set the time origin .t = 0 when the component waves interfere completely
destructively. It is clear that we can set the time origin freely by adjusting constant phase
.φ in (1.31). Here we use the same initial condition as Fig. 1.8, i.e., at .t = 0, the standing
1.2 Mathematics of Waves 17

wave is null. Substituting .t = 0 into (1.46) and setting the value of the function zero, we
find .Tc = 0. So, now the wave function takes the following form.
[ ( ) ( )]
c c
. f (t, x) = Ts sin(ct) X c cos x + X s sin x (1.47)
vp vp

Boundary condition 1

Use the following boundary conditions for . X (x). This condition represents the situation
where the standing wave is null at .x = 0 and .x = L.

. X (0) = 0, X (L) = 0 (1.48)

Substituting the first condition of (1.48) into the spatial function in (1.47), we obtain the
following equation.
. X (0) = X c cos 0 + X s sin 0 = 0 (1.49)
Thus, we find. X c = 0. The second condition of (1.48) and (1.57) yield the following equation.
( )
c
. X (L) = X s sin L =0 (1.50)
vp

It follows that the following condition holds.

c
. L = nπ (1.51)
vp

where .n is an integer.
Since the constant .c is in the form of .sin(ct) in the time function .T (t), we can interpret
this constant as the angular frequency of the wave. Call it the eigen frequency .ωe . From
(1.51) it follows that .ωe satisfies the following condition.
nπ
.ωe = c = v p , n = 1, 2, 3, . . . (1.52)
L
Note that the eigen frequency .ωe is derived from the boundary condition (1.48), which
indicates that the standing wave has a spatial periodicity whose integer multiple is equal to
. L. The physical meaning of this statement becomes clear shortly.

Now substitute (1.52) into the spatial function . X (x) using the above-found condition
. X c = 0.
( ) ( nπ )
ωe
. X (x) = X s sin x = X s sin x (1.53)
vp L
Here (1.52) is used in going through the last equal sign.
So in this case the standing wave function takes the following form.
( nπ ) ( nπ )
. f (t, x) = A sin v p t sin x (1.54)
L L
Here . A = Ts X s .
18 1 General Discussions of Waves

Fig. 1.9 Phase condition for a

passive resonator. In the case of
an active resonator, a source is
placed inside the resonator

Figure 1.9a illustrates standing waves under Boundary condition 1 in a scenario where
a pair of reflectors generate the forward-going and backward-going waves that form the
standing waves. The reflection at each of the reflectors is referred to as fixed-end reflection.

Boundary condition 2

Now consider the following boundary condition instead of (1.48) using the same initial
condition .T (0) = 0. This condition represents the situation where the standing wave has
crests or troughs at .x = 0 and .x = L.

. X (0) = A2 , X (L) = ±A2 (1.55)

Substituting the first condition of (1.55) into the spatial function in (1.47), we obtain the
following equation.
. X (0) = X c cos 0 + X s sin 0 = A 2 (1.56)
Thus, we find . X s = 0. The second condition of (1.55) and (1.56) yields the following
equation. ( )
c
. X (L) = X c cos L = ±A2 (1.57)
vp
It follows that the following conditions hold.
c
. L = nπ, X c = A2 (1.58)
vp

where .n is an integer. Using (1.58) , we find the spatial function . X (x) in this case as follows.
( ) ( nπ )
ωe
. X (x) = A 2 cos x = A2 cos x (1.59)
vp L
1.2 Mathematics of Waves 19

Here whether the standing wave is at a crest or trough in the boundary condition (1.55)
depends on the integer .n; when it is odd, .x(L) = −A2 , and when it is even .x(L) = A2 .
Repeating the same procedure as above using the first condition of (1.58), we find the
same eigenfrequency .ωe as Boundary condition 1. Thus, in this case, the standing wave
function takes the following form.
( nπ ) ( nπ )
. f (t, x) = A sin v p t cos x (1.60)
L L
Here . A = Ts X c = Ts A2 .
Figure 1.9b illustrates standing waves under Boundary condition 2 in a scenario where
a pair of reflectors generate the forward-going and backward-going waves that form the
standing waves. The reflection at each of the reflectors is referred to as open-end reflection.

Boundary condition 3

The last boundary condition considered here represents the situation where the standing
wave is null at .x = 0 and a crest or trough at .x = L.

. X (0) = 0, X (L) = ±A3 (1.61)

Here . A3 represents the peak amplitude when the time function .sin(ct) = 1. Substituting the
first condition of (1.61) into the spatial function term . X (x) in (1.47) and obtain the following
equation. ( ) ( )
c c
. X c cos x + X s sin x = X c cos (0) + X s sin (0) = 0 (1.62)
vp vp
It follows that . X c = 0. Substituting this condition and the second condition of (1.61), we
obtain the following equation.
( )
c
. X s sin L = ±A3 (1.63)
vp

It follows that
c π
. L = nπ − , X s = A3 (1.64)
vp 2
Condition (1.64) leads to the following expressions of the wave under Boundary condition 3.
( )
(2n − 1)π
. f (t, x) = A 3 sin (ωe t) sin x (1.65)
2L

where the eigen frequency is as follows

(2n − 1)π
ωe =
. vp (1.66)
2L
20 1 General Discussions of Waves

Figure 1.9c illustrates standing waves under Boundary condition 3 in a scenario where
a pair of reflectors generate the forward-going and backward-going waves that form the
standing waves. The reflection at the left reflector is open-end reflection and that at the right
reflector is fixed-end reflection.

Resonance of wave

The situation depicted in Fig. 1.9 is associated with a phenomenon called Resonance [8].
Imagine that a longitudinal wave traveling in air is incident to a medium, e.g., a tube of length
. L. Consider that the frequency of the incident wave is equal to one of the eigen frequencies

discussed in the preceding section. We can easily imagine that one of the situations depicted in
Fig. 1.9 is established. Due to the difference in acoustic impedance (see Sect. 2.3.1), the inci-
dent wave experiences partial reflection at the left end of the tube. Similarly, reflection occurs
at the right end of the tube. The wave reflected at the right end travels back toward the left end,
and there it experiences reflection. In this fashion, reflected waves go back and forth in the tube.
It is instructive to consider the resonant condition for each of the three boundary conditions
discussed in the preceding section. First, rewrite (1.47) using the wavenumber or wavelength.
Since.c represents the angular frequency.ωc and the phase velocity is given as.v p = ωc k, we can
express .c/v p with the wavelength.λ and thereby rewrite the spatial function term as follows.

. f (t, x) = Ts sin(ωe t) [X c cos (kx) + X s sin (kx)] (1.67)

[ ( ) ( )]
2π 2π
. = Ts sin(ωe t) X c cos x + X s sin x (1.68)
λ λ

Using .v p = λ(ωe /(2π )) and (1.51) we find the following equality under Boundary con-
ditions 1 and 2.
c ωe 2π
. L= L= L = nπ (1.69)
vp vp λ
Solving (1.69) for .λ or . L we find the following equation.
2L nλ
λ=
. , or, L = n = 1, 2, 3, . . . (1.70)
n 2
Similarly from (1.64) under Boundary condition 3 we find as follows.
( nπ π )
.ωe = − vp (1.71)
L 2L ( )
2L n 1
.λ = ( ) , or, L = − λ n = 1, 2, 3, . . . (1.72)
n − 21 2 4

Using (1.68), the initial condition . f (0, x) = 0, and other conditions used in the above
section we find the following expressions of standing waves under the three Boundary
conditions.
References 21

Boundary condition 1

( nπ )
. f (t, x) = B sin(ωe t) sin x
L
nπ 2L nλ
ωe = vp, λ = , L= , n = 1, 2, 3, . . . (1.73)
L n 2
Boundary condition 2

( nπ )
. f (t, x) = B sin(ωe t) cosx
L
nπ 2L nλ
ωe = vp, λ = , L= n = 1, 2, 3, . . . (1.74)
L n 2
Boundary condition 3

( )
(2n − 1) π
. f (t, x) = B sin(ωe t) sin x
2L
(2n − 1)π 4L 2n − 1
ωe = vp, λ = , L= λ, n = 1, 2, 3, . . . (1.75)
2L 2n − 1 4
The integer .n represents the order of harmonics. Sometimes this number is called the
resonator mode number. As will be discussed in Sect. 1.2.2, the phase velocity .v p is deter-
mined by the medium. The eigen frequency expressions (1.71)–(1.75) indicate that for a
given resonator length . L and a medium inside it, the eigen frequency increases with the
mode number .n. The frequency determines the oscillatory cycle of particles. We can easily
imagine that the higher the frequency the greater the kinetic energy of the particles as they
move faster. Generation of higher-order modes usually requires higher input energy [11].
For example, if you apply an acoustic transducer to a steel structure and monitor the resul-
tant frequency with an acoustic sensor, higher-order modes are observed as the transducer’s
energy is increased.

References

1. Komen GJ, Cavaleri L, Donelan M, Hasselmann K, Hasselmann S, Janssen PAEM (1994)

Dynamics and modeling of ocean waves, Ch. 1. Cambridge University Press, pp 1–68
2. Roithmayr CM, Hodges DH (2016) Dynamics: theory and application of Kane’s method, Append.
I. Cambridge University Press
3. Harper PG, Weaire DL (2009) Introduction to physical mathematics, Ch. 28. Cambridge Uni-
versity Press
4. Prof. Brad Osgood (2014) Lecture notes for EE 261 the Fourier transform and its applications.
CreateSpace Independent Publishing Platform
22 1 General Discussions of Waves

5. Bracewell RN (1999) The Fourier transform and its applications, 3rd edn. McGraw-Hill, Boston,
New York
6. Rienstra SW, Hirschberg A (2023) An introduction to acoustics. https://www.win.tue.nl/~sjoerdr/
papers/boek.pdf (accessed on July 17, 2023)
7. Fletcher NH, Rossing TD (1998) The physics of musical instruments, 2nd edn. Springer Link
8. Standing waves and resonance in university physics vol. 1 (British Columbia/Yukon open
authoring platform) https://pressbooks.bccampus.ca/universityphysicssandboxbook1/chapter/
standing-waves-and-resonance/ (accessed on July 17, 2023)
9. King GC (2009) Vibrations and waves. Wiley, Chichester, UK, pp 137–158
10. Borden B, Luscombe J (2017) Essential mathematics for the physical sciences, vol. 1 homo-
geneous boundary value problems, Fourier methods, and special functions, Ch. 2 Separation of
variables. Morgan and Claypool, San Rafael, CA, USA
11. https://courses.lumenlearning.com/suny-osuniversityphysics/chapter/16-4-energy-and-power-
of-a-wave/
Wave Dynamics
2

In this chapter, we focus on the physics behind wave dynamics. After reviewing the harmonic
oscillation of point mass-spring systems, we discuss how oscillatory motions transition into
a wave. We derive wave equations from the equation of motion for oscillatory dynamics in
air and solids. Considering some solutions to the wave equation, we discuss the propagation
of waves as a flow of energy, along with related topics such as acoustic impedance and
acoustic wave intensity. Finally, after briefly reviewing the laws of reflection and refraction,
we discuss the propagation of acoustic waves through the boundary of media having different
elastic properties.

2.1 Oscillations and Waves

A vibrating object generates sounds. A tuning fork [1] (used by a musician) generates sound
because it vibrates at a specific frequency. The physical mechanism underlying this vibration
is the elasticity of the material of the tuning fork. If you hit a tuning fork made of steel against
something rigid (like an edge of a table), the steel atoms near the contact with the table edge
shift from their equilibrium positions. Due to the elasticity of steel, these shifted atoms return
to the equilibrium locations after they reach the farthest point. Due to inertia, they pass the
equilibrium location and swing in the opposite direction from the initial shifts. This motion
pushes the atoms of the neighboring lattice, shifting them from their equilibrium locations.
In this fashion, the vibratory motion transfers through lattices reaching the other surface
of the tuning fork. Due to the difference in acoustic impedance from the air, the vibratory
motion reflects toward the initial surface hit by the table edge. This process is an example
of the resonant phenomenon associated with the reflection of the open-open ends discussed
in the last chapter. The size (corresponding to . L in Fig. 1.9b) and the phase velocity of steel
(see (1.74)) determine the frequency (the eigen frequency) of the fork.

© The Author(s), under exclusive license to Springer Nature Switzerland AG 2024 23

S. Yoshida, Fundamentals of Acoustic Waves and Applications, Synthesis Lectures on Wave
Phenomena in the Physical Sciences, https://doi.org/10.1007/978-3-031-48200-7_2
24 2 Wave Dynamics

The sound you hear in this fashion has the following properties. If you hit the fork more
strongly you will hear the sound louder. However, the pitch of the sound will most likely
be the same. Here, I say “most likely” because depending on how strongly you hit the fork,
harmonics can be heard [2]. If you hit another tuning fork of a different dimension, you will
hear a sound of a different pitch from the first tuning fork. It is because the second fork
is different in material and/or size from the first fork. In other words, eigen frequency is
different. This is how you tune a musical instrument using tuning forks. In this section, we
explore the physics underlying these phenomena.

2.1.1 Harmonic Oscillations

In Chap. 1 we discussed that a wave is a spatiotemporal oscillation. The temporal oscillation

discussed in Sect. 1.2 indicates that a recovery force acts on the particle, making it oscillate
about an equilibrium position. We can call this type of recovery force an elastic force and
the resultant oscillation is a harmonic oscillation [3].
A harmonic oscillation occurs generally in two modes. In the first mode, the system
oscillates after the external disturbance is removed. The above example of a tuning fork falls
in this category. This type of oscillation is referred to as unforced oscillation [4, 5]. In the
second mode, the external agent keeps applying an oscillatory force and the system oscillates
in response to the external force. This type of oscillation is referred to as forced oscillation [4,
6]. Unforced and forced oscillations occur when the system exerts an elastic force. (Forced
oscillation can occur without an elastic force but in that case, the system simply follows the
oscillatory motion provided by the external agent. This type of oscillation is not a harmonic
oscillation and will not be discussed.)
The elastic force is defined as the force whose magnitude is proportional to the displace-
ment from the neutral (equilibrium) position and the direction is opposite to the displacement.
Naturally, the neutral position is where the elastic force is null. When the oscillating entity
(called the particle) comes back to the neutral position, it passes there due to inertia. Con-
sequently, the particle goes to the other side of the neutral position. As soon as the particle
passes the neutral position, the elastic force changes its direction because it is always toward
the neutral position. We can say that the acceleration of the particle is always toward the
neutral position (according to Newton’s second law). Thus, the acceleration is opposite to
the displacement of the particle.

Point mass-spring system

Spring-mass systems [7–9] are simple yet informative to understand harmonic oscillations.
Consider the above dynamics using a system of a point mass connected to a horizontally
placed spring. In this case, the spring exerts the elastic force that causes the oscillation
of point mass and the point mass is what we called the particle in the above paragraph.
2.1 Oscillations and Waves 25

Fig. 2.1 Spring oscillation. The mass (particle) is on the right side of the neutral point (top), at the
neutral point (middle) and on the left side of the neutral point (bottom). The thicker arrow represents
the spring’s elastic force and the thinner arrow represents the displacement from the neutral position

Figure 2.1 illustrates a sample spring-mass system. Here the object connected to the right
end of the spring is considered to be a point mass. The middle illustration represents the
moment when the point mass passes through the neutral point from the right. The top and
bottom illustrations represent the situation before and after the mass passes through the
neutral point.
The arrows represent the elastic force exerted by the spring. Consider the situation after the
mass passes through the neutral point. Since the magnitude of the elastic force is proportional
to the displacement, the acceleration toward the neutral position keeps increasing. This
causes the velocity of the mass to decrease. Eventually, the velocity becomes null. At that
point, the acceleration is still toward the neutral point. Therefore, the mass starts to move
back towards the neutral point with zero initial velocity (after a momentary stop). From this
nature, the point where the mass stops momentarily is called the turning point. After some
time, the mass comes back to the neutral point from the opposite side from the previous
time. The mass passes the neutral point again due to the inertia and moves toward the other
turning point. There, by the same mechanism as at the first turning point, it switches the
direction of motion after a momentary stop. In this fashion, the mass keeps oscillating back
and forth around the neutral position.
It is instructive to analyze the above dynamics qualitatively. The questions I would like
to ask ourselves are (a) How long does it take the mass to come back to the neutral position?
and (b) How far is the turning point away from the neutral position? The quantity asked by
the first question is called the oscillation period and the one by the second question is the
oscillation amplitude. Let’s consider what determines the oscillation period. Above I said
that the mass moves toward the neutral position due to the elastic force and that it passes
the neutral position due to the inertia. It is easily expected that the stronger the elastic force
the sooner the mass returns to the neutral position and that the greater the inertia the slower
the oscillatory motion of the mass. Thus, it is speculated that the stronger the elastic force
the shorter the oscillation period, and the greater the mass the longer the oscillation period
becomes. Indeed, as we will discuss in the next section quantitatively, the oscillation period
is an increasing function of the elastic constant divided by the mass of the particle. Here the
elastic constant measures the strength of the elastic force for a given displacement.
26 2 Wave Dynamics

Energy of oscillatory systems

As for the oscillation amplitude, neither the elastic constant nor the weight of the mass
plays a role in determining it. It is determined by the initial position of the mass. In fact, the
amplitude is similar to the distance between the initial and neutral positions of the mass. (I
said “similar” for a reason, which will become clear shortly). How can we rationalize it?
The best way is to consider energy conservation. Above, I said, “The mass keeps oscillating
back and forth around the neutral position”. This statement indicates that the oscillation
continues forever with the same period and amplitude. You can easily imagine that such a
motion is unrealistic. Indeed, it is impossible for any system to oscillate forever. Eventually,
the oscillation stops, and the mass becomes stationary at the neutral position. This is because
there is always some mechanism that dissipates the oscillatory energy.
Many oscillatory motions are dissipated by the mechanism known as velocity damping
[4], which will be discussed quantitatively later in this chapter. The velocity damping force
is proportional to the velocity of the mass in magnitude and opposite in direction. In a system
in which the damping effect is low, the rate that the oscillation energy is dissipated is low.
The oscillation energy consists of the elastic potential energy and the kinetic energy of the
mass. At the initial point where the mass starts the oscillatory motion, its velocity is null. All
the oscillatory energy is in the form of the elastic potential energy which is determined by
the elastic constant and the initial elastic force. Since the initial elastic force is proportional
to the initial displacement, the initial mass position determines the maximum elastic energy.
At the neutral point where the elastic potential energy is null, the oscillation energy is in
the form of the kinetic energy. As the system loses the oscillation energy due to the damping
mechanism, both the elastic potential energy and kinetic energy decrease. Consequently, the
displacement at the turning point decreases and the velocity at the neutral point decreases.
This in turn means that the oscillation amplitude keeps decreasing as the oscillation repeats.
If the system is a low damping system, the oscillatory energy diminishes slowly, and
therefore the amplitude decrease slowly. When the damping effect is significant, the ampli-
tude decreases fast.

2.1.2 Equation of Motion of Point Mass-Spring System

Unforced oscillation

A system comprising of a spring and a point mass with a velocity-damping mechanism is a

simple yet very illustrative physical system to understand the decaying oscillatory dynamics.
Let .ξ(t) be the displacement from the equilibrium (neutral point). Being proportional to the
displacement in magnitude and opposite in direction, the elastic force . f sp can be put in the
2.1 Oscillations and Waves 27

following form. (See Fig. 2.1 and note that the displacement vector and spring force vector
are opposite and that the spring force is the only external force acting on the mass.)

. f sp = −ksp ξ(t) (2.1)

Here the constant of proportionality.ksp is called the spring constant or stiffness. It represents
the strength of the elasticity of the spring. From Newton’s second law, the external force
on the point mass, . f ex , is proportional to the acceleration where the mass is the constant of
proportionality. This enables us to write the following equation.

d 2 ξ(t)
. f ex = m (2.2)
dt 2
If the spring force is the only component of the external force, . f ex = f sp and (2.1), and (2.2)
lead to the following equation of motion.

d 2 ξ(t) ksp
.
2
=− ξ(t) (2.3)
dt m
Equation (2.3) indicates that the solution .ξ(t) is a function of time that has the following
property; if differentiated twice the result is proportional to the original function with a
negative proportionality constant. We know that sine and cosine functions have this property.
Thus, it is clear that the motion resulting from a spring force is oscillatory, and can be
represented by a sine or cosine function. The oscillation is called the harmonic oscillation.
When the mass undergoes velocity damping (viscous) force, we need an additional term
on the right-hand side of the equation of motion. The resultant equation of motion looks as
follows.
d 2 ξ(t) dξ(t)
.m
2
= −ksp ξ(t) − b (2.4)
dt dt
Here .b is the damping coefficient.
For simplicity, introduce the following parameters .ω0 and .β.
√
ksp
.ω0 = (2.5)
m
b
.β = (2.6)
2m
Here .ω0 is referred to as the natural (angular) frequency and .β as the decay constant. It will
become clear in the next section why these parameters are called in these ways.
Dividing both-hand sides by.m and using (2.5) and (2.6), rewrite Eq. (2.4) in the following
form.
d 2 ξ(t) dξ(t)
.
2
+ 2β + ω02 ξ(t) = 0 (2.7)
dt dt
28 2 Wave Dynamics

Equation (2.7) is a linear differential equation. The displacement of the point mass is given
as a solution to this linear differential equation. Note that this differential equation has no
source term, i.e., the right-hand side of the equation is zero. Hence, (2.4) is classified as a
linear homogeneous differential equation. Physically, this corresponds to unforced oscilla-
tion. In other words, there is no driving force in the system. Being represented by a linear
differential equation, the system is called a linear system.

Forced oscillation

When an additional force (the driving force) is acting on the mass, the solution.ξ(t) represents
forced oscillation. In this case, the differential equation has a source term as below.

d 2 ξ(t) dξ(t) f dr
. + 2β + ω02 ξ(t) = (2.8)
dt 2 dt m
Here . f dr is the driving force and .m is the mass. It is possible to view the driving force as an
input to the linear system. In this view, the solution .ξ(t) is the output of the linear system.
As long as the coefficient .β and .ω0 are constant, the output can be viewed as the linear
response to the input.
The linear system associated with elastic force can be argued from the viewpoint of
elastic (potential) energy. As a conservative force, the elastic force can be characterized as
the first-order spatial differentiation of the potential energy. When the potential energy has
quadratic dependence, the force has linear dependence.
In an actual physical system, it is possible that the potential energy has a dependence on
the displacement with a third-order or higher polynomial function. In this case, the system
is no more linear. Such a system can be interpreted as the case where the stiffness .ksp is a
function of displacement .ξ . A typical example of such a nonlinear system is an inter-atomic
potential. The famous Lennard-Jones potential curve [10] is approximately quadratic near
the equilibrium (near the potential well). This means that as long as the displacement from
the equilibrium is small, the inter-atomic force can be approximated as a spring-like force.
When the displacement is so large that it exceeds the quadratic part of the potential well,
the dynamics becomes nonlinear. A good example is the acoustic response of atoms in a
solid material having residual stress. Residual stress can lock the atom in a range where
the potential curve is not quadratic. Figure 2.2 illustrates the situation schematically. If the
specimen is free of residual stress, the atom is located near the bottom of the well (the
equilibrium position .ξ0 in Fig. 2.2) where the potential curve has a quadratic dependence
on .ξ − ξ0 . In this case, the inter-atomic force .dU (ξ )/dξ is a linear function of .ξ − ξ0 ,
and therefore the stiffness is constant. (Here .ξ is not defined as the displacement from the
equilibrium. .ξ = 0 is the zero interatomic distance. Therefore, .ξ here is different from .ξ(t)
used in the differential equations (2.3), (2.7) and other equations associated with these two
equations.) If residual stress locks the atom far from the equilibrium position, and therefore
the potential energy cannot be approximated as a quadratic function of .ξ − ξ0 , the stiffness
2.1 Oscillations and Waves 29

Fig. 2.2 Stress energy

potential with residual stress.
The slope of the potential curve
represents inter-atomic force.
Potential curve is
approximately quadratic near
equilibrium .ξ0 . On .ξ < ξ0 side
potential is steeper than
quadratic and on .ξ > ξ0 side it
is less steeper than quadratic

is not a constant anymore; it depends on .ξ − ξ0 . In this case, the atom shows nonlinear
behavior when excited by an acoustic wave. Analysis making use of this non-linearity to
probe residual stress is known as acoustoelasticity [11].

2.1.3 Oscillation to Waves

In the preceding section, we discussed the temporal oscillatory behavior of a wave as a

harmonic oscillation of a point mass. In Chap. 1 we discussed a wave is a spatiotemporal
oscillation of particles. We can view the phase velocity of a wave as the motion of a spatial
oscillatory pattern in time, or the motion of a temporal oscillatory pattern through space. In
other words, if the harmonic oscillation experienced by a portion of a medium has a time lag
from the neighboring portion, the harmonic oscillation travels through the medium forming
a wave. In this section, we look into the physics behind this mechanism.

2.1.4 Wave Equations

One dimensional wave

We first discuss how a harmonic motion of a point mass propagates as a wave in one
dimension. Consider the series of point masses connected with springs in Fig. 2.3. Imagine
you pull the rightmost point mass to the right from its equilibrium point and release it.
When you pull this point mass, the spring connected to it stretches, pulling the next (second
from the right) point mass. In this fashion, the point masses to the left of these two will be
pulled to the right one after another. Now the question is “Are these all point masses pulled
simultaneously?”. Intuitively, we know that the answer is “No”. The ones located to the
left will be pulled after a certain delay, and the more to the left, the longer delay time they
experience.
30 2 Wave Dynamics

Fig. 2.3 Series of point masses. a1 unstretched; a2 rightmost point mass pulled; a3 rightmost point
mass released. b elastic force on .n.th point mass. c elastic force acting on .Δx portion of elastic
continuum

When you release the rightmost point mass, a similar delayed behavior is observed in
other point masses. At the moment you release the rightmost point mass, the next point
mass to the left is still moving to the right. Shortly after, this second point mass changes
its direction being pushed by the spring on its right. Thus, all the point masses change their
direction with a certain delay time. This pattern continues where each point mass oscillates
around its equilibrium point. We can view this oscillatory pattern with the delay time as a
longitudinal wave that propagates the harmonic oscillation of the point masses.
Then, the next question is what is the delay time? Again, intuitively we know that the
weaker the springs the longer the delay time. However, it is not straightforward to express
this delay quantitatively. In fact, this delay time represents the wave’s phase velocity. Below,
we analyze this phenomenon quantitatively.

Equation of motion

Consider the dynamics of the .n.th point mass. This mass receives spring force from the two
springs, one that connects it to the .(n − 1)th point mass and the other that connects it to the
.(n + 1) point mass. The equation of motion that governs this motion is as follows.
th

d 2 ξn
m
. = ksp dξn+1 − ksp dξn−1 = ksp (dξn+1 − dξn−1 ) (2.9)
dt 2
Here .m is the mass of the point mass, .ksp is the spring constant of the two springs, and .dξn+1
and.dξn−1 are the differential displacement of the.n.th point mass relative to the.(n + 1)th and
.(n − 1)
th point masses, respectively. These differential displacements represent the stretch
2.1 Oscillations and Waves 31

of the respective springs, and therefore their products with the spring constant represent the
spring force.
Now we extend the above argument into a limit case where the springs are infinitesimally
small so that the entire series of the point masses can be considered as a continuous elastic
medium. In this limit, we can write the equation of motion (2.9) in the following form.

d 2 ξ(x) ( )
μΔx
. = k sp dξ(x + d x) − dξ(x − d x) (2.10)
dt 2
Here .μ (kg/m) is the linear density of the medium and .Δx is the length of the medium along
the .x-axis that we apply the equation of motion (corresponding to the .n th point mass in
(2.9)). On the right-hand side of (2.10), .dξ(x + d x) and .dξ(x − d x) represent, respectively,
the stretch of the section of the elastic medium neighboring to the section represented by .Δx
on the negative and positive sides. In the infinitesimal limit, we can express the differential
displacement (the stretch of the elastic medium) .dξ(x) as follows.

∂ξ(x)
dξ(x) =
. δx (2.11)
∂x
Here, the quantity .δx represents the length of the section we consider the stretch. Using
(2.11) on the right-hand side, we can rewrite (2.10) as follows.
( ) ( )
d 2 ξ(x) ∂ξ(x + d x) ∂ξ(x − d x) ∂ξ(x + d x) ∂ξ(x − d x)
.μΔx = ksp δx − δx = ksp δx −
dt 2 ∂x ∂x ∂x ∂x
(2.12)

Expressing the spatial derivative of the displacement .ξ(x) as

∂ξ(x)
. = ξ ' (x)
∂x
we can write the quantity inside the parentheses on the right-hand side of (2.12) as follows.

∂ξ(x + d x) ∂ξ(x − d x) ∂ξ ' ∂ 2ξ

. − = ξ ' (x + d x) − ξ ' (x − d x) = dξ ' = Δx = 2 Δx
∂x ∂x ∂x ∂x
(2.13)
Here, we used the following expression on the rightmost-hand side of (2.13).

∂ξ ' ∂ ∂ξ ∂ 2ξ
.dξ ' = Δx = Δx = 2 Δx (2.14)
∂x ∂x ∂x ∂x
The above mathematical process lets us rewrite (2.10) as follows.

d 2 ξ(x) ∂ 2ξ
μ
. = k sp δx (2.15)
dt 2 ∂x2
Note that in deriving (2.15) we canceled .Δx from both sides of the equation.
32 2 Wave Dynamics

Spring constant and elastic constant

Let’s take a moment to consider the physical meaning of the spring constant and elastic
constant. These two constants are similar concepts representing the elasticity of a medium.
However, a spring constant is not a material constant, whereas an elastic constant is a material
constant. Each medium has its unique elastic constant, which characterizes the propagation
of acoustic waves. On the other hand, a spring constant depends on the dimension of the
medium, and thereby it is not a material constant. Below, we consider the difference between
these two constants.
According to Hooke’s law [12], the elastic force at .x is the product of the spring constant
and the local differential displacement .dξ .
∂ξ ∂ξ
. f = ksp dξ = ksp (ξ(x + δx) − ξ(x)) = ksp δx = (ksp δx) (2.16)
∂x ∂x
Here, .dξ represents the elongation between .x and .x + δx. In Fig. 2.3c, the two arrows depict
the elastic force at the two ends of the material having the length of .Δx. Here, the elastic
force is due to the elongation over the segment of thickness .δx.
In (2.16) the quantity .∂ξ/∂ x represents the stretch (the elongation per unit length or
.δx = 1). We call this quantity the normal strain .∈ = ∂ξ/∂ x. Figure 2.4 illustrates the situ-

ation where the same normal force . f acts on a segment of the same material with various
thicknesses and cross-sectional areas causing the same normal strain. In Fig. 2.4a, the seg-
ment has a thickness of .δx and a cross-sectional area of . A; in (b) the thickness is .2δ and the
cross-sectional area is . A; in (c) the thickness is unity (.δx = 1) and the cross-sectional area
is . A; and in (d) the thickness is unity and cross-sectional area is unity (. A = 1).

(a)

(c)
(d)

(b)

Fig. 2.4 Spring constant and Young’s modulus

2.1 Oscillations and Waves 33

' inserted in the figure

According to (2.16), the spring constant for (b) is half of (a), as .ksp
indicates. This is because the segment in (b) is twice as thick as (a), and therefore, the
elongation is double when the same force causes the same strain. This comparison indicates
that the spring constant depends on the thickness of the medium. The medium thickness of
(c) is unity, i.e., .δx = 1. In this case, the quantity corresponding to the elongation in (a) and
(b) becomes .(∂ξ/∂ x)(δx) = (∈)(1) = ∈, i.e. is the strain .∈.
According to continuum mechanics [13], Hooke’s law for a unity thickness is given as
follows.
.σ = E∈ (2.17)
Here .σ is the stress defined as the force per unit area and . E is Young’s modulus. In the
present case, it is the normal stress as the force is normal to the area. Figure 2.4d illustrates
Hooke’s law represented by (2.17).
.f = σA (2.18)
Since . E is defined by the force per unit area and the elongation per unit length, it does not
depend on the dimension of the medium, hence it is a material constant.
From (2.16), (2.17), and (2.18), we obtain the following set of equations.

. f = ∈(ksp δx) (2.19)

. f = σ A = E∈ A (2.20)

From (2.19) and (2.20), we find the following relationship between the spring constant and
Young’s modulus.
ksp δx A
.E = , or ksp = E (2.21)
A δx
The first expression of (2.21) indicates that Young’s modulus is in N/m.2 , and the second
expression tells us the greater the cross-sectional area and the less the thickness, the stronger
the spring constant (stiffness) of the elastic medium becomes.
This situation is similar to the relationship between the electric conductivity .σe and con-
ductance, .G e ; .G e = σe (A/l) where . A is the cross-sectional area of the conductive material
and .l is the length. This relationship tells us the greater the cross-sectional area and the
shorter the length, the same material exhibits higher conductance.

Phase velocity as a material constant

The above discussion lets us state that phase velocity is a material constant. Consider the
wave equation (2.15).
d 2 ξ(x) ∂ 2ξ
.μ = ksp δx 2 (2.15)
dt 2 ∂x
34 2 Wave Dynamics

The quantity .μ on the left-hand side of this equation is linear density. Expressing the cross-
sectional area of the medium with . A, we can relate the linear density and volume density .ρ
as follows.
.μ = ρ A (2.22)
Using (2.22) on the left-hand side of (2.15) and the first expression of (2.21), we can rewrite
the wave equation in the following form.

d 2 ξ(x) ksp δx ∂ 2 ξ E ∂ 2ξ
. = = (2.23)
dt 2 ρ A ∂x2 ρ ∂x2
A comparison with (1.14). we can identify the quantity . E/ρ as the square of the phase
velocity .ω/k.
∂2 f ( ω )2 ∂ 2 f
. = (1.14)
∂t 2 k ∂x2
Thus, we find the phase velocity of the longitudinal displacement wave as follows.
√
long E
.v p = (2.24)
ρ

This discussion tells us that the phase velocity is a material constant. Each elastic material
possesses a unique phase velocity. This causes reflection at a boundary between elastic media
having different phase velocities.
Here (2.25) represents the phase velocity of a longitudinal wave because the underlying
equation of motion represents the normal force acting on the segment of the material and the
resultant elongation (and strain) is parallel to the force. We can repeat the same argument for
shear force and resultant shear strain to derive the phase velocity of the shear (transverse)
wave in the following form. √
G
v shear
. p = (2.25)
ρ
Here .G is the shear modulus.

2.2 Acoustic Wave Equations and Solutions

In Sect. 2.1.4 we discussed that an equation of motion yields a mechanical wave equation.
We also discussed that an elastic force is necessary for the oscillatory behavior that generates
wave dynamics. These statements indicate that an expression representing the elastic force
and an equation of motion associated with the elastic force expression are essential to derive
an acoustic wave equation. Elasticity is intrinsic to a medium and therefore the expression
of the elastic force including the expression of the elastic constant depends on the medium.
In this section, we consider wave equations in air and solids.
2.2 Acoustic Wave Equations and Solutions 35

2.2.1 Acoustic Wave in Air

Hooke’s law in air

We can express Hooke’s law in the following form.

ΔV V' − V
. P = −B = −B (2.26)
V V
Here, . P is the pressure that causes the volume change .ΔV , .V is the initial volume, and
V ' is the volume after compression. . B (N/m.2 ) is the bulk modulus, which is the elastic
.

constant corresponding to Young’s modulus or shear modulus for solids. Note that, unlike
Hooke’s law for solids (see (2.17)), we need to use a negative sign. This is because in solids
conventionally positive stress is expansion whereas in air pressure is compression.
Referring to Fig. 2.5 consider the change in volume due to pressure . P.
'
.V = (Δx + dξx )(Δy + dξ y )(Δz + dξz ) = ΔxΔyΔz + (ΔyΔzdξx + ΔzΔxdξ y + ΔxΔydξz )
+ (dξ y dξz Δx + dξz dξx Δy + dξx dξ y Δz) + dξx dξ y dξz
∼ ΔxΔyΔz + (ΔyΔzdξx + ΔzΔxdξ y + ΔxΔydξz )
=
( )
dξx dξ y dξz
= ΔxΔyΔz + ΔxΔyΔz + + (2.27)
Δx Δy Δz

Here, .ΔxΔyΔz = V . Therefore,

( ( )) ( )
' dξx dξ y dξz dξx dξ y dξz
.ΔV = V − V = V +V + + −V =V + +
Δx Δy Δz Δx Δy Δz
(2.28)
In the infinitesimal limit the fractions .Δξx = ∂ξx /∂ x, etc. can be replaced with the corre-
sponding partial derivatives. Thus (2.26) becomes as follows.
( )
ΔV ∂ξx ∂ξ y ∂ξz
. P = −B = −B + + = −B∇ · ξ (2.29)
V ∂x ∂y ∂z

Fig. 2.5 Change in volume due

to pressure at a single point
36 2 Wave Dynamics

Fig. 2.6 Gradient of volume expansion .∇ · ξ due to pressure gradient

The quantity .∇ · ξ represents the divergence of particles from a single point due to the local
displacement field. In other words, .∇ · ξ , hence the pressure . P is defined at this single point.
The fact that the dimension of .∇ · ξ is m/m, i.e., unity indicates that this quantity is defined
at a single point. We can view .∇ · ξ as the volume expansion at the point.
Now consider that the pressure . P has a gradient, meaning that it is not uniform. From
(2.29) we put Hooke’s law as follows.

.∇ P = −B∇(∇ · ξ ) (2.30)

Consider the meaning of .∇(∇ · ξ ) in Fig. 2.6 that illustrates a cubic unit volume of air. Due
to the pressure gradient, .∇ · ξ has .x, . y, and .z dependence. In other words, at the corners of
the cube, the air experiences different levels of volume expansion illustrated in Fig. 2.5.
The left-hand side of (2.30) represents the forces that apply on this unit volume experi-
encing .∇(∇ · ξ ). By breaking .∇ P into the .x, . y, and .z components, we can visualize that
the differential pressure causes the differential volume expansion .∇ · ξ .
(( ) ( ) ( ) )
∂P ∂P ∂P ∂(∇ · ξ ) ∂(∇ · ξ ) ∂(∇ · ξ )
. î + ĵ + k̂ = −B î + ĵ + k̂ (2.31)
∂x ∂y ∂z ∂x ∂y ∂z

Equation of motion

We can view the left-hand side of (2.31) represents the net force acting on the unit volume
characterized by mass .ρ. Thus, we can write the equation of motion for this unit volume as
follows.
( ) ( ) ( ) (( 2 ) ( 2 ) ( 2 ) )
∂P ∂P ∂P ∂ ξx ∂ ξy ∂ ξz
. î + ĵ + k̂ = −ρ î + ĵ + k̂ (2.32)
∂x ∂y ∂z ∂t 2 ∂t 2 ∂t 2
2.2 Acoustic Wave Equations and Solutions 37

Using (2.30) we can also write this equation of motion expressing with the temporal and
spatial derivatives of the displacement vector .ξ and bulk modulus.

∂ 2ξ
ρ
. = −∇ P = B∇(∇ · ξ ) (2.33)
∂t 2
Equation (2.33) explicitly represents that the pressure gradient causes the acceleration of
the unit volume. Thus, we can view it as the equation of motion governing the unit volume.
Note that we need a negative sign on the right-hand side of (2.32) because the inward
pressure is defined to be positive; if the pressure at .x + d x is greater than at .x, the net force
is in the direction of negative.x (the pressure at .x + d x pushes the unit volume more strongly
inward than at .x).
Taking the divergence of both-hand sides of (2.32), we obtain the following equation.
( ) ( ) ( ) ( ( ) ( ) ( ))
∂ ∂P ∂ ∂P ∂ ∂P ∂ ∂ 2 ξx ∂ ∂ 2ξy ∂ ∂ 2 ξz
. + + = −ρ + +
∂x ∂x ∂y ∂y ∂z ∂z ∂x ∂t 2 ∂ y ∂t 2 ∂ x ∂t 2
( )
∂ 2 ∂ξx ∂ξ y ∂ξz
= −ρ 2 + +
∂t ∂x ∂y ∂z

We can write this equation more compactly as follows.

∂ 2 (∇ · ξ )
∇ 2 P = −ρ
. (2.34)
∂t 2
We can view (2.34) as the equation of motion governing .∇ · ξ where the right-hand side
represents the acceleration (times the mass) and the left-hand side the net force.

Volume expansion (compression) wave equation

Express the equation of motion (2.32) in the following compact form.

∂ 2ξ
∇ P = −ρ
. (2.35)
∂t 2
Hooke’s law is as follows.

∇ P = −B∇(∇ · ξ )
. (2.30)

Equate the right-hand sides of (2.35) and (2.30), and take the divergence of the resultant
equation using the identity .∇ · ∇ = ∇ 2 . Then we obtain the following equation.

∂ 2 (∇ · ξ ) B
. = ∇ 2 (∇ · ξ ) (2.36)
∂t 2 ρ
38 2 Wave Dynamics

We can view (2.36) as an equation that represents a volume expansion .(∇ · ξ ) wave that
√
travels at the phase velocity . B/ρ.

Pressure wave equation

Using Hooke’s law (2.29), we can replace .∇ · ξ on both sides of (2.36) with .−P/B and
obtain the following equation.
∂2 P B
. = ∇2 P (2.37)
∂t 2 ρ
Equation (2.37) is a wave equation that represents that pressure . P travels at the phase veloc-
√
ity of . B/ρ, the same velocity as the volume expansion (compression) wave.

Alternative way to derive pressure wave equation

We can derive the wave characteristics of the pressure . P in the following fashion as well.
Differentiate (2.29) with respect to time twice.
( ( ) ( ) ( ))
∂2 P ∂ ∂ 2 ξx ∂ ∂ 2ξy ∂ ∂ 2 ξz
. = −B + + (2.38)
∂t 2 ∂ x ∂t 2 ∂ y ∂t 2 ∂z ∂t 2

The equation of motion yields the following equations.

∂ 2 ξx ∂P ∂ 2ξy ∂P ∂ 2 ξz ∂P
ρ
. = − , ρ = − , ρ =− (2.39)
∂t 2 ∂x ∂t 2 ∂y ∂t 2 ∂z
Using three relationships in (2.39) on the right-hand side of (2.38), we obtain the following
equation.

( ( ) ( ) ( )) ( )
∂2 P B ∂ ∂P ∂ ∂P ∂ ∂P B ∂2 P ∂2 P ∂2 P
. = + + = + + ,
∂t 2 ρ ∂x ∂x ∂y ∂y ∂z ∂z ρ ∂x2 ∂ y2 ∂z 2

or more compactly, in the following form.

∂2 P B
. = ∇2 P (2.40)
∂t 2 ρ
Apparently, (2.40) is identical to the pressure wave equation (2.37). We derived (2.37)
by considering the spatial differentiation of the equation of motion (2.32), and then using
Hooke’s law (2.29). On the other hand, in deriving the pressure wave equation (2.40), we
differentiated Hooke’s law (2.29) with respect to time first and then used the equation of
motion. In either case, we derived the pressure wave equation based on Hooke’s law and
Newton’s second law (the equation of motion).
2.2 Acoustic Wave Equations and Solutions 39

Density-change wave equation

From the conservation of mass, we find the following equality.

ρ0 V = ρV '
. (2.41)

Here .ρ0 and .ρ are the density before and after the volume expansion. Define the relative
volume change .s as follows.
V' − V V'
.s = = −1 (2.42)
V V
Substituting (2.41) into (2.42),
ρ0 ρ0 − ρ δρ
s=
. −1= = (2.43)
ρ ρ ρ
From Hooke’s law (2.26),
. P = −Bs (2.44)
Substituting (2.44) into pressure wave equation (2.40), we obtain the following wave equa-
tion.
∂ 2s B
. = ∇2s (2.45)
∂t 2 ρ
Using (2.43) we can write wave equation (2.45)
( ) ( )
∂ 2 δρ B 2 δρ
. = ∇ (2.46)
∂t 2 ρ ρ ρ

We can view (2.46) as a wave equation of density change.

Note that all the above four wave-equations have the following same phase velocity.
√
B
.v p = (2.47)
ρ

This phase velocity expression has the same form as the one for solids; the square root of
the ratio elastic constant over density, as will be discussed in the next section.

2.2.2 Longitudinal and Transverse Vibrations in Solid

In elastic solids, longitudinal and transverse vibrations propagate as longitudinal and trans-
verse (shear) waves. We can discuss the dynamics of these waves by considering equations
of motion.
First, consider longitudinal wave dynamics. Figure 2.7 illustrates a block of an elastic
medium between .x and .x + d x (called the central block). This block is part of a long
40 2 Wave Dynamics

Fig. 2.7 Elastic force acting on central block

material extended along the .x axis. At the far-right end of this material, an external agent
exerts a tensile force. The other external agent holds the opposite end (on the negative .x
side). Consequently, the entire material experiences a stretch. Since the external force is
dynamic, the block is not under static equilibrium.
At the face at .x, the central block pulls the neighboring block on the negative .x side in
the positive .x direction. In response, the neighboring block exerts a leftward reaction force
. f el (x) on the central block. This force pair causes a local stretch at the boundary of the

central and left neighboring blocks.

Figure 2.8 illustrates the situation at the boundary in detail. Referring to the upper part of
this figure, we can interpret this local stretch as differential displacement .dξ(x1 ) = ξ(x1 +
δx) − ξ(x1 ). Here .δx represents the thin boundary area between the neighboring and central
blocks, and .x1 is the coordinate of the neighboring block side of the boundary. Thus, using
the spring constant .ke , we can express the leftward force exerted by the neighboring block
on the central block as follows.
( )
∂ξ(x)
. f el (x 1 ) = − (k e dξ ) x = −k e δx = −ke ξ ' (x1 )δx (2.48)
1
∂x x1

Here .ξ ' (x) is the spatial derivative of .ξ(x), i.e., .ξ ' (x) = ∂ξ/∂ x.
On the other face at .x2 = x1 + Δx, the neighboring block located on the positive .x side
exerts a rightward force . f el (x1 + Δx). Repeating the same argument as the face at .x1 , we
can express the elastic force at the .x2 as follows.
( )
∂ξ(x)
. f el (x 1 + Δx) = (k e dξ ) x +Δx = k e δx = ke ξ ' (x1 + Δx)δx (2.49)
1
∂x x1 +Δx

Here we assume that .ke and .δx at .x = x1 + Δx are the same as at .x = x1 .

Thus, the net external force acting on the central block is. f el (x1 + Δx) + f el (x1 ). Accord-
ing to Newton’s second law, the acceleration of the central block is equal to the net external
force. From (2.48) and (2.49), we obtain the following equation of motion.
2.2 Acoustic Wave Equations and Solutions 41

Fig. 2.8 Local stretch at .x1 and .x2 = x1 + Δx

∂ 2ξ ( ) ( ∂ξ ' ) ∂ 2ξ
' ' '
. m = k e ξ (x 1 + Δx) − ξ (x 1 ) δx = k e dξ δx = k e Δx δx = k e Δxδx
∂t 2 ∂x ∂x2
(2.50)

Expressing the mass of the central block with the density .ρ and the cross-sectional area . A
as .m = ρ AΔx, we can rewrite the equation of motion (2.50) as follows.

∂ 2ξ ∂ 2ξ
ρ AΔx
. = k e Δxδx (2.51)
∂t 2 ∂x2
In the infinitesimal limit, we can put .δx = Δx = d x. Thus, canceling .Δx and rearranging
terms, we can write (2.50) as follows.

∂ 2ξ ke d x ∂ 2 ξ
ρ
. = (2.52)
∂t 2 A ∂x2
As discussed with (2.18), we can express the normal stress .σ with elastic force . f el as
follows.
f el ke dξ
.σ = = (2.53)
A A
Normal stress is related to normal strain .∈ with Young’s modulus . E.
dξ
σ = E∈ = E
. (2.54)
dx
Comparing the right-hand side of (2.53) and (2.54), we find the following relation. (This is
the same argument as the derivation of (2.21).)
42 2 Wave Dynamics

. ke d x = E A (2.55)

Using (2.55), we can rewrite (2.52) as follows.

∂ 2ξ E ∂ 2ξ
. = (2.56)
∂t 2 ρ ∂x2
Equation (2.56) is a wave equation that describes the dynamics of a longitudinal wave
traveling through a solid that has Young’s modulus (the longitudinal elastic modulus) . E.
A comparison of (1.14) and (2.56) indicates that the wave equation is, in fact, the equation
of motion, and moreover, the phase velocity of the longitudinal elastic wave is given as
follows. √
long E
.v p = (2.57)
ρ
Similarly, considering the displacement behavior perpendicular to the .x axis and its
variation with .x, we can derive a transverse wave equation. The lower part of Fig. 2.8
illustrates the .x dependence of the displacement in the . y direction, .η(x), at the .x1 and
. x 2 = x 1 + Δx boundaries of the central block.

In this case, the net force acting on the boundary area of thickness .δx is the shear force
in the form of the spring constant times the differential displacement .dη(x). Thus, we can
write the equation of motion for the shear dynamics as follows

∂ 2η ∂ 2η
.m = k e d x d x, (2.58)
∂t 2 ∂x2
and stress-strain equation as follows.
f el ke dη
τ=
. = (2.59)
A A
Here .τ is the shear stress and the other quantities are the same as (2.53). Equations (2.58)
and (2.59) are the shear dynamics versions of (2.50) and (2.53), respectively.
Similarly to (2.54), we can relate the shear stress with shear strain .γ .
dη
. τ = Gγ = G (2.60)
dx
Here.G is the shear modulus and is related to the spring constant as follows. This relationship
is the shear deformation version of (2.55).

ke d x = G A
. (2.61)

With the same procedure as the longitudinal wave case, we can derive a wave equation and
phase velocity for the shear dynamics as follows.
2.2 Acoustic Wave Equations and Solutions 43

∂ 2η G ∂ 2η
. = (2.62)
∂t 2 ρ ∂x2
√
G
.v p =
shear
(2.63)
ρ

Note that as we discussed at the end of Sect. 2.1.3, the phase velocities of the longitudinal
and transverse (shear) wave are material constants. They contain material constants. E,.G, and
.ρ. When an acoustic wave of the frequency (.ν) determined by the source (e.g., the operation

frequency of an acoustic transducer) propagates in an elastic material, the wavelength (.λ) is

determined by the material’s unique phase velocity and the frequency (.λ = v p /ν).

2.2.3 Decaying Acoustic Wave Equations and Solutions

We derived the wave equations (2.36) and (2.37) from the equation of motion (2.35). While
this equation of motion describes the mechanism in which the unit volume of air is accelerated
by the differential pressure (the pressure gradient), it generates a solution that represents
dynamics that lasts forever. In reality, waves decay due to various mechanisms such as
velocity damping associated with collisions between air molecules. (Drag force in viscous
media is proportional to the velocity only for lighter objects and lower speed [14, 15].)
To the first-order approximation, we can represent this effect by including a velocity-
damping force term in the equation of motion (2.33).

∂ 2ξ ∂ξ
ρ
. +b − B∇ (∇ · ξ ) = 0 (2.64)
∂t 2 ∂t
Equation of motion (2.64) leads to the following decay wave equation of volume expansion.

∂ 2 (∇ · ξ ) ∂(∇ · ξ )
. + 2β − v 2p ∇ 2 (∇ · ξ ) = 0 (2.65)
∂t 2 ∂t
Here .β is the decay constant and .v p is the phase velocity defined by (2.47).

b B
.β= , v 2p = (2.66)
2ρ ρ
In the case of longitudinal and transverse waves in solids, we derived the longitudinal
and transverse (shear) wave equations (2.56) and (2.62) from the corresponding equations
of motion (2.50) and (2.58). Similarly to the wave equation in air, we can include the effect
of velocity damping by adding the term proportional to the first-order time derivative of the
displacement.
44 2 Wave Dynamics

∂ 2ξ ∂ξ ( long )2 ∂ 2 ξ
. + 2β − vp =0 (2.67)
∂t 2 ∂t ∂x2
∂ 2η ∂η ( shear )2 ∂ 2 η
. + 2β − vp =0 (2.68)
∂t 2 ∂t ∂x2

As clear from the above arguments, the physical mechanism that generates acoustic waves
in materials in air and solids, in common, is elasticity. The phase velocity has the form of
the square root of the ratio of elastic constant over the density. When the medium is viscous,
the velocity-damping term must be included. Thus, we can express the wave equation and
phase velocity in the following general forms.

∂ 2ξ ∂ξ 2∂ ξ
2
. + 2β − v p =0 (2.69)
∂t 2 ∂t ∂x2
√
κ
.v p = (2.70)
ρ

Here . f represents a quantity that exhibits the wave characteristics in general, .β is the damp-
ing coefficient, and .κ is the elastic constant. In the case of the waves in solids, .ξ is either
longitudinal or shear displacement, and .κ is either Young’s modulus or shear modulus cor-
respondingly. In the case of waves in air, . f can be the volume compression .∇ · ξ , pressure
. p, or density change .δρ/ρ. The other types of elastic waves discussed in Chap. 4 can have

a decay effect in the same fashion.

Decaying acoustic wave solution

Above we discussed that linear acoustic wave equations with a velocity damping mechanism
can be expressed in the form of (2.69) commonly to in air and solids. In this section, we
discuss general solutions to this type of wave equation.
In Chap. 1, we used the cosine function (1.31) as a solution to the one-dimensional wave
equation without a damping term (1.15).

. f (t, x) = A cos(ωt ± kx + φ) (1.31)

The cosine function in (1.31) represents the oscillatory characteristics of the wave. In the
wave equation, the combination of the secondary temporal and spatial derivative of the
wave function represents the oscillatory characteristics. The square of the phase velocity
that appears in front of the spatial secondary derivative term represents the elasticity, which
is the physical origin of the oscillatory behavior. The damping term in (2.69) originates from
the velocity damping effect, and does not cause the oscillatory behavior. The decaying wave
equation (2.69) still has the secondary temporal and spatial derivative terms in the same
form as the non-decaying wave equation.
2.2 Acoustic Wave Equations and Solutions 45

These observations indicate that a decaying wave solution can be expressed with a sine
or cosine function, with the addition of a term that represents the decaying characteristics
of the wave. So, it is natural to assume a solution that has a sinusoidal function as part of it.
In this section, we use the exponential form of a sinusoidal function as it is more convenient
to include the damping effect.
Using Euler’s notation [16, 17], we can express (1.31) as follows.

ξ(t, x) = Aei(ωt±kx) = A {cos(ωt ± kx) + i sin(ωt ± kx)}

. (2.71)

We can view (1.31) as the real part of this exponential expression with initial phase .φ = 0.
We set .φ = 0 because the initial phase does not affect the gist of the discussion.
Substitution of (2.71) into (2.69) leads to the following equation.
{ }
. A (iω) + i2βω − v p (ik) ei(ωt±kx) = 0
2 2 2
(2.72)

It follows that the content of the curly bracket is zero.

.ω2 − i2βω − v 2p k 2 = 0 (2.73)

In order for (2.73) to hold, it is necessary that either .ω or .k is a complex number. Let’s
assume .k is complex, i.e., .k = kr + iki .
( ) ( )
.ω − i2βω − v p (kr + iki ) = ω − v p kr + v p ki − i2 βω + v 2p kr ki = 0 (2.74)
2 2 2 2 2 2 2 2

It follows that the following set of equations holds.

( )
ω2 = v 2p kr2 − ki2
. (2.75)
βω =
. −v 2p kr ki (2.76)

Elimination of .ki from (2.75) and (2.76) yields the following equation for .kr .

(βω)2
.v 2p (kr2 )2 − ω2 (kr2 ) − =0 (2.77)
v 2p

Solve (2.77) as a quadratic equation of .kr2 taking only the real root to obtain the following
expression.
√ ( ⎡ √ ⎤
√ 2) ( )2
ω 2 + ω4 + 4(βω)2 ω2 + ω4 1 + 4β ω ω 2 2β
= 2 ⎣1 + 1 + ⎦
2
.(kr ) = =
2
2v 2p 2v 2p 2v p ω
(2.78)
From (2.75), we find .(ki )2 as follows.
46 2 Wave Dynamics

⎡ √ ⎤
( )2
ω2 ω2 ⎣−1 + 2β ⎦
(ki )2 = (kr )2 −
. = 1+ (2.79)
v 2p 2v 2p ω

Thus, we find it as follows.

⎡ √ ⎤1
( )2 2
ω ⎣1 + 2β ⎦
kr = √
. 1+ (2.80)
2v p ω
⎡ √ ⎤1
( )2 2
ω ⎣−1 + 2β ⎦
. ki = √ 1+ (2.81)
2v p ω

By separating the real and imaginary parts of .k, we can express the solution .ξ(t, x) (2.71)
as follows.

ξ(t, x) = Aei(iki x) ei(ωt±kr x) = Ae−ki x {cos(ωt ± kr x) + i sin(ωt ± kr x)}

. (2.82)

The exponential term .ex p(−iki x) multiplied by amplitude . A represents the decaying effect
as follows. As we discussed in Sect. 1.2.1, when the wave propagates in the positive direction,
the phase term takes the form of .ωt − kx with a positive propagation constant .k. We set the
complex propagation constant as .k = kr + iki . So, for a wave propagating in the positive .x
direction,.ki > 0. This indicates as the wave propagates along the.x axis, i.e., with an increase
in .x, the exponential term .ex p(−iki x) decreases. We can express a wave propagating in the
negative .x direction by assuming .k is negative. In this case, the wave decays as it travels, too.
As an increase of .x in the negative direction (i.e., as .x becomes more negative), the negative
.ki < 0 makes the exponential term decay. So, regardless of the direction of propagation the

wave decays as it travels.

When the damping coefficient .b is negligible, we can put .β = 0 and (2.80) and (2.81)
reduce to the following expressions.

ω [ √ ]1
2 ω 1 ω
kr = √
. 1+ 1+0 = √ [2] 2 = =k (2.83)
2v p 2v p vp
ω [ √ ]1
2 ω 1
.ki = √ −1 + 1 + 0 = √ [−1 + 1] 2 = 0 (2.84)
2v p 2v p

Here (2.84) indicates that when .β = 0, .ki = 0, hence .kr = k. Using this fact, we expressed
the phase velocity .v p as equal to .ω/k in (2.83). Thus, under the condition of .β = 0, the
decay wave solution (2.82) reduces to the exponential form of the non-decaying solution
(1.31).
2.2 Acoustic Wave Equations and Solutions 47

Three dimensional case

Now quickly extend the above discussions regarding the decay wave solution into three
dimensions by solving the three-dimensional wave equation.

∂ 2ξ ∂ξ
. + 2β − v 2p ∇ 2 ξ = 0 (2.85)
∂t 2 ∂t
As clear from the general discussion of extending a one-dimensional wave to a three-
dimensional wave made in Sect. 1.2.2, we can assume the solution in the following form.

ξ (t, r) = Aei(ωt±k·r)
. (2.86)

Use the Laplacian operator for the spatial differentiation of (2.86), as we did in Sect. 1.2.2.
Expressing the magnitude of propagation vector .k with .k, we can express the second-order
spatial differentiation as follows.

∇ 2 eik·r = ∇ 2 ei (k x x+k y y+kz z )

.

= (ik x )2 ei (k x x+k y y+kz z ) + (ik y )2 ei (k x x+k y y+kz z ) + (ik z )2 ei (k x x+k y y+kz z )

= −(k x2 + k 2y + k z2 )ei (k x x+k y y+kz z ) = −k 2 ek·r (2.87)

Substitution of (2.86) into (2.85) with the use of (2.87) yields the following equation.
{ }
. A (iω) + i2βω + v p k ei(ωt±k·r) = 0
2 2 2
(2.88)

Clearly, (2.88) leads to the condition (2.73) for the one-dimensional case.

.ω2 − i2βω − v 2p k 2 = 0 (2.73)

Thus, we find the same expressions of the complex propagation constant as the one-
dimensional case. In other words, except for the direction of propagation, the three-
dimensional wave (2.86) behaves exactly the same as the one-dimensional wave (2.71).
It is instructive to express the spatial part of the phase .k · r that appears in the exponent
of the three-dimensional wave function (2.86) using .î, . ĵ, and .k̂ (the unit vectors for .x, . y,
and .z directions). Referring to Fig. 2.9, express .k · r in the following form.

k · r = k lˆ · (x î + y ĵ + z k̂)
. (2.89)

Here .lˆ is the unit vector in the direction of the propagation vector .k. (I do not use .k̂ for this
unit vector to avoid confusion with the unit vector for the .z direction.)
As discussed in Appendix A, the unit vector .l’s ˆ .x, . y, and .z components are the direction
cosines of the respective axes.

lˆ = cos θx î + cos θ y ĵ + cos θz k̂

. (A.8)
48 2 Wave Dynamics

Fig. 2.9 Propagation vector .k

and its unit vector whose
components are direction
cosines

Using (A.8) we can rewrite (2.89) in the following form.

{ }
k · r = k (cos θx )x + (cos θ y )y + (cos θz )z
.

= (k cos θx )x + (k cos θ y )y + (k cos θz )z = (k x )x + (k y )y + (k z )z (2.90)

We can interpret the quantities .k x , .k y , and .k z that appear in the last term of (2.90) as the .x,
y, and .z components of the propagation constant. Thus, this term indicates that the wave
.

propagates in the direction .cos θx î + cos θ y ĵ + cos θz k̂ gaining the phase (including the
decay factor) along the .x, . y, and .z axes with the respective components of .k.
By separating the propagation constant into the real (oscillatory) and imaginary (decay-
ing) parts, we obtain the following explicit forms.

ξ (t, r) = Ae(i(ωt±k·r )) = Aeiωt eikr ·r ei

2 k ·r
. i = Aei(ωt+kr ·r) e−ki ·r (2.91)

where
{ }
kr · r = kr (cos θx )x + (cos θ y )y + (cos θz )z
.

⎡ √ ⎤1
( )2 2
ω ⎣ 2β ⎦ { }
=√ 1+ 1+ (cos θx )x + (cos θ y )y + (cos θz )z (2.92)
2v p ω
{ }
.ki · r = ki (cos θ x )x + (cos θ y )y + (cos θz )z

⎡ √ ⎤1
( )2 2
ω ⎣ 2β ⎦ { }
=√ −1 + 1 + (cos θx )x + (cos θ y )y + (cos θz )z (2.93)
2v p ω

When the damping coefficient .b is negligible, we can put .β = 0 and (2.92) and (2.93)
reduce to the following expressions.

ω [ √ ]1
2 ω
.kr = √ 1+ 1+0 = =k (2.94)
2v p vp
ω [ √ ]1
2
.ki = √ −1 + 1 + 0 = 0 (2.95)
2v p
2.2 Acoustic Wave Equations and Solutions 49

As is the one-dimensional case, the condition .β = 0 reduces the solution (2.86) to the non-
decaying solution where .k is a real number.

2.2.4 Nonlinear Wave Equations

So far we assumed that the elastic modulus does not depend on the space coordinate. If the
elastic modulus is a function of the space coordinate, the wave equation becomes nonlinear.
Consider this effect for the case of longitudinal waves in a solid.
Reconsider the equation of motion (2.50) allowing the stiffness (spring constant) to
depend on .x. In this case, we need to include differentiation of the stiffness in the evaluation
of the net force on the central block.
∂ 2ξ ( ) ∂(ke ξ ' )
' '
m
. = k e (x 1 + Δx)ξ (x 1 + Δx) − k e (x 1 )ξ (x 1 ) δx = Δxδx (2.96)
∂t 2 ∂x
Using (2.55) allowing . E to depend on .x, we can put (2.97) in the following form.
( ) ( ) ( 2 )
∂ 2ξ ∂(Eξ ' ) ∂ ∂ξ ∂ E ∂ξ ∂ ξ
.ρ = = E = + E (2.97)
∂t 2 ∂x ∂x ∂x ∂x ∂x ∂x2

This situation makes it impossible to put the wave equation in the form of (2.56) where the
secondary differentiation of the wave function (.ξ(t, x)) with respect to time is proportional to
the secondary differentiation of the function with respect to space. Thus, the phase velocity
is not a constant.
The situation is the same as the case when the displacement is a vector or the medium
has viscosity.
∂ 2ξ ∂ξ
.ρ +b = (∇ E) (∇ · ξ ) + E∇ (∇ · ξ ) (2.98)
∂t 2 ∂t
As an example, we can discuss nonlinear wave equations in solids by allowing the spatial
coordinate dependence in the spring constant in (2.50), i.e., Youg’s modulus or shear modulus
depends on the coordinate variable. This type of nonlinearity is caused by the fact that the
restoring (elastic force) has second or higher-order dependence on the space coordinates. A
good example of such a case is the recovery force of residual stress discussed in Fig. 2.2.
When residual stress causes the strain so large that the atom deviates from the bottom of
the inter-atomic potential where the potential curve is approximately a quadratic function
of the inter-atomic distance, its slope becomes nonlinear [18, 19]. Let’s take a moment to
consider this case.
Let .U (r ) be the inter-atomic potential and have a third-order dependence on the distance
from the equilibrium (the bottom of the potential curve) .r .

.U (r ) = ar 3 + br 2 + cr + d (2.99)

Here .a, .b, .c, and .d are constant, and .r corresponds to .ξ − ξ0 in Fig. 2.2.
50 2 Wave Dynamics

The spatial first-order differentiation of potential energy is the restoring force. So, in this
case, the restoring force . fr has a second-order dependence on .r .

U (r )
. fr (r ) = = 3ar 2 + 2br + c (2.100)
dr
Elastic force is expressed as the product of the stiffness (the spring constant) .ke and the
displacement from the equilibrium. From (2.100), we find the following expression of the
stiffness for the potential (2.99).
fr
ke =
. = 3ar + 2b (2.101)
r
Apparently, .ke is not a constant.
Although nonlinear acoustics and related technology [20] are interesting topics, we
mostly discuss linear acoustic waves in this book.

2.3 Wave as a Flow of Energy

In this section, we discuss acoustic waves as a flow of acoustic energy.

2.3.1 Acoustic Energy

Consider in Fig. 2.10 that a cubic block of elastic medium experiences tensile force applied
on the planes at .x and .x + Δx. This pair of forces stretch the block by the differential
displacement in the .x direction .Δξ . We can express the work done by these forces as
follows. ∫ Δξ ∫ Δξ
( ) 1
.W = f dξ = ksp ξ dξ = ksp (Δξ )2 (2.102)
0 0 2
Here .ksp is the stiffness (spring constant) of the elastic medium.
Express the force in two different ways so that we can rewrite (2.102) independent of the
dimension.

Fig. 2.10 Work done by force

f is stored as potential (elastic)
.
energy of the elastic medium
2.3 Wave as a Flow of Energy 51

∂ξ
. f = ksp Δξ = ksp Δx = ksp ∈Δx (2.103)
∂x
. f = σ A = E∈ A (2.104)

Here .∈ is the normal strain, .σ is the normal stress, . E is Young’s modulus, and . A is the
cross-sectional area of the volume element.
By equating (2.103) and (2.104), we find the following relationship between the spring
constant and Young’s modulus.
EA
.k sp = (2.105)
Δx
Substitute (2.105) into (2.102).
( )2
1 1 EA 1 EA ∂ξ 1 2
.W = ksp (Δξ )2 = (Δξ )2 = Δx = E∈ (AΔx) (2.106)
2 2 Δx 2 Δx ∂x 2

Noting that .(AΔx) is the volume of the block, we find the following expression for the
energy density.
W 1
.w p = = E∈ 2 (2.107)
AΔx 2
This energy density represents the elastic potential energy per unit volume. The suffix . p
stands for the potential energy.

Instantaneous acoustic energy

Above, we made the analysis for static equilibrium. Now consider that the force (stress) is
dynamic in the context of an acoustic wave. As we discussed in Sect. 1.2.3, a one-dimensional
wave solution can be put in the following form (see (1.31)).

ξ(t, x) = ξ0 cos(ωt − kx)

. (2.108)

Here .ξ represents the displacement from equilibrium. (In Fig. 2.10 we use .ξ to represent the
particle displacement from the origin, not necessarily from the equilibrium. Here we define
.ξ to be from the equilibrium. This difference does not affect the gist of the discussion here.)

The particle velocity is the time differentiation of (2.108), and the normal strain .∈ is the
spatial differentiation of (2.108).
∂ξ
. v(t, x) = = −ωξ0 sin(ωt − kx) (2.109)
∂t
∂ξ
.∈(t, x) = = kξ0 sin(ωt − kx) (2.110)
∂x

Figure 2.11 illustrates sample snapshots of .ξ(ωt − kx), .∈(ωt − kx), and .v(ωt − kx) at
two instances corresponding to .t = 0 and one-eighth of the period later. Note that the strain
52 2 Wave Dynamics

(au)
(au)
(au)

(au)

Fig. 2.11 Snapshot of displacement (top), strain (middle), and velocity (bottom) waves at two time-
steps. The horizontal axis is common to all plots

wave is a quarter period shifted from the displacement wave, and the velocity wave is .π
out of phase to the strain wave. These phase shifts are explained by (2.110) and (2.109),
respectively.
From (2.109) and (2.110), we find the following relationship between .v(t, x) and .∈(t, x).
k
∈=− v
. (2.111)
ω
As a side note, the negative sign in (2.111) indicates that the velocity wave is .π out of
phase to the strain wave. Also, .k/ω is the reciprocal of the phase velocity. Equation (2.111)
indicates that the ratio of the amplitude of the velocity wave over the strain wave is equal to
the phase velocity and that the velocity and strain waves are out of phase by .π .
Substitution of (2.111) into (2.107) yields the following expression.
( )2
1 2 E k
wp =
. E∈ = v2 (2.112)
2 2 ω

From (1.11) we know .ω/k is the phase velocity and from (2.24) we know the phase velocity
√
of a longitudinal elastic wave is related to the density .ρ as . (E/ρ). Thus, we find the
right-hand side of (2.112) as follows.
2.3 Wave as a Flow of Energy 53

Fig. 2.12 Potential energy

density in a solid (a) and air
(b). In both cases, the external
force increases the acoustic
energy density

( )2
E k E 1 2 E ρ 2 1
. v2 = v = v = ρv 2 ≡ wk (2.113)
2 ω 2 vp
2 2 E 2

Expression (2.113) represents the kinetic energy of particles in the unit volume, .wk .
How about the acoustic potential energy density in air? Fig. 2.12 illustrates the situation
where the external force . f acting normally on the plane of area . A increases the acoustic
energy density. In (a) the external force stretches the volume and in (b) it compresses the
volume. In (a) the stress is .σ/A and in (b) the pressure is . p/A. Let .a and .b represent the
shorter and longer sides of the cube in the direction parallel to the external force (.a < b).
Using Hooke’s law expressions in solids (2.17) and air (2.29), we obtain the following
equations. (For the in-air case, we use a one-dimensional expression of (2.29).) Here, in
both cases, the external force does positive work, hence the change in the acoustic energy
is positive.
∫ ∫
b b E 2b E( 2 )
dw =
. σ d∈ = E∈d∈ =
[∈ ]a = b − a2 > 0 (2.114)
a 2 2
∫ ∫a a
a (−B) 2 a (−B) ( 2 )
dw =
. pd∈ = (−B)∈d∈ = [∈ ]b = a − b2 > 0 (2.115)
b b 2 2
In summary, we find the instantaneous elastic potential energy and the particle’s kinetic
energy as follows.
1 2 1
wp =
. E∈ ; wk = ρv 2 ; w p = wk ; wtotal = w p + wk = E∈ 2 = ρv 2 = B∈ 2
2 2
(2.116)
The above discussions indicate that at a given moment, the acoustic wave has potential
energy density .w p and kinetic energy density .wk , and they are equal to each other. In Sect.
2.3.2, we find that an acoustic wave carries these two types of energy at the phase velocity.
This situation is analogous to that an electromagnetic wave carries electric and magnetic
energy densities that are equal to each other.

Average acoustic energy

What is the actual magnitude of the energy that flows as a wave? It is determined by the
initial condition, i.e., how much the external agent initially pulls the unit volume, i.e., the
54 2 Wave Dynamics

actual value of .ξ0 in (2.109) and (2.110). Putting the amplitude of the velocity and strain
as .∈ p = kξ0 and .v p = ωξ0 , we can express the temporal variation of the velocity and strain
as follows. These expressions correspond the temporal oscillation of .v and .∈ at a fixed .x in
(2.109) and (2.110).

v(t) = v0 sin ωt
. (2.117)
. ∈(t) = ∈0 sin ωt (2.118)

We can express the average values over one period (.τ = (2π )/ω) for these energies as
follows.
∫ 2π ∫ 2π
ω ω ω 2 ω 1 2
. < v 2 (t) > = (v0 sin ωt)2 dt = v sin2 ωtdt = v (2.119)
2π 0 2π 0 0 2 0
∫ 2π ∫ 2π
ω ω ω 2 ω 1 2
. < ∈ 2 (t) > = (∈0 sin ωt)2 dt = ∈ sin2 ωtdt = ∈ (2.120)
2π 0 2π 0 0 2 0
Thus, the average total energy density becomes as follows.
ρ 2 E 2
wtotal
.
av
= wkav + wav
p = v + ∈0 = ρv02 = E∈02 (2.121)
2 0 2
Some note regarding equality of kinetic and potential energy

The equality between the kinetic and potential energy leads to some intriguing insight into
the acoustic field. From (2.116) we obtain the following equality.
E 2 ρ
. ∈ = v2 (2.122)
2 2
Rearranging (2.122) we find the ratio of .v over .∈ as follows.
√
|v| E
. = = vp (2.123)
|∈| ρ
√
Here, from (2.57) we know . E/ρ is the phase velocity of the longitudinal wave character-
ized by the material constants . E and .ρ.
Knowing that .v = dξ/dt and .∈ = dξ/d x, we can view the ratio of the amplitude of the
velocity wave over the amplitude of the strain wave as representing the phase velocity via
the following argument. | | | | | |
|v| || dξ || || dξ || || d x ||
= / = = vp (2.124)
|∈| | dt | | d x | | dt |
.

Note that .v and .∈ in (2.123) represent the amplitude of the respective waves. That is why a
negative sign does not appear in this equality unlike (2.111) which relates the instantaneous
value of the velocity wave and strain wave.
2.3 Wave as a Flow of Energy 55

Acoustic impedance

In the theory of electronic circuits, impedance represents the difficulty of current flow-
ing through a medium. This concept corresponds to the resistance in a DC (direct cur-
rent) circuit, which is defined as the ratio of the voltage drop across a resistor over the
current. In an AC (alternating circuit) transmission line or electromagnetic wave transmis-
sion in air, impedance is significant because when an AC current or electromagnetic wave
passes through a boundary where the impedance varies before and after, reflection occurs.
Consequently, the power is not transmitted as expected. Therefore, various techniques to
match the impedance are used. This operation is called impedance matching [21].
The acoustic impedance.z ac is defined as the ratio of the amplitude of the stress (pressure)
wave to the amplitude of the velocity wave. In the case of a longitudinal wave in a solid, we
obtain the following expression for the acoustic impedance. See (2.123) in going through
the third equal sign in the below equation.

σ E∈ E ρv 2p
. z ac = = = = = ρv p (2.125)
v v vp vp

This definition is analogous to that of the electrical impedance. In electrodynamics, the

scalar potential (voltage) is the spatial integration of the electric field, which exerts an
electric force on the electric charges that flow as an electric current. Here in (2.125), the
acoustic impedance is the ratio of stress (force) over the velocity (flow).
In the case of air, the acoustic impedance takes the following form.
√
B
.v p =
air
(2.126)
ρ

Applying the equality between the potential and kinetic energy density (2.122) to the case
of air, we obtain the following equation.
√
B 2 ρ 2 v B
. ∈ = v , i.e., = = v air
p (2.127)
2 2 ∈ ρ

So, ( )2
−p B∈ B ρ v air
p
.
air
z ac = = = air = = ρv air
p (2.128)
v v vp v air
p

When we excite an ultrasound in a solid with an ultrasound transmitter, it is necessary to

use a buffer, typically distilled water or ultrasound gel, to increase the coupling. The role of
the buffer is to match the impedance. (Tip: If an ultrasound gel is not available, honey may
work well as a buffer. [22])
56 2 Wave Dynamics

2.3.2 Intensity of Acoustic Wave

In the preceding section, we discussed that acoustic energy density is the product of pressure
(stress) and velocity. From this discussion, we can envision that a pair of pressure (stress)
and velocity waves carry acoustic energy. In this section, we discuss this topic quantitatively.
Let’s use the displacement wave in the form of (2.108) for this discussion. In this case,
the velocity and strain waves are as shown in (2.109) and (2.110).
∂ξ
v(t, x) =
. = −ωξ0 sin(ωt − kx) (2.109)
∂t
∂ξ
∈(t, x) = = kξ0 sin(ωt − kx) (2.110)
∂x
Remembering that .ω/k is the phase velocity .v p (see (1.11)), we find the following relation-
ship from (2.109) and (2.110).

v(t, x) −ωξ0 sin(ωt − kx) ω

. = = − = −v p (2.129)
∈(t, x) kξ0 sin(ωt − kx) k
The relationship (2.129) is nothing new. It is an iteration of (2.111) that states that ratio of
velocity wave over the strain wave is equal to the phase velocity with a negative sign. As I
mentioned under (2.111), the negative sign indicates that the velocity and strain waves are
out of phase with each other by .π . In a sinusoidal wave, out of phase by .π is equivalent to
flipping the sign of the wave function.
Viewing .∈ = ∂ξ/∂ x as the one-dimensional version of volume expansion .∇ · ξ and using
(2.30) that relates pressure and volume expansion, we obtain the following expression for
pressure . P.
. P(x, t) = −B∈(x, t) (2.130)
Thus, using (2.129) we can express the product of the pressure and velocity waves in two
ways as follows.

. P(x, t)v(t, x) = −B∈(t, x)v(t, x) = −B∈(t, x)∈(t, x)(−v p ) = B∈(t, x)2 v p

= wtotal (t, x)v p (2.131)
. P(x, t)v(t, x) = Bkξ0 sin(ωt − kx)ωξ0 sin(ωt − kx) = (Bkωξ02 ) sin2 (ωt − kx)
(2.132)

Here going to the second line of (2.131), we use the total energy density expression (2.116).
Dimensional analysis indicates that the . Pv wave is in (N/m.2 )(m/s). = (J/m.3 ) (m/s) =
W/m.2 . This quantity is called intensity often expressed with a symbol . I . Expression (2.131)
literally indicates that the . pv wave carries the total energy density at the phase velocity .v p .
On the other hand, (2.132) indicates that the . Pv wave oscillates at the same frequency as
the displacement wave and its amplitude is . Bkωξ0 . By expressing (2.132) with the pressure
and velocity wave amplitude . B(kξ0 ) = B∈0 = p0 and .ωξ0 = v0 , we can find the average
2.3 Wave as a Flow of Energy 57

intensity as follows. See (2.119) and (2.120) for the equivalent expression for the velocity
and strain waves.
∫ 2π
ω ω 1
.I
av
= p0 v0 sin2 (ωt − kx)dt = p0 v0 (2.133)
2π 0 2
The above discussions indicate that the pressure and velocity wave together carry the
acoustic energy. Figure 2.13 illustrates that the velocity and pressure waves are in phase,
oscillating at the same frequency as the displacement wave.
If we repeat the same procedure to express the product of the normal stress and velocity
waves us .σ = E∈, we obtain the following pair of equations.

.σ (x, t)v(t, x) = E∈(t, x)v(t, x) = E∈(t, x)∈(t, x)(−v p ) = E∈(t, x)2 (−v p )
= wtotal (t, x)(−v p ) (2.134)
.σ (x, t)v(t, x) = Ekξ0 sin(ωt − kx)(−ωξ0 sin(ωt − kx)) = −(Ekωξ02 ) sin2 (ωt − kx)
(2.135)

This time, unlike the product of pressure and velocity wave, we have a negative sign in
front of the phase velocity .v p . Using Fig. 2.13, we can explain this discrepancy between the
cases of the pressure wave and stress wave as follows. The stress wave is out of phase from
the displacement wave by .π/2 and from the velocity and pressure waves by .π . The phase
shift of .±π/2 between the displacement wave and the other three waves originates from
the fact that these three waves are temporal or spatial derivatives of the displacement wave.
(au)
(au)
(au)
(au)

(au)

Fig. 2.13 Displacement, stress, pressure, and velocity waves. Note that the pressure and velocity
waves are in phase. They together carry acoustic energy according to (2.131)
58 2 Wave Dynamics

The.π phase difference between the stress and pressure waves, which have the same physical
dimension of N/m.2 , is simply due to the difference in the definition; a stretch of an elastic
medium corresponds to positive stress and negative pressure, i.e., .σ = E∈ vs . P = −B∇ · ξ .
From this standpoint, we can say that.−σ0 sin(ωt − kx) carries acoustic energy together with
the velocity wave .v0 sin(ωt − kx).
In any event, the intensity is defined as the absolute value of the time average of the
product of the component waves, i.e., the pressure and velocity waves or the stress and
velocity waves. So, it is always a positive value.

2.4 Acoustic Wave Reflection and Transmission

2.4.1 Laws of Reflection and Refraction

Refer to Fig. 2.14. The acoustic wave is incident to the boundary AB from the left. The
acoustic wave travels in the medium on the left of the boundary with a phase velocity .v1 ,
and in the medium on the right of the boundary with .v2 . The frequency of the wave is
constant through the boundary. Therefore, the wavelength on the left of the boundary, .λ1 is
different from that on the right, .λ2 . Lines N.1 A and N.2 B are normal to boundary AB. The
angle made by these lines and AB is called the angle of incidence .θi .
The dashed lines represent the wavefront of three waves, i.e., the incident wave, reflected
wave, and transmitted wave (the crest of the respective waves). Lines S.1 A and S.2 B are
perpendicular to the wavefronts, representing the propagation of the incident wave. Figure
2.14 depicts the moment when the left edge of the wavefront of the incident wave reaches
the boundary at point A. Therefore, line DB is equal to wavelength .λ1 . Similarly, on the
right of the boundary, line AE is equal to the wavelength .λ2 .

N A

S E
C

N
B

Fig. 2.14 Law of reflection and refraction

2.4 Acoustic Wave Reflection and Transmission 59

Consider triangle ABD. Angle ABD is complementary to angle DBN.2 . Angle BAD is
complementary to angle ABD because angle ADB is a right angle. Therefore angle BAD
is equal to angle DBN.2 , which is the angle of incidence .θi . Length BD is equal to the
wavelength in medium 1, .λ1 . We obtain the following equation.

. BD = AB sin θi = λ1 (2.136)

Repeating the same argument for triangle ABC, we find the following equation.

AC = AB sin θr = λ1
. (2.137)

From (2.136) and (2.137) we find the following equality, which is known as the law of
reflection; the angle of reflection is equal to the angle of incidence.

θr = θi
. (2.138)

Now pay attention to triangle ABE. We can easily find the following equation.

. AE = AB sin θt = λ2 (2.139)

From (2.136) and (2.139) we find the following equality, which is known as the law of
refraction, or Snell’s law.
.λ1 sin θt = λ2 sin θi (2.140)
Since the wavelength is proportional to the phase velocity for the same frequency, we can
express Snell’s law in the following form as well.

.v1 sin θt = v2 sin θi (2.141)

Here .v1 and .v2 are the acoustic phase velocity in the medium on the left and right of the
boundary.
Now consider that a wave is incident to a boundary from medium 1 to medium 2 where
the phase velocity in medium 1 is higher than medium 2, i.e., .v1 > v2 . Solving (2.141) for
.sin θt , we obtain the following expression.

v2
. sin θt = sin θi (2.142)
v1
Since the sine function .sin θi cannot take a value greater than unity and .v2 /v1 > 1, (2.142)
does not hold for all .θi . The following condition is necessary.
( )
−1 v1
.θc ≡ θi ≤ sin (2.143)
v2

This angle .θc is referred to as the critical angle.

60 2 Wave Dynamics

2.4.2 Behavior of Acoustic Waves Near Boundary

When an acoustic wave passes through a boundary that separates two elastic media of dif-
ferent elasticity and density, part of the wave energy is reflected and the rest is transmitted.
If the angle of incidence is oblique, the direction of the transmitted wave is determined by
Snell’s law. The amount of reflected and transmitted energy is determined by the boundary
conditions. In this section, we discuss these topics.

Formulation of incident, reflected, and transmitted waves

We first consider acoustic waves in solids. In this section, we use the exponential form as it
avoids complexity in the formulation. Set the incident, reflected, and transmitted displace-
ment in the following form.

ξi (t, x) = ξi0 ei(ωt−k1 x) ; ξr (t, x) = ξi0 ei(ωt+k1 x) ; ξt (t, x) = ξi0 ei(ωt−k2 x)

. (2.144)

Here .ξ0i , .ξ0r , and .ξ0t , are the amplitude of the respective waves. Subscripts .1 and .2 for
Young’s modulus . E and propagation constant .k denote medium 1 (the incident side) and
medium 2 (the transmitted side).
∂ξi ∂ξi
σi (t, x) = E 1
. = E 1 (−ik1 )ξi (t, x); vi (t, x) = = (iω)ξi (t, x) (2.145)
∂x ∂t
∂ξr ∂ξr
.σr (t, x) = E 1 = E 1 (ik1 )ξr (t, x); vr (t, x) = = (iω)ξr (t, x) (2.146)
∂x ∂t
∂ξt ∂ξt
.σt (t, x) = E 2 = E 2 (−ik2 )ξt (t, x); vt (t, x) = = (iω)ξt (t, x) (2.147)
∂x ∂t
Remembering the definition of acoustic impedance (2.125), we obtain the following expres-
sions.
σi E 1 (−ik1 )ξi (t, x) −E 1 k1 E1
. = = =− = −z 1 (2.148)
vi (iω)ξi (t, x) ω v p1
σr E 1 (ik1 )ξr (t, x) E 1 k1 E1
. = = = = z1 (2.149)
vr (iω)ξr (t, x) ω v p1
σt E 2 (−ik2 )ξt (t, x) −E 2 k2 E2
. = = =− = −z 2 (2.150)
vt (iω)ξt (t, x) ω v p2

Notice that the magnitude of the acoustic impedance is the same for the incident and reflected
waves as they are in the same medium. Their signs are opposite because the propagations are
opposite. The acoustic impedance of the transmitted wave is different from the impedance
of the incident or reflected wave as the transmitted beam propagates in a different medium.
2.4 Acoustic Wave Reflection and Transmission 61

Now we make the same discussion as above for acoustic waves in air. According to (2.29)
and (2.54), the only difference between the in-air and in-solids cases is the elastic modulus’s
sign. (You may notice that while (2.29) is expressed in three dimensions for air, (2.54) is in
one dimension for solids. We will discuss Hooke’s law in solids in a three-dimensional form
in Chap. 4.) Replace . E with .−B and .σi with . pi in (2.145)–(2.150) to obtain the following
equations.
∂ξi ∂ξi
. pi (t, x) = −B1 = −B1 (−ik1 )ξi (t, x); vi (t, x) = = (iω)ξi (t, x) (2.151)
∂x ∂t
∂ξr ∂ξr
. pr (t, x) = −B1 = −B1 (ik1 )ξr (t, x); vr (t, x) = = (iω)ξr (t, x) (2.152)
∂x ∂t
∂ξt ∂ξt
. pt (t, x) = −B2 = −B2 (−ik1 )ξt (t, x); vt (t, x) = = (iω)ξt (t, x) (2.153)
∂x ∂t

pi −B1 (−ik1 )ξi (t, x) B1 k1 B1

. = = = = z1 (2.154)
vi (iω)ξi (t, x) ω v p1
pr −B1 (ik1 )ξr (t, x) −B1 k1 B1
. = = =− = −z 1 (2.155)
vr (iω)ξr (t, x) ω v p1
pt −B2 (−ik2 )ξt (t, x) B2 k2 B2
. = = = = z2 (2.156)
vt (iω)ξt (t, x) ω v p2

Similarly to the case in solids, the acoustic impedance for the incident and reflected waves
have the same magnitude and opposite signs, whereas that for the transmitted wave is
different.

Boundary conditions and reflectivity

At the boundary, the following two conditions must hold for the pressure and particle velocity.

.vi cos θi + vr cos θi = vt cos θt (2.157)

σi + σr = σt
. (2.158)

Condition (2.158) represents the force balance at the boundary plane. If the balance is broken,
the boundary plane will have acceleration, and consequently, it moves in the positive or
negative .x direction. Condition (2.157) represents the law of momentum conservation along
the .x axis, i.e., perpendicular to the boundary plane. Per condition (2.158), the net force
acting on the particle along the .x is zero.
Using (2.148)–(2.150) in (2.158) we obtain the following equation.

. − z 1 vi + z 1 vr = −z 2 vt (2.159)

Eliminate .vt from (2.157) and (2.159).

62 2 Wave Dynamics

vi (z 2 cos θi − z 1 cos θt ) + vr (z 2 cos θi + z 1 cos θt ) = 0

. (2.160)

Similarly, eliminate .vr to obtain the following equation.

2vi z 1 cos θi = vt (z 1 cos θt + z 2 cos θi )

. (2.161)

From (2.160) and (2.161), we find the reflection and transmission coefficient for the velocity
wave as follows.
vr z 1 cos θt − z 2 cos θi
rv =
. = (2.162)
vi z 1 cos θt + z 2 cos θi
vt 2z 1 cos θi
.tv = = (2.163)
vi z 1 cos θt + z 2 cos θi
Substituting (2.148)–(2.150) into (2.162) and (2.163), we can express the reflection and
transmission coefficient in terms of the stress waves as follows.
σr vr z 1 −z 1 cos θt + z 2 cos θi
rσ =
. = = (2.164)
σi vi (−z 1 ) z 1 cos θt + z 2 cos θi
σt vt (−z 2 ) 2z 2 cos θi
.tσ = = = (2.165)
σi vi (−z 1 ) z 1 cos θt + z 2 cos θi
Note that (2.162)–(2.165) represent the reflection and transmission of the stress and
velocity waves. Therefore, each of these expressions does not represent the reflection or
transmission of the acoustic energy. The product of the stress (N/m.2 ) and velocity (m/s)
represents the intensity (N/m.2 · m/s.=W/m.2 ), i.e., the area density of acoustic power. As we
discussed in Sect. 2.3.2, it is the product of the pressure (stress) and velocity waves that
represents the flow of acoustic intensity.

Energy conservation at boundary

Now express the intensity of the incident wave striking the boundary plane by considering the
product of the stress and velocity waves. From (2.144) and (2.145), we obtain the following
expression.

. Ii = |σi vi | cos θi = |(−ik1 E 1 )ξi (iω)ξi | cos θi = (k1 E 1 ω)ξi2 cos θi (2.166)

In (2.166), the factor.cos θi represents the component of the intensity parallel to the boundary
plane (see Fig. 2.14). Since we are considering the event on a fixed .x where the boundary
is located, the intensity is a function of time. In other words, the intensity . Ii oscillates
with angular frequency .ω according to .ξi (t, x0 ) = ξi0 ei(ωt−k1 x0 ) . Here .x0 represents the .x
coordinate where the boundary is located.
Similarly, we can formulate the reflected and transmitted intensities.
References 63

. Ir = |σr vr | cos θi = |(rσ σi )(rv vi )| cos θi

|( )( ) |
| −z 1 cos θt + z 2 cos θi z 1 cos θt − z 2 cos θi |
= || (−ik1 E 1 )ξi (iω)ξi || cos θi
z 1 cos θt + z 2 cos θi z 1 cos θt + z 2 cos θi
| |
| −(z 1 cos θt − z 2 cos θi )2 | (z 1 cos θt − z 2 cos θi )2
. = |
2|
| (z cos θ + z cos θ )2 (k1 E 1 ω)ξi | cos θi = (z cos θ + z cos θ )2 Ii (2.167)
1 t 2 i 1 t 2 i
. It = |σt vt | cos θt = |(tσ σi )(tv vi )| cos θt
|( )( ) |
| 2z 2 cos θi 2z 1 cos θi |
=|| (−ik1 E 1 )ξi (iω)ξi || cos θt
z 1 cos θt + z 2 cos θi z 1 cos θt + z 2 cos θi
| |
| 4z 1 z 2 cos θi cos θt | 4z 1 z 2 cos θi cos θt
=|| (k ω)ξ 2|
(z 1 cos θt + z 2 cos θi )2
1 E 1 i | cos θi = (z cos θ + z cos θ )2 Ii (2.168)
1 t 2 i

So, the total reflected and transmitted intensity is

( )
(z 1 cos θt − z 2 cos θi )2 4z 1 z 2 cos θi cos θt
. Ir + It = + Ii
(z 1 cos θt + z 2 cos θi )2 (z 1 cos θt + z 2 cos θi )2
(z 1 cos θt )2 + (z 2 cos θi )2 − 2z 1 z 2 cos θi cos θt + 4z 1 z 2 cos θi cos θt
= Ii
(z 1 cos θt + z 2 cos θi )2
(z 1 cos θt + z 2 cos θi )2
= Ii = Ii (2.169)
(z 1 cos θt + z 2 cos θi )2
Equation (2.169) indicates that through the boundary the acoustic energy is conserved.
By defining the reflectance . R = Ir /Ii and transmittance .T = It /Ii we obtain the follow-
ing simple expression from (2.169).
Ir + It
. R+T = =1 (2.170)
Ii
In the case of acoustic waves in air, .vi , .vr , and .vt are the same as solids (compare (2.145)–
(2.147) and (2.151)–(2.153)). Therefore .rv and .tv are the same as solids. For .r p and .t p , we
need to replace . E 1 and . E 2 with .−B1 and .−B2 in the expression of the corresponding . p
waves. This change results in the reflectivity and transmissivity as follows.
pr vr (−z 1 ) −z 1 cos θt + z 2 cos θi
.r p = = = (2.171)
pi vi z 1 ) z 1 cos θt + z 2 cos θi
pt vt z 2 2z 2 cos θi
.t p = = = (2.172)
pi vi z 1 z 1 cos θt + z 2 cos θi
After all, both coefficients are the same as the solid’s case (2.164) and (2.165), if expressed
with acoustic impedances.

References

1. Addink CCJ (1951) Precise calibration of tuning forks. Philips Tech Rev 12(8):228–232
2. Harvard Natural Sciences Lecture Demonstrations, https://sciencedemonstrations.fas.harvard.
edu/presentations/tuning-forks (accessed on July 17, 2023)
64 2 Wave Dynamics

3. https://www.birmingham.ac.uk/teachers/study-resources/stem/physics/harmonic-motion.aspx
4. Blake RE (1996) Basic vibration theory. In: Harris CM (ed) Shcok and vibration handbook, 4th
edn, Ch. 2. McGraw-Hill, New York
5. Margenau H, Murphy GM (1956) The mathematics of physics and chemistry, 2nd edn. D. Van
Nostrand, Co. Inc., Toronto, New York, pp 50–53
6. King GC (2009) Vibrations and waves. Wiley, Chichester, UK, pp 33–48
7. øAström KJ, Murray RM (2021) Feedback systems, an introduction for scientists and engineers.
Princeton University Press, Princeton, London, pp 27–29
8. Obando L (2018) Mass-spring-damper system, 73 exercises resolved and explained: systems
dynamics and transfer function. Universidad Central de Venezuela, Caracas, Venezuela
9. Yoshida S (2017) Waves; fundamental and dynamics. Morgan & Claypool, San Rafael, CA,
USA, IOP Publishing, Bristol, UK, Chap 1
10. Jones JE (1924) On the determination of molecular fields.—I. From the variation of the viscosity
of a gas with temperature. Proc R Soc Lond Ser A 106(738):441–462
11. Toupin RA, Bernstein B (1961) Sound waves in deformed perfectly elastic materials. Acoustoe-
lastic effect. J Acoust Soc Am 33:216
12. Mavko G, Mukerji T, Dvorkin J (2020) The rock physics handbook, 3rd edn. Cambridge Uni-
versity Press, New York, Chap. 2
13. Reddy JN (2013) An introduction to continuum mechanics, 2nd edn. Cambridge University
Press, Cambridge, UK
14. Tonya Coffey, Drag, https://www.youtube.com/watch?v=xKfZ25yBxu8 (accessed on July 19,
2023)
15. Hooge C. 5.2 drag forces. In: BCIT physics 0312 textbook. https://pressbooks.bccampus.ca/
physics0312chooge/chapter/5-2-drag-forces/
16. Stipp D (2017) A most elegant equation; Euler’s formula and the beauty of mathematics. Basic
Books, New York, USA
17. Boas ML (2006) Mathematical methods in the physical sciences, 3rd edn. Wiley, London, p 61
18. Hughes DS, Kelly JL (1953) Second-order deformation of solids. Phys Rev 92(5):1145–1149
19. Yoshida S, Sasaki T, Usui M, Sakamoto S, Gurney D, Park IK (2016) Residual stress analysis
based on acoustic and optical methods. Materials 9:112
20. Garrett SL (2020) Nonlinea acoustics. In: Understanding acoustics. Graduate texts in physics.
Springer, Cham. https://doi.org/10.1007/978-3-030-44787-8_15
21. Rehman A (2018) Practical impedance matching, 1st edn. Signal Processing Group Inc., Tempe,
AX, USA
22. Miyasaka C (2012) Private communication
Propagation of Acoustic Waves in Air
3

This chapter specifically discusses the propagation of acoustic waves in air. By considering
the source of elasticity of air, we derive the equation of motion and corresponding wave
equations. After these discussions of wave dynamics, we touch on the sounds we hear.
We briefly discuss how human ears conceive and process sound waves. We consider how
musical instruments generate their unique sounds. Toward the end of the chapter, we discuss
amplitude and frequency modulations as methods to transmit sound waves over a distance.

3.1 Dynamics of Air Particles and Acoustic Wave Equation in Air

3.1.1 Particle Motions in Longitudinal Waves

In Sect. 2.2.1 we derived the longitudinal wave equations for volume compression and
pressure from the equation of motion. There we expressed the equation of motion. In this
process, we used the displacement of air particles to derive the equation of motion. These
procedures indicate that the deviation of air particles from the equilibrium is the underlying
mechanism to generate the wave dynamics of air particles. We start this chapter by visualizing
the relationship between displacement and pressure gradient.
Figure 3.1 is a snapshot of air particles where a speaker generates a sound wave in the
air. The diaphragm of the speaker moves sinusoidally. The dots represent the position of
layers of particles. In Fig. 3.1a, the horizontal arrows represent the deviation of layers from
the equilibrium position. In those locations where the dots are dense (space), the pressure is
high (low). Figure 3.1b illustrates the pressure variation as a function of .x, the coordinate
variable along the axis of wave propagation.

© The Author(s), under exclusive license to Springer Nature Switzerland AG 2024 65

S. Yoshida, Fundamentals of Acoustic Waves and Applications, Synthesis Lectures on Wave
Phenomena in the Physical Sciences, https://doi.org/10.1007/978-3-031-48200-7_3
66 3 Propagation of Acoustic Waves in Air

Fig. 3.1 Snapshots of air particles a, pressure wave b, and displacement wave c, .λ and .τ represent
the wavelength and period of the waves

Pay attention to the second positive peak of the pressure graph. In Fig. 3.1a, the horizontal
arrows are inward to this .x location, indicating that the corresponding displacement vector
changes the sign over this location. In fact, on the negative-x (positive-x) side of this location,
the arrow is rightward (leftward), meaning that the displacement is positive (negative). Figure
3.1c explicitly illustrates this change in the sign. Algebraically, we can express the above
relationship between pressure and displacement as . p = −∂ξ/∂ x; the negative slope of .ξ(x)
corresponds to positive pressure (see (2.29)).
The dashed lines in Fig. 3.1b, c are the spatial variation of the respective quantities .Δt
later. Notice that the pressure wave travels to the right so does the displacement wave. Figure
3.1d illustrates this temporal change observed at the leftmost .x position. During the elapse
of .Δt, the pressure at this point decreases while displacement increases.

3.1.2 Wave-Generating Mechanism in Air

Air may not sound like an elastic medium. So, in this section, we discuss what causes
elasticity in air. Here we analyze the dynamics in one dimension to make the analysis easily
visible.
In short, we can say as follows: The pressure gradient is proportional to the density
gradient (gas law). The density gradient is the displacement gradient. Therefore, the pressure
3.1 Dynamics of Air Particles and Acoustic Wave Equation in Air 67

Fig. 3.2 Change in air density due to pressure gradient

gradient is proportional to the displacement gradient. This is how Hooke’s law applies to
air. Below, we will discuss these dynamics quantitatively.
For simplicity, let’s consider the dynamics in one dimension. Figure 3.2 illustrates the
situation where a volume of air expands along the .x-axis. Before the expansion, this volume
has a cross-sectional area of . S and a short length along the .x axis. Due to the pressure
dependence on .x, the cross-sectional plane at .x = x2 is displaced more than that at .x = x1
(.x2 > x1 ). Since we consider a one-dimensional case, the cross-sectional area is unchanged
during the expansion. In the following sections, we discuss this phenomenon in detail.

Density change due to volume expansion

Let .ξ(x, t) be displacement as a function of coordinate variable .x and .l0 be the length of the
volume along the .x-axis before the expansion. Thus, the volume before the expansion (.V0 )
and after (.V ) are given, respectively, as follows.

. V0 = Sl0 (3.1)
. V = S{l0 + (ξ(x2 ) − ξ(x1 )} ≡ S(l0 + Δξ ) (3.2)

So, the volume change .ΔV becomes,

ΔV = V − V0 = SΔξ
. (3.3)

Since the process is an expansion, the mass of the volume is unchanged while the volume
increases. Therefore, we can express the density before the expansion (.ρ0 ) and after (.ρ) as
follows.
m
ρ0 =
. (3.4)
V0
m m
.ρ = = (3.5)
V V0 + ΔV
Here .m is the mass. From (3.4) and (3.5), we find the following expression for the change
in the density.
68 3 Propagation of Acoustic Waves in Air

m m −ΔV
Δρ =
. − =m (3.6)
V0 + ΔV V0 (V0 + ΔV )V0
Now consider the infinitesimal limit where we can express .l0 = d x and assume .ΔV <<
V0 . In this case, we can rewrite (3.3) and (3.6) as follows.

.d V = Sdξ (3.7)
−d V −d V m dV
.dρ = m =m 2 =− (3.8)
(V0 + d V )V0 V0 V0 V0

Using (3.1), (3.7) and replacing .l0 with .d x, we obtain the following equation.
dV Sdξ dξ
. = = (3.9)
V0 Sl0 dx
In the infinitesimal limit, we can express .dξ(x) as follows.
∂ξ
.dξ = dx (3.10)
∂x
From (3.4), (3.8), (3.9), and (3.10), we obtain the following equation.
∂ξ
.ρe ≡ dρ = −ρ0 (3.11)
∂x
Gas law

From the ideal-gas law [1], we know that pressure and density are related as follows.
n
. P= RT (3.12)
V
m
.n = (3.13)
M
Here, . P is the pressure of .n moles of air, . R is the gas constant, .T is the temperature, and . M
is the molar mass. Using (3.12) and (3.13), we can express the pressure as follows.
m/M m RT RT
. P= RT = =ρ
V V M M
So, . P is a function of .ρ, . P(ρ). Noting that the density change is small, i.e., .ρe << ρ0 we
can express the pressure after the expansion as follows.
( ) ( )
dP dP
.P = P0 + Pe = f (ρ0 + ρe ) = f (ρ0 ) + f ' (ρ0 )ρe = f (ρ0 ) + ρe = P0 + ρe (3.14)
dρ 0 dρ 0

From (3.14), we obtain the following relationship between the pressure change and density
change.
. Pe = K ρe (3.15)
3.1 Dynamics of Air Particles and Acoustic Wave Equation in Air 69

where ( )
dP
. K = (3.16)
dρ 0
It is interesting to look into the relationship (3.15) based on Newtonian dynamics. Dimen-
sional analysis on (3.16) indicates that . K is in (N/m.2 )/(kg/m.3 ) = (m/s).2 . Thus (3.15) reads
. Pe (N/m. ) =. K (m/s)(m/s) .ρe (kg/m. ). Interpreting that the product of (m/s) and .ρe (kg/m. ) as
2 3 3

a momentum density (. pe ) and the other (m/s) as the flow velocity of the momentum density,
we can view the right-hand side of (3.15) as the unit volume of air having momentum . pe
passing through a plane perpendicular to its flow with the velocity expressed by the other
(m/s).
Figure 3.3 illustrates the concept. A momentum density (momentum per unit volume)
. pe = ρv passes through a plane. Here .v is the particle velocity. Since the particle velocity

has a gradient .∂v/∂ x / = 0, the momentum of the unit volume changes as it passes through
the plane. Since this is a change in momentum over time, a force is involved in the process
(Newton’s second law). The left-hand side of (3.15) represents this force.
This phenomenon is similar to the so-called optical radiation pressure [2], in which optical
radiation exerts pressure on a reflector. When light is reflected by a reflector, the momentum
carried by the photons changes its direction. Consequently, each of these photons experiences
a momentum change from .h/λ to .−h/λ. Here .h is Planck’s constant and .λ is the wavelength
of the light. This net change of .2h/λ causes the exertion of force on the mirror. The unit
of Planck’s constant is J s = (kg m/s.2 )(ms) = kg (m.2 /s) and .λ is in m. Thus .2h/λ is in
kg(m.2 /s)/m = kg(m/s), i.e., it represents the momentum change. Since this change occurs
during the contact time with the mirror surface, its rate is kg (m/s)/s = N. In other words, we
can explain this phenomenon using the impulse-momentum theorem. Although this force is
in N, not N/m.2 , conventionally it is called radiation pressure.
Radiation pressure is observed in an optical resonator of high intra-resonator power
configured with suspended resonator mirrors employed by a gravitational wave detector [3].
The high optical momentum pushes the resonator mirror out of position, causing the optical
resonator to lose the resonant condition. Consequently, the intra-resonator optical power
decreases, reducing the radiation pressure onto the mirror and making the resonator regain
the operational configuration. Once the resonator operates, the intra-resonator optical power
increases, resulting in high radiation pressure and causing the same problem to repeat. Figure
3.3 illustrates the dynamics. Naively, you may understand this phenomenon by imagining that
you apply a water jet from a hose onto a light plate suspended with a string (like a pendulum).
While the water jet hits the plate perpendicularly (with the normal incident), the plate is
pushed. As the plate swings away, the pushing effect by the water jet reduces and gravity
pulls back the plate making the angle of incidence back to normal. This swinging motion
repeats. (This analogy does not represent the resonant effect observed in a gravitational wave
detector but illustrates the gist of the effect.)
Acoustic waves can cause radiation pressure in conjunction with deformation in the
reflecting object. Cantrell [4] discusses a detailed analysis of this topic.
70 3 Propagation of Acoustic Waves in Air

Fig. 3.3 a Momentum density . pe change causes pressure . Pe . b Radiation pressure pushes the res-
onator mirror. Photon changes its direction as reflected on the mirror surface flipping the sign of the
momentum. .h: Planck’s constant; .λ: wavelength of light

Substitution of (3.11) into (3.15) yields the following expression.

∂ξ
. Pe = K ρe = −K ρ0 (3.17)
∂x
Interpreting .∂∈/∂ x on the right-hand side as the one-dimensional expression of .∇ · ξ , we
find that (3.17) is equivalent to Hooke’s law (2.29).
ΔV
. P = −B = −B∇ · ξ (2.29)
V
From the comparison of (2.29) and (3.17), we find the following relationship between . K
and . B.
B
.K = (3.18)
ρ0
Since . B is in N/m.2 and .ρ0 in kg/m.3 , we find that the quantity . K has a dimension of velocity
squared as (N/m.2 )/(kg/m.3 ) = (kg m/(s.2 m.2 ))/(kg/m.3 ) = (m/s).2 . In fact, as we discussed in
the last chapter (see (2.70), the square root of the ratio of an elastic constant over density is
the phase velocity. So, we can identify . K as the square of the phase velocity of the acoustic
wave in air that we will derive shortly.

Equation of motion

Since . Pe represents the pressure around the plane at .x, we can interpret the quantity .d Pe
as the net force per unit cross-sectional area acting on the volume. Thus, from Newton’s
second law, we obtain the following equation of motion that represents the one-dimensional
dynamics due to pressure gradient along the .x-axis. See Fig. 3.4.

Fig. 3.4 Net force acting on

volume . Sd x due to pressure
gradient = ( ) +
= ( + )
3.1 Dynamics of Air Particles and Acoustic Wave Equation in Air 71

∂ 2ξ ∂ Pe
ρ0 Sd x
. = −Sd Pe = −S dx (3.19)
∂t 2 ∂x
Here, . S and .d x are respectively the cross-sectional area and the thickness of the volume
under consideration, and we replaced .d Pe with .(∂ Pe /∂ x)d x. Thus, eliminating the common
factor . Sd x, we obtain the following equation.

∂ 2ξ ∂ Pe
. ρ0 =− (3.20)
∂t 2 ∂x
From (3.17) and (3.20), we can write the equation of motion (3.19) in the following form.
( )
∂ 2ξ ∂ ∂ξ ∂ 2ξ
.ρ0 = − −K ρ 0 = ρ0 K (3.21)
∂t 2 ∂x ∂x ∂x2

Thus we obtain the following displacement wave equation.

∂ 2ξ ∂ 2ξ
. = K (3.22)
∂t 2 ∂x2
From (3.22), we can derive a wave equation for the pressure. Differentiate (3.22) with
respect to .x.
∂ 2 ∂ξ ∂ 2 ∂ξ
. = K (3.23)
∂t 2 ∂ x ∂x2 ∂x
Use (3.17) in (3.23)
∂ 2 Pe ∂ 2 Pe
.− = −K (3.24)
∂t 2 K ρ0 ∂ x 2 K ρ0
Thus, we obtain the following pressure wave equation.

∂ 2 Pe ∂ 2 Pe
. =K (3.25)
∂t 2 ∂x2

3.1.3 Decaying Pressure Waves in Air

Using (3.18), we can rewrite (2.64) in the following form, which we can interpret as the
equation of motion for the unit volume represented by density .ρ0 .

∂ 2ξ ∂ξ ∂ 2ξ ∂ξ
.ρ + b − B∇ (∇ · ξ ) → ρ0 +b − ρ0 K ∇ (∇ · ξ ) = 0 (3.26)
∂t 2 ∂t ∂t 2 ∂t
By dividing by .ρ and using .β = b/(2ρ0 ) we obtain as follows.

∂ 2ξ ∂ξ
. + 2β − K ∇ (∇ · ξ ) = 0 (3.27)
∂t 2 ∂t
72 3 Propagation of Acoustic Waves in Air

Now take the divergence of (3.27) and obtain the volume expansion (compression) wave
equation.
∂ 2 (∇ · ξ ) ∂(∇ · ξ )
. + 2β − K ∇ 2 (∇ · ξ ) = 0 (3.28)
∂t 2 ∂t
By substituting (2.29) we obtain the pressure wave equation.

∂2 P ∂P
. + 2β − K ∇2 P = 0 (3.29)
∂t 2 ∂t
Remembering that . K = B/ρ represents the square of the phase velocity (see (3.18)), we
find (3.29) takes the general form of decaying elastic wave Eq. (2.69) we derived in the last
chapter.

3.2 Sound We Hear

3.2.1 Our Ear

Structure of our ear

Our ear consists of the outer ear, the middle ear, and the inner ear [5]. Figure 3.5 illustrates
the structure of our ear. The outer and middle ears are filled with air and separated by
the eardrum. The middle ear consists of three bones called the malleus, incus, and stapes.
The middle and inner ears are separated by a tissue membrane called the oval window,
which connects the stapes and the duct called the vestibular duct. As shown in Fig. 3.6,
the vestibular duct lies parallel to the cochlear duct, separated by Reisnner’s membrane.
On the opposite side of the cochlear duct, another duct called the tympanic duct lies. The
cochlear and tympanic ducts are separated by the basilar membrane, which is connected to
the auditory nerves via hair cells. The tympanic duct is terminated with the round window
(another membrane) on the middle ear side.

Fig. 3.5 Anatomy of ear. The

eardrum separates the outer and
middle ear and the oval
window separates the middle
and inner ear. The middle ear
consists of three bones, the
malleus, incus, and stapes. The
stapes is attached to the oval
window. The cochlea has a
spiraled, hollow structure as
the top-right insert illustrates.
(Illustration courtesy of Luc
Allain)
3.2 Sound We Hear 73

Reissner’s

Fig. 3.6 Anatomy of the cochlea. The top right insert is a cross-sectional view of the vestibular,
cochlear, and tympanic ducts. These three ducts inside the cochlea contain fluids. The bottom insert
is a zoomed-in illustration of the basilar membrane. The pair of dashed lines indicate the opposite
directions of fluid motions. The basilar membrane picks up the vibration of the fluid at a specific
frequency depending on its location from the base to the apex. (Illustration courtesy of Luc Allain)

In short, our brain perceives sound by the following mechanism. The sound we capture
vibrates the eardrum. The three bones in the middle ear transfer this vibration to the oval
window. These bones act as an impedance-adjusting mechanism to match the low impedance
of air to the high impedance of the fluid in the inner ear [6]. The vibration of the oval win-
dow moves the fluid that fills the vestibular duct. As the oval window moves in toward the
vestibular duct, the round window moves out from the tympanic duct. These movements
generate mutually out-of-phase motions of the fluids in the respective ducts, which causes
the basilar membrane to vibrate. The vibration of the basilar membrane causes the hair cells
to experience oscillatory displacement, which generates electric impulse signals that are
transmitted through the auditory nerves to the brain. We can say that our ear-brain system
converts a mechanical acoustic signal to an electric signal like a microphone.

Frequency of sound we hear

The frequency band that humans can typically hear ranges from 20 Hz to 20 kHz [7]. In
speeches, the frequency of the vowels ranges from 250 Hz to 2 kHz, and those of voiced
consonants and unvoiced consonants range from 250 Hz to 4 kHz, and 2 kHz to 8 kHz,
respectively [8]. The frequency of the piano ranges from 27.5 Hz to 4.186 kHz over the
leftmost to the rightmost piano keys [9]. The frequency of scratchy sound ranges from 2
kHz to 5 kHz [10]. Table 3.1 lists these frequency ranges.
Human ears process the frequency of sounds in the following fashion [6]. The basilar
membrane varies its width and stiffness along its length. It is the narrowest and stiffest at the
74 3 Propagation of Acoustic Waves in Air

Table 3.1 Frequency range of various types of sound. [8–10]

Type of sound Low frequency (Hz) High frequency (Hz)
Vowel 250 2,000
Voiced consonant 250 4,000
Unvoiced consonant 2,000 8,000
Piano 27.5 4186
Scratchy sound 2,000 5,000

base (closer to the middle ear) end, and is the widest and least stiff at the apex (the other end).
Remember that in Chap. 2 we discussed that the natural frequency of a spring-mass system
increases in proportion to the square root of the ratio of the spring constant (stiffness) over
the mass. The higher the stiffness and lower the mass, the higher the natural frequency of the
system. This structure makes the membrane’s vibration resonant with the highest frequency
near the base side and shifts the resonant frequency towards the apex as the sound frequency
lowers. The hair cell at each location along the length of the basilar membrane dominantly
processes the sound at the corresponding frequency. Thus, complex sounds consisting of
various frequencies (e.g., speech) are resolved into component frequencies and processed
by the corresponding hair cells.

3.2.2 Sounds of Speech and Music

In this section, we look at some sample sounds of speech and musical instruments and make
quick analyses of the signal in the time and frequency domain.

Sample speech signal

Figure 3.7 shows the Fourier spectrum of “a”, “ma”, and “sa” sounds spoken by a female
speaker. The left graphs are logarithmic plots of the sounds, and the right plots are linear
plots zoomed in near the lower ends of the spectra. As indicated in the left graphs, the
respective spectra contain the typical frequency ranges shown in Table 3.1. The spectrum
of the “sa” sound contains both the vowel and unvoiced consonant frequency ranges.
On the right graphs, the “ma” sound indicates somewhat higher peaks near 200 Hz and
400 Hz than the “a” sound, indicating that the “ma” sound is the superposition of the vowel
“a” and the voiced constant “m”. As compared with the spectra of “a” and “ma” sounds, the
“sa” sound has more high-frequency components, which is obviously due to the inclusion
of the unvoiced consonant “s”.
Figure 3.8 is another sample speech signal in the time domain (upper) and the Fourier
spectrum (lower). A female speaker (left) and a male speaker (right) pronounce “chocolate”.
3.2 Sound We Hear 75

Fig. 3.7 Fourier spectra of “a” (vowel), “ma” (voiced consonant plus vowel), and “sa” (unvoiced
consonant plus vowel)

The three envelopes observed in the time-domain signals seem to represent the three syllables
“cho”, “co”, and “late”. The female speaker’s spectrum shows higher amplitude than the
male spectrum in the frequency range of 250 Hz–1000 Hz. This is because females have
higher fundamental voice frequency than males. The frequency ranges marked . f 0 indicate
the typical fundamental voice frequency of 165 Hz–255 Hz (female) and 85 Hz–180 Hz
(male) [11, 12]. The spectral peaks in the further lower range seem to represent syllabic and
fluctuation characteristics of the speech [13].
The spectral shape in the frequency range above the fundamental voice frequency results
from the resonance and filtering characteristics of the vocal tract [14] (the area from the

Fig.3.8 Time (upper) and frequency (lower) domain signals of sound “chocolate” spoken by a female
(left) and male (right). . f 0 indicates typical fundamental voice frequency
76 3 Propagation of Acoustic Waves in Air

nose and the nasal cavity down to the vocal cords) [15]. Since the filtering characteristics
depend on the shape and size of the vocal tract [16], the same sound spoken by different
people with the same fundamental voice frequency sounds differently.
Figure 3.9 zooms in the three enveloped sections of the “chocolate” sound by the female
shown in Fig. 3.8 (left). Here the top row shows the time domain signals of the sections, and
the bottom row plots the Fourier spectrum in a frequency range where most of the peaks
appear in the respective envelopes. These three envelopes correspond to the sound “cho”,
“co”, and “late”. The typical female’s fundamental voice frequency (165 Hz–255 Hz) is
indicated for the three envelopes. It is interesting to note that in the spectra of the “cho” and
“late” sounds, the fundamental frequency range covers prominent peaks. In the case of “co”
sound, higher peaks appear in the frequency range lower than the fundamental frequency.
The spectral shape in this low-frequency range seems to represent the syllabic and fluctuation
characteristics of the speaker. The time domain signal clearly indicates that when the speaker
pronounces “co”, the amplitude fluctuates slower than the other two envelopes.
Figure 3.10 is the male speaker’s version of Fig. 3.9. Similar to the female case, the
fundamental voice frequency covers prominent peaks in the Fourier spectrum. Also, the
sound “co” has high peaks in the frequency range lower than the fundamental frequency. A
comparison of the spectra of “co” and “late” sounds indicates that the energy of the latter

0
0

Fig. 3.9 Time (top) and frequency (bottom) domain signals of three envelopes of sound “chocolate”
pronounced by a female speaker. . f 0 indicates typical female’s fundamental voice frequency
3.2 Sound We Hear 77

0
0

Fig. 3.10 Time (top) and frequency (bottom) domain signals of three envelopes of sound “chocolate”
pronounced by a male speaker. . f 0 indicates typical male’s fundamental voice frequency

is slightly weighted on the higher frequency side. It is possible that this observation can be
explained by the fact that the fundamental frequency of “a” is higher than “o” [17].

Sample sound of music

Figure 3.11 shows the time domain (the left column) and frequency spectrum (middle and
right columns) of c2 sound generated by an electric base (upper row), and an acoustic piano
(lower row). The middle and right columns present the same frequency spectrum at different
frequency ranges.
The frequency of the c2 sound is 65.4 Hz [18]. The right column of Fig. 3.11 indicates
that both musical instruments show a spectral peak at 65.4 Hz. At the same time, they
show overtones [19] (harmonics). While the bass shows a higher peak at the fundamental
frequency of 65.4 Hz, the acoustic piano indicates that the peak at the second harmonic at
65.4 .× 2=130.8 Hz is higher than the fundamental.
The pattern of the overtone spectrum is unique for each musical instrument, and it deter-
mines the characteristic sound. The mechanism of sound generation determines the funda-
mental frequency. For instance, if you place your finger at different positions on the string
78 3 Propagation of Acoustic Waves in Air

Fig. 3.11 Comparison of sound from a Bass and b Acoustic piano. Each instrument is played at
a pitch corresponding to the piano c2 key. Time domain (left), Frequency domain (middle), and
zoomed-in Frequency domain (right) signals

of a guitar, the pitch [20] (frequency) changes. However, it is the body of the instrument that
creates the timbre. Compare the piano and guitar. Both are string instruments, and the length
of the string determines the fundamental frequency. Their distinctively different bodies make
their unique timbre. Figure 3.11 illustrates this situation literally.
The generation of overtones results from the resonance of the standing wave. From this
standpoint, we can say that musical instruments are resonators. Remember the discussion
under “standing wave” in Sect. 1.2.3. The acoustic wave generated by the sound-generation
mechanism goes back and forth inside the instrument, making standing waves. Here, its shape
(the structure) determines the characteristics of the standing wave. Some generate higher
frequency components with a high decay constant, whereas others have sound lasting longer.
Attack, decay, and sustain are the terminologies that characterize the temporal characteristics
of musical sound. The attack represents the rising time to the maximum amplitude, the decay
describes how quickly the sound fades out, and the sustain indicates how long the sound
stays at a constant amplitude. (In Fig. 3.11, for example, the bass exhibits a longer decay
and shorter sustain than the piano).
If instruments have similar shapes, their length determines the fundamental frequency.
For instance, the sound of the flute is lower than the piccolo, and the violin produces a
higher pitch than the viola. The resonant wavelength condition 1.70 precisely describes the
situation. The longer the resonator length, the longer the fundamental wavelength, hence
lower the frequency.

3.3 Transmission of Audible Sound Waves

Sending an acoustic signal such as a speech over a long distance in the form of acoustic
waves is impractical. The speaking person needs to speak very loudly and the listener must
have a very good year even if they are apart by 100 m or so. It is more efficient to convert
3.3 Transmission of Audible Sound Waves 79

the sound signal into an electric signal, transmit the signal through air and receive it with an
antenna.
The size of the antenna is proportional to the wavelength, i.e., inversely proportional to
the frequency of the receiving electromagnetic wave. The optimal length for any antenna
is around 1/4 of the wavelength it is receiving. This type of antenna is known as a quarter
antenna [21]. The audible frequency to humans is 20 Hz to 20 kHz, which is in wavelength
of 1.5.×104 m to 1.5.×107 m. Building an antenna of a quarter of these lengths in size is not
practical.
This is one of the purposes of sending sound information on a carrier frequency that
is high enough to make a practical antenna. For instance, if the carrier frequency is 500
MHz, the wavelength becomes .3 × 108 (m/s) .÷ .5 × 106 (1/s) = 60 (cm), i.e., its quarter
is 15 (cm). In this scheme, the sound information, the signal, is coded in the transmission
signal in the form of modulation. Depending on how to modulate, the modulation method is
referred to as amplitude modulation, phase modulation, and frequency modulation. In this
section, we consider amplitude modulation [22] and frequency modulation [22], along with
demodulations [23].

Amplitude modulation

Consider that we want to transmit signal . S(t) as an amplitude-modulated signal with carrier
frequency . f c . We can perform this operation by multiplying the signal function . S(t) to the
carrier (a sinusoidal function oscillating at the carrier frequency). We assume that . S(t) has
been converted from an acoustic wave to an electric signal. We will discuss this sound-to-
electric signal conversion in Chap. 5.

. Se (t) = S(t) cos ωc t (3.30)

Here .ωc = 2π f c is the angular frequency of the carrier.

Below we continue the discussion by replacing . S(t) with . A cos ωm t, where .ωm is the
angular frequency of the modulation signal. In other words, we simplify the problem by
assuming that the signal is a cosine function with a single, constant frequency. The use
of . A cos ωm t for the signal . S(t) does not lose generality because according to Fourier’s
theorem, we can express . S(t) with a series of cosine and sine functions. We can view
. A cos ωm t as a term of the Fourier series. Note that a term of the sine series can be expressed

by the cosine counterpart of the same component frequency by shifting the phase by .π/2
(.sin ωm t = cos(ωm t − π/2)). In the process of modulation and demodulation discussed
below, we can repeat the same procedure for other terms in the Fourier series as the .m th
term.
Let .sam (t) be the product of . A cos ωm , and .cos ωc t.

sam (t) = A cos ωm t × cos ωc t

. (3.31)
80 3 Propagation of Acoustic Waves in Air

Using the following mathematical identity, putting .α = ωc t and .β = ωm t

1
. cos α cos β = {cos(α + β) + cos(α − β)} (3.32)
2
we can express .sam (t) in the following form.
A
.sam (t) = {cos(ωc + ωm )t + cos(ωc − ωm )t} (3.33)
2
Here we assume that the .ωc >> ωm , which is the condition generally used for amplitude
modulation. The necessity of this condition will be clarified shortly when we discuss ampli-
tude demodulation.
Figure 3.12 illustrates the signal . A cos ωm t, the carrier .cos ωc t, and the modulated signal
.sam (t). The left column shows the carrier, amplitude-modulated signal, and original signal

along with demodulated signal (from top to bottom). The right column shows the Fourier
spectrum of the time domain signal of the same row. The carrier’s frequency is 500 Hz (the
period is 0.002 ms) and the frequency of the signal is 200 Hz (period of 5 ms), as Fig. 3.12
indicates. Notice that the carrier is not visible in the amplitude-modulated signal in the time
domain as the carrier frequency is beyond the resolution on the scale of the signal. The
Fourier spectrum of the amplitude-modulated signal exhibits the carrier frequency of 500

Fig. 3.12 Signals in time (left ) and frequency (right) domains. a carrier .cos ωc t, b modulated signal
.sam (t),
and c signal . A cos ωm t
3.3 Transmission of Audible Sound Waves 81

Fig. 3.13 a . M = 1.0, b . M = 0.5, and c . M = 0.2

kHz at the center and two sidebands at 500 kHz .± 200 Hz (the signal frequency). These
sidebands correspond to the two terms on the right-hand side of (3.33).
In the above case, the signal is fully modulated. It is possible to modulate partially using
the modulation depth . M.

.
M
sam (t) = (1 + M A cos ωm t) × cos ωc t (3.34)

Figure 3.13 illustrates the time series of modulated signals with different modulation
depths. Notice that as the modulation depth decreases, the amplitude modulation becomes
less prominent in the time-domain signals (the left column), and the height of the sidebands
decreases in the frequency domain.

Amplitude demodulation

We need to retrieve the signal . A cos ωt on the receiver side. There are multiple methods
to demodulate amplitude-modulated signals [24]. The following operation of demodulation
consisting of signal mixing (multiplication) and low-pass filtering is one of them.
Consider multiplying the amplitude-modulated signal .sam (t) and the carrier .cos ωc t.
A
.sam (t) cos ωc t {cos(ωc + ωm )t + cos(ωc − ωm )t} cos ωc t
=
2
A[1 1 ]
= {cos(2ωc t + ωm t) + cos(ωm t)} + {cos(ωm t) + cos(2ωc t − ωm t)}
2 2 2
A[1 1 ] A
→ cos(ωm t) + cos(ωm t) = cos ωm t (3.35)
LF 2 2 2 2
82 3 Propagation of Acoustic Waves in Air

Here . L F under the arrow stands for “low-pass filtering”, which eliminates the high-
frequency terms .cos(2ωc t ± ωm ) from the second line of (3.35).

Frequency modulation

Frequency modulation is another common method to transmit audio signals. In this case, the
frequency of the carrier is modulated by the audio signal. Let . Ac and .ωc be the amplitude
and frequency of the carrier and .s(t) be the audio signal. We can express the frequency-
modulated signal .s f m (t) as follows.
( ∫ )
.s f m (t) = A c cos ωc t + s(τ )dτ (3.36)

For the purpose of understanding the principle of frequency modulation and demodula-
tion, consider a simple case where the audio signal is a cosine function.

s(t) = am cos(ωm t)
. (3.37)

In this case, the frequency-modulated signal can be expressed as follows.

( ∫ ) ( )
am
.s f m (t) = A c cos ωc t + am cos ωm tdτ = Ac cos ωc t + sin ωm t
ωm
= Ac cos (ωc t + β sin ωm t) (3.38)

Here, .β is called the modulation depth.

am
β=
. (3.39)
ωm
Figure 3.14 shows a sample of frequency-modulated signal in the time and frequency
domains. Here the carrier frequency is 500 Hz and the audio signal is a cosine function
of frequency 100 Hz. Figure 3.14a1, a2 indicate that the carrier oscillates at the single
frequency of 500 Hz. On the other hand, Fig. 3.14b1 reveals that the frequency-modulated
signal changes the oscillatory pattern with a periodicity of 10 ms; every 10 ms, the frequency
alternates between the high (dense) and low (sparse) oscillations. The periodicity of 10 ms
is equivalent to the modulation frequency of 1/(10.× 10.−3 ) = 100 Hz.
Figure 3.14b2 indicates that the Fourier spectrum of the frequency-modulated signal has
the central peak at the carrier frequency and the side lobes at .± 100 Hz and .± 200 Hz
from the central peak. In principle, these sidebands appear at integer multiples of the signal
frequency (in this case 100 Hz) away from the carrier frequency on both sides. The greater
the modulation depth, the higher the sidelobe peaks towards the far ends of the spectrum.
3.3 Transmission of Audible Sound Waves 83

Fig. 3.14 Frequency modulation time signal (left) and Fourier spectrum (right). The top graphs are
for the carrier and the bottom graphs are for the frequency-modulated signal

Frequency demodulation

Multiple methods are available to retrieve the audio signal on the receiver side by demod-
ulating the frequency-modulated signal [25]. Here, we discuss one of them in which we
retrieve the signal (modulation) frequency as a pattern appears as amplitude modulation. An
envelope detection technique [26] can be used to retrieve the signal frequency.
Differentiate .s f m (t) with respect to time.
( ∫ )
d ( )
. s f m (t) = (−Ac ) [ωc + s(t)] sin ωc t + s(τ )dτ (3.40)
dt

In the case of the simple audio signal (3.37), the result of the differentiation takes the
following form.
d ( )
. s f m (t) = (−Ac ) [ωc + βωm cos ωm t] sin (ωc t + β sin ωm t)
dt
= (−Ac ) [ωc + am cos ωm t] sin (ωc t + β sin ωm t) (3.41)

In going to the second line in (3.41), we used (3.39).

We can view (3.41) as the .sin (ωc t + β sin ωm t) is amplitude modulated by the term
.[ωc + am cos ωm t]. In other words, this sine function’s amplitude oscillates in the range of

.ωc ± am .

Figure 3.15a, b compare the time-differentiated and the original frequency-modulated

signals on the same time axis. Notice that when the frequency-modulated signal has a
high frequency (where the oscillation is dense), the amplitude of the time-differentiated
84 3 Propagation of Acoustic Waves in Air

Fig. 3.15 a Time derivative of the frequency-modulated signal. b The frequency-modulated signal,
c Fourier spectrum of a

signal has a higher amplitude. This observation is something we can easily understand
by examining (3.41); the frequency of the modulated signal becomes highest when .t in
.sin(ωc t + β sin ωm t) makes .sin ωm t swing the largest. With the same .t, the cosine term in

.ωc + am cos ωm t swings the largest from the constant value .ωc .

References

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Propagation of Acoustic Waves in Solids
4

In this chapter, we aim to discuss the propagation of sound waves in solids, clarifying the
dynamical difference in elastic behaviors between air and solids. After briefly reviewing the
theory of elasticity, we derive the equation of motion for isotropic solid media. We derive
the wave equations and discuss the propagation of acoustic waves in solids under various
boundary conditions. The discussions include some nonlinear waves, such as surface waves.

4.1 Acoustic Wave as Solutions to the Equation of Motion

Sound propagating through a solid is an elastic wave. The elasticity of solids causes oscil-
latory motion of the lattice and the oscillation travels as a wave. In Chap. 2, we derived
longitudinal and transverse wave equations from the equation of motion. Here we combine
the longitudinal and transverse effects (i.e., the normal and shear forces) and make a similar
discussion to Chap. 2. We can start with the equation of motion for a unit block in an elastic
medium. We restrict our discussions here to isotropic materials.

4.1.1 Equation of Motion

The equation of motion for a unit volume of an isotropic, elastic medium can be put as
follows. (See Appendix E for the derivation of this equation.)

∂ 2ξ
ρ
. = μ∇ 2 ξ + (λ + μ)∇(∇ · ξ ) (4.1)
∂t 2
Here .λ and .μ are Lamé’s first and second parameters.

© The Author(s), under exclusive license to Springer Nature Switzerland AG 2024 87

S. Yoshida, Fundamentals of Acoustic Waves and Applications, Synthesis Lectures on Wave
Phenomena in the Physical Sciences, https://doi.org/10.1007/978-3-031-48200-7_4
88 4 Propagation of Acoustic Waves in Solids

Equation (4.1) leads to the equations of compression wave and rotation wave as follows.
Take the divergence of (4.1) switching the order of .∇· and .∇ 2 operation for the first term on
the right-hand side and using the identity .∇ · ∇ = ∇ 2 for the second term. These operations
yield the following wave equation of .∇ · ξ (the compression wave equation).

∂ 2 (∇ · ξ ) (λ + 2μ) 2
. = ∇ (∇ · ξ ) (4.2)
∂t 2 ρ
Equation (4.2) indicates that an elastic compression wave travels at the following phase
velocity. /
comp λ + 2μ
.v = (4.3)
ph
ρ
Similarly, taking the curl of (4.1), we obtain the following equation of rotation wave.

∂ 2ω μ
. = ∇2ω (4.4)
∂t 2 ρ

Here, rotation . ω = 21 (∇ × ξ ). (4.4) indicates that an elastic rotation (deformation) wave

travels at the following phase velocity.
/
μ
.v ph =
r ot
(4.5)
ρ
comp
Since .v ph > vrphot , the compression wave is called the Primary Wave (P-wave) and the
rotation wave is called the Secondary Wave (S-wave).

Scalar and vector potentials

It is convenient to express the displacement vector using two potentials [1, 2].

.ξ = ∇φ + ∇ × H (4.6)
.∇ ·H=0 (4.7)

Here .φ and .H are called the scalar potential and vector potential of elastic dynamics. Equa-
tions (4.6) and (4.7) are known as Helmholtz decomposition [3]. As will be clarified shortly,
.φ represents compression (spring) potential energy and .H the rotational potential energy of

elasticity.
Substitution of (4.6) into (4.1) yields the following equation.
4.1 Acoustic Wave as Solutions to the Equation of Motion 89

( )
∂ 2 (∇φ) ∂ 2 (∇ × H)
ρ
. +
∂t 2 ∂t 2
= (λ + μ)∇(∇ · ∇φ) + μ∇ 2 (∇φ)
+ (λ + μ)∇(∇ · (∇ × H)) + μ∇ 2 (∇ × H)
{ }
= ∇ (λ + μ)∇ 2 φ + μ∇ 2 φ
{ }
+ ∇ × (λ + μ)∇(∇ · H) + μ∇ 2 H
{ } { }
= ∇ (λ + 2μ)∇ 2 φ + ∇ × μ∇ 2 H (4.8)

Here, in the last step of (4.8) the two terms in the first curly bracket are combined and (4.7)
is used in the second curly bracket on the right-hand side.
For (4.8) to hold arbitrarily, it follows that the following conditions hold.

∂ 2φ (λ + 2μ) 2
. = ∇ φ (4.9)
∂t 2 ρ
∂ 2H μ
. = ∇2H (4.10)
∂t 2 ρ
Equations (4.9) and (4.10) are wave equations for the two potentials indicating that .φ and
H travel at the following velocity.
.

/
φ λ + 2μ
.v
ph = (4.11)
ρ
/
μ
.v ph =
H
(4.12)
ρ

Notice that wave velocities (4.11) and (4.12) are identical to wave velocities (4.3) and (4.5),
respectively. These identities have physical meanings as will be discussed in the following
section. The .φ wave and .H wave are also called the Primary wave and Secondary wave,
respectively.

Physical meanings of φ and H

.

Split (4.6) into two terms.

.ξ φ + ξ H = ∇φ + ∇ × H (4.13)
where

. ξ φ = ∇φ (4.14)
ξH = ∇ × H
. (4.15)

Equation (4.14) indicates that .ξ φ results from the differential elastic energy, as the fol-
lowing argument explains. Dimensional analysis tells us.φ is in [m.2 ]. If we multiply stiffness
90 4 Propagation of Acoustic Waves in Solids

Fig. 4.1 The product of

stiffness K and vector H
represents torque associated
with the rotational wave.
.∇ × H represents the
z-component of the
displacement due to all torques
along the circumference around
the reference point

K [kg/s.2 ], the dimension of .φ becomes energy as [m.2 kg/s.2 ] .= [kg m/s.2 m] .= [Nm] .= [J].
.

Knowing that the dimension of stiffness . K is in [N/m] .= [kg/s.2 ] and the elastic energy
is . 21 K ξ 2 , we can interpret that .φ is the elastic energy normalized to the unit of [m.2 ] (by
dividing it by the stiffness . K kg/s.2 ).
Dimensional analysis on (4.15) indicates that potential .H is in [m.2 ], because .ξ H is in m
and .∇× is in 1/m. By multiplying the factor . K defined above for the other potential .φ, we
find the dimension of . K H to be [kg m/s.2 m]=[N m]. Being associated with the rotational
(deformation) wave via (4.10), the dimension [N m] can be interpreted as representing
torque.
Material-scientifically, we can interpret the above torque mechanism due to .H and its
association with .ξ H as follows. Figure 4.1 schematically illustrates the situation where a
reference point in an elastic material is surrounded by four infinitesimal segments. The four
segments are on the circumference of an imaginary circle set around the reference point.
The plane of the circle is parallel to the .x y plane and each segment is represented by a
disc perpendicular to the circumference. The external force, . f , generates torque on each
segment as illustrated in Fig. 4.1. Here the torque vector is in the tangential direction of the
imaginary circle, and the torque on each disc is clockwise when viewed from the tail of the
torque vector. Each torque vector .τ is defined by the local position vector .r representing
the radius of the disc and the external force vector, and related to .H with the factor . K as
follows.
.τ = r × f = K H (4.16)
The torques from the four segments all together generate upward displacement at the refer-
ence point and downward displacement outside the imaginary circle. Thus, we can interpret
(4.15) as representing the displacement associated with the rotational (torque) energy stored
in the material surrounding the point of interest.
The above interpretations indicate that the .φ-wave (P-wave) and .H-wave (S-wave) are
generated by different mechanisms and therefore they propagate at mutually different wave
velocities. At the same time, however, both wave-generation mechanisms are due to the
elasticity of the material and therefore they can be converted mutually (see Sect. 4.2.2 under
4.2 Propagation of Acoustic Waves in Solid Under Various Conditions 91

Reflection at the free surface with various incident waves). In the following sections, we
discuss the dynamics of these waves.

4.2 Propagation of Acoustic Waves in Solid Under Various

Conditions

Equations (4.2) and (4.4) indicate that an acoustic wave propagates through an elastic
medium in two forms. The first is the compression wave (P-wave) that is given as a solution
to (4.2) and travels at the wave velocity given by (4.3). The second is the rotational wave
(S-wave) that is given as a solution to (4.4) and travels at the wave velocity given by (4.5).
In this section, we discuss the propagation of these two types of waves in various scenarios.

4.2.1 Plane Waves in an Elastic Medium

In general, a wave solution can be expressed as the product of functions that represent
amplitude and phase, respectively. In a Cartesian coordinate system, let .A(x, y, z) be the
amplitude function and . f be the phase function of a displacement wave .ξ .

.ξ (t, x, y, z) = A(x, y, z) f (ωt − k · r) = A(x, y, z) f (ωt − (k x x + k y y + k z z)) (4.17)

Here .ω is the frequency, .k is the propagation vector and .r is the position vector of coordinate
point .(x.y.z).
When the amplitude .A is a constant, independent of the spatial coordinates, the value of
.ξ is solely determined by the value of the phase function . f . When the spatial dependence

of the phase function is linear, as is the case of (4.17), the three-dimensional space that
shares the same phase (called the wavefront) is planar. If the amplitude is constant and the
wavefront is planar, the wave is called a plane wave.
A plane wave is an idealized solution to a wave equation. Due to diffraction, a wave gen-
erally expands transversely to the direction of propagation. At each point of the wavefront,
the propagation vector is perpendicular to the wavefront. Therefore, if a wave diverges or
converges, the wavefront is curved (curved inward for a divergent wave and outward for
a convergent wave when viewed behind the wavefront). So, a planar wavefront does not
exhibit a diffracting behavior. However, under various conditions, a plane wave solution is
a good approximation (e.g., the light wave from the sun on the Earth). In addition, probably
more importantly, the analysis of plane wave solutions is very informative and helpful to
understand the propagation of acoustic waves.
92 4 Propagation of Acoustic Waves in Solids

Propagation vector

Above, I used the word “propagation vector” without explaining it. Since this is an important
concept in the following sections, take some time to consider it. The phase function varies
with time and space as .ωt − k · r. If the phase function is sinusoidal, for instance, it varies
in the range .−1 < f < 1 as the phase varies with the period of .2π . Since the phase depends
on both time and space, a temporal or spatial change can alter the phase. If the phase change
is an even integral multiple of .π the phase function . f takes the same value, and if the change
is an odd integral multiple of .π , the function takes the same absolute value with the opposite
sign, and so on.
We can understand the concept of the wavefront most easily by freezing the function with
time. We can fix the time at .t = 0 without losing the generality of the argument. With .t = 0,
the phase (argument) of the function . f is .−k · r = −(k x x + k y y + k z z), and the equation
of the plane of a constant phase, .θ0 , is given as follows.

k x x + k y y + k z z = θ0
. (4.18)

Figure 4.2a illustrates this plane. Consider two points in the plane, .(x1 , y1 , z 1 ) and
.(x 2 , y2 , z 2 ). Let .r1 and .r2 represent the position vectors of these two points, and .n a unit

vector normal to this plane.

By definition, the following equations hold.

. k x x1 + k y y1 + k z z 1 = θ0 (4.19)
k x x2 + k y y2 + k z z 2 = θ0
. (4.20)

Using the coordinates expression, we can write .r2 − r1 as follows.

r2 − r1 = (x2 − x1 , y2 − y1 , z 2 − z 1 )
. (4.21)

Using expression (4.21), make the scalar product of vector .r2 − r1 and propagation vector
k = (k x , k y , k z ).
.

.(r2 − r1 ) · k = (x2 − x1 )k x + (y2 − y1 )k y + (z 2 − z 1 )k z

= (k x x2 + k y y2 + k z z 2 ) − (k x x1 + k y y1 + k z z 1 )
= θ0 − θ0 = 0 (4.22)

Equations (4.19) and (4.20) were used in the last step of (4.22). Since vector .r2 − r1 is in
this plane and we can choose the second point .(x2 , y2 , z 2 ) arbitrarily in the plane fixing
the first point .(x1 , y1 , z 1 ), .(r2 − r1 ) · k = 0 derived by (4.22) tells us that the propagation
vector is normal to this plane, i.e., it is parallel the normal vector .n as .k = kn.
Now consider another plane of constant phase .θ0 + 2π . Obviously, this plane is parallel
to the plane of constant phase .θ0 . Therefore, the propagation vector .k is also normal to
4.2 Propagation of Acoustic Waves in Solid Under Various Conditions 93

Fig. 4.2 Wave front of a plane wave

this plane. We can repeat the same argument by adding a constant phase, and say that the
propagation vector is always normal to a plane of constant phase. Figure 4.2b illustrates the
situation.
The fact that the propagation vector of a wave is normal to the wavefront is true even if the
wavefront is curved. At a given point on the wavefront, we can consider an infinitesimal plane
(the plane tangential to the curved wavefront) and apply the above argument. In conclusion,
the propagation vector is always normal to a wavefront.
Now we go back to the discussion of plane wave propagation. Naturally, a plane wave
solution can be found by solving the equation of motion (4.1). For simplicity, we solve
the equation of motion for a two-dimensional case. The same arguments hold for three-
dimensional cases. Define the vectors as follows.

ξ (t, x, y) = ξx (x, y)î + ξ y (x, y) ĵ

. (4.23)
A = A x î + A y ĵ
. (4.24)
k = k x î + k y ĵ
. (4.25)
r = x î + y ĵ
. (4.26)

Substitution of (4.24)–(4.26) into (4.17) yields the following expressions.

.ξx (t, x, y) = A x f (ωt − k x x − k y y) (4.27)

.ξ y (t, x, y) = A y f (ωt − k x x − k y y) (4.28)

Since . A x and . A y are constant, we can differential the phase function to find the derivatives
of .ξx and .ξ y .
94 4 Propagation of Acoustic Waves in Solids

∂f ∂
. = (ωt − k x x − k y y) = (−k x ) f ' (4.29)
∂x ∂x
∂f ∂
. = (ωt − k x x − k y y) = (−k y ) f ' (4.30)
∂y ∂y

Here . f ' represents the derivative of function . f .

Thus, we obtain the following equations.
∂ξx ∂f ∂ξx ∂f
. = Ax = A x (−k x ) f ' , = Ax = A x (−k y ) f ' (4.31)
∂x ∂x ∂y ∂x
∂ξ y ∂f ∂ξ y ∂f
. = Ay = A y (−k x ) f ' , = Ay = A y (−k y ) f ' (4.32)
∂x ∂x ∂y ∂y
∂ξx ∂f ∂ξ y ∂f
. = Ax = Ax ω f ', = Ay = Ay ω f ' (4.33)
∂t ∂t ∂t ∂t
Similarly, the second-order derivatives are given as follows.

∂ 2 ξx ∂2 f ∂ 2 ξx ∂2 f
. = A x 2 = A x (−k x )2 f '' , = A x 2 = A x (−k y )2 f '' (4.34)
∂x 2 ∂x ∂y 2 ∂y
∂ ξx
2 ∂ f
2
. = Ax = A x (−k x )(−k y ) f '' (4.35)
∂ x∂ y ∂ x∂ y
∂ 2ξy ∂2 f 2 '' ∂ ξ y
2 ∂2 f
. = A y = A y (−k x ) f , = A y = A y (−k y )2 f '' (4.36)
∂x2 ∂x2 ∂ y2 ∂ y2
∂ 2ξy ∂2 f
. = Ay = A y (−k x )(−k y ) f '' (4.37)
∂ x∂ y ∂ x∂ y
∂ 2 ξx ∂ ∂ξx 2 '' ∂ 2ξy ∂ ∂ξx
. = A x = A x ω f , = Ay = A y ω2 f '' (4.38)
∂t 2 ∂t ∂t ∂t 2 ∂t ∂t
With (4.34)–(4.38), the derivative terms of (4.1) become as follows.
( ) ( )
∂ ∂ξx ∂ξ y ∂ ∂ξx ∂ξ y
.∇ (∇ · ξ ) = + î + + ĵ
∂x ∂x ∂y ∂ y ∂x ∂y
{} ( ) ( }
= A x k x2 + A y k x k y î + A x k x k y + A y k 2y ĵ f '' (4.39)
( 2 ) ( 2 )
∂ ξx ∂ 2 ξx ∂ ξy ∂ 2ξy
.∇ ξ = + + +
2
î î
∂x2 ∂ y2 ∂x2 ∂ y2
{) ( ) ( }
= A x k x2 + A x k 2y î + A y k x2 + A y k 2y ĵ f '' (4.40)
∂ 2ξ ∂ 2ξ ∂ 2ξ { }
f ''
x y
. = î + ĵ = A x ω2 î + A y ω2 ĵ (4.41)
∂t 2 ∂t 2 ∂t 2
Substitution of (4.39)–(4.41) into the equation of motion (4.1) yields the following system
of equations for . A x and . A y .
4.2 Propagation of Acoustic Waves in Solid Under Various Conditions 95

[ ]
. (λ + μ)k x2 + μ(k x2 + k 2y ) − ρω2 A x + (λ + μ)k x k y A y = 0 (4.42)
[ ]
.(λ + μ)k x k y A x + (λ + μ)k y + μ(k x + k y ) − ρω Ay = 0
2 2 2 2
(4.43)

Here, the common term . f '' is eliminated. Equations (4.42) and (4.43) can be expressed in
the matrix form as follows.
( )
Ax
. (M) =0
Ay
( )
(λ + μ)k x2 + μ(k x2 + k 2y ) − ρω2 (λ + μ)k x k y
(M) = (4.44)
(λ + μ)k x k y (λ + μ)k 2y + μ(k x2 + k 2y ) − ρω2

In order for (4.42) and (4.43) to yield non-trivial solutions, the determinant of matrix . M
must be null. Thus, we obtain the following equations.
[ ][ ]
.|M| = (λ + μ)k x + μ(k x + k y ) − ρω (λ + μ)k 2y + μ(k x2 + k 2y ) − ρω2
2 2 2 2

− (λ + μ)2 k x2 k 2y
[ ]2 [ ]
= μ(k x2 + k 2y ) − ρω2 + μ(k x2 + k 2y ) − ρω2 (λ + μ)(k x2 + k 2y )
+ (λ + μ)2 k x2 k 2y − (λ + μ)2 k x2 k 2y
[ ] {[ ] }
= μ(k x2 + k 2y ) − ρω2 μ(k x2 + k 2y ) − ρω2 + (λ + μ)(k x2 + k 2y )
[ ][ ]
= μ(k x2 + k 2y ) − ρω2 (λ + 2μ)(k x2 + k 2y ) − ρω2 = 0 (4.45)

Equation (4.45) indicates that for displacement .ξ in the form of (4.23) to be a solution to
the differential equation (4.1), one of the following conditions must hold.

. (λ + 2μ)(k x2 + k 2y ) − ρω2 = 0 (4.46)

.μ(k x + k 2y ) − ρω2 =0
2
(4.47)

Since .ω/k is the wave velocity, (4.46) and (4.47) can be interpreted as representing the
following wave velocities.
/
ω λ + 2μ
.c1 = / = (4.48)
(k 2 + k 2 ) ρ
x y
/
ω μ
.c2 = / = (4.49)
(k x2 + k 2y ) ρ

Wave velocities (4.48) and (4.49) are identical to those of the compression wave (P-wave)
(4.3) and rotation wave (S-wave) (4.5), respectively. Equations (4.48) and (4.49) indicate
96 4 Propagation of Acoustic Waves in Solids

that the displacement wave travels in an isotropic elastic medium as a P-wave or S-wave.
In the following sections, we find that these two types of plane waves can coexist in elastic
media and can be converted from one type to the other.
Another feature I would like to point out is that neither velocity (4.48) or (4.49) depends
on the frequency. They simply depend on the material constants .λ, .μ and .ρ. This means that
an acoustic wave propagating through an isotropic media does not exhibit dispersion. Later
in this chapter, we observe that acoustic waves can exhibit dispersion in an isotropic media
when certain boundary conditions are imposed. As we will see in the next section, a free
surface is one of such boundary conditions that cause acoustic waves to exhibit dispersive
behaviors. This type of dispersion can be used in characterizing structural and mechanical
properties of solid objects and is therefore important in various acoustic techniques of
nondestructive evaluation [10].

4.2.2 Plane Waves in a Medium with Free Surface

Consider a plane acoustic wave travels in an isotropic elastic medium towards its infinitely
large free surface in Fig. 4.3. Here we set the free surface on the .zx plane and use the word
“free” to mean that the external force . f acting on the surface exerted by the surrounding
medium (e.g., air) is null. By assuming that the external force is null at . y = 0 for all .(x, z)
coordinate points, we can express this free surface condition as the traction vector .T = f/d S
is null as follows.
.T = σ yx î + σ yy ĵ + σ yz k̂ = 0 (4.50)
Here .σlm represents the .m component of the stress vector on a plane .l. For condition (4.50)
to hold at all .(x, 0, z) coordinate points, it follows that

σ yx = σ yy = σ yz = 0
. (4.51)

Figure 4.4 schematically illustrates the situation where a plane wave is incident to the free
surface at an angle of incidence .θ1 . Here .k is the propagation vector where .k x and .k y are its
. x and . y components. Under these conditions, derive wave equations for the potential .φ and

.H. Assume that the angle of incidence to the free surface is .θ1 and that plane of incidence

is parallel to the .x y plane. This assumption does not lose the generality of the argument. In
the below discussion, we refer to planes parallel to the .zx-plane as a horizontal plane and
planes parallel to the .x y plane as a vertical plane (Fig. 4.3).
From the plane wave condition, we can set .∂/∂z = 0. (Since the plane wavefront is
parallel to the .z axis the displacement components are independent of .z.) Thus, according
to (4.6), we can relate the displacement components to the potentials as follows.
4.2 Propagation of Acoustic Waves in Solid Under Various Conditions 97

Fig. 4.3 Acoustic wave travels towards the free surface at an incident angle of .θ1

Fig. 4.4 Plane wave incidence

to a free surface. .k is the
propagation vector

∂φ ∂ Hz
ξx = (∇φ)x + (∇ × H)x =
. + (4.52)
∂x ∂y
∂φ ∂ Hz
.ξ y = (∇φ) y + (∇ × H) y = − (4.53)
∂y ∂x
∂ Hy ∂ Hx
.ξz = (∇φ)z + (∇ × H)z = − (4.54)
∂x ∂y
We are now in a position to solve wave equations for plane wave solutions of .φ and .H.
Remembering the discussion made in Sect. 4.1, express the wave equations as follows.

∂ 2φ
. = c12 ∇ 2 φ (4.55)
∂t 2
∂ 2H
. = c22 ∇ 2 H (4.56)
∂t 2
98 4 Propagation of Acoustic Waves in Solids

Fig. 4.5 .φ and SV-wave incidence to a free surface. .k is the propagation vector

As observed in the above sections, the acoustic wave can take two modes (the P-wave and
S-wave). (4.52)–(4.54) indicate that .ξx and .ξ y components depend on .φ and . Hz , whereas .ξz
component depends on . Hx and . Hy and independent of .φ. In this situation, it is convenient to
discuss the wave dynamics for the following two types; (1) the wave that involves .ξx and .ξ y
and (2) the wave that involves .ξz . We call the first type (1) the plane strain wave because the
displacement components lie in the plane of incidence. (The plane strain is the state where
the non-zero strain components act in one plane only.) From (4.52) and (4.53) we know that
this wave is a mixture of a P-wave and S-wave because it has the .φ and .H components.
(The .φ wave is a P-wave, and the .H wave is an S-wave, as we discussed in the paragraphs
near (4.16).) Here, the S-wave consists of the displacement vector components in the plane
of incidence, above referred to as a vertical plane. So, we call the S-wave that constitutes
the plane strain wave the Shear Vertical (SV) wave. Figure 4.5 illustrates a P-wave (a) and
SV-wave (b) that constitute a plane strain wave.
We call the second type (2) the Shear Horizontal (SH-wave) because it is a shear wave
whose oscillation is polarized in the direction parallel to the plane of the free surface, above
referred to as a horizontal plane. The SH wave is an S-wave. Figure 4.6 illustrates an SH-
wave traveling along .xs axis.

Plane strain wave solution

In this case, the displacement vector has .x and . y components (see Fig. 4.3). We can solve
wave equations (4.55) and (4.56) for .φ and . Hz using (4.52) and (4.53). Remember that from
the plane wave condition we know that .∂/∂z = 0 and therefore .φ and .H are independent
of .z. Wave functions .φ and . Hz have harmonic time dependence. It is reasonable to assume
that they have a harmonic dependence on .x as well. Thus, using the method of separation
of variables we can put .φ and .H in the following form and start solving the wave equations.
4.2 Propagation of Acoustic Waves in Solid Under Various Conditions 99

Fig. 4.6 SH wave incidence to a free surface. .k is the propagation vector

φ = f (y)ei(ωt−k1x x)
. (4.57)
. Hz = h z (y)e i(ωt−k2x x)
(4.58)

Substitution of (4.57) and (4.58) into wave equations (4.55) and (4.56) yields the following
differential equations for the functions . f and .h z .

d2 f
. = −k1y
2
f (4.59)
dy 2
d2hz
. = −k2y
2
hz (4.60)
dy 2
where
ω2
.
2
k1y = − k1x
2
(4.61)
c12
ω2
.
2
k2y = − k2x
2
(4.62)
c22

Equations (4.59) and (4.60) indicate that the functions . f and .h z are also harmonic functions
of . y. Thus, absorbing these dependencies of . f and .h z on . y in expressions (4.57) and (4.58),
we obtain the following forms of the potentials.

.φ(x, y, t) = A1 ei(ωt−k1x x+k1y y) + A2 ei(ωt−k1x x−k1y y) (4.63)

. Hz (x, y, t) = B1 ei(ωt−k2x x+k2y y) + B2 ei(ωt−k2x x−k2y y) (4.64)

Here, before proceeding further, consider the physical meanings of .k1x etc. in (4.61) and
(4.62). In these equations, .ω/c1 and .ω/c2 represent the wave numbers of the respective
waves. Referring to (4.17) and the discussions under “Propagation vector”, we can interpret
.k 1x and .k 1y as the . x and . y components of the propagation vector .k1 , and .k 2x and .k 2y as the

components of .k2 .
100 4 Propagation of Acoustic Waves in Solids

ω
k1 = k1x î + k1y ĵ, |k1 | =
. (4.65)
c1
ω
.k2 = k 2x î + k 2y ĵ, |k2 | = (4.66)
c2
If we call the wavelength of the respective waves as .λ1 and .λ2 , we find the following
relations (See Fig. 4.3).
2π
.λ1 = (4.67)
k1
λ1 2π 1 2π
.λ1x = = = (4.68)
sin θ1 k1 sin θ1 k1x
λ1 2π 1 2π
.λ1y = = = (4.69)
cos θ1 k1 cos θ1 k1y

From the second and third relations, we find

ω
. k1x = k1 sin θ1 = sin θ1 (4.70)
c1
ω
. k1y = k1 cos θ1 = cos θ1 (4.71)
c1
The same argument holds for the other wave and
ω
k2x = k2 sin θ2 =
. sin θ2 (4.72)
c2
ω
k2y
. = k2 cos θ2 = cos θ2 (4.73)
c2
Using (4.70)–(4.73), we can rewrite (4.63) and (4.64) as follows.

.φ(x, y, t) = A1 eik1 (c1 t−sin θ1 x+cos θ1 y) + A2 eik1 (c1 t−sin θ1 x−cos θ1 y) (4.74)
ik2 (c2 t−sin θ2 x+cos θ2 y) ik2 (c2 t−sin θ2 x−cos θ2 y)
. Hz (x, y, t) = B1 e + B2 e (4.75)

At this point, we can consider the boundary conditions. First, express the stresses in (4.51)
in terms of strain using constitutive relations.
( )
∂ξ y ∂ξx
.σ yx = μ + (4.76)
∂x ∂y
∂ξx ∂ξ y ∂ξz ∂ξx ∂ξ y
.σ yy = λ + (λ + 2μ) +λ =λ + (λ + 2μ) (4.77)
∂x ∂y ∂z ∂x ∂y
( )
∂ξz ∂ξ y ∂ξz
.σ yz = μ + =μ (4.78)
∂y ∂z ∂y
4.2 Propagation of Acoustic Waves in Solid Under Various Conditions 101

Above, we used the plane wave condition.∂/∂z = 0 in the right-hand side of (4.77) and (4.78).
Substituting (4.52) and (4.53) into (4.76) and (4.77) we obtain the following conditions for
. y = 0. Note that for the analysis of the plane strain wave, the boundary condition (4.78) is

irrelevant because the wave does not have the .z component, .ξz .
( ) ( 2 )
∂ξ y ∂ξx ∂ φ ∂ 2 Hz ∂ 2 Hz
.σ yx = μ + =μ 2 − + =0 (4.79)
∂x ∂y ∂ x∂ y ∂x2 ∂ y2
∂ξx ∂ξ y
.σ yy = λ + (λ + 2μ)
∂x ∂y
( 2 ) ( 2 )
∂ φ ∂ 2φ ∂ φ ∂ 2 Hz
= (λ + 2μ) + 2 − 2μ + =0 (4.80)
∂x2 ∂y ∂x2 ∂ x∂ y

By substituting (4.74) and (4.75) into (4.79) and (4.80) and setting . y = 0, we obtain the
following equations.

k12 sin 2θ1 (A2 − A1 )eik1 (c1 t−sin θ1 x)

.

+ k22 cos 2θ2 (B1 + B2 )eik2 (c2 t−sin θ2 x) = 0 (4.81)

ik1 (c1 t−sin θ1 x)
.k 1 (2 sin θ1 − κ )(A2 + A1 )e
2 2 2

− k22 sin 2θ2 (B1 − B2 )eik2 (c2 t−sin θ2 x) = 0 (4.82)

Here in (4.82), .κ is defined as the ratio of .c1 to .c2 as follows.

λ + 2μ
κ2 =
. (4.83)
μ
Using .k1 c1 = k2 c2 = ω, we can factor out the time-dependent term from (4.81) and (4.82).
In order to make these two equations hold for any .x, it is necessary that the space-dependent
term be factored out as well. It follows that the following condition holds.

k1 sin θ1 = k2 sin θ2
. (4.84)

From (4.70) and (4.72), we can interpret the condition (4.84) as the spatial frequency along
the free surface is the same for the .φ and .H waves so that the boundary conditions (4.79)
and (4.80) hold.
With (4.84) we can rewrite (4.74) and (4.75) as follows.
) (
.φ(x, y, t) = A1 eik1 cos θ1 y + A2 e−ik1 cos θ1 y ei(ωt−k1 sin θ1 x) (4.85)
) (
ik cos θ2 y
. Hz (x, y, t) = B1 e 2 + B2 e−ik2 cos θ2 y ei(ωt−k1 sin θ1 x) (4.86)

The .φ wave is called the compression wave (P-wave) and . Hz wave is called Shear-Vertical
(SV) wave because it is a shear wave confined in the vertical plane.
102 4 Propagation of Acoustic Waves in Solids

Shear-Horizontal wave solution

The Shear-Horizontal wave is a shear wave oscillating parallel to the plane of the free surface.
In the present case, a Shear -Horizontal wave has only the z-component of the displacement
vector. We can repeat the same type of argument and analyze the shear mode wave. In this
case, the displacement vector has only the .z component. This means that of the expression
of the three components of the displacement vector (4.52)–(4.54) only the expression for .ξz
(4.54) is relevant. Thus, the shear mode wave is characterized by . Hx and . Hy . Similarly to
(4.58), we can put the solutions in the following forms.

. Hx = h x (y)ei(ωt−k̃2x x) (4.87)
. Hy = h y (y)e i(ωt−k̃2x x)
(4.88)

Note that we use the plane wave condition .∂/∂z = 0 in (4.87) and (4.88).
Substitution of (4.87) and (4.88) into the wave equation (4.56) yields the following
differential equations.

d2hx
. = −k̃2y
2
hx (4.89)
dy 2
d2h y
. = −k̃2y
2
hy (4.90)
dy 2
where
ω2
.
2
k̃2y = − k̃2x
2
(4.91)
c22
Notice that unlike the plane strain case discussed in the preceding section, we use only one
spatial frequency .k̃2y to characterize .h x and .h y waves. This is because both waves share the
same velocity .c2 .
Equations (4.87), (4.88) and (4.91) let us put the solution . Hx and . Hy in the following
forms.

. Hx (x, y, t) = C1 ei(ωt−k̃2x x+k̃2y y) + C2 ei(ωt−k̃2x x−k̃2y y) (4.92)

. Hy (x, y, t) = D1 ei(ωt−k̃2x x+k̃2y y) + D2 ei(ωt−k̃2x x−k̃2y y) (4.93)

There is an additional difference between the plane strain and shear-horizontal cases.
Whereas the compression (.φ) wave and Shear-Vertical (. Hz ) waves are uncoupled in the
plane strain case, the two components of the Shear-Horizontal wave are coupled via (4.7).
Substitution of solutions (4.92) and (4.93) into (4.7) gives us the following equation.
{ }
−i k̃2y y
. −i k̃ 2x (C 1 e 2y + C 2 e
i k̃ y
) + i k̃2y (D1 ei k̃2y y − D2 e−i k̃2y y ) ei(ωt−k̃2x x) = 0
4.2 Propagation of Acoustic Waves in Solid Under Various Conditions 103

Hence,
(−k̃2x C1 + k̃2y D1 )ei k̃2y y + (−k̃2x C2 − k̃2y D2 )e−i k̃2y y = 0
. (4.94)
Since (4.94) holds for any . y it follows that

k̃2x C1 = k̃2y D1
. (4.95)
k̃2x C2 = −k̃2y D2
. (4.96)

Thus we can rewrite solutions (4.92) and (4.93) as follows.

. Hx (x, y, t) = C1 ei(ωt−k̃2x x+k̃2y y) + C2 ei(ωt−k̃2x x−k̃2y y) (4.97)

k̃2x k̃2x
. Hy (x, y, t) = C1 ei(ωt−k̃2x x+k̃2y y) − C2 ei(ωt−k̃2x x−k̃2y y) (4.98)
k̃2y k̃2y

Similarly to (4.72) and (4.73) we can put the spatial frequencies using the angle of
incidence .θ3 as follows.
ω
.k̃2x = sin θ3 = k2 sin θ3 (4.99)
c2
ω
.k̃2y = cos θ3 = k2 cos θ3 (4.100)
c2
With (4.99) and (4.100), (4.97) and (4.98) become
) (
ik cos θ3 y
. Hx (x, y, t) = C 1 e 2 + C2 e−ik2 cos θ3 y ei(ωt−k2 sin θ3 x) (4.101)
) (
ik cos θ3 y
. H y (x, y, t) = tan θ3 C 1 e 2 − C2 e−ik2 cos θ3 y ei(ωt−k2 sin θ3 x) (4.102)

The fact that . Hx and . Hy are coupled is not surprising if we remember that the SH-wave
has only one component of the displacement vector, .ξz , to oscillate. In other words, . Hx and
. H y cooperatively generate .ξz .

It may be a good idea to express the SH-wave in terms of .ξz . By substituting (4.97)
and (4.98) into (4.54), we can easily find that the displacement component .ξz can take the
following form. ) (
−i k̃2y y
.ξz (x, y, t) = E 1 e 2y + E 2 e
i k̃ y
ei(ωt−k̃2x x) (4.103)
where

iC1 ) 2 ( k2 k2
. E1 = − k̃2y + k̃2x
2
= −iC1 2 = −iC1 (4.104)
k̃2y k̃2y cos θ3
iC2 ) 2 ( k2 k2
. E2 = k̃2y + k̃2x = iC2 2 = iC2
2
(4.105)
k̃2y k̃2y cos θ3

Of course, the above solution satisfies the governing (wave) equation for .ξz .
104 4 Propagation of Acoustic Waves in Solids

1 ∂ 2 ξz
. ∇ 2 ξz = (4.106)
c22 ∂t 2

where
ω2
. = k̃2x
2
+ k̃2y
2
(4.107)
c22
Now we consider the boundary condition. In the case of the SH-wave, the relevant bound-
ary condition at the free surface is .σ yz (4.78). Expressing .ξz with the potentials, (4.78)
becomes
( 2 )
∂ξz ∂ Hy ∂ 2 Hx
.σ yz = μ =μ −
∂y ∂ x∂ y ∂ y2
) (
= k22 cos2 θ3 C1 eik2 cos θ3 y + C2 e−ik2 cos θ3 y ei(ωt−k2 sin θ3 x)
) (
+ k22 sin2 θ3 C1 eik2 cos θ3 y + C2 e−ik2 cos θ3 y ei(ωt−k2 sin θ3 x) (4.108)

Setting (4.108) to null for . y = 0 we obtain the following condition.

. (C1 + C2 ) k22 (cos2 θ3 + sin2 θ3 ) = 0 (4.109)

Hence,
C2 = −C1
. (4.110)
Equation (4.110) indicates that the free surface makes the forward-going and backward-
going SH waves have mutually opposite signs. In other words, upon reflection, the SH-wave
undergoes a phase shift of .π .

Reflection at the free surface with various incident waves

In this section, we discuss reflection at the free surface when the incident wave approaches
the surface from the inside of the elastic medium with an oblique angle of incidence. The
behavior at the free surface differs among the three types of wave, i.e., the P-wave, SV-wave,
and SH-wave, as discussed in the preceding section. In the following paragraphs, we will
briefly discuss the reflection for each of these waves. For more detailed discussions of these
topics, see Ref. [4].

(1) Incident SH wave

This is the simplest case of the three wave types. Substitute the SH-wave expression with
the displacement (4.103) into the boundary condition (4.78).
∂ξz
σ yz = μ
. = μ(i k̃2y )(E 1 − E 2 )ei(ωt−k̃2x ) = 0 (4.111)
∂y
4.2 Propagation of Acoustic Waves in Solid Under Various Conditions 105

It follows that
. E 2 = −E 1 (4.112)
Condition (4.112) indicates that the SH-wave undergoes a phase shift of .π on the reflection.
This is easily understood by noting that the boundary condition (4.51) forces the oscilla-
tion along the .z axis to be null at the surface. The situation is similar to the reflection of
an electromagnetic wave whose electric field is parallel to the conductive boundary. Being
conductive, the total electric field at the boundary must be null. When there is no transmitted
wave, the amplitude of the reflected wave negates that of the incident wave so this condition
holds. In the present case, the transmission of the acoustic wave can be neglected because
the acoustic impedance of the air is much less than the elastic medium.

(2) Incident P-wave

In this case, it is convenient to use the .φ and . Hz -wave expressions (4.85) and (4.86). Since
we consider that the incident wave is a P-wave and an H-wave is an S-wave, the first term
on the right-hand side of (4.86) must be null. This term represents a wave traveling in the
negative . y-direction, i.e., toward the boundary (see Fig. 4.3). Thus, we can set . B1 = 0 in
(4.86).
) (
.φ(x, y, t) = A1 eik1 cos θ1 y + A2 e−ik1 cos θ1 y ei(ωt−k1 sin θ1 x) (4.113)

(x, y, t) = B2 e−ik2 cos θ2 y ei(ωt−k1 sin θ1 x)

ref l
. Hz (4.114)

Here (4.114) represents the . Hz -component in the wave reflected at the boundary. Using
B1 = 0, k1 c1 = k2 c2 = ω, and (4.84), we can express the boundary conditions (4.81) and
.

(4.82) as follows.
{ }
. k12 sin 2θ1 (A2 − A1 ) + k22 cos 2θ2 (B2 ) ei(ωt−k1 sin θ1 x) = 0 (4.115)
{ 2 } i(ωt−k sin θ x)
. k 1 (2 sin θ1 − κ )(A 1 + A 2 ) + k 2 sin 2θ2 (B2 ) e =0
2 2 2 1 1 (4.116)

By solving (4.115) and (4.116) for . A1 and . B2 , we can evaluate the reflection coefficients for
the P- and SV-waves. Before proceeding further it is a good idea to express .2 sin2 θ1 − κ 2
term as follows. Remember .κ = c1 /c2 = k2 /k1 ,
1 1
. 2 sin2 θ1 − κ 2 = (2k12 sin2 θ1 − k22 ) = 2 (2k22 sin2 θ2 − k22 )
k12 k1
k22 k22
= 2
(2 sin2 θ2 − 1) = − cos 2θ2 (4.117)
k1 k12

Here, in going through the second equal sign, (4.84) was used. Using (4.117) the two
boundary conditions (4.115) and (4.116) become as follows.

k12 sin 2θ1 A2 − k12 sin 2θ1 A1 + k22 cos 2θ2 B2 = 0

. (4.118)
. − k 2 cos 2θ2 A 2 − k22 cos 2θ2 A1 + k22 sin 2θ2 B2 =0
2
(4.119)
106 4 Propagation of Acoustic Waves in Solids

Fig. 4.7 Reflection coefficient as a function of incident angle for several Poisson’s ratios when the
incident wave is a P-wave

Thus,

A2 k 2 sin 2θ1 sin 2θ2 − k22 cos2 2θ2

. = 12 (4.120)
A1 k1 sin 2θ1 sin 2θ2 + k22 cos2 2θ2
B2 2k12 sin 2θ1 cos 2θ2
. = 2 (4.121)
A1 k1 sin 2θ1 sin 2θ2 + k22 cos2 2θ2

Here .k1 and .k2 are related through Poisson’s ratio.

k22 λ + 2μ λ 2ν 2 − 2ν
.κ2 = = = +2= +2= (4.122)
k12 μ μ 1 − 2ν 1 − 2ν

Figure 4.7 plots the reflection coefficient (4.120) and (4.121) as a function of incident
angle for some Poisson’s ratio. Note that the incident P-wave generates the reflected SV-
wave. This phenomenon is referred to as the mode conversion. When the incident wave is
an SH-wave, there is no mode conversion as we discussed in the preceding section.
The insert in Fig. 4.7 shows the rays of the incident and reflected waves. Notice that when
the incident P-wave is converted to the S-wave, the angle of reflection is smaller than the
angle of incidence. We can explain this phenomenon as follows. As indicated by (4.11) and
(4.12), the P-wave has a higher velocity than the S-wave. From (4.65) and (4.66) this means
that .k1 < k2 , and from (4.84), that .sin θ1 > sin θ2 . It follows that .θ2 < θ1 .
From the law of reflection, the angle of reflection for the P-wave is the same as the angle
of incidence. These arguments all together indicate that no matter how great the angle of
incidence may be the incident P-wave is reflected back to the elastic medium. This situation
4.2 Propagation of Acoustic Waves in Solid Under Various Conditions 107

is contrastive to the case when an incident SV-wave is converted to a P-wave, as discussed

below.

(3) Incident SV wave

We can repeat the same type of argument as the preceding section to discuss the case when
the incident wave is an SV-wave. In this case, we can put . A1 = 0 in (4.85) as there is no
forward-going component in the P-wave. Consequently, we obtain the following expressions.

φ r e f l (x, y, t) = A2 e−ik1 cos θ1 y ei(ωt−k1 sin θ1 x)

. (4.123)
ik2 cos θ2 y i(ωt−k1 sin θ1 x) −ik2 cos θ2 y i(ωt−k1 sin θ1 x)
. Hz (x, y, t) = B1 e e + B2 e e (4.124)

The boundary conditions (4.81) and (4.82) become as follows.

. k12 sin 2θ1 A2 + k22 cos 2θ2 (B1 + B2 ) = 0 (4.125)

. − k 2 cos 2θ2 A 2 − k22 sin 2θ2 (B1 − B2 ) = 0
2
(4.126)

By solving (4.125) and (4.126) we obtain the following reflection coefficients.

B2 k 2 sin 2θ1 sin 2θ2 − k22 cos2 2θ2

. = 12 (4.127)
B1 k1 sin 2θ1 sin 2θ2 + k22 cos2 2θ2
A2 2k22 sin 2θ2 cos 2θ2
. =− 2 (4.128)
B1 k1 sin 2θ1 sin 2θ2 + k22 cos2 2θ2

Figure 4.8 plots the reflection coefficients (4.127) and (4.128). Here the right vertical
axis indicates the reflection coefficient when the mode is unconverted on reflection whereas
the left vertical axis indicates the reflection coefficient when the SV mode is converted to
the P mode.
The insert in Fig. 4.8 illustrates the rays of the incident and reflected waves. Note that in
this case, unlike the case when a P-wave is converted to an SV-wave, the angle of reflection
is greater than the angle of incidence on the SV- to P-wave conversion. This indicates that
if we increase the angle of incidence, at a certain point the angle of reflection reaches .π/2.
Beyond this point, according to (4.84) .sin θ2 exceeds unity and it becomes impossible to find
the angle of reflection from this equation. The angle of incidence that makes the angle of
reflection be .π/2 is called the critical angle .θc . (See (2.143).) When the angle of incidence
is greater than .θc , the resulting P-wave becomes confined in the subsurface region of the
elastic material and decaying exponentially away from the surface. We will discuss this
phenomenon in the next section.
108 4 Propagation of Acoustic Waves in Solids

Fig. 4.8 Reflection coefficient as a function of incident angle for several Poison’s ratios when the
incident wave is a SV-wave

4.2.3 Surface Waves

At the end of the last section, we touched upon the existence of a wave confined near the
free surface in association with the SV- to P-wave mode conversion occurring at an angle
of incidence greater than a critical angle. In this section, we discuss such surface waves in
more general.
Let .θ2c be the critical angle for an SV-wave. By definition, the corresponding angle
of refraction for the transmitting P-wave is .π/2. According to (4.84) the situation can be
expressed for the critical angle and angle .θ2' > θ2c as follows.
π
k2 sin θ2c = k1 sin
. = k1 (4.129)
2
k2 sin θ2' > k1
. (4.130)

From (4.70) and (4.84) we know

ω2
k12 =
. (4.131)
c12
.k1x = k1 sin θ1 (4.132)
k2
. sin θ1 = sin θ2 (4.133)
k1
Substituting (4.131)–(4.133) into (4.61) and using condition (4.130), we find as follows.
4.2 Propagation of Acoustic Waves in Solid Under Various Conditions 109

( ( )2 )
ω2 k 2 sin θ2
.
2
k1y = 2 − k1x
2
= k12 (1 − sin2 θ1 ) = k12 1 − <0 (4.134)
c1 k1

Equation (4.134) indicates that when the angle of incidence .θ2 is greater than the critical
angle, .k1y is not a real number. Remember that .k1y originally appeared in the differen-
tial equation (4.59) for the . f (y) part of the potential .φ(x, y, t) = f (y)ex p{i(ωt − k1x x)}.
Expressing .k1y with a real number .k̄1y as .k1y = i k̄1y , we can rewrite (4.59) as follows.

d2 f
. = −k1y
2
f = −(i k̄1y )2 f = k̄1y
2
f (4.135)
dy 2
We can solve (4.135) and obtain the following form of solution.

. f = F1 e−k̄1y y + F2 ek̄1y y (4.136)

Thus, we can put the potential .φ (4.113) in the following form.

) (
−k̄1y y
.φ(x, y, t) = F1 e + F2 ek̄1y y ei(ωt−k1 sin θ1 x) = F1 e−k̄1y y ei(ωt−k1 sin θ1 x) (4.137)

Here we discard the second term in the parenthesis because it is not physical as representing
an exponentially increasing amplitude. Solution (4.137) represents a wave traveling along
the .x-axis with the amplitude exponentially decreasing with the depth from the surface. We
call this type of wave surface waves.
We can extend the same idea to the other differential equation (4.60) to put the function
. Hz (x, y, t) in the following form.

. Hz (x, y, t) = Hz1 e−k̄2y y ei(ωt−k1 sin θ1 x) (4.138)

It is clear that displacement wave characterized by the two potentials in the form of
(4.137) and (4.138) represents a plane, surface wave. Express the .x and . y components of
such a displacement wave and analyze its feature.
∂φ ∂ Hz ) (
ξx =
. + = − ik1 sin θ1 F1 e−k̄1y y + k̄2y Hz1 e−k̄2y y ei(ωt−k1 sin θ1 x) (4.139)
∂x ∂y
∂φ ∂ Hz ) (
.ξ y = − = − k̄1y F1 e−k̄1y y − ik1 sin θ1 Hz1 e−k̄2y y ei(ωt−k1 sin θ1 x) (4.140)
∂y ∂x
The boundary conditions .σ yx (4.79) and .σ yy (4.80) give us the following equations.
) (
.σ yx = μ 2(−ik 1 sin θ1 )(−k̄ 1y )φ + (k 1 sin θ1 + k̄ 2y )Hz =0
2 2 2
(4.141)
σ yy = (λ + 2μ)(−k12 sin2 θ1 + k̄1y
.
2
)φ − 2μ(−k12 sin2 θ1 φ + ik1 sin θ1 k̄2y Hz )
= κ 2 μ(−k12 sin2 θ1 + k̄1y
2
)φ + 2μ(k12 sin2 θ1 φ − ik1 sin θ1 k̄2y Hz ) = 0 (4.142)

In going through the second equal sign in (4.142), .κ 2 μ = λ + 2μ ((4.83)) is used.

110 4 Propagation of Acoustic Waves in Solids

Here
ω2
. = k12 sin2 θ1 − k̄1y
2
(4.143)
c12
ω2
. = k12 sin2 θ1 − k̄2y
2
(4.144)
c22

So,
k12 sin2 θ1 − k̄2y
2
c12
. = = κ2 (4.145)
k12 sin2 θ1 − k̄1y
2 c22
Use (4.145) to rewrite (4.142) as follows.

σ yy = μ(−k12 sin2 θ1 + k̄2y

.
2
)φ + 2μ(k12 sin2 θ1 φ − ik1 sin θ1 k̄2y Hz )
= μ(k12 sin2 θ1 + k̄2y
2
)φ − 2iμk1 sin θ1 k̄2y Hz = 0 (4.146)

Using (4.137) and (4.138) to express .φ and. Hz at . y = 0 with . F1 and . Hz1 , we obtained the
following set of equations from (4.141) and (4.146).

2ik1 sin θ1 k̄1y F1 + (k12 sin2 θ1 + k̄2y

.
2
)Hz1 = 0 (4.147)
(k12 sin2 θ1 + k̄2y
.
2
)F1 − 2ik1 sin θ1 k̄2y Hz1 = 0 (4.148)

From (4.147) and (4.148), we obtain the following equality.

F1 k12 sin2 θ1 + k̄2y

2
2ik1 sin θ1 k̄2y
. = = 2 2 (4.149)
Hz1 −2ik1 sin θ1 k̄1y k1 sin θ1 + k̄2y2

From the second equality in (4.149) it follows that

. (k12 sin2 θ1 + k̄2y

2 2
) − 4k̄1y k̄2y k12 sin2 θ1 = 0 (4.150)

hence, /
k12 sin2 θ1 + k̄2y
.
2
= 2 k̄1y k̄2y k1 sin θ1 (4.151)
Now compare the amplitude of .ξx (x, 0, t) and .ξ y (x, 0, t) to find out what the actual
displacement vectors look like at the surface. At . y = 0 (on the free surface), (4.139) and
(4.140) become
} (
.ξx (x, 0, t) = − ik1 sin θ1 F1 + k̄2y Hz1 ei(ωt−k1 sin θ1 x) (4.152)
} ( i(ωt−k sin θ x)
.ξ y (x, 0, t) = − k̄ 1y F1 − ik 1 sin θ1 Hz1 e 1 1 (4.153)

Use (4.149) in (4.152) and (4.153) to eliminate . Hz1 .

4.2 Propagation of Acoustic Waves in Solid Under Various Conditions 111

( )
k12 sin2 θ1 + k̄2y
2
.ξx (x, 0, t) = −i k1 sin θ1 − F1 ei(ωt−k1 sin θ1 x)
2k1 sin θ1
( / )
= −i k1 sin θ1 − k̄1y k̄2y F1 ei(ωt−k1 sin θ1 x) (4.154)
( )
k12 sin2 θ1 + k̄2y2
.ξ y (x, 0, t) = − k̄ 1y − F1 ei(ωt−k1 sin θ1 x)
2k̄2y
/ (/ )
k̄1y
=− k̄1y k̄2y − k1 sin θ1 F1 ei(ωt−k1 sin θ1 x)
k̄2y
/
k̄1y
= −i ξx (x, 0, t) (4.155)
k̄2y

Here (4.151) was used in going through the second equal sign for (4.154) and (4.155). From
(4.143) and (4.144),

ω2
.
2
k̄1y = k12 sin2 θ1 − (4.156)
c12
ω2
.
2
k̄2y = k12 sin2 θ1 − (4.157)
c22

Since .c1 > c2 , it follows .k̄1y > k̄2y . Thus (4.155) indicates that

|ξx | < |ξ y |
. (4.158)

The type of surface wave expressed by (4.154) and (4.155) is known as the Rayleigh sur-
face wave [5]. Figure 4.9 illustrates a sample Rayleigh surface wave traveling in the positive
. x-direction. The spatial periodicity is seen to move to the right. Here, .c = ω/(k 1 sin θ1 ) can

be interpreted as the wave velocity. (See the exponential term of (4.154) and (4.155).) Note
that Rayleigh surface waves are non-dispersive. The insert to the right illustrates the ellipti-
cal pattern of the displacement vector as a function of time. Notice that when the Rayleigh
wave travels in the positive .x direction, the trajectory of the displacement vector along the
elliptical line is counterclockwise.
The above discussions cover only the gist of the Rayleigh wave. For more information
about Rayleigh waves, see Refs. [6, 7].

4.2.4 Waves Propagating in Plates

In this section, we extend the discussion on acoustic waves in an elastic medium with a free
surface that we made in Sect. 4.2.2. The elastic medium has two free surfaces, instead of
one. The governing equation (the wave equation) is the same as the one-free surface case
112 4 Propagation of Acoustic Waves in Solids

Fig. 4.9 Rayleigh wave traveling in positive .x-direction. The illustration on the right represents the
elliptical pattern of the displacement vector as a function of time

but the other free surface gives an additional boundary condition, which makes the resultant
displacement vector behave considerably different from the one free-surface case. The most
important difference here is that the displacement wave exhibits dispersion. This feature has
important engineering applications [8]. The aim of this section is to understand the mech-
anism that generates dispersion through some simple cases. For more general discussions
about the dispersion of waves in plates, see Ref. [9, 11].

SH waves in a plate

As is the case of the one free-surface case, we start with the SH-wave case since it is the
simplest. The governing equation is the same as (4.106).

1 ∂ 2 ξz
∇ 2 ξz =
. (4.106)
c22 ∂t 2

As before, we consider a plane wave with the .∂/∂z = 0 condition. The solution can be put
as follows.
) (
−i k̃2y y
.ξz (x, y, t) = E 1 e 2y + E 2 e
i k̃ y
ei(ωt−k̃2x x)
) (
= E 1 (cos k̃2y y + i sin k̃2y y) + E 2 (cos k̃2y y − i sin k̃2y y) ei(ωt−k̃2x x)
) (
= (E 1 + E 2 ) cos k̃2y y + i(E 1 − E 2 ) sin k̃2y y ei(ωt−k̃2x x)
) (
≡ Ẽ 1 cos k̃2y y + Ẽ 2 sin k̃2y y ei(ωt−k̃2x x) (4.159)
4.2 Propagation of Acoustic Waves in Solid Under Various Conditions 113

Fig. 4.10 Plate elastic medium with two free surfaces

where
ω2
. = k̃2x
2
+ k̃2y
2
(4.160)
c22
Now continue the analysis with a plate of thickness.2b having infinitely large free surfaces
at . y = ±b. The free surfaces are parallel to the .zx plane. Figure 4.10 illustrates the plate
geometry. With the geometry shown in Fig. 4.10, the free surface boundary condition relevant
to the SH-wave can be expressed as follows.
( ) ( )
∂ξz ∂ξ y ∂ξz
.σ yz | y=±b = μ − | y=±b = μ | y=±b = 0 (4.161)
∂y ∂z ∂y

Here .∂/∂z = 0 is used for the plane wave solution. Substituting solution (4.159) into (4.161)
we obtain the following equations.
{ }
.σ yz | y=b = k̃ 2y − Ẽ 1 sin k̃ 2y b + Ẽ 2 cos k̃ 2y b e =0
i(ωt−k̃2x x)
(4.162)
{ }
.σ yz | y=−b = k̃ 2y Ẽ 1 sin k̃ 2y b + Ẽ 2 cos k̃ 2y b e =0
i(ωt−k̃2x x)
(4.163)

For the above conditions hold for any .x, the inside the curly brackets must be zero. This
leads to the following condition.

Ẽ 1 cos k̃2y b cos k̃2y b

. = =− (4.164)
Ẽ 2 sin k̃2y b sin k̃2y b

It follows that .(cos k̃2y b) (sin k̃2y b) = −(cos k̃2y b) (sin k̃2y b) = 0, or

(cos k̃2y b) (sin k̃2y b) = 0

. (4.165)

leading to the following condition.

nπ
.k̃2y b = ± , n = 0, 1, 2, 3, 4, . . . (4.166)
2
114 4 Propagation of Acoustic Waves in Solids

Fig. 4.11 SH symmetric waves with even mode numbers and anti-symmetric waves with odd mode
numbers in the plate

Condition (4.166) means as follows. When an SH-wave of a given frequency .ω travels

through a plate in the .x-direction undergoing numerous reflections at the two free surfaces,
the spatial frequency in the . y-direction (along the thickness) must satisfy condition (4.166).
For a given thickness .b, the spatial frequency can be any of .k̃2y = ±(nπ )/(2b). Here, the
integer .n is called the mode number and the corresponding wave is called the mode .n wave.
When .n is an odd integer, .cos k̃2y b = 0 and .sin k̃2y b = 1 in (4.165). From (4.162) it
follows that . Ẽ 1 = 0. This makes the cosine term null in the .ξx (x, y, t) solution (4.159).
Similarly, when .n is an even integer, the sine term of the .ξx (x, y, t) solution becomes null.
Thus, we find the following expressions for .ξx (x, y, t) under the condition of free surfaces
at . y = ±b.
) nπ (
.ξz (x, y, t) = Ẽ 2 sin y ei(ωt−k̃2x x) , n = 1, 3, 5, 7, . . . (4.167)
2b
) nπ (
.ξz (x, y, t) = Ẽ 1 cos y ei(ωt−k̃2x x) , n = 0, 2, 4, 6, . . . (4.168)
2b
Apparently, solution (4.167) presents anti-symmetric waveforms with respect to . y-axis,
and solution (4.168) presents symmetric waveforms. Figure 4.11 illustrates several anti-
symmetric and symmetric waves for low mode numbers. The displacement vector is polar-
ized along the .z-axis and the waves travel along the .x-axis (normal to the page).
Now consider the velocity of the SH plate wave. The wave solution (4.159) indicates
that the SH plate-wave travels in the positive .x-direction at velocity .c = ω/k̃2x . From the
frequency equations (4.160) and (4.166), the spatial frequency along the .x-axis takes the
following form.
4.2 Propagation of Acoustic Waves in Solid Under Various Conditions 115

/
( )2 ) nπ (2
ω
.k̃2x = − (4.169)
c2 2b
Note that .c2 represents the phase velocity of the SH-wave incident to and reflected from the
free surfaces parallel to the .x-axis, and is a constant. On the other hand, the phase velocity
of the SH-wave traveling along the .x-axis .(c ≡ ω/k̃2x ) depends on .ω as shown below.
⎧/ ⎫−1
⎨ ( )2 )
ω 1 nπ ⎬
( 2
.c = = − . (4.170)
k̃2x ⎩ c2 2bω ⎭

In other words, this SH-wave exhibits dispersion. We can find the group velocity as .vg =
∂ω/∂ k̃2x [12]. Below, we consider this dispersion in some detail.
First, normalize the spatial and temporal frequencies as follows so that they become
dimensionless.
2bk2x
ξ̄ =
. (4.171)
π
2bω
.Ω = (4.172)
c2 π
By substituting (4.171) and (4.172) into (4.169), we obtain the following dispersive relation.
√
.ξ̄ = ± Ω2 − n 2 (4.173)

Here, the dimensionless frequency .Ω and the mode number .n are real. Therefore, if .Ω < n,
the dimensionless spatial frequency .ξ̃ becomes imaginary, and if .Ω > n, .ξ̃ is real.
Since .b is real, (4.171) indicates that an imaginary .ξ̃ and a real .ξ̃ correspond to an
imaginary .k2x and a real .k2x , respectively. According to (4.167) and (4.168), the anti-
symmetric and symmetric SH plate wave travels when .ξ̃ is real. If .ξ̃ is imaginary, the
oscillation from the source is not transferred as a wave, and instead, it decays exponentially
as a function of .x.
The condition .Ω > n or .Ω < n, which determines whether .ξ̃ is real or imaginary, tells us
that for a given mode number the oscillation from the source travels as a wave if the oscillation
frequency is higher than a certain value. We can intuitively understand this situation by
imagining oscillating one end of an elastic string at various frequencies with the other end
fixed to the wall. As we increase the oscillation frequency, the motion travels toward the
fixed end with the shorter wavelength, i.e., the higher mode number. In other words, if we
want to excite a wave of a certain mode, we need to oscillate the open end at a certain
frequency. We can imagine that if we oscillate the open end very slowly the entire string
oscillates up and down.
Figure 4.12 shows the dispersion curve for various mode numbers .n. Here, the solid lines
are for symmetric waves and the dashed lines are for anti-symmetric waves, and the right
half is for the case when .ξ̃ is real and the left half is when it is imaginary. Notice that the
116 4 Propagation of Acoustic Waves in Solids

Fig. 4.12 Dispersion curves

for SH-waves traveling through
a plate

dispersion curve depends on the plate thickness via (4.171) and (4.172). This fact can be
used to find a thickness of a layer from the dispersion curve [13].

P and SV waves in a plate and Rayleigh-Lamb frequency equation

In this case, it is more convenient to use the potentials .φ and .H. Recall the discussions made
under Sect. 4.2.2, Plane strain wave solution. The governing equations and the solutions
we should deal with are as follows.
1 ∂ 2φ 1 ∂ 2 Hz
∇2φ =
. , ∇ 2
Hz = (4.174)
c1 ∂t 2 c2 ∂t 2
and
) (
. φ = A1 eik1y y + A2 e−ik1y y ei(ωt−k1x x )
) (
= Ã1 cos k1y y + Ã2 sin k1y y ei(ωt−k1x x ) (4.175)
) (
−ik2y y
. Hz = B1 e 2y + B2 e ei(ωt−k1x x )
ik y
) (
= B̃1 cos k2y y + B̃2 sin k2y y ei(ωt−k1x x ) (4.176)

where . Ã1 = A1 + A2 , Ã2 = i(A1 − A2 ) and . B̃1 = B1 + B2 , B̃2 = i(B1 − B2 ). Similarly

to the case of the free surface condition discussed in the Plane strain wave solution section,
the boundary conditions relevant to the P and SV waves are as follows.

σ yy = σx y = 0, at y = ±b
. (4.177)
4.2 Propagation of Acoustic Waves in Solid Under Various Conditions 117

Here, the boundaries are assumed to be traction free and the relevant stress are expressed as
follows.
( 2 )
∂ φ ∂ 2 Hz ∂ 2 Hz
.σ x y = μ 2 + − (4.178)
∂ x∂ y ∂ y2 ∂x2
( 2 ) ( )
∂ φ ∂ 2φ ∂ 2φ ∂ 2 Hz
.σ yy = (λ + 2μ) + − 2μ +
∂x2 ∂ y2 ∂x2 ∂ x∂ y
( ( 2 ) ( 2 2 H ))
∂ φ ∂ 2 φ ∂ φ ∂ z
= μ κ2 + 2 −2 + (4.179)
∂x2 ∂y ∂x2 ∂ x∂ y

At this point, express the displacement vector components .ξx and .ξ y using potentials
(4.175) and (4.176)
∂φ ∂ Hz
ξx (x, y, t) =
. +
∂x ∂y
) (
= −ik1x Ã1 cos k1y y + k2y B̃2 cos k2y y ei(ωt−k1x x)
) (
+ −ik1x Ã2 sin k1y y − k2y B̃1 sin k2y y ei(ωt−k1x x) (4.180)
∂φ ∂ Hz
ξ y (x, y, t) =
. −
∂y ∂x
) (
= −k1y Ã1 sin k1y y + ik1x B̃2 sin k2y y ei(ωt−k1x x)
) (
+ k1y Ã2 cos k1y y + ik1x B̃1 cos k2y y ei(ωt−k1x x) (4.181)

Of the two terms enclosed by the parenthesis in expressions (4.180) and (4.181), the first
one is symmetric about the .x-axis and the second one is anti-symmetric. The symmetric and
anti-symmetric expressions of .ξx and .ξ y are shown below.

Symmetric displacement components:

) (
sym
ξx
. (x, y, t) = −ik1x Ã1 cos k1y y + k2y B̃2 cos k2y y ei(ωt−k1x x) (4.182)
) (
sym
.ξ y (x, y, t) = −k1y Ã1 sin k1y y + ik1x B̃2 sin k2y y ei(ωt−k1x x) (4.183)

Anti-symmetric displacement components:

) (
asym
.ξx (x, y, t) = −ik1x Ã2 sin k1y y − k2y B̃1 sin k2y y ei(ωt−k1x x) (4.184)
) (
asym
.ξ y (x, y, t) = k1y Ã2 cos k1y y + ik1x B̃1 cos k2y y ei(ωt−k1x x) (4.185)

Figure 4.13 illustrates the symmetry.

118 4 Propagation of Acoustic Waves in Solids

Fig. 4.13 Symmetric and anti-symmetric parts of .ξx and .ξ y components of plane P and SV waves
traveling in a plate of thickness .2b

Substitute (4.182) and (4.183) into expressions (4.79) and (4.80) and find the stress due
to the symmetric waves as follows.
{
sym
.σ x y = μ − 2ik1x k1y (− Ã1 sin k1y y)
}
+ (k1x
2
− k2y
2
)(+ B̃2 sin k2y y) ei(ωt−k1x x) (4.186)
{
sym
.σ yy = μ (2k1x − κ 2 (k1x + k1y ))( Ã1 cos k1y y)
2 2 2
}
+ 2ik1x k2y (+ B̃2 cos k2y y) ei(ωt−k1x x)
{
= μ (k1x
2
− k2y
2
)( Ã1 cos k1y y)
}
+ 2ik1x k2y (+ B̃2 cos k2y y) ei(ωt−k1x x) (4.187)

Here the following substitution was made for .σ yy .

4.2 Propagation of Acoustic Waves in Solid Under Various Conditions 119

2 + k2
c12 ω2 /c22 k2y 1x
κ2 =
. = =
c22 ω2 /c12 2 + k2
k1y 1x
κ 2 (k1x
2
+ k1y
2
) = k2y
2
+ k1x
2

2
2k1x − κ 2 (k1x
2
+ k1y
2
) = k1x
2
− k2y
2
(4.188)

Substitution of (4.186) and (4.187) into boundary conditions (4.177) yields the following
equations.

. ± {2ik1x k1y ( Ã1 sin k1y b) + (k1x

2
− k2y
2
)( B̃2 sin k2y b)} = 0 (4.189)
.(k 1x − k2y )( Ã1 cos k1y b) + 2ik1x k2y ( B̃2 cos k2y b) =0
2 2
(4.190)

Here the .± sign comes from the boundary condition .σx y = 0 at . y = ±b.
We can view (4.189) and (4.190) as a system of equations for . Ã1 and . B̃2 , and express
them in the form of a matrix as follows.
( )
2 − k 2 ) sin k b (
2ik1x k1y sin k1y b (k1x
) ( )
2y Ã1 0
.
2y
= (4.191)
2 − k 2 ) cos k b 2ik k cos k b
(k1x 2y 1y 1x 2y 2y B̃2 0

For the matrix equation (4.191) to hold for a nontrivial pair of . Ã1 and . B̃2 , it follows that
the determinant of the matrix must be null. This observation yields the following frequency
equation.
tan k2y b 4k 2 k1y k2y
. = − 21x 2 2 (4.192)
tan k1y b (k1x − k2y )
Equation (4.192) is known as the Rayleigh-Lamb frequency equation for the propagation of
symmetric waves in a plate. Remember that .k1x , .k1y and .k2y are related through angle .θ1 as
follows.

ω
k1x = k1 sin θ1 =
. sin θ1 (4.70)
c1
ω
. k1y = k1 cos θ1 = cos θ1 (4.71)
c1
ω
k2y
. = k2 cos θ2 = cos θ2 (4.73)
c2
. k1 sin θ1 = k2 sin θ2 (4.84)

For a given elastic material (.c1 and .c2 ) with a plate thickness (.2b), once frequency .ω is
determined (4.192) tells us the angle .θ1 and .θ2 for the symmetric waves (4.182) and (4.183).
From the system of equations (4.189) and (4.190), we can derive the amplitude ratio as
follows.
Ã1
2 − k 2 ) sin k b
(k1x 2y 2y 2ik1x k2y cos k2y b
. =− =− 2 (4.193)
B̃2 2ik1x k1y sin k1y b (k1x − k2y2 ) cos k b
1y
120 4 Propagation of Acoustic Waves in Solids

Repeating the same argument for the anti-symmetric case, we obtain the following equa-
tions. ( )
2 − k 2 ) cos k b (
−2ik1x k1y cos k1y b (k1x
) ( )
2y Ã2 0
.
2y
= (4.194)
2 − k 2 ) sin k b −2ik k sin k b
(k1x 2y 1y 1x 2y 2y B̃1 0

tan k2y b
2 − k 2 )2
(k1x 2y
. =− 2 (4.195)
tan k1y b 4k1x k1y k2y
Equation (4.195) is the Rayleigh-Lamb frequency equation of anti-symmetric waves.

Ã1
2 − k 2 ) cos k b
(k1x 2y 2y 2ik1x k2y sin k2y b
. =− =− (4.196)
B̃2 2ik1x k1y cos k1y b (k1x
2 − k 2 ) sin k b
2y 1y

Equation (4.196) is the amplitude ratio for the anti-symmetric wave case.

A note on vibrations and waves in a thin plate

When a longitudinal wave is to be excited in a thin plate with the application of an acoustic
transducer normally to the surface, we need to be careful about the relation between the
wavelength and the thickness. If the thickness is significantly smaller than the wavelength,
say less than a quarter of the wavelength, the phase difference between the leading and
trailing edge of the longitudinal wave is less than .π/4. The plate is more or less oscillating
as a whole. Recall that the wave velocity is intrinsic to the material. Therefore the wavelength
is determined by the frequency. To make the wavelength shorter for a given wave velocity
(material), it is necessary to increase the frequency.
The same can be said about the spatial resolution. When we try to detect a void in a
material, if the length of the void in the direction of the longitudinal wave’s propagation,
the void falls in a small portion of the wavelength. It is necessary to increase the frequency.
However, increase of frequency is not straightforward as the attenuation can be proportional
to the square of the frequency [14–19].

References

1. Graff KF (1975) Wave motion in elastic solids. Oxford University Press, Oxford, UK
2. Rose JL (1999) Ultrasonic waves in solid media. Cambridge University Press, Cambridge, UK
3. Rose JL (1999) Ultrasonic waves in solid media. Cambridge University Press, Cambridge, UK,
p 25
4. Graff KF (1975) Wave motion in elastic solids. Oxford University Press, Oxford, UK, Chap. 5
5. Rayleigh JWS (1887) On waves propagated along the plate surface of an elastic solid. Proc Lond
Math Soc 17:4–11
6. Rose JL (1999) Ultrasonic waves in solid media. Cambridge University Press, Cambridge, UK,
chap. 7
7. Graff KF (1975) Wave motion in elastic solids. Oxford University Press, Oxford, UK, Chap. 6
References 121

8. Pilarski A, Rose JL (1988) A transverse wave ultrasonic oblique incidence technique for inter-
facial weakness detection in adhesive bonds. J Appl Phys 63(2):300–307
9. Graff KF (1975) Wave motion in elastic solids. Oxford University Press, Oxford, UK, Chap. 8
10. Maev RG (2008) Scanning acoustic microscopy. Physical principle and methods. Current devel-
opment in R. G. Maev Acoustic microscopy: fundamentals and applications, Ch1. Wiley-VCH,
Verlag. https://doi.org/10.1002/9783527623136.ch1 (accessed on July 28, 2023), pp 9–19
11. Rose JL (1999) Ultrasonic waves in solid media. Cambridge University Press, Cambridge, UK,
chap. 8
12. Yoshida S (2017) Waves; Fundamentals and dynamics, Chap. 4. Morgan & Claypool, San Rafael,
CA, USA
13. Parmon W, Bertoni H (1979) Ray interpretation of the material signature in the acoustic micro-
scope. Electron Lett 15(21):684–686
14. Szabo TL, Wu J (2000) A model for longitudinal and shear wave propagation in viscoelastic
media. J Acoust Soc Am 107(5):2437–2446
15. Szabo TL (1994) Time domain wave equations for lossy media obeying a frequency power law.
J Acoust Soc Am 96(1):491–500
16. Chen W, Holm S (2003) Modified Szabo’s wave equation models for lossy media obeying
frequency power law. J Acoust Soc Am 114(5):2570–2574
17. Chen W, Holm S (2004) Fractional Laplacian time-space models for linear and nonlinear lossy
media exhibiting arbitrary frequency power-law dependency. J Acoust Soc Am 115(4):1424–
1430
18. Carcione JM, Cavallini F, Mainardi F, Hanyga A (2002) Time-domain modeling of constant-Q
seismic waves using fractional derivatives. Pure Appl Geophys 159:1719–1736
19. D’astrous FT, Foster FS (1986) Frequency dependence of ultrasound attenuation and backscatter
in breast tissue. Ultrasound Med Biol 12(10):795–808
Electrical-Mechanical Transduction
5

This chapter discusses acoustic transduction. We focus on the operation principle of typical
acoustic transducers to understand how the devices work. To this end, we discuss the physics
underlying the operation.

5.1 General Argument

Acoustic transduction generally means the conversion of an electric signal to an acoustic

signal or an acoustic signal to an electric signal. The former is acoustic transmission, and
the latter is acoustic sensing. A loudspeaker is an example of an acoustic transmitter, and a
microphone is an example of an acoustic sensor. In the application to experimental mechan-
ics dealing with solid objects, often acoustic sensors are vibration sensors, and acoustic
transmitters are vibration actuators.
Thus acoustic transduction is the interaction of an electric system and a mechanical
system. The following pair of equations conveniently represent the process of the acoustic
transaction [1].

v = z e0 i + Tem u
. (5.1)
. f = Tme i + z m0 u (5.2)

Here .v is the voltage on the electric system, .u is the velocity of the mechanical load, .z e0
is the electrical impedance, .Tem is the mechanical to electrical transduction coefficient,
. f is the force on the mechanical system, . Tme is the electrical to mechanical transduction

coefficient, and.z m0 is the mechanical impedance. (5.1) and (5.2) are the governing equations
that describe the relationship between the electrical and mechanical systems. Call the former
the electrical governing equation and the latter the mechanical governing equation. When
the transducer acts as an acoustic sensor, the electrical governing equation expresses the

© The Author(s), under exclusive license to Springer Nature Switzerland AG 2024 123
S. Yoshida, Fundamentals of Acoustic Waves and Applications, Synthesis Lectures on Wave
Phenomena in the Physical Sciences, https://doi.org/10.1007/978-3-031-48200-7_5
124 5 Electrical-Mechanical Transduction

acoustic (mechanical) input represented by the velocity of the sensing mechanism (e.g., a
microphone’s diaphragm) represented by .u and the resultant current and voltage .i and .v
detected by the electric system. When the transducer operates as a transmitter (actuator),
voltage .v represents the input electric signal, and .i and .u represent the resultant current
flowing through the electric system and the velocity of the actuation mechanism (e.g., a
loudspeaker’s diaphragm). The mechanical governing equation describes the force acting on
the vibrating part (e.g., the diaphragm) where the first term on the right-hand side represents
the electric force and the second term represents the mechanical force.
As these equations indicate, the conversion from acoustic to electric or electric to acoustic
energy is not 100% efficient. For instance, when an acoustic signal oscillates the diaphragm
in a microphone, part of the generated electric energy is lost in the circuit. The first term on
the right-hand side of (5.1) represents this loss. Similarly, in the opposite process, part of
the generated mechanical energy is lost by mechanical impedance expressed by the second
term on the right-hand side of (5.2).
Thus, the coupling factor . K is defined as follows.
Tem Tme
. K2 = (5.3)
z e0 z m0
Here the numerator is the product of the active terms on the right-hand side of (5.1) and
(5.2), whereas the denominator is the product of the ineffective terms.

5.2 Reciprocal Transduction

This is the case when the following condition (reciprocal condition) holds.

. Tem = Tme = T (5.4)

Define the transduction factor .φ as follows.

T
.φ= (5.5)
z e0
(5.1) and (5.2) become as follows.

v = z e0 i + φz e0 u
. (5.6)
. f = φz e0 i + z m0 u (5.7)

In this case, the coupling factor takes the following form.

Tem Tme T2 1 T 2 z e0 z e0
. K2 = = = 2 = φ2 (5.8)
z e0 z m0 z e0 z m0 z e0 z m0 z m0
5.2 Reciprocal Transduction 125

5.2.1 Electrostatic Transducer

An electrostatic transducer is an example of a reciprocal transducer. Figure 5.1 schemati-

cally illustrates a sample electrostatic transducer. This transducer consists of two stationary
electrodes connected to an AC voltage supply and one movable plate (diaphragm) located
in between the two stationary electrodes. Typically, the diaphragm consists of a thin plastic
film coated with a conductive material. Initially, the diaphragm is positively charged with .q0
at a DC voltage .v0 . Throughout the operation, this DC voltage remains the same. (In other
words, the diaphragm is AC-isolated.) The diaphragm forms two capacitors with each of
the stationary electrodes, respectively.
When this transducer operates as an acoustic transmitter, the AC voltage supply applies
a sinusoidal voltage (an electric input) to the system. As explained below, the applied AC
signal is proportional to the output acoustic signal. Figure 5.1 illustrates the moment when
the applied AC voltage is negative on the left stationary electrode and positive on the right
stationary electrode. The absolute value of these negative and positive voltage changes are
the same. This application of the AC voltage causes two things; first, positive charges .q
flow into the right electrode, and the same amount of positive charges, .q, flow out of the left
electrode; second, these changes in the two electrodes generate a leftward electric field in
the space between the electrodes. This electric field exerts Coulomb force on the positively
charged diaphragm, causing it to move to the left by .x. Here, the charge .q caused by the
applied AC voltage is much smaller than the total charge on the diaphragm, .q < q0 , and the
corresponding displacement of the diaphragm is much smaller than the initial gap distance
between the stationary electrode and diaphragm, .x << x0 .
Below, we analyze the operation of this transducer as an acoustic transmitter. When the
same transducer operates as an acoustic sensor, the oscillatory displacement of the diaphragm

Fig. 5.1 Electrostatic

transducer
126 5 Electrical-Mechanical Transduction

due to the acoustic pressure (input) causes an AC voltage change that is read out as an output
electric signal.
In the following analysis, we use the interaction between the left stationary electrode and
the diaphragm for simplicity. The same argument holds for the interaction between the right
electrode and the diaphragm. If the total charge on the diaphragm is positive, the change in
the charge .q makes the right electrode repel the diaphragm and the left electrode attracts it.
Consequently, the diaphragm moves to the left.

5.2.2 Electrical Governing Equation of Electrostatic Transducer

First, derive the electrical governing equation. Our goal here is to find the expression of
voltage change due to the application of AC voltage. Consider a sinusoidal voltage .v is
superposed to the bias voltage, and consequently, the charge on the capacitor changes by .q0
and the gap length changes by .x (.q << q0 and .x << x0 ).

q (q0 + q)(x0 + x) q0 x 0 + q0 x + q x 0
.v + v0 = = ≈
C ∈S ∈S
q x 0 + q0 x
= v0 + (5.9)
∈S
Here, . S is the area of the capacitor, .∈ is the electric permittivity of the space between the
electrodes and the diaphragm, .x0 and .q0 are the initial gap length and charge, and .q x << 1
is neglected. We can relate .∈ S and .q0 with the initial capacitance .C0 as follows.
∈S
C0 =
. (5.10)
x0
.q0 = C 0 v0 (5.11)

From (5.9), (5.10) and (5.11), we can express the change in the voltage across the capacitor
as follows.
q x 0 + q0 x x0 C 0 v0 1 v0
v=
. = q+ x= q+ x (5.12)
∈S ∈S ∈S C0 x0
On the right-hand side of (5.12), the first term represents the effect due to the change in the
stored charge (with the initial capacitance), and the second term represents the effect due to
the change in the capacitor’s gap length (with the initial voltage).
Since the superposed voltage is sinusoidal, the resultant displacement is sinusoidal with
the same frequency. Let .ω be the frequency. We can express the current to the capacitor and
the velocity of the diaphragm as follows.

i = q̇ = jωq
. (5.13)
u = ẋ = jωx
. (5.14)
5.2 Reciprocal Transduction 127

Here . j is the imaginary unit . j 2 = −1.

Substitution of (5.13) and (5.14) into (5.12) yields
1 v0
.v= i+ u (5.15)
jωC0 jωx0
Comparison of (5.6) and (5.15) with the use of (5.5) identifies the expressions for the
electrical impedance and the mechanical to electrical transduction coefficient as follows.
1
. z e0 = (5.16)
jωC0
v0
. Tem = T = (5.17)
jωx0
The physical meaning of the electric governing equation (5.15) is as follows. The acoustic
signal (input) oscillates the diaphragm when the electrostatic transducer acts as a sensor.
The resultant velocity .u generates current (the flow of charges) via the second term on the
right-hand side of (5.15) with the transduction coefficient (5.17). The generated current
causes the voltage drop via the first term on the right-hand side of (5.15). The left-hand side
of (5.15) represents the overall voltage that the transducer outputs.
When the transducer acts as a transmitter (actuator), the applied voltage represented by the
left-hand side of (5.15) causes charges to flow into or from the stationary capacitor (depend-
ing on the polarity). This flow of charges oscillates the diaphragm via the corresponding
electric force, as discussed above.

5.2.3 Mechanical Governing Equation of Electrostatic Transducer

Now we consider the mechanical governing equation (5.7). The first term on the right-hand
side of (5.7) represents the electric force on the diaphragm due to current .i. The second term
represents the mechanical force on the diaphragm, called . f cm , which causes the diaphragm
to oscillate harmonically. Below, we consider the second term first.
The explicit form of . f cm is not readily available. So, we find it out from the equation of
motion for the diaphragm. Let .m be the mass, .x be the displacement from the equilibrium
position of the diaphragm, .γ and .k be the damping coefficient and the stiffness of the elastic
mechanism that pulls the diaphragm to the equilibrium position. We can put the equation of
motion as follows.
.m ẍ = −γ ẋ − kx + f cm (5.18)
Expressing the velocity of the diaphragm .u = ẋ and rearranging the order of the terms, we
can rewrite the equation of motion (5.18) as follows.
∫
. f cm = γ ẋ + m ẍ + kx = γ u + m u̇ + k udt (5.19)
128 5 Electrical-Mechanical Transduction

Assuming that the motion of the diaphragm is sinusoidal with frequency .ω, we can express
∫
u̇ = jωu, . udt = u/( jω) and put the equation of motion (5.19) in the following form.
.

( )
1 k
. f cm = γ u + m jωu + k u = γ u + j mω − u (5.20)
jω ω

Next, consider the electric force on the diaphragm, the first term on the right-hand side
of (5.7). Call this force . f ce . The electric field inside the capacitor, . E, acting on the change
in the stored charge, .q, generates . f ce . Since .q << q0 , we can use the initial charge .q0 to find
. E (with the application of Gauus’ law [2]).

q0 v0
. E= = (5.21)
∈S x0
Here (5.10) and (5.11) are used in going through the second equal sign. Note that the
rightmost-hand side of (5.21) represents the uniform electric field inside the capacitor with
the initial gap length .x0 . When charge .q << q0 flows into the capacitor, the capacitor’s
electrodes feel electric force . f ce as follows.
v0 v0
. f ce = q E = q= i (5.22)
x0 jωx0
Here .i is the current flowing into the capacitor and can be expressed as .i = q̇ = jωq.
Using (5.20) and (5.22) we can express the total force on the diaphragm, . f , as
( ( ))
v0 k
. f = f ce + f cm = i + γ + j mω − u (5.23)
jωx0 ω

Comparison of (5.7) and (5.23) identifies the expressions for the mechanical to electrical
transduction coefficient and the mechanical impedance as follows.
v0
. Tme = T = (5.24)
jωx0
( )
k
. z m0 = γ + j mω − (5.25)
ω

Equations (5.17) and (5.24) confirm that the electrostatic transducer is reciprocal, .Tem =
Tme = T .

5.2.4 Physics Behind Electrostatic Transducer’s Operation

Consider first the physics behind the mechanical governing equation. The mechanical force
expression (5.20) tells us the following features of the force. First, the greater the damping
coefficient.γ and the mass.m, the stronger force is necessary to excite the oscillation at a given
velocity .u. This makes sense because higher .γ means more mechanical energy dissipation
5.2 Reciprocal Transduction 129

in the mechanical system. Second, the complex unit . j in front of the parenthesis on the
rightmost-hand side indicates the elastic force represented by .(ω/k) is .π/2 in phase behind
the damping force. Finally, setting the quantity inside the parenthesis to zero represents
the situation where the mechanical force . f cm is null regardless of the velocity .u. This
corresponds to the resonant condition where the frequency of the mechanical input matches
the resonant frequency of the elastic mechanism (the diaphragm and its support). Setting
√
the inside the parenthesis to zero provides us with angular frequency .ω0 = k/m. This is
the natural angular frequency of a harmonic system, see (2.5) in Sect. 2.1.2.
Now consider the electro-mechanical interaction from the viewpoint of electrodynamics.
Figure 5.2 illustrates the above electrostatic transducer as two capacitors. Here the left
capacitor represents the capacitor formed by the left stationary electrode and the diaphragm,
and the right one represents the capacitor formed by the diaphragm and the right stationary
electrode. The diaphragm’s voltage is fixed at .v0 as mentioned above (Fig. 5.1).
First, consider the behavior of the charges stored in both capacitors when the transducer
acts as an acoustic actuator. Assume that initially the diaphragm is in the middle of the space
between the two stationary electrodes. Thus, the two capacitors’ capacitance is the same as
.C 1 = C 2 = C 0 . The charge stored in the two capacitors is .q0 , hence

q 0 = C 0 v0
. (5.26)

Consider that the current source causes charge .q << q0 to flow from the left capacitor to the
right capacitor. We can implement this operation by raising the voltage at the top-left corner
of the circuit and lowering the top-right corner of the circuit and turning on the switch.
Charge .q will flow as an electric current due to the potential difference. At this moment

Fig. 5.2 a Electrostatic acoustic actuator and b sensor as two capacitors connected in series
130 5 Electrical-Mechanical Transduction

the voltage across the two capacitors commonly remains at .v0 . However, the stored charges
become different between the two capacitors. The left electrode of the left capacitor loses .q,
so the new charge will be .−q0 − q. The right electrode of the right capacitor’s new charge
will be .−q0 + q. Since the voltage across both capacitors is .v0 , this situation leads to the
following expressions for the new capacitance .C1 and .C2 .
q0 + q q0 q0 q0
C1 =
. = + = C0 + (5.27)
v0 v0 v0 v0
q0 − q q0 q0 q0
.C 2 = = − = C0 − (5.28)
v0 v0 v0 v0
Equations (5.27) and (5.28) indicate that .C1 > C2 , meaning that after the flow of charge .q,
the left capacitor’s gap length decreases and the right capacitor’s gap length increases. In
other words, if charges flow rightward through the current source, the diaphragm moves to
the left.
Now consider that the transducer acts as an acoustic sensor. Assume that the mechanical
force pushes the diaphragm to the left by.Δx. The new capacitance of each capacitor becomes
as follows.
S
C1 = ∈
. > C0 (5.29)
x − Δx
S
.C 2 = ∈ < C0 (5.30)
x + Δx
At the moment when the diaphragm moves, the charges on the two capacitors are still .q0
whereas .C1 > C0 and .C2 < C0 . This situation leads to the following expressions for the
new voltage across the electrodes.
q0
v 0 − v1 = ∈
. (5.31)
C1
q0
.v0 − v2 = ∈ (5.32)
C2
Since .C1 > C2 , (5.31) and (5.32) lead to the following equations.

v 0 − v 1 < v 0 − v2 → v 1 > v 2
. (5.33)

Equation (5.33) indicates that when the diaphragm moves to the left, current will flow to the
right as .v1 > v2 .
It is interesting to discuss the role of the electric force . f ce in the force governing equation
when the transducer acts as a sensor. (When the transducer is an actuator, obviously the elec-
tric force drives the diaphragm’s oscillation, as we discussed above.) Consider the situation
where the mechanical force acts on the diaphragm while the AC circuit is open (.i = 0). The
right illustration in Fig. 5.2b with the terminals labeled .v1 and .v2 open corresponds to this
5.2 Reciprocal Transduction 131

situation. Under this condition, the electric force term in the mechanical governing equation
(5.23) is null. Call the total force on the diaphragm under this condition . f o . By setting .i = 0
in (5.23) we find the following expression for . f o .
( )
k
. f o = f cm = γ u + j mω − u (5.34)
ω

We can define the mechanical impedance under this condition as

( )
fo k
. z mo = = γ + j mω − (5.35)
u ω

Now consider that the AC circuit is short-circuited, and the electric force term becomes
active in the mechanical governing equation. At the same time, the voltage in the electric
governing equation (5.6) becomes zero (.v = 0). From the latter condition, we obtain the
following equation.
.i = −φu (5.36)
Substituting (5.36) into the first term on the right-hand side of (5.23), and using (5.5), (5.16)
and (5.17) under the reciprocal condition (5.4) to find .v0 /( jωx0 ) = φ/( jωC0 ), we obtain
the following expression.
( ) ( )
v0 k φ2 k
. fs = (−φu) + γ u + j mω − u=− u + γ u + j mω − u
jωx0 ω jωC0 ω
(5.37)
Then defining .z ms as . f s /u and using (5.35), we obtain the following expression.
( ) ( 2 )
k − φ 2 /C0 φ /C0
. z ms = γ + j mω − = z mo + j (5.38)
ω ω

We can view the effect of short-circuiting the AC circuit as the decrease in the stiffness by
φ 2 /C0 (i.e., from .k to .k − φ 2 /C0 ). What does this observation mean? It means as follows.
.

When the mechanical force shifts the diaphragm toward the left stationary electrode, the
capacitance of the space between the diaphragm and the left electrode increases. At the
same time, the capacitance of the space between the diaphragm and the right stationary
electrode decreases because the gap distance increases. Consequently, the electric energy
stored in the left capacitor increases, and the energy stored in the right capacitor decreases.
Nature is not fond of such energy imbalance, and therefore it forces the left capacitor
to discharge by carrying an electric current flowing through the wire connecting the two
stationary electrodes. If we short-circuit the circuit under this condition, the charges on the
capacitor’s electrodes flow with no resistance to the other electrode. If we open the circuit,
there is no way for the charges to flow. In this case, nature tries to oppose the movement of
the diaphragm. The mechanical force feels more resistance from the system as compared
132 5 Electrical-Mechanical Transduction

with the short-circuited case. This explains why the stiffness is lower when the AC circuit
is short-circuited.
We can argue this extra stiffness based on the electric force on the diaphragm as well.
Under the “open circuit” condition, when the diaphragm moves to the left the left electrode
has higher voltage than the right electrode (.v1 > v2 .) This means that the electric field
vector is directed from the left stationary electrode towards the right stationary electrode
through the diaphragm. Since the diaphragm is positively charged, this electric field exerts a
rightward electric force on the diaphragm. This is a resistive force for the mechanical force.
The external agent exerting the mechanical force feels a higher stiffness.

5.3 Anti-reciprocal Transducers

In this case, the mechanical-to-electrical and the electrical-to-mechanical transduction coef-

ficients have mutually opposite signs.

. Tem = −Tme (5.39)

Define the transformation factor .φ M as follows.

φ M = Tem = −Tme
. (5.40)

The subscript . M denotes “magnetic” because the magnetic field plays the main role in
this transduction, as will be discussed later in this section. The electrical and mechanical
governing equations take the following form.

v = z e0 i + φ M u
. (5.41)
. f = −φ M i + z m0 u (5.42)

Similarly to the reciprocal case, consider short-circuiting the electric circuit to make
v = 0. From (5.41), the current under this condition becomes .i = −φ M u/z e0 . Substituting
.

this into (5.42) yields . f s and .z ms = f s /u as follows.

1
. f s = φ 2M u + z m0 u (5.43)
z e0
φ2
. z ms = z m0 + M (5.44)
z e0

5.3.1 Moving-Coil Transducer

The most typical example of an anti-reciprocal transducer is the moving-coil transducer. In

this case, .z e0 in (5.43) and (5.44) is the electrical impedance consisting of the DC resistance
5.3 Anti-reciprocal Transducers 133

. R0 and inductance . L 0 .
. z e0 = R0 + jωL 0 (5.45)
Unlike the electrostatic case, the open-circuited condition makes the electric force inactive.
This is because while in the electrostatic case, the electric force is the Coulomb force
proportional to the voltage across the capacitor, in the moving-coil case the electric force is
the magnetic force generated in the coil in proportion to the current.
According to electrodynamics [2], the magnetic force due to the magnetic field . B acting
on current .i flowing through a conductive wire of length .l is . f B = i Bl. On the other hand,
when the electric current moves in the magnetic field with velocity .u, the magnetic induction
induces an opposing voltage of . Blu.
Consider the above-mentioned mechanism in more detail by referring to Fig. 5.3. In this
figure, a permanent magnet applies a constant magnetic field into the page. Two conductive
wires are connected to either an AC power supply (Fig. 5.3a) or a resistor (Fig. 5.3b). A
conductor bar is placed on the two wires. Here (Fig. 5.3a) models an acoustic actuator and
(b) models an acoustic sensor. The conductor bar represents a diaphragm in both cases.
In the acoustic actuator configuration, the transducer induces a force on the diaphragm
(represented with . f ) in response to an input electric current, whereas in the acoustic sensor
configuration, the transducer induces an electric current in response to the motion of the
diaphragm (represented with velocity .u) due to mechanical force. In Fig. 5.3, the dashed
quantity depicts the induced quantity.
When the transducer operates as an actuator, the AC power supply generates current.
Figure 5.3a illustrates the moment when the current from the power supply flows upward in
the conductor bar (length .l). Here, .x̂ etc. are the unit vector of the direction. The magnetic
field due to the permanent magnet (.B = B ŷ) exerts magnetic force leftward. The diaphragm
moves to the left. In the next cycle when the current flips the direction, the magnetic force
exerts rightward force on the diaphragm. In this fashion, the diaphragm oscillates. The
magnetic force at the moment when the electric current is .i = i ẑ, the magnetic force .f B is
.−Bli x̂

.f B = i ẑ × B ŷl = Bli(ẑ × ŷ) = −Bli x̂ (5.46)

When the transducer operates as a sensor, the mechanical force causes velocity .u. Mag-
netic induction states that the induced voltage .v is the change in the magnetic flux .φ B = B A
over time. Here . A is the area of the magnetic field. Since the area . A changes over time as
the conductor bar moves at velocity .u while the magnetic field is constant, we can express
the temporal change in the magnetic flux as follows.
dφ B dA
v=
. =B = Blu (5.47)
dt dt
According to Faraday’s law, the induced current resulting from magnetic induction flows in
such a way that the resultant magnetic field opposes the temporal change in the magnetic
134 5 Electrical-Mechanical Transduction

Fig. 5.3 Conceptual (a) Acoustic actuator

illustration of a magnetic force
generated by an acoustic
actuator and b magnetic
induction occurring in an
acoustic sensor

diaphragm
(b) Acoustic sensor

flux. When the magnetic force causes the rightward velocity as shown in Fig. 5.3b, the
magnetic force increases the magnetic flux. Hence, the induced current flows in the direction
that generates a magnetic field in the out-of-page direction inside the loop formed by the
conductor bar and the two wires.
Figure 5.4 illustrates a more realistic structure of a moving-coil transducer. Typically, a
moving-coil transducer consists of a solenoid wound around a cylindrical core placed inside
the inner space of a doughnut-shaped permanent magnet. The diaphragm (not shown in Fig.
5.4 to avoid confusion) is attached to the solenoid in such a way that the solenoid’s motion
parallel to the axis of the core cylinder (called the vertical motion) oscillates the diaphragm.
In this configuration, as the solenoid experiences vertical motion, the magnetic flux varies.
In the situation illustrated in Fig. 5.3, the area that penetrates the permanent magnetic field
changes as the conductor bar moves parallel to the plane. In the case of Fig. 5.4, the (active)

Fig. 5.4 Schematic view of a

moving-coil transducer

AC power supply

Electric magnet
(Solenoid)
N
Permanent magnet
S
Core

Magnetic field by
permanent magnet
5.3 Anti-reciprocal Transducers 135

area changes as the conductor (the solenoid) moves perpendicular to the plane that the
permanent magnet penetrates. In this situation, the circular area formed by each turn of
the solenoid defines the active area involved in the magnetic flux. When the solenoid is
moving out of the vertical field line of the permanent magnet, the induced voltage decreases
as the circular area of each turn goes out of the effective zone of the permanent magnet.
This is the phase when the induced voltage decreases. In the opposite phase, as each turn of
the solenoid enters the effective zone of the permanent magnetic field, the induced voltage
increases because the number of circular areas that contributes to the magnetic induction
increases.
Figure 5.5 illustrates another way of understanding the magnetic induction in the con-
figuration illustrated in Fig. 5.4. The top illustration of this figure represents the situation
where a solenoid carries an electric current provided by an AC voltage supply. A permanent
magnet is partially located inside the cylindrical space of the solenoid with two poles as
shown in the figure. In the phase when the current increases in the direction depicted by
solid arrows (the current increases clockwise when the solenoid’s cross-sectional area is
viewed from the left side of the solenoid), the solenoid increases the magnetic field vector
rightward along the axis of the cylindrical space. Electromagnetic induction reduces this
change. Since the magnetic field due to the permanent magnet is leftward, going deeper into
the solenoid performs this task. Thus, the permanent magnet is pushed to the right. The solid
big arrow in Fig. 5.5 indicates the direction in which the permanent magnet is pushed.
We can understand the above effect in terms of the interaction between two magnets.
When the current flows in the direction indicated by the solid line, the solenoid forms
an electric magnet with the S-pole on the left and the N-pole on the right. (Consider this
imaginary permanent magnet in the space of the solenoid’s core on the right of the permanent
magnet.) On the other hand, the polarity of the permanent magnet is opposite, as the lower
illustration of Fig. 5.5 indicates. The two S-poles facing each other repel the magnets from
one another. Hence, the permanent magnet is pushed to the left. When the solenoid current
increases, the electric magnet’s repelling force increases. Nature does not like this abrupt
change, so it exerts a counterforce. This force corresponds to the above-mentioned opposing
effect due to electromagnetic induction.

Fig. 5.5 Explanation of

temporal change in magnetic
flux in moving-coil transducer N S

N S S N
136 5 Electrical-Mechanical Transduction

In the other phase, when the current flows in the opposite direction (depicted by the
thin dashed line in the top illustration), the direction of the magnetic field generated by
the solenoid flips, whereas the direction of the permanent magnetic field remains the same.
Consequently, the permanent magnet is pulled into the solenoid, as the thick arrow with the
dashed line indicates.
The above description illustrates the dynamics when the solenoid carries current. Hence
it applies to the explanation of an acoustic actuator. Instead of supplying current to the
solenoid, consider that a mechanical force pushes the permanent magnet into the solenoid.
According to the law of magnetic induction, it will induce an electric current. This situation
corresponds to the case when the transducer acts as an acoustic sensor.
In the above explanation, we assumed that the solenoid is stationary in space, and the
permanent magnet is movable. Since all the interactions between the permanent magnet and
the solenoid are relative, we can argue the same effect by making the permanent magnetic
field stationary and the solenoid movable. The dynamics illustrated in Fig. 5.4 corresponds
to that case.
The magnetic force and induced voltage yield the following two governing equations. The
above argument indicates that for a common direction of the electric current, the direction
of the induced force . f and the velocity .u caused by the magnetic force are opposite to each
other. This explains the negative sign appearing in the first term on the right-hand side of
the mechanical governing equation (5.49).

v = z e0 i + Blu
. (5.48)
. f = −Bli + z m0 u (5.49)

Comparison of the set of equation (5.48) and (5.49) with (5.41) and (5.42) indicates the
following expression for .φ M
.φ M = Bl (5.50)
Repeating the same type of argument as the electrostatic transducer case, we find the
following force expression when the electric circuit is open. Note that, unlike the electro-
static case, the electromagnetic force is inactive when the electric circuit is open. This is
because, in this case, the electromagnetic force is magnetic force, which is proportional to
the current. Thus, calling this force . f o we obtain the following expressions for the force and
the corresponding impedance .z m0 .
( )
k
. fo = γ u + j mω − u (5.51)
ω
( )
k
. z m0 = γ + j mω − (5.52)
ω

When the electric circuit is closed and therefore .v = 0, from (5.48) and (5.50) we find
i = −Blu/z e0 = −φ M u/z e0 . Substitution of this current into (5.49) with the use of (5.25)
.

yields the expressions for the corresponding force . f s and impedance .z ms as follows.
5.3 Anti-reciprocal Transducers 137

( )
φ 2M k + φ 2 /L 0
. fs = u + z m0 u = γ u + j mω − u (5.53)
z e0 ω
( )
k + φ 2 /L 0
. z ms = γ + j mω − (5.54)
ω

Here assuming . R0 << jωL 0 , we use .z e0 ≈ jωL 0 ((5.45). As expected in the case of the
moving coil transducer the short-circuited impedance has greater stiffness than the open-
circuited case (i.e., .k vs. .k + φ 2 /L 0 ).

5.3.1.1 Piezoelectric Transducer

As indicated by the name, the piezoelectric transducer uses the phenomenon referred to as
the piezoelectric effect [3] and the inverse piezoelectric effect [4, 5]. The piezoelectric effect
refers to the property of certain materials that exhibits voltage in proportion to the applied
stress. The materials that exhibit the piezoelectric effect are called piezoelectric materials.
The inverse piezoelectric effect refers to the reversed effect. In response to applied electric
voltage, piezoelectric materials are compressed or stretched in proportion to the applied
voltage. The applied stress or electric voltage can be in high frequency, such as the ultra-
sonic frequency. Thus, a variety of ultrasonic transducers are commercially available. Here,
ultrasonic stress sensors use the piezoelectric effect, and ultrasonic transmitters (actuators)
use the inverse effect.
The piezoelectric effect is essentially the induction of electric dipole moment in the
material. The electric dipole is defined as a pair of unlike electric charges, .q and .−q,
separated by a small distance .d. [2] The electric dipole moment .p is defined as the product
of them, .p = qd. It is a vector quantity whose magnitude is .qd, and the direction is the
displacement vector .d drawn from the negative charge to the positive charge of the pair. The
polarization is defined as the density of the electric dipole moment.
We can explain the essence of the piezoelectric and inverse effects based on the above
explanation of dipole moment and polarization vectors. Take quartz (SiO.2 ) as an example
since it is a well-known solid that exhibits piezoelectric and inverse effects. Figure 5.6 illus-
trates a simplified, two-dimensional configuration of quartz crystal. Figure 5.6a represents
the stress-free situation where in each pair of neighboring atoms the electrons are closer to
the oxygen atoms but the entire crystal is electrically neutral. The electrons are closer to the
oxygen atom because it has higher electron affinity than the silicon atom. The entire crystal
is neutral because the electron distribution makes it so.
Now imagine that you stretch the crystal vertically as indicated by Fig. 5.6b. In the
enclosed region near the top of the crystal, the silicon atom has moved away from the
two oxygen atoms. Consequently, the polarity of this region has become more positive as
compared with the stress-free state. In other words, the contribution of the two relatively
negative oxygen atoms is lower in the stretched state. On the other side of the crystal,
the exact opposite phenomenon occurs. The contribution of the relatively positive silicon
138 5 Electrical-Mechanical Transduction

(a) (b) (c)

Fig. 5.6 Unit cell of quartz a stress-free, b under vertical tensile stress, c under vertical compressive
stress

atoms is reduced and consequently, this end becomes negative. Overall, the entire crystal is
polarized with the top side positive and the bottom side negative.
If the crystal is compressed vertically, the situation is reversed (Fig. 5.6c.) On the top side
of the crystal, the contribution of the two oxygen atoms increases and therefore this side
becomes negative. On the bottom side, the contribution of the silicon atom increases, and this
side becomes positive. If the stretching and compressing stresses are applied alternatively, the
polarity changes accordingly. If we place a quartz crystal on the surface of a solid specimen
undergoing compression and stretching, the quartz crystal exhibits alternative positive and
negative electric polarities. This quartz crystal behaves as a stress sensor. It outputs the
electric signal sensing the stretching and compressive stress caused by the specimen.
It is now easy to understand the same quartz crystal as an ultrasonic stress transmitter
(actuator). When an alternating electric voltage is applied to the crystal, the dipole moment
is induced. Consequently, the crystal is stretched and compressed, following the change in
the electric polarity. This demonstrates the quartz crystal as an ultrasonic transmitter.
The efficiency of the piezoelectric effect is indicated by the piezoelectric coefficient
(usually denoted by .d33 ) in the unit of m/V (meter/volt). The piezoelectric coefficient of
quartz is approximately 3.×10−12 m/V [6].
Many ceramics that have a perovskite structure (ABX.3 ) [8] exhibit the piezoelectric effect.
Here A atoms are larger metal cations, B atoms are smaller metal cations, and X atoms are
anions (usually oxide). PZT (lead zirconate titanate Pb [.Zrx Ti1−x ].O3 .(0 ≤ x ≤ 1) is one of
the most used piezoelectric materials. It is a binary solid-solution of PbZrO.3 and PbTiO.3
having a perovskite structure with Pb.2+ for A, O.2− for B, and Ti.4+ or Zr.4+ for B.
Figure 5.7a illustrates the structure of a unit cell of PZT when the cell is above the Curie
point. With the Ti.4+ or Zr.4+ ion at the center of the cubic cell, the polarization is null.
Below the Curie temperature, the Ti.4+ or Zr.4+ ion shifts from the center generating a dipole
moment as illustrated in Fig. 5.7b. When external force stretches or compresses the unit cell,
the position of the Ti.4+ or Zr.4+ changes, exhibiting the piezoelectric effect. When the unit
References 139

(a) (b)

Pb2+

O2-

Ti4+, Zr4+

Fig. 5.7 Unit cell of lead zirconate titanate PZT; a above Currier temperature and b below Currier
temperature

cell is under an external electric field, the electric force changes the location of the Ti.4+ or
Zr.4+ ion, exhibiting the inverse piezoelectric effect.
The piezoelectric coefficient of PZT is two orders of magnitude higher than quartz [6].
Normally, PZT cell units are randomly oriented. To make an efficient acoustic transducer,
it is necessary to pole the crystal by applying a high electric field [9].

References

1. Kinsler LE, Frey AR, Coppens AB, Sanders JV (1980) Fundamentals of acoustics, 3rd edn, Chap.
14. Wiley, New York, USA
2. See, for example, Griffiths DJ (1999) Introduction o electrodynamics, 3rd edn. Prentice Hall,
Upper Saddle River, NJ, p 74
3. Curie J, Curie P (1880) Bulletin de la Société Minérologique de France 3(4):90–93
4. Lippmann G (1881) Principe de la conservation de l’électricité (Principle of the conservation of
electricity). Annales de chimie et de physique (in French) 24:145
5. Curie J, Curie P (1881) Contractions et dilatations produites par des tensions dans les cristaux
hémièdres á faces inclinées (Contractions and expansions produced by voltages in hemihedral
crystals with inclined faces). Comptes Rendus (in French) 93:1137–1140
6. Bottom VE (1970) Measurement of the piezoelectric coefficient of quartz using the Fabry-Perot
dilatometer. J Appl Phys 41:3941
7. Kholkim AL, Kiselev DA, Kholkine LA, Safari A (2008) Piezoelectric and electrostrictive ceram-
ics transducer and actuators. In: Schwartz M (ed) Smart materials, 1st edn, Chap. 9. CRC Press,
London, New York
8. Tilley RJD (2016) Pervoskites, 1st edn. Wiley, Chichester, UK
9. Waller D, Iqbal T, Safari A (1989) Poling of lead zirconate titanate ceramics and flexible piezo-
electric composites by the corona discharge technique. J Am Ceramic Soc 72(2):322–324
Appendix A
Direction Cosines
A

Assume .lˆ be a unit vector in the direction that makes angles of .θx , .θ y , and .θz to the three
coordinate axes.
.lˆ = l x î + l y ĵ + l z k̂ (A.1)
Consider the following scalar products based on the definition .A · B = AB cos θ AB . Here
A and .B are arbitrarily chosen vectors, . A and . B are their magnitude, and .θ AB is the angle
.

between the two vectors.

lˆ · î = |l||
. ˆ î| cos θx = cos θx (A.2)
.lˆ · ĵ = |l||
ˆ ĵ| cos θ y = cos θ y (A.3)
.lˆ · k̂ = |l||
ˆ k̂| cos θz = cos θz (A.4)

ˆ .î, . ĵ, and .k̂ are all unit vectors, hence their magnitudes are all
Here we used the fact that .l,
one.
Now substitute the component expression (A.1) into the left-hand sides of (A.2)–(A.4).
( )
.lˆ · î = l x î + l y ĵ + l z k̂ · î = l x (A.5)
( )
.lˆ · ĵ = l x î + l y ĵ + l z k̂ · ĵ = l y (A.6)
( )
.lˆ · k̂ = l x î + l y ĵ + l z k̂ · k̂ = l z (A.7)

Comparing (A.2)–(A.4) with (A.5)–(A.7), we find that the components of .lˆ are the direction
cosines.
.lˆ = cos θ x î + cos θ y ĵ + cos θz k̂ (A.8)

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Appendix B
Gradient Operation
B

Consider the differentiation of . f (t, x, y, z) referring to Fig. B.1. The total differential .d f is
given by the following expression.
∂f ∂f ∂f
.df = dx + dy + dz (B.1)
∂x ∂y ∂z
As indicated in Fig. B.1, the addition expressed by the right-hand side of (B.1) consists of
three portions each corresponding to one of the following three steps; finding the change in
the value of . f associated with the parallel movement along the .x-axis by .d x, the change
associated with the parallel movement along the . y-axis by .dy, and that associated with
.dz. While we can find the scalar quantity .d f with (B.1), each of these three steps has

a directionality. The short answer to the question “What is the three-dimensional version
of .d f /d x?” is the gradient operation (vector) given by (1.22). Since this concept is not
straightforward, at least for me, I would like to take a moment to consider it.
( )
∂ ∂ ∂ ∂f ∂f ∂f
.∇ f = î + ĵ + k̂ f = î + ĵ + k̂ (B.2)
∂x ∂y ∂z ∂x ∂y ∂z
Here the operator .∇ is called the gradient operator, and .î, . ĵ, and .k̂ are the unit vector for .x,
.y, and .z directions. This operator applies to a scalar function and produces a gradient vector.
We use the case of a Cartesian coordinate here but we can express the gradient operator for
other coordinate systems.
Let’s start the discussion with a simple case where the wave function . f (t, x) depends
on a single spatial coordinate variable .x. For this discussion, we assume that the function
. f represents the gravitational potential to make the argument intuitive. The slope indicates

the path of an object to fall by gravity if it is released. Figure B.2a illustrates an example of
such a case where the potential curve is expressed as follows.

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144 Appendix B: Gradient Operation

Fig. B.1 Total differential .d f consisting of partial differential with respect to .x, . y and .z

( )1
. f (x) = 102 − x 2 2 (B.3)

The potential curve represents an arc of a circle with a radius of 10. Consider the slope at
two representative points, .x = ±0.5. As the arrows in this figure indicate, the slope at these
two points has the same magnitude and mutually opposite signs. Thus, even in this case, the
slope shows a vector-like behavior. Since we are dealing with only one-spatial coordinate,
the slope appears to be a signed scalar.

Fig. B.2 a One-dimensional gravitational potential; b Two-dimensional potential depending on .x

and independent of . y; c Two-dimensional potential having quadratic dependence on .x and . y

Figure B.2b is the case where the wave function has two independent spatial variables as
. f (t, x, y). Its . x-dependence of the wave function is the same as Fig. B.2a and . y-dependence

is null, i.e., .∂ f /∂ y = 0. Imagine you put a tennis ball at a representative point .(x, y) =
(−0.2, 0) and release it from your hand. It will fall in a direction parallel to the .x-axis. If
you repeat the same thing on the other side at .(x, y) = (0.2, 0), the tennis ball will fall in
the opposite direction but still parallel to the .x-axis. This observation is the same as the one
we made for Fig. B.2a, i.e., the slope has an opposite sign (direction). In this case, since the
Appendix B: Gradient Operation 145

function . f has two independent spatial variables (.x and . y), the slope is literally a vector
quantity.1
Consider the situation in Fig. B.2b in terms of (B.2). Substituting (B.3) into (B.2), we
obtain the following gradient vector.
∂ ( 2 )1 ∂ ( 2 )1 −x
∇f =
. 10 − x 2 2 î + 10 − x 2 2 ĵ = √ î (B.4)
∂x ∂y 102 − x 2
As expected from the above intuitive observation of the tennis ball, the gradient vector has
only .x-component, corresponding to the ball falling parallel to the .x-axis. The two arrows in
Fig. B.2b indicate this coordinate dependence of the gradient. The arrow pointing downward
represents the .x-component, and the one on the ridge of the . f (x, y) indicates the zero . y-
component. Equation (B.4) also reveals the fact that the slope (gradient) is proportional to
the coordinate value .x with a negative sign; if the reference coordinate is positive (.xr > 0),
it is negative. This situation corresponds to the fact that the gradient decreases with .xr , i.e.,
its positive value decreases as .xr increases positively; consequently, the ball falls faster as
. xr increases. Similarly, we can explain the observation that the falling speed of the ball

decreases as the negative reference coordinate increases toward .xr = 0, i.e., when .xr < 0,
the gradient is positive.
Now in Fig. B.2c, the gravitational potential has the same dependence on .x and . y.
Consider the motion of a tennis ball when you release if at .(x, y) = (0, −1). You can easily
imagine that the ball falls along a longitudinal line because in the tangential direction,
the height does not change. As is the case of Fig. B.2b, let’s consider the gradient vector
quantitatively. In this case, the function . f (x, y) has a dependence on .x and . y and can be
expressed as follows.
( 2 )1
. f (x, y) = 10 − x − y
2 2 2
(B.5)
Substituting (B.5) into (B.2), we obtain the following expression for the gradient vector.
−x −y
∇f = √
. î + √ ĵ (B.6)
10 − x − y
2 2 2 10 − x 2 − y 2
2

This time, the gradient vector has both .x and . y components. This fact is consistent with the
above qualitative observation that the ball should fall along a longitudinal line. In a top view,
a longitudinal line contains both .x and . y components. However, (B.6) is not convenient to
find the longitudinal direction that the ball falls from a reference point. It is better to use
spherical coordinates.
Using.r̂ ,.θ̂ and.ψ̂ to represent the unit vector for the radial, polar, and azimuthal component
of . f (r , θ, ψ) we can express (B.2) and (B.5) as follows.

1 You may wonder what if we place the tennis ball at a point where the . x coordinate is 0. It is unclear
if the ball stays there forever or fall in either the positive or negative .x direction. In theory, the ball
does not fall. In reality, it will fall either direction as another force, such as air pressure acts on it.
146 Appendix B: Gradient Operation

Fig. B.3 Coordinate point expressed with a Cartesian and b Spheric coordinate systems

. f (r , θ, ψ) = r cos θ = 10 cos θ (B.7)

∂f 1∂f 1 ∂f
.∇ f = r̂ + θ̂ + ψ̂ (B.8)
∂r r ∂θ r sin θ ∂ψ
Here we use that the surface shown in Fig. B.2c is part of the surface of a sphere with a
radius of 10.
Substituting (B.7) into (B.8), we find the gradient vector as follows.
1
.∇f = (−r sin θ )θ̂ = (− sin θ )θ̂ (B.9)
r
The arrows in Fig. B.2c represent a tangential vector (the one parallel to the .x y-plane),
and .∇ f expressed by (B.9) at a reference point. The tangential vector corresponds to the
one on the ridge seen in Fig. B.2b.
Referring to Fig. B.3 we find the following relationship between the spheric and Cartesian
coordinate variables.

. x = r sin θ cos ψ, y = r sin θ sin ψ (B.10)

. cos θ θ̂ = cos ψ î + sin ψ ĵ (B.11)

Using .r = 10 and noting that from (B.10) .x 2 + y 2 = r 2 sin2 θ (cos2 ψ + sin2 ψ) = 102
.sin θ , we can write (B.6) as follows.
2

−x −y −10 sin θ cos ψ −10 sin θ sin ψ

∇ f =√
. î+ √ ĵ = î + ĵ
102 − (x 2 + y 2 ) 102 − (x 2 + y 2 ) 10 cos θ 10 cos θ
(B.12)
sin θ ( )
. =− cos ψ î + sin ψ ĵ (B.13)
cos θ
Appendix B: Gradient Operation 147

Now use (B.11) in (B.9) to derive the following equation.

1 ( )
∇ f = (− sin θ )θ̂ = (− sin θ )
. cos ψ î + sin ψ ĵ (B.14)
cos θ
Comparison of (B.12) and (B.14) indicates that these two equations are identical.
Appendix C
Orthogonality of Sine and Cosine Functions
C

According to Fourier’s theorem [1, 2] a periodic function can be expanded into a series of
cosine and sine functions.
a0
. f (t) = + a1 cos(ω0 t) + a2 cos(2ω0 t) + · · · + an cos(nω0 t)
2
+ b1 sin(ω0 t) + b2 sin(2ω0 t) + · · · + bn sin(nω0 t) (C.1)

Here .ω0 is the lowest angular frequency. If function . f (t) is defined on .[0 T ], the lowest
frequency . f 0 and the corresponding angular frequency are as follows.
1
. f0 = (C.2)
T
2π
ω0 = 2π f 0 =
. (C.3)
T
With this notation, we can express (C.1) as follows.
n ( ( ) ( ))
a0 ∑ 2π k 2π k
. f (t) = + ak cos t + bk sin t
2 T T
k=1

a0 ∑
n
= + {ak cos (ωk t) + bk sin (ωk t)} (C.4)
2
k=1

The next step is to determine the coefficients .an and .bn . We can use the following property to
determine the coefficients. This property is known as the orthogonality of the trigonometric
functions.

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150 Appendix C: Orthogonality of Sine and Cosine Functions

∫ t1 +T
. cos(nω0 t) cos(mω0 t)dt =0 (n / = m) (C.5)
t1
T π
. = = (n = m) (C.6)
2 ω0
Here the following identities are used.
1
. cos(nω0 t) cos(mω0 t) = {cos((m − n)ω0 t) + cos((m + n)ω0 t)} (C.7)
2
1
. sin(nω0 t) sin(mω0 t) = {cos((m − n)ω0 t) − cos((m + n)ω0 t)} (C.8)
2
1
. cos (nω0 t) =
2
{1 + cos(2nω0 t)} (C.9)
2
1
. sin (nω0 t) =
2
{1 − sin(2nω0 t)} (C.10)
2
Multiply .cos(2ω0 t) to both sides of Eq. (C.1) and integrate for one period (from .t = t1 to
t = t1 + T ). From the orthogonality, all terms vanish except .a2 cos(2ω0 t) on the right-hand
.

side. Thus, using Eq. (C.6), we find that coefficient .a2 can be expressed as follows.
∫ t1 +T
2
.a2 = f (t) cos(2ω0 t)dt (C.11)
T t1

It is clear that other coefficients can be evaluated with the same procedure. Generally,
coefficients .an and .bn can be expressed as follows.
∫ t1 +T
2
.an = f (t) cos(nω0 t)dt (C.12)
T t1
∫ t1 +T
2
. bn = f (t) sin(nω0 t)dt (C.13)
T t1
Appendix D
Notes on Standing Waves
D

When the amplitude of the forward-going wave (. A f ) is different from that of the
backward-going wave (. Ab ), the superposed wave can be expressed as follows.

A f sin(ωt − kz) + Ab sin(ωt + kz)

. (D.1)
= 2 Ab cos(kz) sin(ωt) + (A f − Ab ) sin(ωt − kz)

Equation (D.1) indicates that in this case the expression of the superposed wave consists of the
standing-wave term (the first term on the right-hand side of Eq. (D.1)), and the traveling-wave
term (the second term on the right-hand side of Eq. (D.1)). Because of this traveling-wave
term, the superposed wave is behavior differs from the case when the forward-going and
backward-going waves have the same amplitude. First, the superposed wave is not zeroed
out even when the phase difference between the forward-going and backward-going waves
is an odd integer multiple of .π . Figure D.1 illustrates the situation. Here, Fig. D.1a shows
the forward-going (top), the back-ward-going (middle) and superposed (bottom) waves for
three representative moments when the interference is destructive (left), constructive (right)
and intermediate (center). Notice that under the destructive interference the superposed
wave is completely zeroed out whereas under the constructive interference, the peak of the
superposed wave is doubled as compared with the forward-going wave. The superposed
wave oscillates as a standing wave. Figure D.1b exhibits the situation when the amplitude
of the backward-going wave is 25% smaller than the forward-going beam (. Ab = 0.75A f ).
It is seen that even when the two wave are completely out of phase (the phase difference is
.π ), the superposed wave is not zeroed-out.

Second, the superposed wave travels. According to the traveling-wave term of Eq. (D.1),
the superposed wave travels forward if the amplitude of the forward-going wave is greater
than the backward-going, and it travels backward if the amplitude of the forward-going wave
is smaller than the backward-going wave. The superposed wave in Fig. D.1b travels but is

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152 Appendix D: Notes on Standing Waves

Fig. D.1 Superposition of two waves propagating in mutually opposite directions. a When forward-
going and backward-going waves have the same amplitude. b When the forward-going wave has
greater amplitude

not obvious. This is because the amplitude of the traveling-wave term in Eq. (D.1) increases
in proportion to the difference in amplitude between the forward-going and backward-
going waves. Figure D.2 shows that the superposed wave travels forward clearly when the
amplitude of the backward-going wave is 25% of the forward-going wave (. Ab = 0.25A f ).
With the increase in the difference between . A f and . Ab the amplitude of the traveling-
wave term increases. Thus, the traveling feature of the superposed wave is much clearer as
compared with . Ab = 0.75A f case shown in Fig. D.1b.
Appendix D:Notes on StandingWaves 153

Fig. D.2 Superposition of two waves where backward-going wave’s amplitude is 25% of the forward-
going wave
Appendix E
Generalized Hooke’s Law and Equation of Motion
E
for Isotropic Media

E.1 Strain and Stress Matrices

The strain matrix consists of normal strain components and shear strain components.
⎛ ∂ξ ∂ξ ∂ξ ⎞
x x x
⎛ ⎞
⎜ ∂ x ∂ y ∂z ⎟ ∈x x ∈x y ∈x z
⎜ ∂ξ y ∂ξ y ∂ξ y ⎟ ⎝
.[∈] = ⎜ ⎟ = ∈ yx ∈ yy ∈ yz ⎠ (E.1)
⎝ ∂ x ∂ y ∂z ⎠
∂ξz ∂ξz ∂ξz ∈zx ∈zy ∈zz
∂x ∂y ∂z

Here the normal strain and shear strain components are defined as follows.
( )
1 ∂ξi ∂ξi
.∈ii = + (E.2)
2 ∂ xi ∂ xi
( )
1 ∂ξ j ∂ξi
.∈i j = + (E.3)
2 ∂ xi ∂x j

where .i, j = 1, 2, 3 and .x1 = x, x2 = y, x3 = z and .ξ1 = ξx , ξ2 = ξ y , ξ3 = ξz .

Similarly, we define the stress matrix as follows.
⎛ ⎞
σx x σx y σx z
.[σ ] = ⎝σ yx σ yy σ yz ⎠ (E.4)
σzx σzy σzz

Here .σi j represents the stress on the plane .i in the direction of . j as indicated in Fig. E.1.
The plane .i is the plane normal to the .i axis. Stress is defined as a force acting on a plane
divided by the area of the plane. For instance, .σx x (x) is the force applied in the positive .x
direction on the plane located at .x perpendicular to the .x axis. .σ yx is the force acting on the
plane perpendicular to the . y axis at . y in the direction of positive .x.
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156 Appendix E: Generalized Hooke’s Law and Equation of Motion for Isotropic Media

Fig. E.1 Normal and shear

stress components. The first
subscript denotes the plane and
the second the direction of the
stress (force) vector

E.2 Compliance and Stiffness Matrices

The compliance or stiffness matrix constitutes the basis of generalized Hooke’s law. These
two matrices are mutually the inverse of the other and essentially represent the same stress-
strain relationship.
The compliance matrix . S and stiffness matrix .[C] are defined as follows.

[∈] = [S][σ ]
. (E.5)
[σ ] = [C][∈]
. (E.6)

Here .[∈] and .[σ ] are the strain and stress matrices in Voigt’s notation [3]. In this notation
the 3.×3 matrices (E.2) and (E.4) become a vector with 6 components, respectively. Voigt’s
notation is applicable to a symmetric matrix in general.
( )†
. [∈] = ∈x x ∈ yy ∈zz ∈ yz ∈zx ∈x y (E.7)
( )†
.[σ ] = σ x x σ yy σzz σ yz σzx σ x y (E.8)

Here .† represents transpose.

For later uses, we express the compliance and stiffness matrices in the following compo-
nent forms. ⎛ ⎞
S11 S12 S13 S14 S15 S16
⎜S S S S S S ⎟
⎜ 21 22 23 24 25 26 ⎟
⎜ ⎟
⎜ S31 S32 S33 S34 S35 S36 ⎟
.[S] = ⎜ ⎟ (E.9)
⎜ S41 S42 S43 S44 S45 S46 ⎟
⎜ ⎟
⎝ S51 S52 S53 S54 S55 S56 ⎠
S61 S62 S63 S64 S65 S66
⎛ ⎞
C11 C12 C13 C14 C15 C16
⎜C C C C C C ⎟
⎜ 21 22 23 24 25 26 ⎟
⎜ ⎟
⎜C31 C32 C33 C34 C35 C36 ⎟
.[C] = ⎜ ⎟ (E.10)
⎜C41 C42 C43 C44 C45 C46 ⎟
⎜ ⎟
⎝C51 C52 C53 C54 C55 C56 ⎠
C61 C62 C63 C64 C65 C66
Appendix E: Generalized Hooke’s Law and Equation of Motion for Isotropic Media 157

Fig. E.2 When expressed with the coordinate system of principal axes .x yz the shear stress is zero

E.3 Principal Axes

Before proceeding to further analysis, take a moment to consider the coordinate system to
express stress and strain. Refer to Fig. E.2 and consider that an external force is acting on
an elastic cubic object. In three dimensions, we can set up an .x yz coordinate system in
any direction. In this figure, we consider two coordinate systems, .x yz and .x ' y ' z ' . Notice
that the .x yz has the three coordinate axes normal to the corresponding plane of the cube.
Consequently, there is no shear stress component. The coordinate axes that satisfy this
condition are called principal axes [4]. The axes of the other coordinate system .x ' y ' z ' are
not principal axes because when decomposed with this coordinate system, the force vector
has shear components.

E.4 Isotropic Case

Now we are in a position to express the strain tensor components with the .x yz and .x ' y'z'
coordinate systems. We restrict ourselves in an isotropic case. Here the isotropy means
that the elasticity is independent of the orientation. Under this condition, the elasticity is
plane-symmetric.

E.4.1 Axis Inversion

Using plane symmetry, we can reduce the number of independent compliance matrix ele-
ments. First, consider axis inversions. The elastic response is symmetric about the .x y plane.
This means that if the .z axis is inverted the displacement components behave as follows.

ξx' = ξx , ξ y' = ξ y , ξz' = −ξz

. (E.11)
158 Appendix E: Generalized Hooke’s Law and Equation of Motion for Isotropic Media

On the other hand, the spatial derivatives behave as follows.

∂ ∂ ∂ ∂ ∂ ∂
. = , = , =− (E.12)
∂x' ∂ x ∂ y' ∂ y ∂z ' ∂z
Consequently, the strain components behave as follows through this inversion operation.

∂ξx' ∂ξx
.∈x' x = '
= = ∈x x (E.13)
∂x ∂x
'
∂ξ y' ∂ξ y
. ∈ yy = = = ∈ yy (E.14)
∂ y' ∂y
' ∂ξz' ∂(−ξz ) ∂ξz
∈zz
. = =− = = ∈zz (E.15)
∂z ' ∂z ∂z
( ) ( )
' 1 ∂ξz' ∂ξ y' 1 ∂ξz ∂ξ y
. ∈ yz = + = − − = −∈ yz (E.16)
2 ∂ y' ∂z ' 2 ∂y ∂z
( ) ( )
' 1 ∂ξx' ∂ξz' 1 ∂ξx ∂ξz
. ∈zx = + ' = − − = −∈zx (E.17)
2 ∂z ' ∂x 2 ∂z ∂x
( ) ( )
1 ∂ξ y' ∂ξx' 1 ∂ξ y ∂ξx
. ∈x' y = + = + = ∈x y (E.18)
2 ∂x' ∂ y' 2 ∂x ∂y

Repeating the same procedure for the stress matrix components, we obtain the following
equations.

σx' x = σx x , σ yy
.
' '
= σ yy , σzz '
= σzz , σ yz '
= −σ yz , σzx = −σzx , σx' y = σx y (E.19)

Now express the first row of (E.5) using the components of the compliance matrix in
both .x yz and .x ' y ' z ' systems, respectively.

.∈x x = S11 σx x + S12 σ yy + S13 σzz + S14 σ yz + S15 σzx + S16 σx y (E.20)
∈x' x = S11 σx' x + '
S12 σ yy + '
S13 σzz + '
S14 σ yz + '
S15 σzx + S16 σx' y
. = S11 σx x + S12 σ yy + S13 σzz − S14 σ yz − S15 σzx + S16 σx y (E.21)
Appendix E: Generalized Hooke’s Law and Equation of Motion for Isotropic Media 159

⎛ ⎞ ⎛ ⎞⎛ ' ⎞ ⎛ ⎞⎛ ⎞
∈x' x S11 S12 S13 S14 S15
S16 σx x S11 S12 S13 S14 S15 S16 σx x
⎜∈ ' ⎟ ⎜ S S26 ⎟ ⎜ ' ⎟ ⎜ S26 ⎟ ⎜ ⎟
⎜ yy ⎟ ⎜ 21 S22 S23 S24 S25 ⎟ ⎜σ yy ⎟ ⎜ S21 S22 S23 S24 S25 ⎟ ⎜ σ yy ⎟
⎜ ∈' ⎟ ⎜ ⎟⎜ ' ⎟ ⎜ ⎟⎜ ⎟
⎜ zz ⎟ ⎜ S31 S32 S33 S34 S36 ⎟ ⎜ σzz
S35 ⎟ ⎜ S31 S32 S33 S34 S35 S36 ⎟ ⎜ σzz ⎟
.⎜ ' ⎟ = ⎜ ⎟⎜ ' ⎟ = ⎜ ⎟⎜ ⎟
⎜ ∈ yz ⎟ ⎜ S41 S42 S43 S44 S46 ⎟ ⎜ σ yz
S45 ⎟ ⎜S S S S44 S45 S46 ⎟ ⎜−σ yz ⎟
⎜ ' ⎟ ⎜ ⎟ ⎜ ' ⎟ ⎜ 41 42 43 ⎟⎜ ⎟
⎝ ∈zx ⎠ ⎝ S51 S52 S53 S54 S55
S56 ⎠ ⎝ σzx ⎠ ⎝ S51 S52 S53 S54 S55 S56 ⎠ ⎝−σzx ⎠
∈x' y S61 S62 S63 S64 S65
S66 σx' y S61 S62 S63 S64 S65 S66 σx y
⎛ ⎞⎛ ⎞ ⎛ ⎞
S11 S12 S13 −S14 −S15 S16 σx x ∈x x
⎜S −S24 −S25 S26 ⎟ ⎜ ⎟ ⎜ ⎟
⎜ 21 S22 S23 ⎟ ⎜σ yy ⎟ ⎜ ∈ yy ⎟
⎜ ⎟⎜ ⎟ ⎜ ⎟
⎜S S32 S33 −S34 −S35 S36 ⎟ ⎜ σzz ⎟ ⎜ ∈zz ⎟
= ⎜ 31 ⎟⎜ ⎟ = ⎜ ⎟ (E.22)
⎜ S41 S42 S43 −S44 −S45 S46 ⎟ ⎜ σ yz ⎟ ⎜−∈ yz ⎟
⎜ ⎟⎜ ⎟ ⎜ ⎟
⎝ S51 S52 S53 −S54 −S55 S56 ⎠ ⎝ σzx ⎠ ⎝−∈zx ⎠
S61 S62 S63 −S64 −S65 S66 σx y ∈x y

By changing the sign of the fourth and fifth rows of (E.5) we obtain the following equation.
⎛ ⎞ ⎛ ⎞⎛ ⎞
∈x x S11 S12 S13 S14 S15 S16 σx x
⎜∈ ⎟ ⎜ S ⎟ ⎜ ⎟
⎜ yy ⎟ ⎜ 21 S22 S23 S24 S25 S26 ⎟ ⎜σ yy ⎟
⎜ ⎟ ⎜ ⎟⎜ ⎟
⎜ ∈zz ⎟ ⎜ S31 S32 S33 S34 S35 S36 ⎟ ⎜ σzz ⎟
.⎜ ⎟=⎜ ⎟⎜ ⎟ (E.23)
⎜−∈ yz ⎟ ⎜−S41 −S42 −S43 −S44 −S45 −S46 ⎟ ⎜ σ yz ⎟
⎜ ⎟ ⎜ ⎟⎜ ⎟
⎝−∈zx ⎠ ⎝−S51 −S52 −S53 −S54 −S55 −S56 ⎠ ⎝ σzx ⎠
∈x y S61 S62 S63 S64 S65 S66 σx y

Comparing (E.22) and (E.23), we obtain the following equation.

⎛ ⎞ ⎛ ⎞
S11 S12 S13 −S14 −S15 S16 S11 S12 S13 S14 S15 S16
⎜ S S S −S −S S ⎟ ⎜ S S26 ⎟
⎜ 21 22 23 24 25 26 ⎟ ⎜ 21 S22 S23 S24 S25 ⎟
⎜ ⎟ ⎜ ⎟
⎜ S31 S32 S33 −S34 −S35 S36 ⎟ ⎜ S31 S32 S33 S34 S35 S36 ⎟
.⎜ ⎟=⎜ ⎟ (E.24)
⎜ S41 S42 S43 −S44 −S45 S46 ⎟ ⎜−S41 −S42 −S43 −S44 −S45 −S46 ⎟
⎜ ⎟ ⎜ ⎟
⎝ S51 S52 S53 −S54 −S55 S56 ⎠ ⎝−S51 −S52 −S53 −S54 −S55 −S56 ⎠
S61 S62 S63 −S64 −S65 S66 S61 S62 S63 S64 S65 S66

Note that on the left-hand side, the fourth and fifth columns (corresponding to the non-
inverted axes) the sign is flipped, and on the right-hand side, the fourth and fifth rows
(corresponding to the non-inverted axes) the sign is flipped. From (E.24) we find as follows.

. S14 = S24 = S34 = S15 = S25 = S35 = S46 = S56

= S41 = S42 = S43 = S51 = S52 = S53 = S64 = S65 = 0 (E.25)

Note that the first and second lines of (E.25) list the matrix elements at mutually symmetric
positions. Thus, by considering inverting the .z axis, we find the compliance matrix has the
following form.
160 Appendix E: Generalized Hooke’s Law and Equation of Motion for Isotropic Media

⎛ ⎞
S11 S12 S13 0 0 S16
⎜S S26 ⎟
⎜ 21 S22 S23 0 0 ⎟
⎜ ⎟
⎜ S31 S32 S33 0 0 S36 ⎟
.[S] = ⎜ ⎟ (E.26)
⎜ 0 0 0 S44 S45 0 ⎟
⎜ ⎟
⎝ 0 0 0 S54 S55 0 ⎠
S61 S62 S63 0 0 S66
Similarly, by considering inverting the .x axis, we obtain the following equation. Those
matrix elements found to be 0 from (E.25) are set to 0 in the below equation.
⎛ ⎞ ⎛ ⎞
S11 S12 S13 0 0 −S16 S11 S12 S13 0 0 S16
⎜S S S 0 −S26 ⎟ ⎜ S26 ⎟
⎜ 21 22 23 0 ⎟ ⎜ S21 S22 S23 0 0 ⎟
⎜ ⎟ ⎜ ⎟
⎜ S31 S32 S33 0 0 −S36 ⎟ ⎜ S31 S32 S33 0 0 S36 ⎟
.⎜ ⎟=⎜ ⎟ (E.27)
⎜ 0 0 0 S44 −S45 0 ⎟ ⎜ 0 0 0 S44 S45 0 ⎟
⎜ ⎟ ⎜ ⎟
⎝ 0 0 0 S54 −S55 0 ⎠ ⎝ 0 0 0 −S54 −S55 0 ⎠
S61 S62 S63 0 0 −S66 −S61 −S62 −S63 0 0 −S66

From (E.27), we find as follows.

. S45 = S16 = S26 = S36 = 0; S54 = S61 = S62 = S63 = 0 (E.28)

Substituting (E.27) into (E.26), we can further simplify the compliance matrix as follows.
⎛ ⎞
S11 S12 S13 0 0 0
⎜S S S ⎟
⎜ 21 22 23 0 0 0 ⎟
⎜ ⎟
⎜ S31 S32 S33 0 0 0 ⎟
.[S] = ⎜ ⎟ (E.29)
⎜ 0 0 0 S44 0 0 ⎟
⎜ ⎟
⎝ 0 0 0 0 S55 0 ⎠
0 0 0 0 0 S66

E.4.2 Rotation Around Axis

Next, consider rotating the .x y plane around the .z axis by angle .θ . The plane symmetry
condition indicates that the elastic property is intact by this operation. Call the coordinate
system before this rotational operation .x yz and the one after .x ' y ' z ' . By expressing the same
point on the plane by expressing it with the two coordinate systems, we obtain the following
equation.
⎛ ⎞ ⎛ ⎞ ⎛ '⎞
x cos θ − sin θ 0 x
. ⎝ y ⎠ = ⎝ sin θ cos θ 0⎠ ⎝ y ' ⎠ (E.30)
z 0 0 1 z'
Similarly, we can express the displacement vector components .ξx , .ξ y , and .ξz written in
the old coordinate system with the new coordinate system as follows.
Appendix E: Generalized Hooke’s Law and Equation of Motion for Isotropic Media 161

⎛ ⎞ ⎛ ⎞⎛ '⎞
ξx cos θ − sin θ 0 ξx
. ⎝ξ y ⎠ = ⎝ sin θ cos θ 0⎠ ⎝ξ y' ⎠ (E.31)
ξz 0 0 1 ξz'

Next, express the spatial derivatives of the new coordinate system with the old one using
(E.30) and (E.31).
∂ ∂ ∂x ∂ ∂y ∂ ∂
. = + = cos θ + sin θ (E.32)
∂x' ∂x ∂x' ∂ y ∂x' ∂x ∂y
∂ ∂ ∂x ∂ ∂y ∂ ∂
.
'
= '
+ '
= − sin θ + cos θ (E.33)
∂y ∂x ∂y ∂y ∂y ∂x ∂y
∂ ∂
. = (E.34)
∂z ' ∂z
Using the above relationship between the quantities expressed with .x yz and .x ' y ' z ' sys-
tems, we can express .∈x' x in terms of the strain tensor components expressed with the .x yz
system.
( )
' ∂ξx' ∂ ( ) ∂ ∂ ( )
.∈ x x = = ξ x cos θ + ξ y sin θ = cos θ + sin θ ξx cos θ + ξ y sin θ
∂x' ∂x' ∂x ∂y
∂ξx ∂ξ y ∂ξx ∂ξ y
= cos2 θ + cos θ sin θ + sin θ cos θ + sin2 θ
∂x ∂x ∂y ∂y
= ∈x x cos2 θ + ∈ yy sin2 θ + 2∈x y cos θ sin θ (E.35)

Similarly, we obtain the following equations.

'
.∈ yy = ∈x x sin2 θ + ∈ yy cos2 θ − 2∈x y cos θ sin θ (E.36)
'
. ∈zz = ∈zz (E.37)
'
.∈ yz = ∈ yz cos θ − ∈zx sin θ (E.38)
'
.∈zx = ∈zx cos θ + ∈ yz sin θ (E.39)
'
( )
.∈ x y = 2∈x y cos2 θ − sin2 θ + 2(∈ yy − ∈x x ) cos θ sin θ (E.40)

With the same procedure, we can obtain the following equation for .σx x .

σx' x = σx x cos2 θ + 2σx y cos θ sin θ + σ yy sin2 θ

. (E.41)

We can express .∈x x etc. on the right-hand side of (E.35) with stress matrix components
by using the compliance matrix expression (E.29).
162 Appendix E: Generalized Hooke’s Law and Equation of Motion for Isotropic Media

∈x' x = ∈x x cos2 θ + ∈ yy sin2 θ + 2∈x y cos θ sin θ

.
( ) ( )
= S11 σx x + S12 σ yy + S13 σzz cos2 θ + S21 σx x + S22 σ yy + S23 σzz sin2 θ
+ 2S66 σx y cos θ sin θ
( ) ( )
= σx x S11 cos2 θ + S21 sin2 θ + σ yy S12 cos2 θ + S22 sin2 θ
( )
+ σzz S13 cos2 θ + S23 sin2 θ + σx y (2S66 ) cos θ sin θ (E.42)

Now this time first express .∈x' x with .σi j using (E.22) and the compliance matrix (E.29).
Note that because of the isotropic condition, the elastic behavior does not change by the
rotational operation, hence we can use the same compliance matrix.

∈x' x = S11 σx' x + S12 σ yy

.
' '
+ S13 σzz
( )
= S11 σx x cos2 θ + σ yy sin2 θ + 2σx y cos θ sin θ
( )
+ S12 σx x sin2 θ + σ yy cos2 θ − 2σx y cos θ sin θ + S13 σzz
( ) ( )
= σx x S11 cos2 θ + S12 sin2 θ + σ yy S12 cos2 θ + S11 sin2 θ
+ σzz (S13 ) + σx y (2S11 − 2S12 ) cos θ sin θ (E.43)

Comparing (E.42) and (E.43) for the same .σii terms, we find as follows.

. S12 = S21 (E.44)

. S22 = S11 (E.45)
. S66 = S11 − S12 (E.46)

Repeating the same procedure for rotation around other axis, we find the following condi-
tions.

S11 = S22 = S33 ; S12 = S13 = S23 = S21 = S31 = S32 ; S44 = S55 = S66 = S11 − S12
.

(E.47)
Using (E.44)–(E.46) in (E.29), we obtain the following expression of compliance matrix.
⎛ ⎞
S11 S12 S12 0 0 0
⎜S S S ⎟
⎜ 12 11 12 0 0 0 ⎟
⎜ ⎟
⎜ S12 S12 S11 0 0 0 ⎟
.[S] = ⎜ ⎟ (E.48)
⎜ 0 0 0 S11 − S12 0 0 ⎟
⎜ ⎟
⎝ 0 0 0 0 S11 − S12 0 ⎠
0 0 0 0 0 S11 − S12

Expression (E.48) is the general form of compliance matrix for an isotropic elastic medium.
Appendix E: Generalized Hooke’s Law and Equation of Motion for Isotropic Media 163

E.5 Expression of Compliance Matrix with Elastic Constants

We next express the compliance matrix with elastic modulus. Consider in Fig. E.3 that exter-
nal force acts on a block of an elastic object along the .x axis. The elastic block experiences
stretch in the .x direction and compression in the . y and .z directions according to Poisson’s
effect. Clearly, there is no shear strain if we use the.x yz coordinate system.
As shown in Fig. E.3, the external force along the.x axis causes three normal strains; tensile
strain along the .x axis, and compressive strains along the . y and .z axes. The compressive
strains are due to Poisson’s effect. The compressive strains can be related to the tensile strain
via Poisson’s ratio .ν. The overall strain (the addition of all three strains) constitutes .∈x x .
Thus, we obtain the following equation.
σx x σ yy σzz
.∈x x = −ν −ν (E.49)
E E E
Here the first term on the right-hand side represents that the tensile strain in the .x direction
is related to .σx x via Young’s modulus . E. The other terms represent that the normal strain is
related to the normal stress and that the compressive strain is lower than the tensile strain
by a factor of Poisson’s ratio .−ν where the negative sign indicates the strain is opposite to
the one along the .x axis, which is the direct consequence of the applied force.
Comparison of the compliance matrix (E.48) and (E.49) allows us to identify . S11 and . S12
as follows.
1 −ν
. S11 = , S12 = (E.50)
E E
Using (E.50), we can express the compliance matrix for an isotropic elastic medium with
Youg’s modulus and Poisson’s ratio, and thereby write the strain-stress relationship (E.5) as
follows. ⎛ ⎞ ⎛ ⎞⎛ ⎞
∈x x 1 −ν −ν 0 0 0 σx x
⎜∈ ⎟ ⎜−ν 1 −ν 0 ⎟ ⎜σ ⎟
⎜ yy ⎟ ⎜ 0 0 ⎟ ⎜ yy ⎟
⎜ ⎟ 1 ⎜ ⎟⎜ ⎟
⎜ ∈zz ⎟ ⎜−ν −ν 1 0 0 0 ⎟ ⎜ σzz ⎟
.⎜ ⎟= ⎜ ⎟⎜ ⎟ (E.51)
⎜ ∈ yz ⎟ E ⎜ 0 0 0 1+ν 0 0 ⎟ ⎜ σ yz ⎟
⎜ ⎟ ⎜ ⎟⎜ ⎟
⎝ ∈zx ⎠ ⎝ 0 0 0 0 1 + ν 0 ⎠ ⎝ σzx ⎠
∈x y 0 0 0 0 0 1+ν σx y
By finding the inverse matrix to the compliance matrix in (E.51), we can express the stress-
strain relationship (E.6) in the following form.

Fig. E.3 Elastic block y

experiencing normal strains

z
x
164 Appendix E: Generalized Hooke’s Law and Equation of Motion for Isotropic Media

Table E.1 Various elastic moduli and their relationships

(.λ, μ) (. E, G) (. E, ν) (.G, ν)
μ(3λ+2μ)
.E .
λ+μ .E .E .2(1 + ν)G
.G(μ) .μ .G .
E .G
2(1+ν)
.ν .
λ .
E−2G .ν .ν
2(λ+μ) 2G
.λ .λ .
G(E−2G)
.
νE .
2νG
3G−E (1+ν)(1−2ν) 1−2ν

⎛ ⎞ ⎛ ⎞⎛ ⎞
σx x 1−ν ν ν 0 0 0 ∈x x
⎜σ ⎟ ⎜ ν 1−ν ν 0 ⎟ ⎜ ⎟
⎜ yy ⎟ ⎜ 0 0 ⎟ ⎜∈ yy ⎟
⎜ ⎟ ⎜ ⎟⎜ ⎟
⎜ σzz ⎟ E ⎜ −ν −ν 1 − ν 0 0 0 ⎟ ⎜ ∈zz ⎟
.⎜ ⎟= ⎜ ⎟⎜ ⎟
⎜ σ yz ⎟ (1 + ν)(1 − 2ν) ⎜ 0 0 0 1 − 2ν 0 0 ⎟ ⎜ ∈ yz ⎟
⎜ ⎟ ⎜ ⎟⎜ ⎟
⎝ σzx ⎠ ⎝ 0 0 0 0 1 − 2ν 0 ⎠ ⎝ ∈zx ⎠
σx y 0 0 0 0 0 1 − 2ν ∈x y
(E.52)
Conventionally, the stiffness matrix is expressed with Lamé’s first and second parameters .λ
and .μ. ⎛ ⎞ ⎛ ⎞⎛ ⎞
σx x λ + 2μ λ λ 0 0 0 ∈x x
⎜σ ⎟ ⎜ λ λ + λ ⎟ ⎜∈ ⎟
⎜ yy ⎟ ⎜ 2μ 0 0 0 ⎟ ⎜ yy ⎟
⎜ ⎟ ⎜ ⎟⎜ ⎟
⎜ σzz ⎟ ⎜ λ λ λ + 2μ 0 0 0 ⎟ ⎜ ∈zz ⎟
.⎜ ⎟=⎜ ⎟⎜ ⎟ (E.53)
⎜ σ yz ⎟ ⎜ 0 0 0 2μ 0 0 ⎟ ⎜ ∈ yz ⎟
⎜ ⎟ ⎜ ⎟⎜ ⎟
⎝ σzx ⎠ ⎝ 0 0 0 0 2μ 0 ⎠ ⎝ ∈zx ⎠
σx y 0 0 0 0 0 2μ ∈x y
Table E.1 summarizes the relationship among Lamé’s parameters, Young’s modulus,
shear modulus, and Poisson’s ratio. Notice that we can always express an elastic constant
with a pair of other elastic constants.

E.6 Equation of Motion

Refer to Fig. E.4 and consider the force acting on an elastic block of density .ρ having three
sides of .d x, .dy, and .dz. The net force in the .x, . y, and .z directions is the addition of the
normal and shear force in line with the respective direction. Thus, we obtain the following
equation of motion for this elastic block.

( ) ( )
∂ 2ξ ∂σx x ∂σ yx ∂σzx ∂σx y ∂σ yy ∂σzy
.ρ = + + î + + + ĵ
∂t 2 ∂x ∂y ∂z ∂x ∂y ∂z
( )
∂σx z ∂σ yz ∂σzz
+ + + k̂ (E.54)
∂x ∂y ∂z
Appendix E: Generalized Hooke’s Law and Equation of Motion for Isotropic Media 165

y
+ +

=
= + +

y x

z + +

+ + = + +

= + +

Fig. E.4 All external force acting on an elastic block. The three cubes in this figure represent the same
elastic block. Each of the three cubes illustrates the net force exerted in the corresponding direction

Here, .î, . ĵ, and .k̂ are the unit vector for the corresponding direction. Remember that the first
subscript .i of the stress tensor component .σi j represents the plane on which the force is
acting and the second subscript . j represents the direction of the force. The volume .d xd ydz
is canceled on both-hand sides on deriving (E.54).
Using (E.53) we can substitute the stress tensor components with the pertinent strain
tensor components and obtain the following equation.
( )
∂ 2ξ ∂ [ ] ∂∈ yx ∂∈zx
.ρ = (λ + 2μ)∈x x + λ∈ yy + λ∈zz + 2μ + 2μ î
∂t 2 ∂x ∂y ∂z
( )
∂∈x y ∂ [ ] ∂σzy
+ 2μ + 2λ∈x x + (λ + 2μ)∈ yy + 2λ∈zz + ĵ
∂x ∂y ∂z
( )
∂∈x z ∂∈ yz ∂ [ ]
+ 2μ + 2μ + 2λ∈x x + 2λ∈ yy + (λ + 2μ)∈zz k̂ (E.55)
∂x ∂y ∂z

The next step is to express the right-hand side of (E.55) in terms of displacement.ξ . Remember
the following expressions of the strain tensor components.
∂ξi
. ∈ii = (E.56)
∂ xi
( )
1 ∂ξ j ∂ξi
.∈i j = + (E.57)
2 ∂ xi ∂x j
166 Appendix E: Generalized Hooke’s Law and Equation of Motion for Isotropic Media

Using (E.56) and (E.57), consider expressing the .x component of the right-hand side of
(E.55) expressing with the displacement vector components.

∂ [ ] ∂∈ yx ∂∈zx
. (λ + 2μ)∈x x + λ∈ yy + λ∈zz + 2μ + 2μ
∂x ∂y ∂z
[ ] ( ) ( )
∂ ∂ξx ∂ξ y ∂ξz ∂ ∂ξ y ∂ξx ∂ ∂ξx ∂ξz
= (λ + 2μ) +λ +λ +μ + +μ +
∂x ∂x ∂y ∂z ∂ y ∂x ∂y ∂z ∂z ∂x
( 2 ) ( )
∂ ξx ∂ ξx
2 ∂ ξx
2 ∂ ∂ξx ∂ξ y ∂ξz
=μ + + + (λ + μ) + + (E.58)
∂x 2 ∂y 2 ∂z 2 ∂x ∂x ∂y ∂z

We can view the first term on the last line of (E.58) as the .x component of .∇ 2 ξ , and the
second term as the .x component of .∇(∇ · ξ ). By repeating the same term rearrangement for
the. y and.x components on the right-hand side of (E.55), we obtain the following expression.

∂ 2ξ
ρ
. = μ∇ 2 ξ + (λ + μ)∇(∇ · ξ ) (E.59)
∂t 2
Equation (E.59) is the equation of motion we wanted to derive.

References

1. Prof. Brad Osgood (2014) Lecture notes for EE 261 the Fourier transform and its applications.
CreateSpace Independent Publishing Platform
2. Bracewell RN (1999) The Fourier transform and its applications, 3rd edn. McGraw-Hill, Boston,
New York
3. Brannon RM (2018) Rotation, reflection, and frame changes, Chap. 26, Voigt and Mandel com-
ponents. IOP, Bristol, UK
4. Chou PC, Pagano NJ (1967) Elasticity: tensor, dyadic, and engineering approaches. Dover Pub-
lications, New York
Index

A Decay constant, 27, 43

Acoustic energy conservation at boundary, 62 Decaying acoustic wave solution, 44
Acoustic impedance, 55, 60, 63 Decaying pressure wave in air, 71
Acoustic radiation pressure, 69 Decay wave equation of volume expansion, 43
Acoustic wave intensity, 56 Destructive interference, 14
Acoustoelasticity, 29 Direction cosines, 9
Amplitude demodulation, 81 Dispersion, 112
Amplitude modulation, 79
Angular spatial frequency, 7
Angular temporal frequency, 7 E
Audible frequency, 73 Eigen frequency, 17, 23
Average acoustic energy, 53 Elastic constant, 32
Average intensity of acoustic wave, 57 Elastic force, 24
Elasticity, 23
Elasticity of air, 66
B
Elastic potential energy, 26
Boundary condition, 17–19
Elastic potential energy density, 51
Bulk modulus, 35
Electrical impedance, 55, 123, 127, 133
Electric conductivity, 33
C Equation of motion due to pressure gradient, 70
Complex propagation constant, 45 Equation of motion for longitudinal vibration,
Constructive interference, 14 40
Critical angle, 59, 108 Equation of motion governing series of point
masses, 30
Equation of motion of air compression, 36
D Equation of motion of isotropic solids, 87
Damping coefficient, 27 Equation of motion of spring-point mass system,
Deaying, longitudinal displacement wave 26
equation, 44 Equation of motion of unit volume, 37
Deaying, transverse displacement wave Equation of motion of unit volume with
equation, 44 velocity-damping force term, 43

© The Editor(s) (if applicable) and The Author(s), under exclusive license 167
to Springer Nature Switzerland AG 2024
S. Yoshida, Fundamentals of Acoustic Waves and Applications, Synthesis Lectures on Wave
Phenomena in the Physical Sciences, https://doi.org/10.1007/978-3-031-48200-7
168 Index

Euler’s notation, 45 M
Magnetic force, 133
Mechanical impedance, 123
F
Faraday’s law, 134
Fixed-end reflection, 18 N
Forced oscillation, 24, 28 Natural frequency, 27
Fourier series, 12 Normal strain, 32
Fourier’s theorem, 12 Normal stress, 33
Frequency demodulation, 83
Frequency modulation, 82
Frequency ranges of vowel and consonants, 74 O
One-dimensional wave equation of series of
point masses, 29
G Open-end reflection, 19
Gas law, 68 Overtones, 77
Gradient operation, 11
P
H Particle velocity, 2
Harmonic oscillation, 24, 27 Period, 2
Harmonics, 13 Phase velocity, 1, 8
Hooke’s law, 32 Phase velocity as a delay of oscillation in a
Hooke’s law in air, 35 series of point masses, 30
Hooke’s law of continuum, 33 Phase velocity as a material constant, 33, 43
Human ear, 72 Phase velocity of air expansion (compression)
wave, 38
Phase velocity of air pressure wave, 38
I Phase velocity of compression wave in isotropic
Impedance matching, 55 media, 88
Initial condition, 16 Phase velocity of longitudinal displacement
Inverse piezoelectric effect, 137 wave, 34
Phase velocity of rotation wave in isotropic
media, 88
K Phase velocity of transverse displacement wave,
Kinetic energy density, 53 34
Piezoelectric effect, 137
Plane strain wave solution, 98
L Plane wave, 15
Laplacian, 11 Plane wave in elastic media, 91
Law of reflection, 59 Primary (P-) wave, 88, 89
Law of refraction, 59 Propagation constant, 9
Linear differential equation, 28 Propagation vector, 9, 92
Longitudinal and transverse vibrations in solid, P-wave velocity, 96
39
Longitudinal elastic wave equation, 42
Longitudinal elastic wave velocity, 42 R
Longitudinal wave, 4 Radiation pressure, 69
Rayleigh-Lamb frequency equation, 116
Rayleigh surface wave, 111
Index 169

Reflectance, 63 Transverse elastic wave equation, 42

Reflection at free surface with various incident Transverse elastic wave velocity, 42
waves, 104 Transverse waves, 5
Reflection coefficient, 62 Traveling waves, 8
Resonance, 14
Resonance of wave, 20
Resonator, 14 U
Resonator mode number, 21 Unforced oscillation, 24, 26
Rotation wave equation of solids, 88
V
Variable separation, 15
S
Vector potential, 88
Scalar potential, 88
Vector potential wave equation, 89
Scalar potential wave equation, 89
Velocity damping, 26
Secondary (S-) wave, 88, 89
Volume compression wave equation of solids,
Shear-Horizontal wave solution, 102
88
Shear modulus, 34
. S H waves in a plate, 112
Snell’s law, 59 W
Spatial frequency, 7 Wave equation, 9
Spring constant, 27 Wave equation of air density change, 39
Spring-mass system, 24 Wave equation of air pressure, 38
Standing wave, 13 Wave equation of air volume expansion
Stiffness, 27 (compression), 37
S-wave velocity, 96 Wave equation of density change, 39
Wavefront, 92
Wavelength, 2
T
Temporal frequency, 7
Transmission coefficient, 62 Y
Transmittance, 63 Young’s modulus, 33