Introduction to mechanical waves

1
PHYSICS
MECHANICAL WAVE - AN INTRODUCTION
Arun Umrao
https://sites.google.com/view/arunumrao
DRAFT COPY - GPL LICENSING

2 Mechanical Waves
Contents
1 Mechanical Waves 3
1.1 Mechanical Waves . . . . . . . . . . . . . . . . . . . . . . . . . . 3
1.1.1 Wave Parameters . . . . . . . . . . . . . . . . . . . . . . . 5
1.2 One Dimensional Wave Propagation . . . . . . . . . . . . . . . . 7
1.2.1 Travelling Wave . . . . . . . . . . . . . . . . . . . . . . . 9
1.2.2 Motion of Oscillating Particles . . . . . . . . . . . . . . . 11
1.2.3 Wave Equation in Form of Power . . . . . . . . . . . . . . 12
1.2.4 Angular Wave Number . . . . . . . . . . . . . . . . . . . . 13
1.2.5 Phase Distance Relation . . . . . . . . . . . . . . . . . . . 14
1.2.6 Properties of One Dimensional Travelling Waves . . . . . 14
1.2.7 Velocity Of Travelling Wave . . . . . . . . . . . . . . . . . 15
1.2.8 Superposition of Waves . . . . . . . . . . . . . . . . . . . 16
1.2.9 Reflection of Waves . . . . . . . . . . . . . . . . . . . . . 18
1.3 Standing Wave . . . . . . . . . . . . . . . . . . . . . . . . . . . . 21
1.3.1 Properties of Standing Waves . . . . . . . . . . . . . . . . 22
1.3.2 Speed of Wave in String . . . . . . . . . . . . . . . . . . . 22
1.3.3 Reflection of String Wave . . . . . . . . . . . . . . . . . . 23
1.3.4 Refraction of String Wave . . . . . . . . . . . . . . . . . . 25
1.3.5 Transverse Waves in String . . . . . . . . . . . . . . . . . 26
1.3.6 Frequencies in One End Closed Pipe . . . . . . . . . . . . 28
1.3.7 Frequencies in Open Ends Pipe . . . . . . . . . . . . . . . 30
1.3.8 Standing Wave Components . . . . . . . . . . . . . . . . . 32
1.4 Beats . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 33
2 Sound Waves 37
2.1 Sound Wave . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 37
2.1.1 Why Sound Travels Faster in Solids? . . . . . . . . . . . . 37
2.1.2 Speed of Sound in Solids . . . . . . . . . . . . . . . . . . . 39
2.1.3 Speed of Sound in Liquid . . . . . . . . . . . . . . . . . . 39
2.1.4 Velocity of Sound in Atmosphere . . . . . . . . . . . . . . 40
2.1.5 Velocity of Sound in Gaseous System . . . . . . . . . . . . 41
2.1.6 Thermodynamic Effect . . . . . . . . . . . . . . . . . . . . 41
2.2 Reflection And Refraction . . . . . . . . . . . . . . . . . . . . . . 44
2.2.1 Reflection of Sound Wave . . . . . . . . . . . . . . . . . . 45
2.2.2 Refraction of Sound Wave . . . . . . . . . . . . . . . . . . 45
2.2.3 Echo . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 45
2.3 Doppler Effect (Relative Theory of Sound) . . . . . . . . . . . . . 50
2.3.1 Observer Stationary & Source Moving Away . . . . . . . 50
2.3.2 Observer is Moving Away From Stationary Source . . . . 51
2.3.3 Observer & Source Moving in Same Direction & Source is
Leading Observer . . . . . . . . . . . . . . . . . . . . . . . 52

1.1. MECHANICAL WAVES 3
1Mechanical Waves
“Mechanical Waves” word is assigned with those waves those require ma-
terial to propagate from one point to other. For example, sound wave, water
wave, sonar, waves in rope or string are examples of mechanical waves. Though
mechanical waves need medium to propagate yet they do not carry medium
themselves. Energy of mechanical waves propagate from one place to other
place by transferring of energy by using principle of particle oscillation and
energy transfer by particle collisions.
Example One of the most common example of mechanical waves is observed
in water bodies. For example, in ponds or in lakes, when a stone is dropped,
mechanical waves start moving away from the point of disturbance concentri-
cally.
Figure 1.1: Concentric circles of mechanical wave generated by disturbance
caused by stone thrown in a pond.
1.1 Mechanical Waves
When a medium is disturbed by applying an external energy source then dis-
turbing energy moves in all directions in the medium. This disturbing energy
caused “disturbance” in medium particle. This disturbance is known as mechan-
ical wave. Though energy propagated through medium yet medium particles
remains stationary about their equilibrium position. Mechanical waves are of
two types:
Longitudinal Waves Longitudinal waves have oscillatory motion of medium
particles about their equilibrium position is parallel to the direction of propa-
gation of the waves.

4 Mechanical Waves
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Wave Propagation Direction
b
Figure 1.2: Longitudinal motion of wave. Particles oscillates parallel to the
wave propagation direction.
Sound wave is main example of the longitudinal waves. Compression and
rarefaction in medium is observed during the longitudinal wave propagation.
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Compression Rarefraction
Figure 1.3: Longitudinal motion of wave.
Transverse Waves Transverse waves have oscillatory motion of medium
particles about their equilibrium position is perpendicular to the direction of
wave propagation.
t
y
b
A
b
B
Wave Propagation Direction
Figure 1.4: Transverse wave moving from left to right while its particles are
oscillating vertically.
Light wave, electro-magnetic wave and seismic wave are main examples of
the transverse waves. Trough and Crest are observed in transverse wave prop-
agation.

1.1. MECHANICAL WAVES 5
t
y
b
A
b
B
Figure 1.5: Transverse Motion of Wave. Point ‘A’ is crest while point ‘B’ is
trough.
1.1.1 Wave Parameters
The lateral geometrical shape of mechanical wave is same in all medium. Thus
their some characteristics are same and unique. These unique characteristics
are called wave parameters.
Periodic Wave Periodic waves are those waves which replicate their shape
after a fixed interval of time. For example, all mechanical waves which are
represented by sine or cosine functions replicates themselves after each 2π angle.
y = sin(ωt) = sin(2π + ωt)
“Time period” is that time in which a mechanical sinusoid wave covers a distance
equal to its wavelength. Hence we can say that a sinusoidal wave is periodic to
its wavelength or its time period. So,
y = sin(ωt) = sin (ω(t + T ))
If sinusoidal wave has travelled much distance, then
y = sin(ωt) = sin (ω(t − T ))
Or
y = sin(ωt) = sin (ω(t ± T ))
t
y λ
λ, T
Figure 1.6: We see that in sinusoidal wave, transverse position of particles is
same in interval of time period (T ). This is main characteristics of periodic
wave.

6 Mechanical Waves
Linear Wave Velocity It is linear distance covered by a mechanical wave in
one second. It is represented by v.
v =
d
1s
In terns of frequency and wavelength, linear wave velocity is given as
v = fλ
Wavelength It is distance between two consecutive crests or trough of the
mechanical wave. It is represented by λ.
t
y λ
λ
Figure 1.7: Wavelength and Time Period.
Time Period Time required by mechanical wave to cover a distance equal
to its wavelength. It is represented by T . Its relation with frequency is
T =
1
f
and it is related to linear wave velocity and wavelength as
v =
λ
T
t
y T
T
Figure 1.8: Wavelength and Time Period.

1.2. ONE DIMENSIONAL WAVE PROPAGATION 7
Frequency It is number of oscillations in one second by oscillating particle
of medium in which mechanical wave is propagating. It is represented by n in
mechanical science, f in electronics and atomic physics. It is related with time
period as
f =
1
T
1.2 One Dimensional Wave Propagation
Wavelength λ of a wave is the distance travelled by wave in its time period T .
Mechanical wave is replica of waveform of wavelength λ. After each interval
of time T , wave repeats itself. In figure 1.9, a waveform of a mechanical wave
is shown. A line is drawn parallel to the axis of time. This line intersects to
mechanical wave in multiple points. Points, where tangent is in same direction,
are said to be in same phase.
t
y
y(t)
b
A1
b
A2
b
Ar
b
An
T1 Tn
t
y
y(t − nT )
b
A1
b
An
t
Figure 1.9: Periodic waveform.
Initially, observation point is at A1 and after each interval of time period
T , it reaches to A2, . . ., An successively. Number in suffix shows the number
of time periods are elapsed. Finally, assume that wave is at point An. If this
point is at time distance of t from its origin then its wave equation will be
y(t) = A sin(ωt)
This equation does not contain the time period or wavelength. To include it,
we transform this equation for previous points of same phase to the point An.
At previous point A1, that is lagging by ‘n’ multiple of time period T to the
point An, the wave equation will be
y(t) = A sin ω(t − nT )
This equation is known as general wave equation of mechanical wave. If leading
point is taken then
y(t) = A sin ω(t + nT )

8 Mechanical Waves
Solved Problem 1.1 A transverse mechanical wave of time period T is travelling
in a string of length l. At t = 0, wave starts propagating from its origin. After
time t, the wave is at distance x from the origin. Find the vertical displacement
of the wave just after 0.25T .
Solution A mechanical transverse wave equation in form of time period is
given by
y(t) = A sin ω(t + nT )
t
y
y(t)
b
A1
b
An
t
T/4
Here nT = 0.25T , so the vertical displacement of the wave will be
y(t) = A sin ω(t + 0.25T )
Solved Problem 1.2 An observation is started at time t = 0. After 4 seconds,
the displacement of the wave is 0.25 cm. Find the amplitude of the wave. Time
period of the wave is 0.5 second. What was the vertical position of the wave
particle from time axis just before the 0.5T second from the current time, i.e.
from time instant of 4 seconds?
Solution In this problem, we have to compute amplitude of the wave and
transverse position of wave particle at a given time.
For Amplitude From the equation of wave propagation y = A sin ωt, ampli-
tude of wave can be found by substituting the values of unknowns, i.e. t = 4
second, T = 0.5 second and y = 0.25 cm. This equation will compute the
amplitude of the wave as
0.25 = A sin

2 × 3.14
0.5
× 4

On simplification amplitude will be 9.8cm.
t
y
y(t)
b
A1
b
An
4s
0.25s

New Vertical Distance Just before the time t = 0.5T , vertical position of
the waveform will be
y′
= A × sin (ω × (T − t))
Substituting the values of unknown parameters, we have
y′
= 9.8 × sin

2 × 3.14
0.5
× (4 − 0.5 × 0.5)

It gives y′
= 0.25 cm.
1.2.1 Travelling Wave
A wave continuously propagated in a homogeneous medium along one dimen-
sion, it is called one dimensional travelling wave. Let y0 is the displacement
of medium particles from their equilibrium position at any time t0, a is the
maximum lateral displacement of medium particles about equilibrium position
and x is distance travelled by the wave after time t − t0, then displacement of
particles, y, at distance x about their equilibrium position is given by
x
y
t0
φ0
b
y
t
x
λ
Figure 1.10: Travelling Wave
t
y
t0
φ0
b
y
t
x
λ
Figure 1.11: Travelling Wave
y = a sin(ωt − φ) (1.1)

10 Mechanical Waves
Where φ is initial phase of wave and ω is angular velocity. The maxima of wave
is called crest and minima of the wave is called trough. The phase difference
at time t0 is φ0 and at distance x phase is φ′
then from definition of phase
difference
φ =
2π
λ
x
And equation (1.1) becomes
y = a sin

ωt −
2πx
λ

(1.2)
Substituting the value of ω = 2π/T the equation (1.1) becomes
y = a sin

2πt
T
−
2πx
λ

(1.3)
Equation (1.3) is one-dimensional travelling wave.
Amplitude of a wave is maximum lateral displacement of medium particles
about their equilibrium position when wave propagate through themselves.
The angular displacement of wave from the origin along +x axis, when it
starts propagation is called phase φ of wave. Normally, phase is the angular
distance of the point from origin where wave first intersects to +x-axis.
Second Thought for Travelling Transverse Wave
A travelling wave is propagating in a homogeneous medium upto infinite
distance taking sufficient large time. If time taken for wave propagation is
equals to its ‘Time Period’ then wave equation will be
y = a sin(ωT )
As travelling wave is a period function, hence plane wave propagating in a
medium is replica of this equation. This equation is principle equation of the
travelling wave. Assume at time ‘t’, displacement of medium particle from
their equilibrium position normal to the direction of wave propagation is y
and its wave equation is
y = a sin(ωt)

t
y
b
A1
(
t
−
T
,
y
)
b
A2
(
t
,
y
)
T
A wave is replica of principle wave, displacement y is given by the above
equation will be same even if time t is replaced by t − T .
y = a sin ω(t − T )
If this particle is at distance ‘x’ from the origin and velocity of wave is ‘v’,
then time taken by wave to reach at this distance is
t − T =
x
v
or
T = t −
x
v
Substituting ‘Time Period’ in principle equation of wave
y = a sin ω

t −
x
v

It is genera equation of travelling transverse wave.
1.2.2 Motion of Oscillating Particles
y is displacement of medium particles normal to the direction of wave propaga-
tion about their equilibrium and equation for this displacement is given by
y = a sin ω

t −
x
vx

Here vx is velocity of wave in medium i.e. velocity of the wave along the direction
of propagation. Instantaneous vertical velocity of particle is dy/dt
vy =
dy
dt
= aω cos ω

t −
x
v

(1.4)
If above equation is derivated second time with respect to time then acceleration
of oscillating particle will be obtained.
ay =
d2
y
dt2
= −aω2
sin ω

t −
x
v

(1.5)

12 Mechanical Waves
Here equations 1.4 and 1.5 are knows as velocity and acceleration of oscillating
particles in transverse wave.
Solved Problem 1.3 A transverse wave is travelling in a medium with horizontal
velocity vx = 10m/s along the +x direction. Frequency is 10Hz. Find the
equation of wave. Find the velocity and acceleration of an oscillating particle
at x = 6m when t = 5.1 seconds.
Solution The wave equation is given by
y = A sin ω

t −
x
vx

Substituting the values of parameters
y = A sin

2πn × t − 2πn ×
x
10

Or
y = A sin (20πt − 2πx)
Velocity of the wave at x = 6m when t = 5.1 second is
dy
dt
= vy = A20π cos (20πt − 2πx)
Or
vy = 20πA cos (20π × 5.1 − 2π × 6)
on solving vy = 62.15A metre per second. Again acceleration of the wave at
x = 6m when t = 5.1 second is
d2
y
dt2
= ay = −(20π)2
A sin (20πt − 2πx)
Or
ay = −(20π)2
A sin (20π × 5.1 − 2π × 6)
on solving ay = −563.37A meter per square second.
1.2.3 Wave Equation in Form of Power
y = A sin ωt represents a sinusoidal transverse wave where y is instantaneous
vertical displacement of medium particles at time t. Sinusoidal wave has am-
plitudes in both positive and negative axes. It gives zero average value over the
time period T = 2π/ω. But in terms of power, average value is not zero over
the period of T as P ∝ y2
. Negative displacement sinusoidal wave is converted
into positive magnitude.

t
f(x)
A
y
P =
1
T
Z T
0
y2
dt
It gives the solution for power
P =
A2
2
Wave equation in terms of power is
y =
√
2P sin ωt
1.2.4 Angular Wave Number
From the equation (1.3), at initial observation time t = 0, one dimensional
travelling wave becomes
y = −a sin

2πx
λ

which represents to a new one dimensional travelling wave. Here 2πx/λ is a
constant and known as angular wave number and 2π/λ is known as wave
number, k.
Solved Problem 1.4 Equation of a travelling wave is y = 1.25 sin(4πt − 1.75x).
Find wave number of the travelling wave.
Solution Wave number of a transverse mechanical wave is given by 2π/λ.
Comparing given equation with one dimensional wave equation for position
y = A sin

2π
T
t −
2π
λ
x

Wave number will be 1.75.

14 Mechanical Waves
1.2.5 Phase Distance Relation
S
X Y
P1
P2
2π 4π 6π 8π
φ
λ
dx
K
To understand the relation between path difference and phase difference, a point
source S, is illuminated by a monochromatic light source. Two waves are ema-
nating in same phase from source and propagate towards the points P1 and P2
which are in same vertical plane. Distances of SP1 and SP2 are not equal. The
wave-fronts of two waves of equal phase are shown by arcs at 2π, 4π, 6π and
8π radians. Assume that when first wave reach at P1, its phase is φ. An arc
of phase φ is drawn as shown in the figure. This arc intersects the second wave
at point K. It means that the second wave at point K has phase φ. To reach
the point P2, second wave has to travel an extra path. Consequently, there is
an extra angular displacement in second wave when it reaches to the point P2.
From figure, 2π angular distance is equal to the wavelength of wave λ. Hence
dx distance will be equal to angular distance
dφ =
2π
λ
dx (1.6)
It is phase distance relation in mechanical waves. This relation is also applicable
to electro-magnetic waves.
1.2.6 Properties of One Dimensional Travelling Waves
The properties of waves are given below.
1. The wave is sort of disturbance that travels through the medium.
2. The medium particles oscillates about their mean position without leaving
their position along the direction of motion of wave.
3. Only energy produced by disturbance is passes through the medium.
4. There is continuous change in the phase and position of medium particles.
Hence at different time, amplitude of the particle is different.
5. Compression and rarefaction is visible in longitudinal waves while in trans-
verse wave, trough and crest are visible.

1.2.7 Velocity Of Travelling Wave
Let a one dimensional travelling wave is propagating in +x direction with ve-
locity v. Two consecutive particles at distance dx are oscillating about their
equilibrium position attains their maximum displacement at time difference dt
then velocity of wave is dx/dt. From the principle of one-dimensional travel-
ling wave propagation, particles are not travel along +x direction then their
displacement is zero and their oscillation have no effect in wave motion. Now
differentiating the equation (1.3) with respect to t for +x direction and taking
y as constant. Hence
v
y
b
A
b
B
dx
b b
dx
Figure 1.12: Travelling wave with velocity v.
v
y dx
dy
dt
= a

cos

2πt
T
−
2πx
λ

2π
T
−
2π
λ
dx
dt

Here dy/dt = 0 hence
2π
T
−
2π
λ
dx
dt

= 0
dx
dt
= vx =
λ
T
(1.7)
Equation (1.7) is the velocity of one dimensional travelling wave.
Solved Problem 1.5 A wave propagating in a medium. After time t, wave length
of wave is measured that equals to 4m. If time period of the wave is 1.5 second
then find the velocity of wave.
Solution As we know that the wave velocity is ratio of wavelength to time
period. Hence vx = 4/1.5 = 2.66 meter per second.

16 Mechanical Waves
1.2.8 Superposition of Waves
Two waves, which are propagating in same medium, interact with each-other
and after some time their resultant wave of variable amplitude is generated
which propagate in the same medium with parental wave velocity. The intensity
of resultant wave is algebraic sum of two waves. This phenomenon is known as
superposition of waves. Let two waves
y1 = a sin 2π

t
T
−
x
λ

(1.8)
y2 = a sin 2π

t
T
−
x
λ
+ φ

(1.9)
are propagating in the same medium then due to superposition the resultant
wave y is y = y1 + y2 which give
y = a sin 2π

t
T
−
x
λ

+ a sin 2π

t
T
−
x
λ
+ φ

(1.10)
v
y
y1
y2
y
Using algebraic identity sin α + sin β = 2 sin(α + β)/2 cos(α − β/2)
y =

2a cos2π

1
2
φ

sin 2π

t
T
−
x
λ
+
1
2
φ

(1.11)
This is the equation of resultant wave of two waves. If phase difference between
two waves is zero i.e. φ = 0 then
y = 2a sin 2π

t
T
−
x
λ

(1.12)
The amplitude is twice of the initial amplitudes. Again if waves are in opposite
phase i.e. φ = π then
y = 0 (1.13)

i.e. the amplitude is zero.
Solved Problem 1.6 Two mechanical waves represented by mathematical equa-
tions y1 = 2 sin(ωt) and y2 = sin(ωt + π/4) respectively are travelling in same
medium. Find the amplitude of wave generated after superposition.
Solution The mathematical equations of two waves are
y1 = 2 sin(ωt)
and
y2 = sin(ωt + π/4)
On superposition, displacement equation of resultant wave will be y = y1 + y2.
Now
y = 2 sin(ωt) + sin(ωt + π/4)
1
2
−1
−2
−3
1 2 3 4 5 6 7 8 9
t
y
y1
y2
y
A
A
A

18 Mechanical Waves
t
y
y1
y2
y
Amplitude is maximum transverse displacement of medium particle when
wave propagate through it. As two partial wave equations are sine function
hence resultant wave will also be a sinusoidal wave. So, at amplitude point,
there will be either local maxima or local minima of super-positioned wave y.
So,
dy
dt
= 0 ⇒ 2 cos(ωt) × ω + cos(ωt + π/4) × ω = 0
It will give
2 cos(ωt) + cos(ωt + π/4) = 0
Or
2 cos(ωt) + cos(ωt) cos(π/4) − sin(ωt) sin(π/4) = 0
Or
ωt = tan−1
(
√
2 + 1) = 67.5◦
Here, we do not need to check (where second order derivative is negative or
positive at ωt for local maxima or local minima respectively) whether there is
local maxima or local minima at ωt value as amplitude of sinusoidal wave is
equal in both side about time axis as shown in the above figure. Now amplitude
of the super-positioned wave is
y = 2 sin(67.5◦
) + sin(67.5◦
+ 45◦
) = 2 sin(67.5◦
) + sin(112.5◦
) = 2.77
This is amplitude of the resultant wave.
1.2.9 Reflection of Waves
Medium are not infinitely large. They have boundaries. The boundaries of
medium are either rigid (closed) or non-rigid (open). Example of closed bound-
aries are pool walls, container walls etc. Example of open boundaries are inter

medium surface, rubber surfaces etc. When mechanical waves collides with
these boundaries wave energy reflects back into the same medium but in differ-
ent phases and direction. It is known as reflection of wave.
Reflection From Closed Boundary When a propagating wave (primary wave)
collides with boundary, colliding point at boundary starts acting as a wave
generating source itself for secondary wave.
t
y
S
Figure 1.13: Primary wave is propagating with speed of vf to a rigid boundary.
After reflection it returns with speed vr in opposite polarity. Boundary S act
source for secondary waves, i.e. reflected wave.
When one-dimensional travelling wave collides with closed boundary (rigid
or rare-to-dense medium boundary) then it returns to the same medium with
same velocity but with a opposite phase difference. This happens due to action-
reaction between rigid boundary and wave. This is similar to the reflection of
elastic ball from rigid surface.
t
y
~
vr
~
vf
After reflection it returns with speed vr in opposite polarity.
Assume one dimensional mechanical wave
y1 = A sin 2π

t
T
−
x
λ

20 Mechanical Waves
when collides with closed boundary then it reflects back as
y2 = A sin 2π

t
T
−
x
λ
+ π

= −A sin 2π

t
T
−
x
λ

(1.14)
Reflection From Open Boundary Similarly when one dimensional mechanical
wave collides with open boundary (elastic or dense-to-rare medium boundary)
then it return to the same medium with same velocity and without changing its
phase. This happens because of absorption of wave energy by open boundary.
When open boundary fully absorbs the wave energy, boundary starts acting
similar to the wave source.
t
y
~
v
t
y
~
v
Figure 1.15: Wave propagating to non-rigid boundary.
t
y
~
v
t
y
~
v
Figure 1.16: Wave reflected from non-rigid boundary.
And this wave source generates identical wave of same phase and amplitude.
For one dimensional mechanical wave
y1 = A sin 2π

t
T
−
x
λ

1.3. STANDING WAVE 21
reflected wave from open boundary will be
y2 = A sin 2π

t
T
−
x
λ

(1.15)
From the principle of superposition, superposition intensity of resultant wave of
propagating and reflected wave from closed boundary is zero while it is twice
when propagating wave superposition with reflected wave from open boundary.
1.3 Standing Wave
When a travelling wave collides with closed boundary then it return to the
same medium and under superposition a standing wave produced. It is known
as standing wave as amplitude of wave at a fixed distance remains constant with
time. At any point two waves travels in opposite side. Let along to +x direction
a wave is travelling according to
y1 = a sin 2π

t
T
−
x
λ

which returns to same medium after reflecting with closed boundary. The equa-
tion of returned wave is
y2 = a sin 2π

t
T
+
x
λ
+ π

Applying superposition theory the resultant wave is y = y1 + y2 and
y = a sin 2π

t
T
−
x
λ

− a sin 2π

t
T
+
x
λ

= −2a cos 2π

t
T

sin 2π
x
λ

(1.16)
Equation (1.16) is standing wave. As standing wave travels between two closed
boundaries at distance L, hence displacement of oscillating particles will be zero
at x = 0 and x = L. Now, equation (1.16) can be separated into amplitude of
standing wave and its wave form as
A = 2a cos2π

t
T

(1.17)
and waveform equation is
Y = −A sin 2π
x
λ

(1.18)
As the amplitude of wave varies from point to point hence points where am-
plitude is zero are called nodes and where amplitude is maximum are called
anti-nodes. For nodes, the value of cos(2πx/λ) must be zero i.e.
sin

2π
x
λ

= 0 = sin

2m
π
2

(1.19)

22 Mechanical Waves
Where m = 0, 1, 2, 3, . . .. Above equation would give
x = m
λ
2
i.e. at
x = 0,
λ
2
,
2λ
2
,
3λ
2
, . . .
Amplitude of standing wave would be zero. For maximum amplitude i.e. anti-
node, the value of sin(2πx/λ) must be maximum i.e. 1.
sin

2π
x
λ

= 1 = sin
h
(2m + 1)
π
2
i
(1.20)
Where m = 0, 1, 2, 3, . . .. Equation (1.20) would give
x = (2m + 1)
λ
4
i.e. at
x =
λ
4
,
3λ
4
,
5λ
4
, . . .
Amplitude is maximum. It is also noted that at m = 0 the value of frequency
is called root harmonics and at m = 1, frequency is called first harmonic or
fundamental frequency, at m = 2 is called second harmonic or first overtone
and so on.
1.3.1 Properties of Standing Waves
Followings are the properties of the standing waves.
1. Standing waves travel between two fixed boundaries.
2. The amplitude of a medium particle remains constant through out the
motion.
3. The amplitude of the medium particles varies from zero to the maximum
amplitude of the standing wave.
4. Nodes and anti-nodes are formed in standing wave.
1.3.2 Speed of Wave in String
A wave pulse travelling in a string is of transverse in nature. The speed of wave
in string depends on its elastic restoring force (tension T ) and inertia (mass per
unit length µ say). The mathematical relation is
v =
s
T
µ

1.3.3 Reflection of String Wave
String wave has reflection from two types of boundary.
Reflection From a Rigid Boundary When a propagating wave (primary
wave) collides with boundary, colliding point at boundary starts acting as a
wave generating source itself for secondary wave.
b
S
vf b
S
vf
On incident of approaching wave at point S, it will start acting as source of
secondary wave i.e. reflected wave.
These secondary waves may be smaller or equal or larger than primary waves.
Assume a string wave is moving from left to right towards the end which is rigidly
clamped. Due to rigid clamping of this end, it can not move. When internal
restoring force of string wave approaches to the fixed end, it applied force in
upward direction.
b
S
vf b
S
vf
F
Figure 1.18: Force exerted by primary wave at point S.
Fixed end acts as source of secondary wave. Due to rigidity, fixed end exerted
equal and opposite force on the string in downward direction according to the
Newton’s third law.
b
S
F′
b
S
vr
Figure 1.19: Force exerted by source S and generation of secondary at point S
with opposite polarity, velocity vr and amplitude A.
This downward force creates a wave pulse that propagates from right to left,
with the same speed and amplitude as the incident wave (if there is no loss of
energy at boundary), but with opposite polarity (upside down).

24 Mechanical Waves
Reflection From a Soft/Free Boundary When a propagating wave (primary
wave) collides with boundary, colliding point at boundary starts acting as a
wave generating source itself for secondary wave.
S
vf
S
vf
On incident of approaching wave at point S, it will start acting as source of
secondary wave i.e. reflected wave.
These secondary waves may be smaller or equal or larger than primary waves.
Assume a string wave is moving from left to right towards the end which is
loosely clamped. Due to loose clamping of this end, it can move vertically.
Hence vertical force acting at loosely fixed end must be zero.
S
vf
S
vf
F
Figure 1.21: Force exerted by string wave at loose end.
When internal restoring force of string wave approaches to the fixed end, it
can apply pushing force into wall in horizontal direction. Loosely fixed end acts
as source of secondary wave. Due to loose fixing at wave incident end, it exerted
equal and opposite force on the string in horizontal direction according to the
Newton’s third law.
S
F′
S
vr
Figure 1.22: Force exerted by source on string and generation of secondary wave
with same phase, velocity vr and amplitude A.
This horizontal force creates a wave pulse that propagates from right to left,

with the same speed and amplitude as the incident wave, and in same polarity.
1.3.4 Refraction of String Wave
String waves are also undergo refraction at junction of strings of different mass
density.
Refraction From a Rigid Boundary When a propagating wave is moving in
a thin string (low mass density) and it passes to the thick string (high mass
density) it suffers refraction at the string junction. The wave transition take
place from low to high mass density boundary. This boundary acts as rigid/hard
boundary but energy is transferable from one medium to other medium. Thus
propagating wave energy is partially refracted and partially reflected. Reflected
wave suffers a π phase difference while refracted wave moves with same phase
to the incident wave. Amplitude and velocities of both reflected and refracted
waves are different.
Refraction From a Non Rigid Boundary When a propagating wave is mov-
ing in a thick string (high mass density) and it passes to the thin string (low
mass density) it suffers refraction at the string junction. The wave transition
take place from high to low mass density boundary. This boundary acts as
non-rigid/soft boundary but energy is transferable from one medium to other
medium. Thus propagating wave energy is partially refracted and partially re-
flected. Reflected wave does not suffers phase difference while refracted wave
moves with same phase to the incident wave. Amplitude and velocities of both
reflected and refracted waves are different.

26 Mechanical Waves
1.3.5 Transverse Waves in String
Let a string is bound with two rigid boundaries and having length of L. Nodes
in string are form at bounded ends and anti-node forms at center of the string.
Assume one end at x = 0 and other end at x = L. From figure 1.23 L = mλ/2
which gives
λ =
2L
m

λ/2
Fig: 1
m = 1
b
N
b
N
b
A
λ/2
Fig: 2
m = 2
b
N
b
N
b
N
b
A
b
A
λ/2
Fig: 3
m = 3
b
N
b
N
b
N
b
N
b
A
b
A
b
A
λ/2
Fig: 4
m = 4
b
N
b
N
b
N
b
N
b
N
b
A
b
A
b
A
b
A
P Q
x = 0 x = L
L
Figure 1.23: Transverse Wave in String.
If velocity of wave in string is v = nλ then
v = n
2L
m
Where m = 1,2,3, . . . (1.21)
Where n is the frequency of the wave. The frequency of wave in string can be
find by rearranging equation (1.21)
n = m
v
2L
(1.22)
As the velocity of wave increases, number of nodes and anti-nodes increases.
Formation of nodes and anti-nodes is called harmonics1
of the wave. There is
one fundamental frequency and other harmonic frequencies.
1
An oscillation of a periodic quantity whose frequency is an integral multiple of the fun-
damental frequency.

28 Mechanical Waves
First Harmonic Frequency or Fundamental Frequency Substituting m = 1
in equation (1.22) which gives
n1 =
v
2L
(1.23)
This forms one anti-node and two nodes.
Second Harmonic Frequency or First Overtone Substituting m = 2 in equa-
tion (1.22) which gives
n2 = 2
v
2L
(1.24)
This forms two anti-node and three nodes.
Third Harmonic Frequency or Second Overtone Substituting m = 3 in
equation (1.22) which gives
n3 = 3
v
2L
(1.25)
This forms three anti-node and four nodes and so on.
1.3.6 Frequencies in One End Closed Pipe
In a pipe whose one end is closed and other is open then node is form at closed
end and anti-node is form at open end. If L is the length of pipe and closed end
is at x = 0 then open end would be at x = L. For an anti-node

λ/4
Fig : 0
m = 0
b
N
b
A
λ/4
Fig : 1
m = 1
b
N
b
A b
N
b
A
λ/4
Fig : 2
m = 2
b
N
b
A b
N
b
A b
N
b
A
λ/4
Fig : 3
m = 3
b
N
b
A b
N
b
A b
N
b
A b
N
b
A
L
Figure 1.24: Frequency in pipe open at one end.
L =

m +
1
2

λ
2
λ =
2L
m + 1
2
(1.26)
Where m is 0, 1, 2, 3, . . .. If velocity of wave is v = nλ then
v = n
2L
m + 1
2
(1.27)
Where n is the frequency of the wave. Equation (1.27) gives the frequency is
one end closed pipe
n =

m +
1
2

v
2L
(1.28)
Formation of nodes and anti-nodes is called harmonics of the wave. There is
one fundamental frequency and other harmonic frequencies. This is fundamental
frequency of one end closed pipe. One anti-node is formed.

30 Mechanical Waves
n1 =

1
2

v
2L
(1.29)
This forms one anti-node and one nodes.
n2 =

3
2

v
2L
(1.30)
This forms two anti-node and two nodes.
Third Harmonic Frequency or Second Overtone Substituting m = 2 in
equation (1.28) which gives
n3 =

5
2

v
2L
(1.31)
This forms three anti-node and three nodes and so on.
Solved Problem 1.7 A wave y = 0.25 sin(2πt − 4x) is entered into a pipe of
length 75cm, closed at one end. Find (i) how many nodes will be there in the
pipe at resonance, (ii) position of third anti-node from open end if possible and
(iii) maximum possible sum of nodes and anti-nodes.
Solution
1.3.7 Frequencies in Open Ends Pipe
In a pipe which is open at both end form anti-node at open end. If L is the
length of pipe and closed end is at x = 0 then open end would be at x = L. For
a anti-node
L = m
λ
2

λ/2
Fig : 0
m = 0
b
N
b
A
b
A
λ/2
Fig : 1
m = 1
b
N
b
N
b
A
b
A
b
A
λ/2
Fig : 2
m = 2
b
N
b
N
b
N
b
A
b
A
b
A
b
A
λ/2
Fig : 3
m = 3
b
N
b
N
b
N
b
N
b
A
b
A
b
A
b
A
b
A
L
Figure 1.25: Frequency in pipe open at both ends.
λ =
2L
m
(1.32)
Where m = 1, 2, 3, . . .. . .. If velocity of wave is v = nλ then
v = n
2L
m
(1.33)
Where n is the frequency of the wave. Equation (1.33) gives the frequency is
one end closed pipe
n = m
v
2L
(1.34)
Formation of nodes and anti-nodes is called harmonics of the wave. There is
one fundamental frequency and other harmonic frequencies.
n1 = 1
v
2L
(1.35)
This forms two anti-node and one nodes.

32 Mechanical Waves
n2 = 2
v
2L
(1.36)
This forms three anti-node and two nodes.
Third Harmonic Frequency or second Overtone Substituting m = 3 in equa-
n3 = 3
v
2L
(1.37)
This forms four anti-nodes and three nodes and so on.
1.3.8 Standing Wave Components
A standing wave is generated by superposition of the two propagating waves in
same medium. Two simple sinusoidal waves y1 and y2 are superimposed and
form a single wave by using summation relation as
y = y1 + y2 (1.38)
Assume a standing wave
y = −2a cos

2π
t
T

sin

2π
x
λ

which is generated by using relation 1.38. This standing wave can be re-factored
into its generating waves by using trigonometric relation
2 cos A sin B = sin(A + B) − sin(A − B)
By using this relation, standing wave is re-factored as
y = −a

sin 2π

t
T
+
x
λ

− sin 2π

t
T
−
x
λ

Now the standing wave is re-factored into its two constituents components.
Where
y1 = −a sin 2π

t
T
+
x
λ

and
y2 = a sin 2π

t
T
−
x
λ

By this way, we can re-factored a standing wave into its constituent parts.
Solved Problem 1.8 A mechanical wave y = 0.25 sin(2πt − 4x) is entered into a
pipe of length 22cm, opened at both ends. All units in the wave equation are in
centi-meters. Find (i) how many nodes will be there in the pipe at resonance,
(ii) position of third anti-node from open end if possible and (iii) maximum
possible sum of numbers of nodes and anti-nodes.

1.4. BEATS 33
Solution
1. Comparing this equation with propagating waves, it gives ω = 2π, hence
frequency of the wave in pipe is 1Hz. Again
2π
λ
= 4 ⇒ λ =
π
2
Now maximum number of wavelength possible inside the pipe of 22cm is
n =
22 × 2
π
= 14
At open end there is always anti-node in both end open pipe. Now, number of
nodes in the pipe will be 14 × 2 = 28.
2. Third anti-node will be at distance λ from the open end.
3. In both end open pipe, number of anti-nodes is more than number of
nodes by one. Hence, sum of total number of anti-nodes and nodes is 29 + 28 =
57.
Solved Problem 1.9 A standing mechanical wave is represented by y =
2 sin(2.5πt) cos(5x). Find the waves that are generating this standing wave.
Solution Using trigonometric relation, 2 sin A cos B = sin(A+ B)+ sin(A−
B),
y = sin(2.5πt + 5x) + sin(2.5πt − 5x)
Segregating the two equations in right hand side
y1 = sin(2.5πt + 5x)
and
y2 = sin(2.5πt − 5x)
Here y1 and y2 are the super-positioning waves.
1.4 Beats
When two waves, having negligible frequency difference travels in same medium
then frequency of resultant wave varies with time continuously. This phe-
nomenon is called beats. If frequencies of two waves are n1 and n2 then num-
ber of beats in one second is n1 ∼ n2. Let two waves are y1 = a sin ω1t
and y2 = a sin ω2 and they produces resultant wave under superposition is
y = y1 + y2. So,
y = a sin ω1t + a sin ω2t

34 Mechanical Waves
f1
y
F : 1
f2
y
F : 2
(f1 − f2)
y
F : 3
(ω1 − ω2) (ω1 + ω2)
△f
y
F : 4 A B
A = (ω1 − ω2) B = (ω1 + ω2)
Figure 1.26: Beats.
y = 2a sin

ω1 − ω2
2
t

cos

ω1 + ω2
2
t

(1.39)
If difference between frequency is negligible then

ω1 + ω2
2

≈ ω1 ≈ ω2
The intensity of frequency of resultant wave is near to original wave hence
identification of wave is not possible. But if ω1 ≈ ω2 then

ω1 − ω2
2

≈ 0
and identification of this frequency is very easy.
Solved Problem 1.10 A long pipe having radius 4cm is filled with water from
the base tap at a rate of quantity 20cm3
/s. A fork having frequency of 250Hz is
vibrating at the open end of pipe. Find the estimated time after which resonance
frequency would be sound.

1.4. BEATS 35
Solution Frequency of the fork is f = 250Hz and quantity of water entered
into pipe is 20cm3
/s. Area of cross-section is 16πcm2
. Now the rate of height
at which water is filled in the pipe in centi-meter per second is
h =
20
16π
(1.40)
Wavelength of the sound wave in meter is
λ =
vs
f
=
332
250
(1.41)
Distance between two successive harmonics is
d = 3 ×
λ
4
−
λ
4
= 2 ×
λ
4
(1.42)
Now the time required for second harmonics if first harmonic is observed at
t = 0
t =
d
h
=
2 × λ
4
20
16π × 10−2
(1.43)
Substituting the value of λ we have
t = 167 (1.44)
This is the estimated time in second.

2.1. SOUND WAVE 37
2Sound Waves
2.1 Sound Wave
The speed of sound is the distance travelled during a unit of time by a sound
wave propagating through an elastic medium. At normal temperature and pres-
sure the speed of sound is 343.2 meters per second. This is 1,236 kilometres per
hour (768 mph), or about one kilometre in three seconds or approximately one
mile in five seconds. Speed of sound in a medium depends on the properties of
material of medium. Speed of sound in a medium is given by
v =
s
Ks
ρ
(2.1)
Where Ks is a coefficient of stiffness for solid, bulk modulus for liquid and the
isentropic bulk modulus (or modulus of bulk elasticity) for gases.
2.1.1 Why Sound Travels Faster in Solids?
The speed of sound in a material is depends on the stiffness of the material.
Stiffness means resistance to deformation of material on applying of external
forces. For example, iron is more stiff than aluminium because iron opposes
more strongly when it is deformed by either hammering, or stretching or on
compression than aluminium. Similarly, aluminium is more stiff than water.
Atomic arrangement of stiff material is highly packed and geometrically ar-
ranged therefore inter-atomic forces are too stronger. In other words, we can
say that stiff material have larger Young’s Modulus.
b
b
b
b
b
b
b
b
b
b
b
b
b
b
b
b
b
b
b
b
b
b
b
b
b
b
b
b
b
b
b
b
b
b
b
b
b
b
b
b
b
b
b
b
b
b
b
b
b
b
(a) (b)
Figure 2.1: Atomic arrangement of gas (fig:a) and solid (fig:b).

38 Sound Waves
Sound waves are not materialistic subjects. They are energies those travel
through a medium. Sound waves need medium to propagate. Sound energy
moves from one point to other either by convection or conduction. In convection,
energised material particles moves from one place to other while in conduction,
energy is transferred by inter-atomic collision by oscillating atomic particles.
Contrast to light wave, sound wave can not travel in space (i.e. via phenomenon
of radiation).
b
b
b
b
b
b
b
b
b
b
b
b
b
b
b
b
b
b
b
b
b
b
b
b
b
b
b
b
b
b
b
b
b
b
b
b
b
b
b
b
b
b
b
b
b
b
b
b
b
b
b b
Energy
Movement
Path
A B
d
(a)
A B
d
(b)
Figure 2.2: Movement of sound energy by gas particle (fig:a) and by solid atoms
(fig:b) between two points A and B at distance d.
In gases, atomic particles are at distances and they move randomly. Hence,
gas particles need to travel long path to transfer the sound energy from one
place (say A) to other place (say B). Thus gas particle needs more time on
reaching from A to B. In solids and liquids, movement of atoms (either by
oscillation or by convection) is highly directional.
b
b
b
b
b
b
b
b
b
b
b
b
b
b
b
b b
b
b
b
b
b
b
b
b
b
b
b
b
b
b
b
b
b
b
b
b
b
b
b
b
b
b
b
b
b
b
b
b
b
b b
Random
Particle
Path
A B
(a)
Highly Directional
(b)
Figure 2.3: Movement of gas particle is random (fig:a) while oscillation collision
between solid atoms is highly directional (fig:b).
Hence atoms of solid or liquid material needs much less time to transfer the
energy between between A and B. This is why, in per unit time, sound energy

2.1. SOUND WAVE 39
propagate larger distance in solid and liquid in comparison to the gases. This
is why, sound wave moves with larger velocity in solids than gases.
Now, the distance between points A and B is d. So, sound velocities in gas
and solid are
vgas =
d
t1
and
vsolid =
d
t2
Here, ti and t2 are time required by atoms to transfer sound energy from point
A to B. From above explanation, t1 t2 therefore vgas vsolid.
2.1.2 Speed of Sound in Solids
The speed of sound for pressure waves in stiff materials such as “long rod of
metals” is sometimes given by
v =
s
Y
ρ
Where Y is Young’s modulus. This is similar to the expression for shear waves
where Young’s modulus replaces the shear modulus.
Solved Problem 2.1 Find the speed of sound in a solid medium whose density
is 4.65 gram per cubic centimetre and its Young’s modulus is 45 mega-newton
per square meter.
Solution Velocity of sound in solid medium is given by
v =
s
T
ρ
We have given, ρ =
4.65 × 10−3
10−2 × 10−2 × 10−2
kilogram per cubic meter and T =
45 × 106
newton per square meter. Now
v =
s
45 × 106
4.65×10−3
10−2×10−2×10−2
On solving it, we have speed of sound v = 98.37 meter per second.
2.1.3 Speed of Sound in Liquid
. In a fluid, the only non-zero stiffness is to volumetric deformation. Hence the
speed of sound in a fluid is given by
v =
s
K
ρ

40 Sound Waves
Where K is the bulk modulus of the fluid. The speed of sound in solid as well
as in liquid, increases with the stiffness1
.
Solved Problem 2.2 A uniform string of length 20m is suspended from a rigid
support. A short wave pulse is introduced at its lowest end. It starts moving
up the string. What is the time taken by it to reach at the support?
Solution In the string when a small disturbance is created at the lower end of
the string the wave propagate towards the support. Taking wave disturbance as
small as possible and string as rigid medium, velocity of the wave is v =
p
T/µ,
where T is tension on the string and µ is mass per unit length of the string.
Velocity of wave in this string is is given by
v =
s
T
µ
Take an experimental element of length x in the string string. For this element,
T = µ × x × g. For this element, using above velocity relation, we have velocity
of the wave as v =
√
gx. Derivative it w.r.t. x,
dv
dx
=
√
g ×
1
2
√
x
From the velocity and acceleration relation
d2
x
dt2
= v
dv
dx
Substituting the values of v and dv/dx, we get the acceleration of the wave as
a = g/2. Applying the third relation of linear motion in vertical plane with
initial velocity zero at lower end of the string. So
s = ut +
1
2
at2
⇒ 20 =
1
2
×
10
2
× t2
The time is t = 2
√
2s.
2.1.4 Velocity of Sound in Atmosphere
Atmosphere is like an open gaseous systems are those systems in which gaseous
substances are free to expand, free to entry into the system and free to exit
from system. Speed on sound in atmosphere or atmospheric like conditions is
approximately 330 meter per second approximately at 0◦
C and 0% humidity
(dry air).
1
the property of elastic material by which it resists its deformation when external forces
acts to deform it.

2.1. SOUND WAVE 41
2.1.5 Velocity of Sound in Gaseous System
Similar to the relation of sound in solids, the speed of sound in gaseous system
is given by
v =
s
Kg
ρ
(2.2)
Where Kg is a coefficient of stiffness, the bulk modulus (or the modulus of bulk
elasticity for gases), ρ is the density of the gas. It is to be noted that it is true
if propagation of sound wave does not losses energy or medium temperature
does not change during compression and rarefaction of the medium. When
sound moves in atmosphere, air compressed and rarefied continuously. Thus,
air medium behaves like an adiabatic process system. For an ideal gas, Kg is
given by
Kg = γ · P
Where γ is the ratio of the two specific heats of gases at constant pressure and
constant volume. This equation includes the adiabatic changes of the medium
due to sudden compression and rarefaction of longitudinal waves. Now the
velocity of sound becomes
v =
s
γP
ρ
(2.3)
Where P is pressure on gas.
2.1.6 Thermodynamic Effect
The speed of sound is variable and depends on the properties of the substance
through of which the wave is travelling. In solids, the speed of longitudinal waves
depend on the stiffness to tensile stress, and the density of the medium. In fluids,
the medium’s compressibility and density are the important factors. In gases,
compressibility and density are related, making other compositional effects and
properties important, such as temperature and molecular composition. In other
words, speed of sound in gases can be increased (i) either by increasing density
of the gas by means of compressing or by means of increasing humidity so that
gas particles travel comparatively smaller path between two points (ii) or by
increasing particle’s speed by means of thermal heating so that sound energy
can be transferred from one point to other point speedily by gas particles.
Effect of Temperature
Velocity of sound in a gas is given by
v =
s
γP
ρ

42 Sound Waves
Now, using the ideal gas law to replace P with
nRT
V
, we have
v =
s
γnRT
ρV
=
s
γRT
ρV/n
Here, ρV/n is molar mass (mass of one mole gas) of the air and it is represented
by M.
v =
r
γRT
M
For air, R = 8.3144 Joule per Kelvin per Mole, molar mass is 0.0289 kilogram
per mole and γ = 1.403 at 273K. Substituting these values, we have
v =
r
1.403 × 8.3144 × T
0.0289
Or
v = 20.08
√
T (2.4)
This is velocity of sound in terms of absolute temperature scale. Substitute
T = 273 + t, where t is temperature in Celsius scale. 273K is equal to the 0◦
C.
It gives
v = 20.08
√
273 + t = 20.08 ×
√
273 ×
s
1 +
t
273

Expanding right hand side upto two terms only. Leaving higher order terms as
they are negligible. So
v = 331

1 +
t
273
1/2
≈ 331

1 +
1
2
×
t
273
+ . . .

Or
v = 331 +
331
2
×
t
273
= 331 + 0.606t
This is velocity of sound at t◦
C. When t = 0◦
C, v0 = 331 meter per second. So,
above relation can be written as
vt = v0 + 0.606t (2.5)
Assume vr is reference speed of sound at r◦
C in a gaseous medium then speed
of sound in gas at t◦
C is
vt = vr + 0.606 × (t − r)
Here, 0.606 is increase of speed (∆v = 0.606) of sound in meter at rise of
temperature of medium by one degree Celsius (∆t = 1◦
C). In air, at 0◦
C, speed
of sound is 331.3 meter per second. For given ideal gas the sound speed depends
only on its temperature.

2.1. SOUND WAVE 43
Solved Problem 2.3 Velocity of sound in air at 0◦
C is 330 meter per second.
Find the velocity of sound at 45◦
C.
Solution We know that
vt = v0 + 0.606 × ∆t
Substituting the values, we have
v45◦ = 330 + 0.606 × 45
Or
v45◦ = 330 + 27.27
And it gives v45◦ = 357.27 meter per second.
Solved Problem 2.4 Velocity of sound in air at standard temperature, i.e. 0◦
C
is 330 meter per second. In summer, the temperature of atmosphere raised to
50◦
C. Find the velocity of sound in summer.
vt = v0 + 0.606 × ∆t
v45◦ = 330 + 0.606 × 50
Or
v45◦ = 330 + 33.30
And it gives v45◦ = 363.30 meter per second.
Solved Problem 2.5 Velocity of air at temperature of 10◦
C is 336 meter per
second. In summer season, velocity of sound is measured as 345 meter per
second by a student. Find the temperature of the surroundings in summer
season.
vt = vr + 0.606 × ∆t
345 = 336 + 0.606 × (t − 10)
Or
9 = 0.606 × (t − 10) ⇒ t − 10 = 14.85
And it gives t = 24.85◦
C.

44 Sound Waves
Effect of Pressure
At a constant temperature, the ideal gas pressure has no effect on the speed of
sound, because pressure and density (also proportional to pressure) have equal
but opposite effects on the speed of sound, and the two contributions cancel
out exactly. In a similar way, compression waves in solids depend both on
compressibility and density. In non-ideal gases, such as a Vander Waals gas, the
proportionality is not exact, and there is a slight dependence of sound velocity
on the gas pressure.
Mathematical Derivative Consider the sound wave propagating at speed
v through a pipe having a cross-sectional area A. In time interval dt it moves
length dx = v dt. In steady state, the mass flow rate ṁ = ρvA must be the same
at the two ends of the tube, therefore the mass flux is constant and v dρ = −ρ dv.
Pressure-gradient force provides the acceleration:
dv
dt
= −
1
ρ
dP
dx
dP = (−ρ dv)
dx
dt
v2
=
dP
dρ
Therefore:
v =
s
∂P
∂ρ

s
where P is the pressure and ρ is the density and the derivative is taken isentrop-
ically at constant entropy s. Thus sound velocity in air is depends on pressure
and consequently change in density.
Effect of Humidity
Humidity has a small but measurable effect on sound speed (causing it to in-
crease by about 0.1% to 0.6%), because oxygen and nitrogen molecules of the
air are replaced by lighter molecules of water. This is a simple mixing effect.
2.2 Reflection And Refraction
When a sound wave reached to boundary, it suffers reflection or refraction ac-
cordingly the boundary is reflective or transparent. If boundary is rigid, then
reflected ray returns back without changing of phase. This reflected sound wave
interfere with incident wave and produces standing sound wave. This inter-
ference generates node and anti-node in the standing sound wave. At closed
boundary, there are always node and at open boundary there are always anti-
node.

2.2. REFLECTION AND REFRACTION 45
2.2.1 Reflection of Sound Wave
Reflection of sound is takes place similar to the reflection of light and it also
follows the laws of reflection. The reflected wave can interfere with incident
waves and produce patterns of constructive and destructive interference. This
leads to the resonance called standing waves. It is also seen that sound intensity
enhances because reflected wave added some pressure to the incident wave. This
behaviour is similar to the change in momentum of the rebounding ball from
rigid wall. This pressure enhancement is very useful in increasing microphone
sensitivity. From the Huygen’s law of reflection of wave, when a wave front
collides with an opaque object, secondary waves emanates from the each point
of the surface. The velocity of secondary wave is similar to the mother wave
and its frequency and wavelength do not change after reflection.
Pressure Zone The sound intensity near a hard surface is enhanced because
the reflected wave adds to the incident wave, giving a pressure amplitude that
is twice as great in a thin “pressure zone” near the surface.
2.2.2 Refraction of Sound Wave
When sound wave enters from one medium to other medium, it suffers refraction
at inter-medium boundary. The speed of refracted sound depends on the type
of medium from which it enters into the other medium. The velocity of sound
is greater in larger stiff materials, in contrast to the refraction of light.
2.2.3 Echo
Echo is reflected sound from a right surface. Sound propagating from a source
towards a rigid boundary, gets reflected back and returns to the location of
source. The reception of reflected sound is called Echo.
Solved Problem 2.6 A person is standing in front of a hight. He shouted and
hears echo after 2.5s. Find the distance of hight from the person. Take speed
of sound 332m/s.
Solution Let the person is at distance d from the hight. When we heard echo
of own sound, sound travels double distance of d. Now, from velocity, distance
and time relation, we have
332 =
2d
2.5
It gives, d = 415 meter.
Solved Problem 2.7 If speed of sound is 330 meter per second then find the time
taken by it to travel 2.3 kilo meter.
Solution The time taken by sound to travel a distance of 2.3 kilometres is
t =
2300
330

46 Sound Waves
It will give t = 6.97 seconds.
Solved Problem 2.8 A sound source is generating sound waves of 290 Hz at
medium temperature 25◦
C. Find the (i) wavelength of sound and (ii) time period
of the sound wave.
Solution
i. The speed of sound at 25◦
C is given by
v25◦C = 330 + 0.606 × 25
Or, v = 345.15 meter per second. The frequency of sound is 290Hz. Therefore,
wavelength of the sound wave is
λ =
v
f
=
345.15
290
It gives, wavelength of sound wave as λ = 1.19 meter.
ii. Time period of sound wave is given by
T =
1
f
=
1
290
It gives time period of sound wave as T = 3.44 milliseconds.
Solved Problem 2.9 A sound wave is produced with 340Hz and 1m wavelength.
Find (i) speed of sound wave, (ii) time period of the sound wave and (iii) tem-
perature of atmosphere.
Solution
i. Speed of sound is given by v = fλ. It gives
v = 340 × 1
It gives, v = 340 meter per second.
ii. Time period of the sound wave is given by
T =
1
f
=
1
340
i.e. T = 2.94 milliseconds.
iii. Temperature of environment where sound wave is travelling is found by
relation
vt = v0 + 0.606 × t
Substituting the values
340 = 330 + 0.606t ⇒ t =
340 − 330
0.606
i.e. temperature of the environment is t = 16.50◦
C.

Solved Problem 2.10 A person is standing between two cliffs, separated by
1000m. One of the two cliffs is at 300m from him. He shouted and hears echoes.
He hears first echo after 2s. Now, find (i) speed of sound, (ii) temperature of
the medium at that location and (iii) time after that he hears second echo.
Solution Assume scene scenario is like given below. Cliff at front side of
person is 300 meter away while cliff at his back side is 700 meter away. Simply,
the firs echo he hears would come from his front side.
b b
b
1km
0.3km
i. Now, from the given parameters in question, speed of sound in cliff side
is
v =
300 + 300
2
= 300m/s
ii. As the speed of sound in cliff side is not equal to standard speed of sound.
It means temperature at cliff side is different. We can find the temperature of
the cliff side as
vt = v0 + 0.606t
300 = 330 + 0.606t ⇒ t =
−30
0.606
It gives the temperature of cliff side t = −49.50◦
C.
iii. Second echo he will hear when sound would reflect from his backside
cliff. Secondary echoes will be heard by him after two primary echoes. Time
after that he will hear second echo is
t =
700 + 700
300
It gives, t = 4.667 seconds.

48 Sound Waves
Solved Problem 2.11 A man is standing in front of huge wall at 700m. Tem-
perature of atmosphere is 30◦
C. man fires gun. Find the time after which he
hears echo. If he approaches to wall by 300m and again fires his gun. Find the
time after that he hears echo.
Solution First we find the speed of sound in atmosphere whose temperature
is not at 0◦
C. For this purpose, we shall use relation
vt = v0 + 0.606t
Assuming that speed of sound at 0◦
C is 330 meter per second. So,
v30◦C = 330 + 0.606 × 30 ≈ 348m/s
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700m
Man is at 700 meter away from wall as shown in above figure. Sound will
travel 700 × 2 meters from the time of firing of gun and hearing of echo of gun
firing by man. So, time lapsed is
t =
2 × 700
348
It gives, t = 4.023 seconds.
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400m 300m
If man approaches to wall by 300 meter then distance remains between man
and wall is 400 meter. For this distance, time after man hears echo of gun firing
is
t =
2 × 400
348
It gives, t = 2.30 seconds. These are desired results.

Solved Problem 2.12 A man hears echo of own shout after 3s when atmospheric
temperature is 30◦
C. Find the distance of the sound reflector from man. If
temperature of the atmosphere rises to 45◦
C then find the time after that he
would hear echo. Take speed of sound 330m/s at 0◦
C.
Solution First we find the speed of sound in atmosphere when temperature
is 30◦
C. For this purpose, we shall use relation
vt = v0 + 0.606t
Given that speed of sound at 0◦
C is 330 meter per second. So,
v30◦C = 330 + 0.606 × 30 ≈ 348m/s
Using echo relation for distance d (distance between man and sound reflector),
we have
t =
2d
v
⇒ 3 =
2d
348
It gives d = 522 meters. It is distance between man and reflector. If temperature
rises to 45◦
C then speed of sound is
v45◦C = 330 + 0.606 × 45 ≈ 357m/s
Now, the time after which man would hear echo is
t =
2 × 522
357
It gives t = 2.92 seconds approximately.
Solved Problem 2.13 Two cliffs are in same side of a girl, who is shouting loudly.
First cliff is at 495m away from her. She hears two consecutive echoes after 3s
and 4s respectively. Find the distance between two cliffs. If temperature of
atmosphere rises to 45◦
C, then find the times after that she hears first and
second echoes. Take speed of sound 330m/s at 0◦
C.
Solution Using first echo, we get the speed of sound in atmosphere.
v =
2 × 495
3
= 330m/s
Using this velocity of sound, we can get the distance of second cliff from the
position of girl. So,
330 =
2d
4
⇒ d = 660m
Now, the distance between two cliffs is 660m − 495m = 165 meters. At 45◦
C,
speed of sound is
v45◦C = 330 + 0.606 × 45 ≈ 357m/s

50 Sound Waves
Now, the time after which man will hear echoes are
t1 =
2 × 495
357
= 2.77s
and
t2 =
2 × 660
357
= 3.70s
These are desired results.
2.3 Doppler Effect (Relative Theory of Sound)
Phenomenon of change in real frequency of source observed by observer due to
relative motion of source and observer is known as Doppler Effect. There are
three possible conditions for Doppler effect (i) observer is stationary and source
is moving, (ii) observer is moving and source is stationary and (iii) when both,
observer and stationary are moving. Here it should be note that the speed of
sound does not depend on the the velocity of the emitting source but depends on
the medium in which it travels. Only observed frequency depends on the speed
of emitting source as well as speed of observer.
2.3.1 Observer Stationary Source Moving Away
Let observer is at O and sound source is at S1 at distance L and moving away
with velocity vs. Source produces a sound wave which is observed by observer at
time t1 = L/v. Where v velocity of the sound wave produced by the source. As
observer is stationary, therefore relative velocity of sound wave to the observer
is v − 0 = v. The source reaches at S2 after time T0 and produces second sound
wave which reaches at observer after time (L + vsT0)/v. If observer observes
second wave at time t2 then
S1 S2
bcb
O
vo
vs
v
L vsT0
Figure 2.4:
time t2 is sum of time taken by source between S1 to S2 and time taken by
sound wave between S2 to O.
t2 = T0 +
L + vsT0
v
(2.6)
Here v is the velocity of wave in air medium and T0 is the real time period of
the sound wave. Simplifying above equation
t2 = T0 + t1 +
vsT0
v

2.3. DOPPLER EFFECT (RELATIVE THEORY OF SOUND) 51
Or
t2 − t1 = T0

1 +
vs
v

The difference of time period between two wave pulse observed by observer is
the virtual time period observed by observer hence
T = T0

1 +
vs
v

Substituting T = 1/n in above equation
1
n
=
1
n0

1 +
vs
v

On simplification above result becomes
n = n0

v
v + vs

(2.7)
Where v (v + vs), i.e. the observed frequency by observer is less than real
frequency of source. If source is moving toward observer then
n = n0

v
v − vs

(2.8)
and observed frequency would be greater than the real frequency.
2.3.2 Observer is Moving Away From Stationary Source
Let observer is moving away from source S with velocity vo and at any instant
it is at O′
. Source produces a sound wave which is observed by observer at
time t1 = L/(v − vo). Where v − vo is the relative velocity of the sound wave
with respect to observer. The observer reaches at O after time T0 and observes
sound wave which reaches at observer after time (L+voT0)/(v−vo). If observer
observes second wave at time t2 then
S
bcb
O′
bcb
O
vo v
L
voT0
Figure 2.5:
t2 = Time taken by observer between O′
to O + time taken by sound wave
between S to O
t2 = T0 +
L + voT0
v − vo
(2.9)

52 Sound Waves
t2 = T0 + t1 +
voT0
v − vo
t2 − t1 = T0

1 +
vo
v − vo

T = T0

v
v − vo

1
n
=
1
n0

v
v − vo

n = n0

v − vo
v

(2.10)
Where (v − vo) v, i.e. the observed frequency by observer is less than real
frequency of source. If observer is moving toward source then
n = n0

v + vo
v

(2.11)
and observed frequency would be greater than the real frequency.
2.3.3 Observer Source Moving in Same Direction
Source is Leading Observer
Let at any instant source is at S1 and observer is at O′
which are L distance
apart from each-other. They are moving with velocities vs and vo respectively
in same direction. The observer observes wave pulse in time t1 = L/(v + vo).
Here v+vo is the velocity wave with respect to observer. After time T0 observer
and source reaches at O and S2 and wave pulse produces by source is observed
after time t2 by observer then
S1 S2
bcb
O
bcb
O′
vo v
L
voT0 vsT0
vs
Figure 2.6:

t2 = T0 +
L − voT0 + vsT0
v + vo
(2.12)
t2 = T0 + t1 +
−voT0 + vsT0
v + vo
t2 − t1 = T0

1 +
−vo + vs
v + vo

T = T0

v + vs
v + vo

1
n
=
1
n0

v + vs
v + vo

n = n0

v + vo
v + vs

(2.13)
Where (v + vo) (v + vs) −→ vo vs, i.e. if the velocity of observer is less
than source then observed frequency by observer is less than real frequency of
source. If observer leading to source then
n = n0

v − vo
v − vs

(2.14)
and observed frequency would be less than the real frequency. If source and
observer is moving towards each-other then
n = n0

v + vo
v − vs

(2.15)
and observed frequency would be greater than the real frequency of source.
Image of Moving Sound Source
Assume a condition where source of sound is moving towards the non-
penetrable surface (wall) with speed of v. When sound waves are collide with
a surface, secondary waves with same frequency and wavelength, but may be
of lesser amplitude, are generated and move in all directions. When sound
waves emitted from the moving source collide with a surface, an image of

54 Sound Waves
real source is produced at opposite side of the surface that have an imaginary
velocity equal to the velocity of real source but in opposite direction of the
direction of real source. This is the similar to the phenomenon of a person
moving towards a plain mirror and seeing ones image.
Observer Moving Away To Source
In figure, observer is moving away from the standing source. Sound pulses
are shown in color bands, while frequencies of a pulse shown in vertical lines.
Assume that source and observers are standing initially and source emits sound
pulse in each interval of time T . Assume first pulse just reaches to observer
and starts passing it and time is taken as t = 0. Pulse of first interval of time
T1, starts passing it and after time t = T1 it completely passes to the observer.
vo
bc
bc
bcb
T3 T2 T1
voT2
(3)
bc
bcb
T2 T1
(2)
bc
bcb
n
T1
(1)
As the first pulse passes to the observer, second pulse of time interval
t = T2 is just reaches to the observer. If observer is further standing, second
pulse would pass him in time T2. But if he starts moving away from source
when second pulse is just reaches to observer with velocity vo and reaches a
distance voT2 rightward. Now in the same time second pulse passes completely
the origin of observer in time T2. After time T1 + T2 pulse frequency that
passes to the observer are those frequencies that are in the right hand side
of the observer. This is also clear that standing observer hears more pulse
frequencies in the time T1 +T2 rather than he is moving away from the source.
Observer Moving Towards Source

In figure observer is moving towards the standing source. Sound pulses
As the first pulse passes to the observer, second pulse of time interval t = T2
is just reaches to the observer. If observer is further standing, second pulse
would pass him in time T2.
vo
bc bc
bcb
T3 T2 T1
voT2
(3)
bc
bcb
T2 T1
(2)
bc
bcb
n
T1
(1)
But if he starts moving towards source when second pulse is just reaches
to observer with velocity vo and reaches a distance voT2 leftward. Now in
the same time second pulse passes completely the origin of observer in time
T2. Observer not only hears pulse frequencies of time T2 but he also swept
some pulse frequencies of pulse emitted by source in time interval of T3 that
just reaches to the observer’s origin. After time T1 + T2 pulse frequency that
passes to the observer are those frequencies that are in the right hand side
of the observer. This is also clear that standing observer hears lesser pulse
frequencies in the time T1 + T2 rather than he is moving towards the source.
Source Moving Away To Observer
In figure, source is moving away from the standing observer. Sound pulses

56 Sound Waves
As the first pulse passes to the observer, second pulse of time interval t = T2 is
just reaches to the observer. If source is further standing, second pulse would
pass him in time T2.
vs bcb
bcb bc
T3 T2 T1
vsT2
(3)
bc
bcb
T2 T1
(2)
bc
bcb
n
T1
(1)
But if source starts moving away from the observer when second pulse is
just reaches to observer with velocity vs and reaches a distance vsT2 leftward.
Now in the same time second pulse does not passes completely the origin of
observer in time T2. Observer does not hears pulse frequencies of time T2
completely as the sound pulse shifted leftward in second second. After time
T1 + T2 pulse frequency that passes to the observer are those frequencies that
are in the right hand side of the observer. This is also clear that observer
hears lesser pulse frequencies in the time T1 + T2 when source is in motion
away from observer rather than source and observer are standing.
Source Observer Moving Towards Each Other
In figure, source and observer are moving towards each-other. Sound
pulses are shown in color bands, while frequencies of a pulse shown in vertical
lines. Assume that source and observers are standing initially and source
emits sound pulse in each interval of time T . Assume first pulse just reaches
to observer and starts passing it and time is taken as t = 0. Pulse of first
interval of time T1, starts passing it and after time t = T1 it completely passes
to the observer. As the first pulse passes to the observer, second pulse of time
interval t = T2 is just reaches to the observer.

vs vo
bcb bcb bc bc
T3 T2 T1
voT2
vsT2
(3)
bc
bcb
T2 T1
(2)
bc
bcb
n
T1
vw
(1)
If source and observer are further standing, second pulse would pass ob-
server in time T2. But if source and observer starts moving towards each other
when second pulse is just reaches to observer with velocity vs and v0 respec-
tively and covers a distance vsT2 rightward and voT − 2 leftward respectively.
In time T2 observer swept frequencies of pulse of time interval T3 and source
shifted frequencies of time interval T3 rightward. Thus observer not only hears
pulse frequencies of time interval T3 due to his leftward motion but also hears
some frequencies of shifted source of time interval T3 during time T2. This
is clear that observer hears larger pulse frequencies in the time T1 + T2 when
source and observer are in motion towards each other.
Source Observer Moving Away To Each Other
In figure, source and observer are moving away from each-other. Sound
pulses are shown in color bands, while frequencies of a pulse shown in vertical
lines. Assume that source and observers are standing initially and source
emits sound pulse in each interval of time T . Assume first pulse just reaches
to observer and starts passing it and time is taken as t = 0. Pulse of first
interval of time T1, starts passing it and after time t = T1 it completely passes
to the observer. As the first pulse passes to the observer, second pulse of time
interval t = T2 is just reaches to the observer. If source and observer are
further standing, second pulse would pass observer in time T2.

58 Sound Waves
vs vo
bcb
bcb bc bc
T3 T2 T1
vsT2 voT2
(3)
bc
bcb
T2 T1
(2)
bc
bcb
n
T1
(1)
But if source and observer starts moving away from each other when
second pulse is just reaches to observer with velocity vs and v0 respectively
and covers a distance vsT2 leftward and voT −2 rightward respectively. In time
T2 observer misses some frequencies of pulse of time interval T2 and source
shifted frequencies of time interval T3 leftward. Thus observer hears lesser
pulse frequencies of time interval T2. This is clear that observer hears lesser
pulse frequencies in the time T1 + T2 when source and observer are in motion
away from each other.

Introduction to mechanical waves

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Introduction to mechanical waves