or gas) for its propagation.

These waves are also

known as elastic waves because their
SOUND AND WAVES propagation depends upon the elastic
WAVE is a form of disturbance, which travels properties of the medium through which they
through a material medium due to the repeated pass. Examples for mechanical waves are sound
periodic motion of the particles of the medium waves and water waves. In these waves, the
about their position. particles of the medium just vibrate to and fro
about their mean position.
WAVE MOTION
A wave motion is the transmission of energy
from one place to another through a material or Requisites of the medium to propagate
a vacuum. Wave motion may occur in many mechanical waves
forms such as water waves, sound waves, radio
waves and light waves. The medium must possess the following
properties for the propagation of the waves:
The main characteristics of wave motion are:
• The medium should be able to return to its
• Particles of the medium vibrate about their original condition after being disturbed, i.e.,
mean position while the wave moves the medium must possess elasticity.
forward.
• The medium must be capable of storing
• Each particle of the medium vibrates, takes energy.
energy from its preceding particle and
• The frictional resistance must be negligible
transmits it to the next particle.
so as not to damp the oscillatory
• During a wave motion, the medium does movement.
not move as a whole. Only the disturbance
travels through the medium. Electromagnetic waves

Waves may be broadly grouped under two main An electromagnetic wave is a disturbance,
headings: Mechanical and Electromagnetic
which does not require any material medium
waves.
for its propagation and can travel even through
Mechanical and Electromagnetic waves vacuum. They are caused due to varying electric
and magnetic fields.
Depending upon their propagation in a Radio waves and light waves are examples for
medium, the waves are classified as follows: electromagnetic waves.
• Mechanical waves or elastic waves and
Let us now tabulate the differences between
• Electromagnetic waves.
the mechanical waves and electromagnetic
Mechanical Waves waves.

A mechanical wave is a periodic disturbance,

which requires a material medium (solid, liquid
 Electromagnetic
Mechanical Waves
Waves
Mechanical waves Electromagnetic waves
need a material do not need any
medium for their material medium for
propagation. their propagation.
These waves are
produced due to the They are caused due to
vibrations of the change in electric and
particles of the magnetic fields.
medium.
They have low speed They have high speed
or frequency or frequency

In general all waves : mechanical and

electromagnetic (non-mechanical) waves are
classified into two types. These are

• longitudinal waves
• transverse waves
Fig (a) – (d) Propagation of sound waves in air

Longitudinal waves A wave motion in which the particles of the

medium oscillate about their mean positions in
the direction of propagation of the wave, is
called longitudinal wave.
Sound waves are classified as longitudinal
waves. Let us now see how sound waves
propagate. Take a tuning fork, vibrate it and
concentrate on the motion of one of its prongs,
say prong A. The normal position of the tuning
fork and the initial condition of air particles is
shown in the fig (a). As the prong A moves
towards right, it compresses air particles near it,
forming a compression as shown in fig (b). Due
to vibrating air layers, this compression moves
forward as a disturbance. As the prong A moves
back to its original position, the pressure on its
right decreases, thereby forming a rarefaction.
This rarefaction moves forward like
compression as a disturbance. As the tuning
fork goes on vibrating, waves consisting of the lowest point, i.e. the position of maximum
alternate compressions and rarefactions spread displacement is called trough. Thus in a
in air as shown in fig (d). The direction of transverse wave crests and troughs appear
motion of the sound waves is same as that of alternatively.
air particles, hence they are classified as
longitudinal waves. The longitudinal waves Definition of some terms
travel in the form of compressions and
Amplitude of a wave, A
rarefactions. It is the maximum displacement of particles
from their mean position. S.I unit is the metre
Transverse waves (m).

A wave motion, in which the particles of the Period of a wave, T

medium oscillate about their mean positions at It is the time taken for one complete cycle or
vibration. It is measured in seconds (s)
right angles to the direction of propagation of
the wave, is called transverse wave. Frequency, f
It is the number of complete cycles or vibrations
These waves can propagate through solids and in one second. The S.I unit is Hertz (Hz).
liquids but not through gases, because gases do
not possess elastic properties. Examples of 1
f =
these waves are: vibrations in strings, ripples on T
water surface and electromagnetic waves.
In a transverse wave the particles of the
Wavelength, λ
medium oscillate in a direction perpendicular to It is the distance between any two successive
the direction of propagation as shown in the crests or troughs. It can also be defined as the
figure. distance covered by the wave after one
complete cycle. It is measured in metres (m)
Velocity of a wave, V
It is the distance travelled by the wave in one
second. The S.I unit is ms-1.

Wave front
It is a line connecting points that are in the
same phase.

Particles of the medium oscillate in a direction

Relationship between V, f and λ
perpendicular to the direction of propagation
Velocity, V = distance travelled (λ)
Thus, during their oscillations, the particles may
Time taken (T)
move upwards or downwards from the plane
passing through their mean positions. The 
V =
uppermost point of the wave, i.e., the position T
of maximum positive displacement is crest and
 1 V = f
T=
f V 400
But  f = = = 1000Hz
 0.4

V = = f 4. A travelling wave has a frequency of 400KHz
1
f and a speed of 3.0 x 106m/s. Calculate the
wavelength of the wave.

General representation of a wave Solution

f = 400KHz = 400 x 103Hz
displacement (y) V = 3.0 x 106m/s

a V = f
V 3x106
t, x  = = = 7.5m
f 400x103
T, λ
5. The frequency of a sound wave is 300Hz.
Calculate the
(i) period (ii) wavelength
Examples
[velocity of sound in air = 340m/s]
1. The speed of a radio wave is 3 x 108m/s.
Calculate the wavelength of an FM station
Solution
whose frequency of transmission is 100MHz.
1 1
T= = = 0.0033s
Solution (i) f 300
V = 3 x 108m/s , f = 100MHz = 100 x 106Hz
V = f (ii)
V 3x108 V = f
= = = 3m
f 100x106 V 340
= = = 1.13m
f 300

2. The period of a note is 0.4s. Calculate the 6. The diagram below shows the profile of a
frequency of the note. wave.
Solution y/cm
1 1
f = = = 2.5 Hz 5
T 0.4
0 4 8 12 16 20 24 28 t/s
3. A wave of wavelength 0.4m travels 900m in
2.25s. Calculate the frequency of the wave. The wave moves with a velocity of 20m/s.
Determine the
Solution (a) amplitude
900 (b) period
V = = 400m / s
2.25 (c) frequency
(d) wavelength
Solution
(a) amplitude = 5cm = 0.05m y/m
(b) period, T = 8s (time for 1 cycle) dxn of the wave
1 1 A
f = = = 0.125Hz
(c) T 8
O x P t

T, λ
(d)
V = f
The displacement at any instant of time is given
V 20
 = = = 160m by
f 0.125

7. The diagram below shows the profile of

a wave moving in the negative x - direction. Where A is the amplitude, w is the angular
frequency of the wave. Consider a particle P at
y/cm a distance x from the particle O on its right. Let
the wave travel with a velocity v from left to
2
right. Since it takes some time for the
0 2 4 6 8 10 12 14 x/cm disturbance to reach P, its displacement can be
written as
-2

The wave moves with a velocity of 1000cm/s. Where φ is the phase difference between the
Determine the
(a) maximum amplitude particles O and P.
(b) wavelength
(c) frequency
(d) period
Substituting equation (1.5) in equation (1.4)
Solution We get,
(a) maximum amplitude = 2cm = 0.02m
(b) wavelength, λ = 4cm = 0.04m
(c)
V = f
V 1000cms −1
 f = = = 250Hz
 4cm

1 1
T= = = 0.04s
(d) f 250

Progressive Wave Equation

Consider the wave profile below:

  x
y = A sin 2  ft + 
 
Similarly, for a particle at a distance x to the left  x
y = 5 sin 2 100t + 
of 0, the equation for the displacement is given (b)  10 
by
2. A wave profile is represented by
  
y = a sin 200t − x 
 17 
Hence the following are the various progressive
Calculate its
wave equations:
(a) wavelength
(b) velocity
2  x
y = A sin(2ft − x) = A sin 2  ft −  (c) frequency
   (d) period
 2v 2  2
y = A sin t− x  = A sin (vt − x ) Solution
    
 

y = a sin 200t − x 
If the wave is moving from right to left or in the Comparing  17  to the
negative x-direction then the equations 2
become: y = a sin(2ft − x)
general equation 
200 = 2f
2  x
y = A sin(2ft + x) = A sin 2  ft +  200
    f = = 100Hz
2
 2v 2  2
y = A sin t+ x  = A sin (vt + x )
      2
=
Also 17 
Examples
2 (17)
1. A progressive wave of amplitude 5m, = = 34m
frequency 100Hz and wavelength 10m. 
Write down the wave equation of the wave
if it travels in velocity, v = f = 34x100 = 3400m / s
(a) positive x-direction
(b) negative x-direction 1 1
T= = = 0.01s
Solution Period, f 100
(a)
 x (a) wavelength = 34m
y = A sin 2  ft −  (b) velocity = 3400m/s
 
(c) frequency = 100Hz
 x (d) period = 0.01s
y = 5 sin 2 100t − 
 10 

3. A progressive wave is represented by the

    (i) amplitude, A = 10m
y = 0.2 sin100t − x 
equation  10 
2
where y is measured in centimeters and 4 =
(ii) 
t in seconds. Determine the
(a) amplitude
(b) frequency 2
= = 0.5m
(c) wavelength 4
(d) velocity of the wave
(iii)
Solution 60 = 2f
   60
y = 0.2 sin100t − x   f = = 30Hz
Comparing  10  to the 2
2
y = A sin(2ft −
(iv) velocity, v = f = 0.5 x30 = 15m / s
x)
general equation 
(a) amplitude, A = 0.2cm = 0.002m
5. A plane progressive wave is represented by
(b)  x
y = 0.2 sin 2 100t + 
100 = 2f  15 
the equation
100 where y and x are in centimeters and t in
 f = = 50Hz
2 seconds.
(c) (a) sketch a graph of y against x over one
 2 complete cycle
= (b) Calculate the
Also 10 
(i) frequency
(ii) wavelength
2 (10)
= = 20m (iii) velocity of the wave
 (iv) phase difference between two points
20cm apart
(d) velocity, v = f = 20x50 = 1000m / s

4. A progressive wave is represented by the Solution

equation y = 10sin 4 (15t + x ) . Determine

(a) y/cm
0.2
its
(i) amplitude x/cm
(ii) wavelength λ = 15
(iii) frequency
(iv) velocity  x
y = 0.2 sin 2 100t + 
Solution (b) Comparing  15  to
y = 10sin 4 (15t + x ) = 10sin(60t + 4x ) x
y = A sin 2 ( ft + )
Comparing y = 10sin(60t + 4x ) to the the general equation 
2 (i) f = 100Hz
y = A sin(2ft + x)
general equation  (ii) λ =15cm = 0.15m
 Reflection of waves involves a change in
(iii) velocity, v = f = 0.15x100 = 15m / s direction of waves when they bounce off a
barrier.
2 2(3.142)
= x= ( 20cm ) Reflection does not change the velocity,
(iv)  15cm frequency or the amplitude of the wave.

  = 8.38rad

6. (a) What is a transverse wave?

(b) The equation of motion of a wave is

 2x 
y = A sin t − 
represented by   
(i) Identify the quantities A, ω, x and λ

= m
(ii) If 10 and ω = 6.6 x 103s -1,
calculate the velocity of the wave.

Solution
(a) Refer to notes

(b) (i) A = amplitude Refraction of waves

ω = angular velocity Refraction of waves involves a change in the
x = distance travelled by the wave direction of waves as they pass from one
λ = wavelength medium to another. Refraction, or the bending
of the path of the waves, is accompanied by a
(ii)  = 2f change in speed and wavelength of the waves.
 6.6 x103 3300
 f = = =
2 2 

 3300
v = f = x = 330m / s
10 

Properties of Waves
The properties are:
- reflection
- refraction
- diffraction
Diffraction of waves
- interference
Diffraction involves a change in direction of
waves as they pass through an opening or
around a barrier in their path. Water waves
Reflection of waves
have the ability to travel around corners,
around obstacles and through openings.

The smaller the opening (or aperture), the

smaller the wavebands and vice versa.

wide aperture (gap) small aperture (gap)

Interference STATIONARY OR STANDING WAVES

It is the superposition of two waves of the same Stationary waves are produced by superposition
amplitude and frequency. of two progressive waves of equal amplitude
and frequency, travelling with the same speed
When two groups of waves (called wave trains) in opposite directions.
meet and overlap they interfere with each
other. The resulting amplitude will depend on
the amplitudes of both the waves at that point. Properties of a stationary wave
Stationary waves have nodes (N) where
If the crest of one wave meets the crest of the there is no displacement at any time.
other, the waves are said to be in phase and the
resulting intensity will be large. This is known as In between the nodes are positions
constructive interference. If the crest of one called antinodes (A), where the
wave meets the trough of the other (and the displacement has maximum amplitude.
waves are of equal amplitude) they are said to
be out of phase by p then the resulting intensity
will be zero. This is known as destructive
interference.
y
A A A A A

N N N N N t, x

λ/2

[NA] = λ/4 [NN] or [AA] = λ/2

Fig. constructive and destructive interference
Stationary wave equation

 2 2  2003 Q5a
y1 = a sin t − x A transverse wave is represented by
T  
The equation  2t 2x 
y = 2a sin cos
  where the symbols
represents a progressive wave in the positive
 T
 2 2 
y2 = a sin t + x have their usual meanings.
x -direction and T   in (i) State with reasons whether it is a travelling
the negative x-direction. or stationary wave.
(ii) Sketch and label two cycles of the wave
y = y1 + y2 profile.
(iii) Calculate the amplitude of the wave after
 2 2   2 2  t = ⅓T seconds.
y = a sin t − x  a sin t + x
T  + T   Solution
(i) It is a stationary wave because the equation
  2t 2x   2t 2x  is made up of the sum of two sine functions.
y = a sin −  + sin + 
  T    T  
(ii) y
A A A A
From trigonometry,
sin( A − B) + sin( A + B) = 2 sin A cos B
N N N N N t, x
 2t 2x 
 y = a 2 sin cos
 T   λ/2

λ
 2t 2x  2λ
 y = 2a sin cos
 T   2x
A = 2a cos
(iii) Amplitude, 
If t = ⅓T, x = ⅓λ
2x 2 (1 / 3 ) 2
2a cos = A = amplitude  A = 2a cos = 2a cos
But   3
 = 180
2t
 y = A sin 2(180)
T  A = 2a cos = 2a cos120
3
maximum displacement = 2a  A = 2a (−0.5) = − a

NB 2006 Q25 (Obj)

If x = λ, t = T A wave of frequency 30Hz forms a stationary
x = 2λ, t = 2T wave such that the distance between adjacent
x = λ/2 , t = T/2 antinodes is 3cm. Calculate the speed of the
wave.
In general, if x = kλ, t = kT
Solution

= 3cm Experiment to show that sound waves needs a
[AA] = 2
material medium for propagation OR To show
  = 2 x3 = 6cm = 0.06m that sound waves cannot be propagated
through vacuum.
v = f = 0.06x30 = 1.8m / s
Diagram
Difference between progressive and stationary
Battery
Progressive wave Standing wave
Switch (key)
Particles have the Amplitudes are
same amplitude different
Particles are not Particles at the
permanently at rest nodes are
permanently at rest
Bell Jar
energy is transmitted no transfer of energy
Electric Bell

SOUND WAVES
Sound waves are longitudinal waves
propagated through a material by the transfer to vacuum pump
of kinetic energy from one molecule to another. Method
The diagram for the experiment is as shown
- in solids and liquids by intermolecular forces above. When the key is closed, the bell is heard
and collisions, ringing. While the bell is ringing, air is gradually
- in gases by intermolecular collisions alone. removed from the jar by the vacuum pump. The
sound from the electric bell gradually dies down
Hence the velocity of sound is greater in solids as the air becomes less. No sound is heard
and liquids than in gases. when all the air in the jar is removed.

Conclusion
This shows that sound cannot be propagated in
Sound waves occur in the following musical vacuum, it needs a material medium for
instruments: propagation.
- Wind instruments : whistle, mouth organ,
trumpet etc 2002 Q5b
- Percussion instruments : drums, xylophone Describe an experiment to demonstrate that
- Stringed instruments : guitar, violin etc sound waves require a material medium for
their propagation.
Propagation of Sound waves
Sound needs a material medium for its Musical Note and Noise
transmission. Gases, liquids and solids are all A Note is a sound produced from vibrations at
material medium to transmit sound waves but one or regular frequency.
vacuum cannot. This explains why Astronauts
(spacemen) can only speak to each other in
space through radio because there is no air.
Noise is a sound produced from vibrations of
irregular or no fixed frequency and is Velocity of sound by echo
unpleasant to the ear.
wall or cliff
Characteristics of a musical note
The characteristics of a musical note are : Source of
- pitch sound
- quality
- loudness
x
Pitch
It is the characteristic of a note which enables a
high note to be differentiated from a low note. The sound hits the wall and reflects so it travels
It depends on the frequency. The higher the a distance of 2x
frequency, the higher the pitch and vice versa.
Velocity, V = distance travelled = 2x
Loudness time taken t
It is a sensation on the mind of the individual
observer depending on the intensity of the Examples
sound. Intensity depends on the following: 1. A boy stands in an empty room and shout.
- amplitude of vibration If he is 10m from the wall, calculate the time
- volume of the sound taken to hear the echo.
- frequency of the sound [ velocity of sound in air = 330m/s ]
- distance away of the sounding body
The unit of loudness is bel.

Quality or Timbre
It is the characteristic of a note that Solution
distinguishes it from another note of the same 2x
pitch and loudness. Quality depends on v=
t
overtones or harmonics
2 x 2(10)
t = = = 0.061s
Reflection of sound (Echo) v 330
Echo is the sound heard when sound waves are
reflected from hard surfaces. 2. An observer stands at a distance from a tall
Echoes are undesirable in big halls or cliff and blows a whistle. If an echo is heard 6.2s
auditoriums because they interfere with the later, how far is the observer from the cliff.
original sound thereby making hearing difficult. [speed of sound in air = 330m/s]

To reduce echo, soft pads or wood are used for Solution

the ceilings and floors of big halls or 2x
v=
auditoriums. t
vt (330)(6.2)
Applications of Echo x= = = 1023m
It is used 2 2
- to determine the depth of sea or ocean beds
3. A boy shouts in front of a cliff and hears an
- for exploration of oil
echo after 0.8s. He then moves 15m closer
- to determine the velocity of sound in air
to the cliff and shouts and receives the echo
after 0.5s. Calculate the initial position of the Since v is constant,
boy from the cliff.
 10x = 5( x + 10)
Solution  10x = 5 x + 50
Let his initial position from the cliff = x
The time of echo, t1 = 0.8s  10x − 5 x = 50
2x 2x  5 x = 50
v= = = 2.5 x
t1 0.8 50
x= = 10m
5
As he moves 15m closer to the cliff, the new The initial distance from the wall = 10m
position = x – 15
The time of echo, t2 = 0.5s 5. As a ship approaches a cliff, its siren is
2( x − 15) sounded and the echo is heard in the ship
v= = 4( x − 15) after 12s. 2.1 minutes later the siren is
0.5
Since v is constant, sounded again and the echo is heard 8s
later. If the speed of sound in air is 340m/s,
 2.5 x = 4( x − 15) calculate the velocity at which the ship is
approaching the cliff.
 2.5 x = 4 x − 60
 4 x − 2.5 x = 60 Solution
 1.5 x = 60 Let the initial distance of the ship = x1
v = 340m/s , t1 = 12s
60
x= = 40m v=
2 x1
1.5 t1
The initial position = 40m vt1 340(12)
 x1 = = = 2040m
2 2
Let x2 = the second position of the ship from
4. A man shouts in front of a wall and hears an the cliff.
echo after 0.2s. He then moves back 10m and v = 340m/s , t2 = 8s
shouts and receives the echo after 0.4s. 2 x2
Calculate the initial distance of the boy from v=
t2
the wall.
vt2 340(8)
 x2 = = = 1360m
Solution 2 2
Let his initial distance from the wall = x The time taken between the two
The time of echo, t1 = 0.2s instants = 2.1mins = 2.1 x 60 = 126s
2x 2x
v= = = 10x The distance between the two
t1 0.2 instants = x1 – x2 = 2040 – 1360 = 680m
As he moves 10m back, the new Hence the velocity at which the ship approaches
position = x + 10 the cliff = 680 = 5.40m/s
The time of echo, t2 = 0.4s 126

2( x + 10) Modes of vibration in Stationary Wave

v= = 5( x + 10)
0.4
A stationary wave occurs between two 
boundaries. l=
2
  = 2l
Stationary waves occur in
- stringed instruments in which the
boundaries are two fixed ones. v = f = 2lf
- open tubes or pipes in which the two
boundaries are free. v
 f0 =
2l
- closed tubes or pipes in which one boundary
is fixed and the other free. But the speed of sound in stretched string
T
The modes of vibration depend on the type of v=
boundaries involved. The first mode of vibration m
is the fundamental which has the longest T
wavelength and the lowest frequency. The next
mode of vibration is the first overtone and so  f = m
2l
on.

Overtone has a frequency which is an integral 1 T

multiple of the fundamental frequency. The  f0 =
2l m
integral multiple of the frequency of the
fundamental mode of vibration is called
harmonic. T = tension in the string
m = mass per unit length of the string
NB
1st harmonic = fundamental mode
M
m=
Vibration in stringed instruments NB: l
Waves in stretched strings are stationary and M = mass of the string
transverse. l = length of the string

2nd harmonic (1st overtone)

A A
Fundamental mode (1st harmonic)
N N N
A  
N N l= + =
2 2

l=
2


l = NN =
NB: 2
l=
 v = f = fl f =
1 T
=
1 40
v v 2l m 2(1) 7 x10− 3
 f1 = = 2 
l  2l 
1
f == 5714.29 = (0.5)(75.593)
 f1 = 2 f 0 2

f = 3f f = 4f f = 37.8Hz
Similarly, 2 0 , 3 0

Hence the harmonics in stretched string are :

f0 , 2 f0 , 3 f0 , 4 f0 , 5 f 0
etc
2. The length of a sonometer wire is 1.2m.
Stretched strings have both odd and even Calculate the wavelength if it resonates in
harmonics. (a) the 1st overtone
(b) the 2nd overtone
Factors affecting the frequency of stretched (c) the 4th harmonic
string
Solution
The factors affecting the frequency are:
- Length of the string. (a) 1st overtone
1 A A
f
l
N N N
 f1l1 = f 2l2  
l= + =
2 2
- Tension in the string.
f T   = l = 1.2m
f f
 1 = 2
T1 T2
(b) 2nd overtone
- Mass per unit length A A A
1
f
m N N N N
   3
 f1 m1 = f 2 m2 l= + + =
2 2 2 2

3
Examples l=
1. A wire of length 1.0m is kept under a 2
constant tension of 40.0N. Calculate the 2l 2(1.2)
frequency of the tuning fork which sets the = = = 0.8m
3 3
wire in a resonant vibration.
[mass per unit length = 7.0 x10-3Kgm-1 ] (c) 4th harmonic = 3rd overtone

Solution
 and released.
N N N N N [ speed of sound in air =330m/s ]

l = 1.2m
Solution
    (a) (i) Stationary wave is a wave obtained when
l= + + + = 2 two progressive waves of equal amplitude and
2 2 2 2
frequency travelling in the opposite directions
l 1 .2 are superimposed.
= = = 0 .6 m
2 2 (ii) Amplitude is the maximum displacement
of a wave from its equilibrium position.
3. A sonometer resonates in the fundamental A node is a point on a stationary wave of
mode with a tuning fork of frequency 440Hz,
zero amplitude whereas antinode is a
calculate the resonant length. point of maximum amplitude.
[ v = 330m/s)
Solution (b) (i) Characteristics of a musical note are :
(fundamental mode) Pitch, Loudness and Quality
N N (ii) Pitch is the position of a musical note on

l= musical scale. It depends on frequency of
2 sound.

v = f Loudness is a measure of the intensity

v 330 of sound. It depends on the amplitude
= = = 0.75m of sound.
f 440
Quality enables one to distinguish between
 the same notes played on different
l=
But 2 instruments. It depends on overtones.

0.75 (c) (i) Factors affecting the frequency are:

l = = 0.375m - length of the string
2
- tension in the string
- mass per unit length of the string
2006 Q13
(a) (i) What is a stationary wave?
(ii) Define amplitude and use it to distinguish (ii) l = 50cm = 0.5m
between the node and antinode of a stationary
wave.
(b) (i) List the characteristics of a musical note
(ii) Explain each of the characteristics listed
(c) (i) List the factors on which the frequency of
vibration in a stretched string depends.
(ii) A wire of length 50cm and mass
5.0 x10-3 Kg is stretched under a tension
T = 40N. Calculate the wavelength of the
sound emitted when the wire is plucked
 Diagram
1 T wire movable bridge
f =
2l m
paper rider
−3
M 5.0 x10
m= = = 0.01kgm−1 T
l 0.5

1 40 W
 f = = 1 4000 = 63.25Hz
2(0.5) 0.01
Method
The set up for the experiment is as shown
v = f above. The wire is kept stretched or taut by
v 330 hanging a weight W from it. A tuning fork of a
 = = = 5.22m known frequency, f is sounded and the wire is
f 63.25
plucked at the same time. The vibrating tuning
THE SONOMETER
fork is placed above the paper rider and the
The sonometer consists of a wire stretched
length of the wire is adjusted by moving one of
between two supports or bridges on a wooden
the bridges until the paper rider falls off. The
sounding box.
length l of the sonometer wire between the
bridges and the frequency of the tuning fork are
The tension of the string may be varied either
recorded. The procedure is repeated for four
by a screw or by hanging weights on one end of
other tuning forks of different frequencies and
it.
the corresponding lengths are recorded.
1
The length of the vibrating wire can be changed
using a small wooden support that can be A graph of f is plotted against l to obtain a
moved along under the wire. 1
f
The purpose of the sounding box is to make a l.
straight line indicating that
larger mass of air vibrate and so amplify the
very small sounds produced by the vibrating
To verify the dependence of the frequency on
string itself. The same principle is applied in
tension, the length of the wire between the
stringed instruments such as the guitar.
bridges is kept constant and different weights
are used to vary the tension T.
A vibrating tuning fork is held above the paper
rider and the wire is plucked. Weights are
added until the paper rider falls off.
The procedure is repeated using four tuning
forks of different frequencies and the
corresponding weight, W are recorded.

A graph of f is plotted against W to obtain a

straight line indicating that f  T since
Experiment to show the dependence of the
W =T
frequency emitted by a wire on its length and
tension using a sonometer
Precautions
- the tuning fork must be sounded gently
- the sonometer wire must be firmly fixed 3rd harmonic (2nd overtone)
- the experiment must be conducted in a
quite environment. A N N N A

l
Vibrations in Pipes or Tubes     3
Waves in pipes or tubes are both stationary and l= + + + =
4 2 2 4 2
longitudinal.
3
Open Pipe l=
2
Fundamental mode (1st harmonic) 2l
 =
3
A N A
2 fl
   v = f =
l= + = 3
4 4 2 3v v
 f2 = = 3  = 3 f 0
 2l  2l 
[NA] or [AN] = 4 Hence the harmonics in open pipes are:

l= f0 , 2 f0 , 3 f0 , 4 f0 , 5 f 0
etc
2
  = 2l
Open pipes have both odd and even
harmonics.
v = f = 2 fl
Closed Pipes

v Fundamental mode (1st harmonic)

 f0 =
2l
N A

 
2nd harmonic (1st overtone) l=
4 [NA] = 4
A N N A 
l=
4
l
  = 4l
  
l= + + = v = f = 4 fl
4 2 4

v = f = fl v
 f0 =
v v 4l
 f1 = = 2  = 2 f 0
l  2l 
2nd harmonic ( 1st overtone)
 When air in a pipe is vibrating, small amount of
N N A the air beyond the pipe also vibrate. The
distance of the air column from the open end is
  3 known as end correction.
l= + =
2 4 4

3 Closed Pipe (fundamental mode)

l=
4
4l N A
 =
3
l c
4 fl
v = f = 
3 l+c =
4
  = 4(l + c )
3v v
 f1 = = 3  = 3 f 0
 4l 
v = f = 4 f (l + c )
4l

3rd harmonic (2nd overtone) v

 f0 =
4(l + c )
N N N A

l Open Pipe (fundamental mode)

   5 A N A
l= + + =
2 2 4 4
c l c
5
l=   
4 l + 2c = + =
4l 4 4 2
 =
5
  = 2(l + 2c )
4 fl
v = f =
5 v = f = 2 f (l + 2c )

5v v v
f2 = = 5  = 5 f 0  f0 =
4l  4l  2(l + 2c )
Hence the harmonics in open pipes are:
f0 , 3 f0 , 5 f 0 7 f0 Examples
, etc 1. A tube of length 27.5cm is closed at one end.
If it resonates at the fundamental mode with a
Closed Pipes have only odd harmonics. tuning fork of frequency 262Hz, calculate the
end correction of the tube.
End correction, c [velocity of sound in air = 330m/s]
 l = 30cm = 0.30m
Solution c = 2.25cm = 0.0225m

N A 3
l+c =
4
l c 3
 0.30 + 0.0225 =
v = f 4
3
v 330  0.3225 =
= = = 1.26m 4
4(0.3225)
f 262
= = 0.43m
3

l+c =
4 v = f
l = 27.5cm = 0.275m
v 330
 f = = = 767.44Hz
1.26  0.43
 0.275 + c = = 0.315
4
3. 2007 Q13c
A pipe closed at one end is 100cm long. If the
 c = 0.315 − 0.275 = 0.04m = 4cm air in the pipe is set into vibration and a
fundamental note is produced, calculate the
frequency of the note.
2. A 30cm tube closed at one end resonates at [velocity of sound in air = 340m/s]
its first overtone with a tuning fork of
unknown frequency. If the end correction of Solution
the tube is 2.25cm, determine the frequency
of the fork.
[velocity of sound in air = 330m/s] N A

Solution l = 100cm = 1m


N N A l=
4
  = 4l = 4(1) = 4m
l c

  3 v = f
l+c = + = v 340
2 4 4  f = = = 85Hz
 4

4. A tube of length 55cm is opened at both

ends. If it resonates at the fundamental mode
with a tuning fork of frequency 215Hz,
calculate the end correction of the tube.   
[ velocity of sound in air = 330m/s ] l + 2c = + + =
4 2 4
Solution
 0.285 + 2(0.012) = 
A N A
  = 0.285 + 0.024 = 0.309m
c l c

l = 55cm = 0.55m
v = f
v = f
v 330
v 330  f = = = 1068Hz
= = = 1.53m  0.309
f 215

   6. (a) Explain the term end correction.

l + 2c = + =
4 4 2 (b) An open pipe 30cm long and a closed pipe
23cm long, both of the same diameter,
are each sounding its first overtone, and
1.53
 0.55 + 2c = = 0.765 they are in unison. What is the
2 end-correction of the pipes.

Solution
 2c = 0.765 − 0.55 = 0.215
(a) Refer to notes
0.215
c= = 0.1075m
2 (b) open pipe

5. A 28.5cm tube opened at both ends

resonates at its first overtone with a A N N A
tuning fork of unknown frequency.
If the end correction is 1.2cm, determine c l1 c
the frequency of the tuning fork.
[ velocity of sound in air = 330m/s ]   
l1 + 2c = =
+ +
Solution 4 2 4
 1 = 30 + 2c − − − − − 1
A N N A
closed pipe
c l c

l = 28.5cm = 0.285m N N A
c = 1.2cm = 0.012m
l2 c

  3
l2 + c = + =
2 4 4
 3
 23 + c = Experiment to determine the velocity of sound
4
4(23 + c )
in air and the end correction of a tube
 2 = −−−−−2
3 Diagram
In unison, 1 = 2
Vibrating tuning
4(23 + c ) fork
 30 + 2c =
3 
 3(30 + 2c ) = 4(23 + c ) l+c =
4
 90 + 6c = 92 + 4c
 6c − 4c = 92 − 90
 2c = 2
 c = 1cm
Resonance Tube
RESONANCE
It is said to occur whenever a body is set in
vibration at its own natural frequency. water
It is as a result of impulses received from some
other body vibrating at the same frequency.

Resonance in Tubes or Pipes Beaker

When a vibrating tuning fork is placed over a
column of air and the volume of air is gradually
varied, a stage will be reached when the natural
frequency of the air in the tube becomes equal
to the frequency of the tuning fork. At this stage Method
resonance occurs and a very loud sound is The set up for the experiment is as shown
heard. above.
Tuning fork A tuning fork of a known frequency is sounded
and brought very close to the open end of the
resonance tube without touching it.The water
l in the tube is drained gradually until a loud
sound is heard.
The length of the vibrating air column l is
measured and the frequency of the tuning fork
recorded.
The experiment is repeated for four other
tuning forks of different frequencies and the
corresponding values of l noted.

Theory
  Precautions
l+c = - the vibrating tuning fork must not touch the
4
 = 4(l + c )
tube.
- the tuning fork must be sounded gently.
= 4(l + c )
v - the environment must not be noisy
=
But f

 4(l + c ) =
v
f
v
 4l + 4c =
f
First and second Resonance
v
 4l = − 4c
f First resonance

c
v
l= −c
4f l1

1
A graph of l is plotted against f to give a
v
straight line with 4 as the slope and − c as 
l1 + c =
the intercept on the l - axis. 4 − − − −1

l
Second resonance

1
c
f

−c l

v
Slope =
4

  3
 v = 4 xSlope l2 + c = + = − − − − − −2
2 4 4
c = − int ercept on the l - axis
3 
2 − 1  l2 + c − (l1 + c ) = −
4 4
  2. The first position of resonance in a resonance
 l2 − l1 = tube is 25cm from the open end. Neglecting end
2
correction, calculate the distance from the open
end to the next position of resonance.
  = 2(l2 − l1 )
Solution

l=
But v = f = 2 f (l2 − l1 ) For 1st resonance, 4
  = 4l = 4(25) = 100cm = 1m

For 2nd resonance,

3 3(1m )
l= = = 0.75m = 75cm
4 4
Applications of resonance
It is used
- to determine the velocity of sound in air
- in tuning radio receivers
3. In determining the velocity of sound using
a resonance tube, the position of resonance
is found to be 31.0cm. If the frequency of
Examples
the tuning fork is 256Hz, calculate the
1. In a resonance tube experiment, the first
end-correction.
position of resonance is found to be 20.0cm
[ velocity of sound in air = 330m/s ]
and the end correction of the tube is 2.0cm. If
the velocity of sound in air is 330m/s, calculate
Solution
the frequency of the tuning fork.
c
Solution
c
l
l

v 330
= = = 1.29m
 f 256
l +c =
4
 = 4(l + c ) = 4(20 + 2) = 88cm = 0.88m
l = 31cm = 0.31m

v 330 l+c =
f = = = 375Hz 4
 0.88 1.29
 0.31 + c = = 0.3225
4
 c = 0.3225 − 0.31 = 0.0125m = 1.25cm
4. 1994 Q5b v = 2 f (l2 − l1 )
In a resonance tube experiment, the first and
second resonant positions have lengths l1 and l2
v 340
Derive an expression that relates velocity, v of  f = =
2(l2 − l1 ) 2(0.40 − 0.10)
the sound produced to the lengths l1 , l2 and
the frequency, f of the tuning fork.
340
f = = 566.67Hz
Solution 0.60
Refer to notes - v = 2 f (l2 − l1 )
v 340
= = = 0.6m
f 566.67
5. In a resonance tube experiment, the first and
second resonant lengths are 0.34m and Using the first resonant length,
1.05m respectively. Calculate the frequency
of the tuning fork used. 
[ velocity of sound in air = 340m/s ] l1 + c =
4
0.6
 0.10 + c = = 0.15
4
Solution  c = 0.15 − 0.10 = 0.05m = 5cm

v = 2 f (l2 − l1 )

v 340
 f = =
2(l2 − l1 ) 2(1.05 − 0.34) 7. A resonance tube of total length 1.1m is filled
with water to a depth of 1.0m. When a
vibrating tuning fork is positioned over the
340 end of the tube, a loud sound is heard when
f = = 239.4 Hz 9.0cm of the water has been drained.
1.42
Another loud sound is heard again when
40.2cm of the column of water has been
6. 2004 Q6c drained. If the frequency of the tuning fork
The first and second resonant lengths of an air is 500Hz, calculate the
column in a resonance tube are 0.10m and (α) end correction of the tube
0.40m respectively. Calculate the (β) speed of sound in air
(i) frequency of sound in air
(ii) end correction of the tube Solution
[ velocity of sound in air = 340m/s ] Length of air column before
draining = 1.1 – 1.0 = 0.1m = 10cm
Solution
First resonant length, l1 = 9 + 10 = 19cm
l1 = 0.19m
Second resonant length, 90
l2 = 19 + 40.2 = 59.2cm 80
l2 = 0.592m 70
60

(β) v = 2 f (l2 − l1 ) = 2(500)(0.592 − 0.19) 50

l/cm
40
v = 1000(0.402) = 402m / s 30
20
v 402 10
= = = 0.804m 0
(α) f 500
0 0.002 0.004 0.006 0.008 0.01 0.012
1/f
Using the first resonant length,

 v 78.5 − 12.5
l1 + c = =
4 Slope = 4 0.01 − 0.002
0.804
 0.19 + c = = 0.201
4 v 66cm 0.66m
 = = = 82.5m / s
 c = 0.201 − 0.19 = 0.011m = 1.1cm 4 0.008s 0.008s
8. In a resonance tube experiment to determine
the speed of sound in air, the following were  v = 4(82.5) = 330m / s
the results:
BEATS
f/Hz 100 200 300 400 500 It is the periodic rise and fall in the intensity of
l /cm 78.5 37.3 23.5 16.6 12.5 sound produced by the interference of two
waves of nearly equal frequency.
Plot a suitable graph to determine the
velocity of sound in air. Conditions for the occurrence of Beats
The conditions are:
Solution - the two waves must have nearly equal
frequency
- the two waves must interfere.
f/Hz 100 200 300 400 500
l /cm 78.5 37.3 23.5 16.6 12.5
Uses or applications of Beats
1/ f 0.01 0.005 0.0033 0.0025 0.002 Beats are used in
- tuning musical instruments to a given tone
- determining unknown frequencies

How audible beats arise when two tuning forks

of slightly different frequencies are sounded
together

When the two tuning forks are sounded

together, at one instant the waves interfere
constructively to produce a loud note. At
another instant the waves interfere 2005 Q6 (Nov)
destructively to produce a low note. Hence (a) (i) Explain the phenomenon of beats as
alternate loud and low notes called beats are obtained in sound.
produced. (ii) State two practical applications of beats.

f b = f 2 − f1 (b) Two strings each of length 1.0m and masses

Beat frequency, 4.5g and 5.0g respectively are kept under a
constant tension of 80.0N.
If the strings are plucked at the same time in
f1 + f 2 their mid-points, calculate the
f =
Frequency of note produced, 2 (i) frequency of the resulting note emitted
(ii) number of beats produced per second

How a tuning fork is used in tuning a piano (c) Describe an experiment to determine the
The tuning fork is first sounded together with a end correction of a resonance tube.
note to hear the number of beats heard per
second. The string of the piano is tightened or
loosened to increase or decrease the frequency Solution
(number of beats per second) respectively until (a) Refer to notes
no beats are heard.
(b) M1 = 4.5g = 0.0045kg
2009 Q3 M 2 = 5.0 g = 0.005Kg
A piano tuner hears 3 beats per second when a
tuning fork and a note are sounded together
M 1 0.0045
and 6 beats when the string is tightened. What m1 = = = 0.0045kgm−1
should the tuner do next, tighten or loosen the l 1
string?
M 2 0.005
m2 = = = 0.005kgm−1
Solution l 1
The string should be loosened to decrease the
frequency. The procedure is repeated until no 1 T
beats are heard. f =
2l m

Examples 1 T 1 80
1. Two tuning forks of frequencies 254Hz and  f1 = =
2l m1 2(1) 0.0045
258Hz are sounded together. Calculate the
(a) beats frequency
(b) frequency of the note produced f1 = (0.5)(133.33) = 66.67Hz

Solution
1 T 1 80
 f2 = =
fb = f 2 − f1 = 258 − 254 = 4 Hz 2l m2 2(1) 0.005
(a)
f1 + f 2 254 + 258
f = = = 256Hz f 2 = (0.5)(126.49) = 63.25Hz
(b) 2 2
 f1 + f 2 The waves are equally spaced in all directions
f = around the source.
(i) Frequency of the note , 2

66.67 + 63.25
 f = = 64.96Hz
2

(ii) Number of beats produced per second

fb
= beat frequency ,

fb = f1 − f 2 = 66.67 − 63.25 = 3.42Hz

(c) Refer to notes

DOPPLER EFFECT
It is the apparent change in the frequency or
pitch of sound wave as a result of a relative
motion between a source of sound and an
observer.

You may have heard the Doppler effect in

sound as a police car with its siren on. As the
car comes towards you the pitch of the siren Source moving towards an observer
goes up, only to fall as it passes you and goes If the source is moving towards the observer
away. the wave is "squashed up" so the wavelength is
made smaller. The faster the source moves the
Stationary source more the waves are squashed.
With sound this means that the pitch of the
sound that you hear is increased.

Source moving away from an observer

If the source is moving away from the observer
the wave is "stretched out" and so the
wavelength is made larger. The faster the
source moves the more the waves are
stretched. With sound this means that the pitch
of the sound that you hear is decreased.

Practical situations in which Doppler effect is

experienced

- When an observer moves towards a

stationary sound source
- When an observer moves away from a
stationary sound source
- When a sound source moves towards a
stationary observer
- When a sound source moves away from Source moving away from stationary observer
a stationary observer
- When sound source and observer are  v 
moved towards each other. f 1 =   f
- Red shift of the galaxies  v + u s 

- Siren of a train passing through a station

- Change of pitch of a car horn as it moves f 1 = apparent frequency
towards you and then away.
f = frequency produced by the source
v = velocity of sound in air
Application or uses of Doppler effect us =
velocity of the source
It is used to
- determine the speed of a moving car
- calculate the relative velocity of the stars

Example
Calculation of apparent frequency
An ambulance moving with a speed of 30m/s
sounds its siren as it moves away from a
Source moving towards stationary observer
stationary observer. If the frequency of the note
emitted is 440Hz, determine the frequency of
 v 
f 1 =   f the sound heard by the observer as the
 v − us  ambulance moves away from him.
[ velocity of sound in air = 340m/s ]
f 1 = apparent frequency
f = frequency produced by the source Solution
v = velocity of sound in air
 v 
 f = 
us = 340 
velocity of the source f 1 =  440
 v + us   340 + 30 
2007 Q5 (Nov)
A train moving with a speed of 40m/s sounds its 340(440)
whistle as it approaches a stationary observer. f1 = = 404.32Hz
If the frequency of the sound produced is 370
500Hz, determine the frequency of the sound
heard by the observer as the train approaches Observer moving towards a stationary source
him.
[ velocity of sound in air = 345m/s ]  v + u0 
f1 = f
 v 
Solution
 v 
 f = 
345  f 1 = apparent frequency
f 1 =  500
 v − us   345 − 40  f = frequency produced by the source
v = velocity of sound in air

f1 =
345(500)
= 565.57Hz u0 =
velocity of the observer
305
 properties, namely:
2008 Q5 (a) it is propagated by varying electric and
A stationary siren emits a note of frequency magnetic fields oscillating at right angles to
400Hz. If a train approaches the siren at a each other;
constant speed of 45m/s, determine the (b) it travels with a constant speed of
frequency of the note that will be received by a 3.0 x108 ms-1 in a vacuum;
passenger inside the train. (c) it is unaffected by electric and magnetic
[velocity of sound in air = 340m/s] fields;
(d) it travels in straight lines in a vacuum;
Solution (e) it may be polarized;
(f) it can show interference and diffraction.
 v + u0   340 + 45 
f1 = f = 400
 v   340 
Electromagnetic spectrum

385(400)  -rays x -rays U.V visible infra micro radio

f1 = = 452.94Hz
340 light red wave wave

Observer moving away from a stationary increasing wavelength

source decreasing frequency

 v − u0 
f1 = f Production and Uses of E.M waves
 v 
Gamma-radiation
f = apparent frequency
1
This radiation is normally produced by
transitions within the excited nucleus of an
f = frequency produced by the source
atom and usually occurs as the result of some
v = velocity of sound in air previous radioactive emission.
u0 =
velocity of the observer
Gamma- radiation can result from fission or
TRY
fusion reactions. It is used in some medical
A car, sounding a horn producing a note of
treatment and also for checking flaws in metal
500Hz, approaches and then passes a stationary
castings, and it may be detected by
observer at a steady speed of 20m/s. Calculate
photographic plates or radiation detectors such
the frequencies of the note received by the
as the Geiger tube or scintillation counter.
observer as the car approaches and moves
away from him.
X- radiation
[ velocity of sound in air = 340m/s ]
This occurs due to electron transitions between
the upper and lower energy levels of heavy
[ Answer = f1 = 531Hz, f2 = 472Hz ]
elements, usually excited by electron
bombardment or by the rapid deceleration of
electrons. X-rays are primarily used in medicine
ELECTROMAGNETIC WAVES and dentistry, and may be detected using
photographic film.
Electromagnetic radiation (wave) is the name
given to a whole range of transverse radiation Ultraviolet radiation
having different wavelengths but six common This is produced by fairly large energy changes
in the electrons of an atom. It may occur with Microwaves may be detected with crystal
either heavy or light elements. The Sun detectors or solid-state diodes.
produces a large amount of ultraviolet
radiation, most of which is absorbed by the Radio waves
ozone layer in the upper atmosphere. These waves have the longest wavelengths of
any region of the electromagnetic spectrum and
Ultraviolet radiation will cause fluorescence and therefore the smallest frequency and hence the
ionization, promote chemical reactions, affect lowest energy per quantum. They are produced
photographic film and produce photoelectric by electrical oscillations and may be detected
emission. Its main uses are in spectroscopy and by resonant circuits in radio receivers. Their use
mineral analysis (some minerals exhibit strong is of course in radio and television
fluorescence under ultraviolet radiation). communications.

Visible light Laboratory experiment to show the reflection of

This is due to electron transitions in atoms. It E.M waves
affects a photographic film, stimulates the meter
retina in the eye and causes photosynthesis in Diagram
plants. Receiver (R)
B
i
Infrared radiation O
Infrared radiation is due to small energy r
changes of an electron in an atom or to
molecular vibrations. It may be detected by a A
thermopile or special photographic film. It is
also used by Earth resource satellites to detect metal Q
healthy crops; most of us are familiar with its sheet
use for heating, both in the home and in Transmitter
hospitals. It may be refracted by rock salt.
Method
A highly polished metal plate is set up in a
stand. A source of electromagnetic wave, Q is
Microwaves set up at the end of line A. A probe or a
These are produced by valves such as a receiver R is plugged into an amplifier with a
magnetron or with a maser. They are used in meter set up to receive the reflected wave.
radar, telemetry and electron spin resonance When the source is turned on the probe is
studies and in microwave ovens. In a microwave moved round until maximum deflection is
oven the food is heated because it contains received on the meter. A line BO is drawn from
water that is a strong absorber of microwaves. where the probe is stopped.
The microwaves excite the water molecules, the The angles I and r are measured. It is observed
velocity of the molecules rises and therefore that I = r and this proves the law of reflection.
the temperature of the food rises. This explains
why the food is heated but the temperature of 1995 Q5c
the containers does not rise very much. (i) Give a labeled sketch of the electromagnetic
Microwave ovens are useful because they spectrum. Use an arrow to indicate the
reduce cooking time considerably since they increasing order of wavelength.
cook the food from within. (ii) State one use each of the following
radiations:
 (α) ultra-violet (β) infra red (γ) radio wave

Solution – Refer to notes