P hysi cs | 12.

13

11.2 Closed pipe

L
The first three modes of vibration of a closed pipe are given as follows:

For fundamental or first harmonic mode in Fig. 12.13.

A N
1
λ1 v
=l ; λ=
1 4l; n=
1 4
4 4l
Figure 12.13

For the first overtone or third harmonic in Fig. 12.14

3λ2 4l 3v
=l ;=
λ2 ;=
n2
4 3 4l A N A N
33
4
For the second overtone or the fifth harmonic, Figure 12.14
5λ2 4l 5v
=l ;=
λ3 ;=
n
4 5 3 4l

For the pth overtone or (2p+1)th harmonic, n =

( 2p + 1) v .
2l A N ANA N
43
At the open end of the pipe, the antinode is formed at a small distance outside the open end. 4
Thus, the correct length of the closed pipe is l + e and that for an open pipe, it is l + 2 e and
Figure 12.15
e is equal to 0.3D where D is the internal diameter of the pipe.

12. DETERMINATION OF SPEED OF SOUND IN AIR

12.1 Resonance Column Method

Generally, systems have one or more natural
vibrating frequencies. Further, when a
system is driven at a natural frequency, then
there is a maximum energy transfer and the
vibrating amplitude steadily increases till up
to a maximum. However, when a system is
driven at a natural frequency, we say that L=3/4
L=/4
the system is in resonance (with the driven
source) and refer to the particular frequency
at which this occurs as a resonance frequency.
Moreover, from the relationship between the
frequency f, the wavelength λ , and the wave
speed v, which is λf =v , it is very obvious
that if both the frequency and wavelength
are known, then the wave speed can be Figure 12.16
easily determined. Further, if the wavelength
and speed are known, then the frequency can be determined.
We know that air column in pipes or tubes of fixed length has particular resonant frequencies. Moreover, the
interference of the waves travelling down the tube and the reflected waves traveling up the tube produces
(longitudinal) standing wave which must have a node at the closed end of the tube and an antinode at the open
end of the tube.
The resonance frequencies of a pipe or tube usually depend on its length L. As we observe from the Fig. 12.16, only
a certain number of lengths or “loops” can be “fitted” into the tube length with the node–antinode requirements.
1 2 . 1 4 | Sound Waves

However, since each loop corresponds to one-half wavelength, resonance occurs when the tube is nearly equal to
an odd number of quarter wavelength, i.e., L = λ / 4,L =
3λ / 4, L =
5λ / 4, etc
or in general, L = ( 2n + 1) λ / 4 ; =
λ 4L / ( 2n + 1) ; f=
n ( 2n + 1) v / 4L
Hence, an air column (tube) of length L has particular resonance frequencies and therefore will be in resonance with
the corresponding odd harmonic driving frequencies.
As we can observe from the above equation, the three experimental parameters involved in the resonance condition
of an air column are f, V, and L. However, to study the resonance in this experiment, the length L of an air column
will be varied for a given driven frequency. The length of the air column achieved by changing the position of the
movable piston in the tube is as seen in the Fig. 12.17.

Speaker - Connect to power supply

Resonance tube

Movable piston

Microphone - Connect to voltage sensor

Figure 12.17

Further, as the piston is removed, increasing the length of the air column, more wavelength segments will fit into
the tube, consistent with the node–antinode requirements at the ends. Thus, the difference in the tube lengths
when successive antinodes are at the open end of the tube and resonance occurs is equal to a half wavelength; for
example: ∆L = L2 − L1 = 3λ / 4 − λ / 4 = λ / 2
Further, when an antinode is at the open end of the tube, a loud resonance tone is heard. Hence, the tube length
for antinodes to be at the open end of the tube can be determined by moving the piston away from the open
end of the tube and “listening” for resonances. However, no end correction is needed for the antinode occurring
slightly above the end of the tube since in this case, difference in tube lengths for successive antinodes is equal
to λ/2. Further, if we know the frequency of the driving source, then the wavelength is determined by measuring
difference in tube length between successive antinodes, ∆L =λ / 2 or λ = 2∆L , the speed of sound in air, vs = λf .

12.2 Kundt’s Tube Method

In the Kundt’s method, a gas is filled in a long cylindrical tube
closed at both the ends, one by disk and the other by a movable
piston. A metal rod is welded with the disk and is clamped exactly
at the middle point. The length of the tube in this method can be D I
varied by moving the movable piston. Some powder is sprinkled
Figure 12.18
in the tube along its length.
The rod in the setup is set into longitudinal vibrations either electronically or by rubbing it with some cloth or
otherwise. Further, if the length of the gas column is such that one of its resonant frequency is equal to the
frequency of the longitudinal vibration of the rod, then standing waves originate in the gas. Moreover, the powder
particles at the displacement antinodes fly apart due to the inherent violent disturbance there, whereas the powder
at the displacement nodes remain undisturbed because the particles here do not vibrate. Thus, the powder which
was initially dispersed along the whole length of the tube gets collected in a heap at the displacement nodes.
By measuring the seperation ∆l between the successive heaps, we can find the wavelength of the sound in the
enclosed gas. λ = 2∆l
 P hysi cs | 12.15

However, it should be noted that the length of the column is adjusted by moving the piston such that the gas
resonates and wavelength λ is obtained.
The speed of sound is given by ν = λv = 2∆l * v .
Further, if the frequency of the longitudinal vibration in the rod is not known, then the experiment is repeated with
air filled in the tube. Now, the length between the heaps of the powder, ∆l ′ is measured. The speed of sound in air
is then ν = 2∆l' v .  … (i)
ν ∆l ∆l
Now, = or ν = ν'
ν ∆l' ∆l'
By calculating the speed of ν ’ of sound in air, we can find the speed of sound in the gas.

13. BEATS
It two sources of slightly different frequencies produce sound waves in the same direction at the same point, these
waves then superpose to produce alternate loud and feeble sounds. Such variations in loudness are called beats.
The number of times such a fluctuation in loudness from maxima to minima takes place per second is called the
beat frequency.
If two waves y1 = asin(2πn2 t) of respective frequencies n1 and n2 superpose at the same place
asin(2πn1t) and y 2 =
x = 0, then y = y1 + y=
2 a[sin(2πn1t) + (sin2πn2 t)]

 2π (n1 − n2 ) t   2π (n1 + n2 ) t 
∴y 2 acos   × sin  =  2 acos 2π (n1 − n2 ) t  × sin 2π (n1 + n2 ) t 
 2   2 

y A sin  π (n1 + n2 ) t  ;=
= A 2acos  π (n1 − n2 ) t 

n +n 
The resultant wave is a harmonic wave with a frequency  1 2  but its amplitudes vary harmonically as a function
 2 
of the difference in the frequency n1 − n2 . The beat frequency nB is nB= n1 − n2 .

If n1 − n2 is small, i.e., the number of times the intensity of sound fluctuates between maxima and minima per
second is small, i.e., less than about 10 to 15, then the beats can be heard distinctly.

Illustration 7: Suppose that a string of length 25 cm and 2.5 g is under tension. A pipe closed at one end is 40 cm
long. When the string is set vibrating in its first overtone and the air in the pipe in its fundamental frequency, then
8 beats per second are heard. It is observed that decreasing the tension in the string decreases the beat frequency.
If the speed of sound in air is 320 m/s, then find the tension in the string  (JEE ADVANCED)
1 T v
Sol: The fundamental frequency of the string and the closed organ pipe are νs = and νp = . When two
2 m 4
waves of equal amplitude and slightly different frequencies superimpose with each other, phenomenon called
beats take place. Number of beats n = ∆ν where ∆ν is the difference in the frequencies of superimposed waves.

1 T 1 T
Fundamental of the string=
νs = = 20 T
2 m 2 × 0.25 10−2
v 320
The fundamental frequency of a closed pipe νp= = = 200 Hz
4 4 × 0.40

The frequency of the first overtone of the string = 2νs =40 T

Since there are 8 beat per second, 2νs − νp = 8 or 40 T − 200 =

8
1 2 . 1 6 | Sound Waves

Since on decreasing the tension, the beat frequency decreases, 2νs is definitely greater than νp
∴ 40 T − 200
= 8 or =
T 27.04N

Illustration 8: A sonometer wire of 100 cm in length has a fundamental frequency of 330 Hz. Find
(a) The velocity of propagation of transverse waves along the wire and
(b) The wavelength of the resulting sound in air if velocity of sound in air is 330 ms–1. (JEE ADVANCED)

Sol: As the wave travelling on the sonometer wire is the standing wave, the wavelength of the wire is λ =2L . And
the velocity of the wave is given by v = f λ = 2f L .
(a) In the case of transverse vibration of a string for fundamental mode: L =( λ / 2 ) ⇒ λ =2 L =2 × 1 ==2m
i.e., the wavelength of transverse wave propagating on the string is 2 m. Now, as the frequency of the wire is given
to be 330 Hz, so from v = f λ , the velocity of transverse wave along the wire will be v= 330 × 2= 660m / s
(b) Here, the vibrating wire will act as a source and produce sound, i.e., longitudinal waves in air: Now, as the
frequency does not change with change in medium so f = 330Hz, and as velocity in air is given to be = 330 m/s
v 330
so from relation v =f λ we get λair = air = =1m
f 330
i.e., for sound (longitudinal mechanical waves) in air produced by vibration of wire (body),
f = 330 s−1 , λair = 1 m and v = f × λ = 330m / s

14. DOPPLER EFFECT

We are familiar with the fact that when a source of sound or an observer or both are Velocity of sound, v
moving relative to each other, then there is an apparent change in the frequency of
sound as heard by an observer and this is called Doppler Effect. Further, the apparent Source Observer
frequency increases if the source is moving toward the observer or the observer is vs v0
moving toward the source. On the contrary, the apparent frequency decreases if either
the source is moving away from the observer or the observer is moving away from the Figure 12.19
source. This apparent change in the frequency is principally due to the basic effect of
motion of source to change the effective wavelength and the basic effect of motion of
observer is the change in the number of waves received per second by the observer.
However, if both the source and the observer are moving in the positive direction of x-axis, then sound of
frequency ‘n’ propagating in air with velocity in still air will result in an apparent frequency n′ heard by observed
 v − v0 
as n' =   n
 v − vs 
Moreover, if the direction of motion of source or observer is changed, then the signs of v0 and vs are accordingly
changed from negative to positive. Thus, the frequency n′, in still air for the different cases is obtained as follows:
(a) Both the source and observer are moving toward right when the source is approaching
vS v0
 v − v0 
a receding observer toward right n' =   n Figure 12.20
 v − vs 
 v − v0 
(b) Both the source and observer are receding from each other n' =   n vS v0
 v + vs 
Figure 12.21
 v + v0 
(c) Both the source and observer are moving toward each other n' =   n
 v − vs 
Figure 12.22
 P hysi cs | 12.17

 v + v0 
(d) When the observer is approaching a receding source, n' =   n If the wind is
 v + vs  vS v0
blowing with a velocity ω in the direction of sound, then ω is added to ν and if the
Figure 12.23
wind is blowing with velocity ω opposite to direction of wind, then ω is subtracted
 ν ± ω  ν0 
from ν . The general formula for the apparent frequency n′ due to Doppler effect is, n' =   n
 ν ± ω  νs 

Illustration 9: Assume that a siren emitting a sound of frequency 2000 Hz moves away from you toward a cliff at
a speed of 8 m/s.
(a) What is the frequency of the sound you hear coming directly from the siren?
(b) What is the frequency of sound you hear reflected off the cliff? Speed of sound in air is 330 m/s.  (JEE MAIN)

Sol: As the siren being source is moving away from you the observer on cliff, the apparent frequency is given by
 v 
 v + v  . Where f0 is natural frequency of the sound wave. The intensity of the sound wave appears to be
f ' = f0 
 s 
decreasing. When sound reflects from cliff it moves towards observer (cliff) and hence the frequency of the sound
 v 
wave is f ' = fo   . When source moves towards the observer, the intensity of sound wave appears to be
 v − vs 
increasing.
(a) The frequency of sound heard directly is given by

 v   330 
=f1 f0 
v+v
; v s 8m / s; ∴
 = = f1   × 2000
 s   330 + 8 
(b) The frequency of the reflected sound is given by

 v   330  330
f2 = f1   ; ∴ f2 =   × 2000 ; f2 = × 2000 = 2050Hz.
 v − vs   330 − 8  322

Illustration 10: Let us suppose that a sound detector is placed on a railway platform. A train, approaching the
platform at a speed of 36 km h–1, sounds its horn. The detector detects 12.0 kHz as the most dominant frequency
in the horn. If the train stops at the platform and sounds the horn, what would be the most dominant frequency
detected? The speed of sound in air is 340 ms–1.  (JEE MAIN)
Sol: In the first case, when train is moving towards the stationary observer on the platform, the intensity of the
 v 
wave appears to be increasing. And the frequency is given by f ' = fo   . In the second case both the train
 v − vs 
and the observer are stationary so we hear the natural frequency f0 of the sound wave.
Here, the observer (detector) is at rest with respect to the medium (air). Suppose that dominant frequency as
emitted by the train is v0. When the train is at rest at the platform, the detector will select the dominant frequency
as v0. When this same train was approaching the observer, then frequency detected would be

ν ν − us  u 
v=' v ; or v 0= v=' 1 − s  v '
ν − us 0 ν  ν 
−1 36 × 103 m
The speed of the source
= is us 36kmh
= = 10ms−1
3600s
 10 
Thus v 0 =−
1  × 12.0kHz = 11.6kHz
 340 
 1 2 . 1 8 | Sound Waves

14.1 Change in Wavelength

If a source moves with respect to the
medium, then wavelength becomes
’
different from the wavelength observed
when there is no relative motion between   ’’
the source and the medium. Thus, the 4
formula for calculation of apparent 3 4
wavelength may be derived immediately 3
2
from the relation λ = ν / v . It is given as 2
ν −u 1 1
=λ λ.
ν Figure 12.24

15. SONIC BOOM AND MACH NUMBER

A sonic boom is basically the sound associated with the shock waves created by an object travelling through air
faster than the speed of sound. This boom generates a huge amount of energy, sounding much like an apparent
explosion. The crack of a supersonic bullet passing overhead is an excellent example of a sonic boom in miniature.
Mach number is purely a dimensionless quantity representing the ratio of speed of an object moving through a
ν
fluid and the local speed of sound, M = where M is the Mach number,
νsound
ν is the velocity of the source relative to the medium, and νsound is the speed of sound in the medium.

16. MUSICAL SCALE

A musical scale is a sequence of frequencies which has a particularly pleasing effect on our ear. A widely used
musical scale, called diatonic scale, has eight frequencies covering an octave. We call each frequency as a note.

17. ACOUSTICS OF A BUILDING

Good concert halls: Good concert halls are so designed to eliminate unwanted reflection and echoes and
to optimize the quality of the sound perceived by the audience. This is accomplished by suitably engineering
the shape of the room and the walls, as well as to include sound-absorbing materials in areas that may cause
echoes.

Lecture hall: Similar consideration such as the one made in the above must be made particularly in a college
lecture hall, so that the professor can be heard by all of the students in the session. Although the sound quality
need not be as good as in a concert hall where music is being played, it still must be good enough to prevent
echoes and other things that will distort the audio quality of the speech delivered by the professor.

Work buildings: In an office building where there are cubicles with a divider in a large work area, there is often the
problem of noise from conversation and activities. However, in this case the quality of the sound is not an issue as
much as suppressing unwanted noise.

17.1 Echo
An echo (plural echoes) is a reflection of sound, arriving back at the listener particularly sometime after the direct
sound.
 P hysi cs | 12.19

17.2 Reverberation and Reverberation Time

Reverberation is the persistence of sound in a particular space after the original sound is produced. A reverberation,
or reverb, is generated when sound is produced in an enclosed space causing a large number of echoes to build
up and then slowly decay as the sound is absorbed by the wall and air.
Reverberation time: The interval between the initial direct arrival of a sound wave and the last reflected wave is
called the reverberation time.

18. APPLICATION OF ULTRASONIC WAVES

Biomedical application: Ultrasound has very good therapeutic applications, which can be highly beneficial when
used with appropriate dosage precautions. Relatively high power ultrasound can eliminate stony deposits or tissue,
accelerate the effect of drugs in a targeted area, assist in the measurement of the elastic properties of tissue, and
can also be used to sort cells or small particles for research.
Ultrasonic impact treatment: Ultrasonic impact treatment (UIT) is a technique wherein ultrasound is used to
enhance the mechanical and physical properties of metals. Basically, it is a metallurgical processing technique in
which ultrasonic energy is applied to a metal object.
Ultrasonic welding: In ultrasonic welding of plastics, high frequency (15 kHz to 40 kHz) low amplitude vibration is
used to create heat by way of friction between the materials to be joined. The interface of the two parts is specially
designed so as to concentrate the energy for the maximum weld strength.
Sonochemistry: Power ultrasound in the 20–100 kHz range alone is used in chemistry. The ultrasound does
not interact directly with molecules to induce the chemical change, as its typical wavelength (in the millimeter
range) is too long compared to the molecules. Instead, the energy causes cavitation, which generates extremes of
temperature and pressure in the liquid where the reaction takes place.

19. SHOCK WAVES

A shock wave is one type of propagating disturbance. Similar to an ordinary wave, it carries energy and can
propagate through a medium (solid, liquid, gas or plasma) or in some cases even in the absence of a material
medium, through a field such as an electromagnetic field. Generally, shock waves are characterized by an abrupt,
nearly discontinuous change in the characteristics of the medium.

PROBLEM-SOLVING TACTICS
1. Most of the questions are naturally related with the concepts of wave on a string. Therefore, one must
be thorough with the concept of that particular topic. (E.g., standing waves formed in open pipe here are
analogous to string tied at both ends. Further, many of the cases can be related in the same way.)
2. Questions dealing with physical experiments form another set of questions. Therefore, one must be familiar
with usual as well as unusual (or specific) terminology of each experiment. Mostly, it happens that if we do not
know the term, then we are usually stuck (E.g., end correction is one term used with the resonance column
method, which is directly related with the radius of the tube.)
3. Path difference between two sources form another set of questions and this is the only place where some
mathematical complexity can be involved. Hence, one must take care of them.
4. Questions related to Doppler effect and beats are generally formulae specific; therefore, one must carefully
use the formulae. (It is, however, also advised that one must know about the derivation of these formulae.)
1 2 . 2 0 | Sound Waves

FORMULAE SHEET

S. No. Term Description

1. Wave It is a disturbance, which travels through the medium due to repeated periodic motion of
particles of the medium about their equilibrium position.

Examples include sound waves travelling through an intervening medium, water waves,
light waves, etc.
2. Mechanical waves Waves requiring material medium for their propagation. These are basically governed by
Newton’s laws of motion.

Sound waves are mechanical waves in the atmosphere between source and the listener
and hence require medium for their propagation.
3. Non-mechanical These waves do not require material medium for their propagation.
waves
Examples include waves associated with light or light waves, radio waves, X-rays, micro
waves, UV light, visible light and many more.
4. Transverse These are waves in which the displacements or oscillations are perpendicular to the
direction of propagation of wave.
waves
5. Longitudinal These are those waves in which displacement or oscillations in medium are parallel to the
direction of propagation of wave, for example, sound waves.
waves
6. Equation of At any time t, displacement y of the particle from its equilibrium position as a function of

harmonic wave ( )
the coordinate x of the particle is y(x,y) = A sin ωt − kx where A is the amplitude of the
wave,

k is the wave number,

ω is the angular frequency of the wave,

( )
And ωt − kx is the phase.

7. Wave number Wave length λ and wave number k are related by the relation k = 2n/λ.

8. Frequency
Wavelength λ and wave number k are related by the relation v = ω / k = λ / T = λf.

9. Speed of a wave Speed of a wave is given by v =

ω/k = λf .
λ/T=

10. Speed of a Speed of a transverse wave on a stretched

transverse wave
string depends only on tension and the linear

mass density of the string but not on frequency of the wave, i.e.,
= v T/µ .
11. Speed of a Speed of longitudinal waves in a medium is given by v =
longitudinal wave
B = bulk modulus; ρ = density of medium;

Speed of longitudinal waves in an ideal gas is v = γp / ρ P = pressure of the gas , ρ =

density of the gas and y = Cp/Cv.
12. Principle of super When two or more waves traverse through the same medium, then the displacement
position of any particle of the medium is the sum of the displacements that the individual waves
would give it, i.e., y =∑ ( )
y i x,t
.
 P hysi cs | 12.21

S. No. Term Description

13. Interference of If two sinusoidal waves of the same amplitude and wavelength travel in the same direction,
waves then they interfere to produce a resultant sinusoidal wave travelling in the direction with
resultant wave given by the = ( )
relation y ' x,t ( ) (
2Am cos u / 2  sin ωt − kx + u / 2
  )
where u is the phase difference between the two waves.

If u = 0, then interference would be fully constructive.

If u = π , then waves would be out of phase and their interference would be destructive.
14. Reflection of waves When a pulse or travelling wave encounters any boundary, it gets reflected. However, if
( )
an incident wave is represented by y i x,t ( )
= A sin ωt − kx , then the reflected wave at
( )
rigid boundary is yr x,t ( ) ( )
= A sin ωt + kx + n =−A sin ωt + kx and for reflection at
( )
open boundary, reflected waves is given by yr x,t
= A sin ωt + kx .( )
15. Standing wavesThe interference of two identical waves moving in opposite directions produces
standing waves. The particle displacement in a standing wave is given by
= y x,t ( )
 ( ) ( )
2A cos kx  sin ωt . In standing waves, amplitude of waves is different at
different points, i.e., at nodes, amplitude is zero and at antinodes, amplitude is maximum
which is equal to sum of amplitudes of constituting waves.

16. Normal modes of Frequency of transverse motion of stretched string of length L fixed at both the ends is
stretched string given by f = nv/2L where n = 1, 2, 3, 4. The set of frequencies given by the above relation
is called normal modes of oscillation of the system. Mode n = 1 is called the fundamental
mode with the frequency f1 = v/2L. Second harmonic is the oscillation mode with n = 2
and so on.

Thus, a string has infinite number of possible frequencies of vibration which are harmonics
of fundamental frequency f1 such that fn = nf1.
17. Beats Beats arise when two waves having slightly differing frequency V1 and V2 and comparable
amplitudes are superposed.
18. Doppler effect Doppler effect is a change in the observed frequency of the wave when the source S and
the observer O move relative to the medium.

There are three different ways where we can analyze this change in frequency as listed
hereunder.

(1) when observer is stationary and source is approaching observer

v = v0(1+Vs/V) where Vs = velocity of the source relative to the medium

v = velocity of wave relative to the medium

V = observed frequency of sound waves in terms of source frequency

V0 = source frequency

Change in the frequency when source recedes from stationary observer is

v = V0(1–VS/V)

Observer at rest measures higher frequency when source approaches it and it

measures lower frequency when source recedes from the observer.

(2) observer is moving with a velocity V0 toward a source and the source is at rest is
V = V0(1+V0/V)

(3) both the source and observer are moving, then frequency observed by observer is
V = V0 (V+V0)/(V+VS) and all the symbols have respective meanings as discussed
earlier