NCERT Solutions for Class 11 Physics Chapter 14 Waves

NCERT
SOLUTIONS
CLASS - 11th

aglase .co

Class : 11th
Subject : Physics
Chapter : 15
Chapter Name : Waves

Q15.1 A string of mass 2.50 kg is under a tension of 200 N. The length of the stretched string is 20.0 m. If the transverse jerk is struck at one
end of the string, how long does the disturbance take to reach the other end?

Answer. Mass Of the string, M = 2.50 kg
Tension in the string, T = 200 N
Length of the string, I = 20.0 m
M

Mass per unit length μ=
l =
2.50

20
= 0.125kgm
−1

The velocity (v) of the transverse wave in the string is given by the relation:
T
v = √
μ

200
= √ = √1600 = 40m/s
0.125

¯
¯¯ 20
∴ Time taken by the disturbance to reach the other end, t = v = = 0.50s
40

Page : 391 , Block Name : Exercise

Q15.2 A stone dropped from the top of a tower of height 300 m high splashes into the water of a pond near the base of the tower. When is the
splash heard at the top given that the speed of sound in air is 340 ms ? (g = 9.8ms ) −1 −2

Height of the tower, s = 300m

Initial velocity of the stone, u = 0

2
Acceleration, a = g = 9.8m/s

Speed of sound in air = 340m/s

The time (t1 ) taken by the stone to strike the water in the pond can be calculated using

the second equation of motion, as:

1 2
s = ut1 + gt
2 1

1 2
300 = 0 + × 9.8 × t
2 1

300×2
∴ t1 = √ = 7.82s
9.8

Time taken by the sound to reach the top of the tower, t = t1 + t2

Therefore, the time after which the splash is heard, t = t1 + t2

= 7.82 + 0.88 = 8.7s

Page : 391 , Block Name : Exercise

Q15.3 A steel wire has a length of 12.0 m and a mass of 2.10 kg. VMnat should be the tension in the wire so that speed of a transverse wave on
the wire equals the speed of sound in dry air at 20 C = 343ms ∘ −1

Length of the steel wire, l = 12m

Mass of the steel wire, m = 2.10kg

Velocity of the transverse wave, v = 343m/s

Mass per unit length μ = m
=
2.10
= 0.175kgm
−1

l 12

For tension T, velocity of the transverse wave can be obtained using the relation:
T
v = √
μ

2
∴ T = v μ

2 4
= (343) × 0.175 = 20588.575 ≈ 2.06 × 10 N

Page : 391 , Block Name : Exercise

γP
Q15.4 Use the formula v = √ p
to explain why the speed of sound in air

(a) is independent of pressure,
(b) increases With temperature,
(c) increases with humidity.

γP
M = √
ρ

v here,

Mass M
Density, ρ = =
Volume V

M = Molecular weight of the gas

V = Volume of the gas

Hence, equation ( i ) reduces to:

γP V
v = √
M

Now from the ideal gas equation for n = 1 :

P V = RT

For constant T , P V = Constant

since both M and y are constants, v = Constant

Hence, at a constant temperature, the speed of sound in a gaseous medium is

independent of the change in the pressure of the gas.

(b) Take the relation:

γP
v = √
ρ

For one mole of an ideal gas, the gas equation can be written as:

P V = RT
RT
P = … (ii)
V

Substituting equation (ii) in equation (i), we get:

γRT γRT
v = √ = √
Vρ M

where,
Mass, M = ρV is a constant

γ and R are also constants

We conclude from equation (iv) that v ∝ √T .

Hence, the speed of sound in a gas is directly proportional to the square root of the

temperature of the gaseous medium, l..e., the speed of the sound increases with an

increase in the temperature of the gaseous medium and vice versa.

(c) Let vm and v4 be the speeds of sound in moist air and dry air respectively.

Let ρm and ρs be the densities of moist air and dry air respectively.

Take the relation:

γP
v = √
ρ

Hence, the speed of sound in moist air is:

γP
vm = √
ρm

And the speed of sound in dry air is:

γP
vd = √
ρd

On dividing equations ( i) and (ii), we get:

vm γP ρd ρd
= √ × = √
vd ρm γP ρm

However, the presence of water vapour reduces the density of air, l.e.,

ρd < ρm

∴ vm > vd

Hence, the speed of sound in moist air is greater than it is in dry air. Thus, in a gaseous

medium, the speed of sound increases with humidity.

Page : 391 , Block Name : Exercise

Q15.5 You have learnt that a travelling wave in one dimension is represented by a function y = f (x, t)where x and t must appear in the
combination x - v t or x + v t, i.e. y = f (x ± vt). Is the converse true? Examine if the following functions for V can possibly represent a travelling
wave:

2
(a) (x − vt)

x+yt
log[ ]
x0

1

(x+vt)

Answer. No;
(a) Does not represent a wave
(b)Represents a wave
(c) Does not represent a wave
The converse of the given statement is not true. The essential requirement for a function to represent a travelling wave is that it should remain
nite for all values Of x and t.
Explanation:
2
(a) For x = 0 and t = 0, the function (x − vt) becomes 0 .

Hence, for x = 0 and t = 0, the function represents a point and not a wave.

(b) For x = 0 and t = 0, the function

x+yt
log( ) = log 0 = ∞
x0

Since the function does not converge to a nite value for x = O and t =O, it represents a
travelling wave.
(c) For x = 0 and t = 0, the function

1 1
= = ∞
x+vt 0

since the function does not converge to a finite value for x = 0 and t = 0, it does not

represent a travelling wave.

Page : 391 , Block Name : Exercise

Q15.6 A bat emits ultrasonic sound of frequency 1000 kHz in air. If the sound meets a water surface, what is the wavelength of (a) the re ected
sound, (b) the transmitted sound? Speed Of sound in air is 340 ms and in water 1486 ms
−1 −1

6
(a) Frequency of the ultrasonic sound, v= 1000kHz = 10 Hz

Speed of sound in air, va = 340m/s

The wavelength (λr ) of the reflected sound is given by the relation:
v
λr =
v

340 −4
= = 3.4 × 10 m
6
10
6
(b) Frequency of the ultrasonic sound, v = 1000kHz = 10 Hz

Speed of sound in water, vw = 1486m/s

The wavelength of the transmitted sound is given as:

1486 −3
λ1 = = 1.49 × 10 m
6
10

Page : 391 , Block Name : Exercise

Q15.7 A hospital uses an ultrasonic scanner to locate tumours in a tissue. What is the wavelength of sound in the tissue in which the speed of
sound is 1.7 km s ? The operating frequency of the scanner is 4.2 MHz.
−1

3
Speed of sound in the tissue, v = 1.7km/s = 1.7 × 10 m/s

6
Operating frequency of the scanner, v = 4.2MHz = 4.2 × 10 Hz

The wavelength of sound in the tissue is given as:
v
λ =
v

3
1.7×10 −4
= 6
= 4.1 × 10 m
4.2×10

Page : 391 , Block Name : Exercise

Q15.8 A transverse harmonic wave on a string is described by
π
y(x, t) = 3.0 sin(36t + 0.018x + )
4

Where x and V are in cm and t in s. The positive direction Of x is from left to right.
(a) Is this a travelling wave or a stationary wave? If it is travelling, what are the speed and direction of its propagation?
(b) What are its amplitude and frequency?
(c) What is the initial phase at the origin?
(d) What is the least distance between two successive crests in the wave?

(a) Yes; Speed = 20m/s , Direction = Right to left

(b) 3cm; 5.73Hz

π
(c)
4

(d) 3.49m

Explanation:

(a) The equation of a progressive wave travelling from right to left is given by the

displacement function:

y(x, t) = a sin(ωt + kx + Φ) … (i)

The given equation is:

π
y(x, t) = 3.0 sin(36t + 0.018x + )
4

On comparing both the equations, we find that equation (ii) represents a travelling wave,

propagating from right to left.

Now, using equations (i) and (ii), we can write:
−1
ω = 36rad/s and k = 0.018m

We know that:

ω 2π
v = λ =
2π and k

Also,

v = vΛ

ω 2π ω 36
∴ v = ( ) × ( ) = = = 2000cm/s = 20m/s
2π k k 0.018

Hence, the speed of the given travelling wave is 20m/s

(b) Amplitude of the given wave, a = 3cm

Frequency of the given wave:

ω 36
v = = = 5.73Hz
2π 2×3.14
π
ϕ=
(c) On comparing equations (i) and (ii), we find that the initial phase angle, 4

(d) The distance between two successive crests or troughs is equal to the wavelength of

the wave.

Wavelength is given by the relation:

2π
k =
λ

2π 2×3.14
∴ λ = = = 348.89cm = 3.49m
k 0.018

Page : 391 , Block Name : Exercise

Q15.9 For the wave described in Exercise 15.8, plot the displacement (V) versus (t) graphs for x = 0, 2 and 4 cm. What are the shapes of these
graphs? In which aspects does the oscillatory motion in travelling wave differ from one point to another: amplitude, frequency or phase?

All the waves have different phases.

The given transverse harmonic wave is:

π
y(x, t) = 3.0 sin(36t + 0.018x + )
4

For x = 0, the equation reduces to:
π
y(0, t) = 3.0 sin(36t + )
4

2π −1
Also, ω = = 36rad/s
T

π
∴ T = s
8

Now, plotting V vs. t graphs using the different values of t, as listed in the given table.

Page : 391 , Block Name : Exercise

Q15.10
For the travelling harmonic wave y(x, t) = 2.0 cos 2π(10t − 0.0080x + 0.35) Where x and y are in cm and t in s. Calculate the phase

motion of two points separated by a distance of

(a) 4m,

(b) 0.5m t

λ
(c)
2

3λ
(d)
4

Equation for a travelling harmonic wave is given as:

y(x, t) = 2.0 cos 2π(10t − 0.0080x + 0.35)

= 2.0 cos(20nt − 0.016nx + 0.70π)

Where,

Propagation constant, k = 0.0160n

Amplitude, a = 2cm

Angular frequency, ω = 20rad/s

Phase difference is given by the relation:
2π
ϕ = kx =
λ

(a) For x = 4m = 400cm

Φ = 0.016n × 400 = 6.4rad

(b) For 0.5m = 50cm

Φ = 0.016n × 50 = 0.8πrad
λ

(c) For
x=
2

2π λ
ϕ = × = πrad
λ 2

(d) For

2π 3λ
ϕ = × = 1.5πrad
λ 4

Page : 391 , Block Name : Exercise

Q15.11
2
The transverse displacement of a string (clamped at its both ends) is given by y(x, t) = 0.06 sin x cos(120πt)
3
−2
Where x and y are in m and t in s. The length of the string is 1.5m and its mass is 3.0×10 kg. Answer the following:

(a) Does the function represent a travelling wave or a stationary wave? What is the wavelength, frequency, and speed of each wave

(c) Determine the tension in the string.

(a) The general equation representing a stationary wave is given by the displacement

function:

y(x, t) = 2a sin kx cos ωt

This equation is similar to the given equation:

2
y(x, t) = 0.06 sin( x) cos(120πt)
3

Hence, the given function represents a stationary wave.

(b) A wave travelling along the positive x -direction is given as:

y1 = a sin(ωt − kx)

The wave travelling along the negative x − direction is given as:

y2 = a sin(ωt + kx)

The superposition of these two waves yields:

y = y1 + y2 = a sin(ωt − kx) − a sin(ωt + kx)

= a sin(ot) cos(kx) − a sin(kx) cos(ωt) − a sin(ωt) cos(kx) − a sin(kx) cos(ωt)

= −2a sin(kx) cos(ωt)

2π
= −2a sin( x) cos(2πvt)
λ

The transverse displacement of the string is given as:

The transverse displacement of the string is given as:

2π
y(x, t) = 0.06 sin( x) cos(120πt)
3

Comparing equations (i) and (ii), we have:

2π 2π
=
λ 3

∴ Wavelength, λ = 3m

It is given that:

120n = 2nv

Frequency, v = 60Hz

Wave speed, v = vλ

= 60 × 3 = 180m/s

(c) The velocity of a transverse wave travelling in a string is given by the relation:

T
v = √ … (i)
μ

Where,

Velocity of the transverse wave, v = 180m/s

−2
Mass of the string, m = 3.0 × 10 kg

Length of the string, / = 1.5m

m
Mass per unit length of the string, μ =
l
3.0 −2
= × 10
1.5

−2 −1
= 2 × 10 kgm

Tension in the string = T

From equation (i), tension can be obtained as:

2 −2
= (180) × 2 × 10

= 648N

Page : 392 , Block Name : Exercise

Q15.12 (i) For the wave on a string described in Exercise 15.11, do all the points on the string oscillate with the same (a) frequency, (b) phase,
(c) amplitude? Explain your answers.
(ii) What is the amplitude of a point 0.375 m away from one end?

Answer. (i)(a) Yes, except at the nodes
(b) Yes, except at the nodes
(c) No

(ii) 0.042 m
Explanation:

(i)

(a) All the points on the string oscillate with the same frequency, except at the nodes

which have zero frequency.

(b) All the points in any vibrating loop have the same phase, except at the nodes.

(c) All the points in any vibrating loop have different amplitudes of vibration.

(ii) The given equation is:

2π
y(x, t) = 0.06 sin( x) cos(120πt)
3

For x = 0.375m and t = 0

2π
Amplitude = Displacement = 0.06 sin( x) cos 0
3

2π
= 0.06 sin( x) cos 75) × 1
3

π
= 0.06 sin(0.25π) = 0.06 sin( )
4

1
= 0.06 × = 0.042m
√2

Page : 392 , Block Name : Exercise

Q15.13
Given below are some functions of x and t to represent the displacement (transverse or longitudinal) of an elastic wave. State which

(c) y = 3 sin(5x − 0.5t) + 4 cos(5x − 0.5t) (d) y = cos x sin t + cos 2x sin 2t

Answer. (a) The given equation represents a stationary wave because the harmonic terms kx and ωt (Dt appear separately in the equation
(b) The given equation does not contain any harmonic term. Therefore, it does not represent either a travelling wave or a stationary wave
(c) The given equation represents a travelling wave as the harmonic terms kx and tot are in the combination kx − ωt.
(d) The given equation represents a stationary wave because the harmonic terms kx and ωt appear separately in the equation. This equation
actually represents the superposition of two stationary waves.

Page : 392 , Block Name : Exercise

Q15.14
−2
A wire stretched between two rigid supports in its fundamental mode with a frequency of 45Hz . The mass of the wire is 3.5 × 10 k

−2
(a) Mass of the wire, m = 3.5 × 10 kg

m −2 −1
μ = = 4.0 × 10 kgm
l

Frequency of vibration, v = 45Hz
−2

Length of wire, l = m

μ
=
3.5×10
−2
= 0.875m
4.0×10

The wavelength of the stationary wave (A) is related to the length of the wire by the
relation:
2l
λ =
n

Where, n = Number of nodes in the wire

For fundamental node, n = 1 :

λ = 2l

λ = 2 × 0.875 = 1.75m

The speed of the transverse wave in the string is given as:

v = vλ = 45 × 1.75 = 78.75m/s

(b) The tension produced in the string is given by the relation:

2
T = v μ

2 −2
= (78.75) × 4.0 × 10 = 248.06N

Page : 392 , Block Name : Exercise

Q15.15 A metre-long tube open at one end, with a movable piston at the other end, shows resonance with a xed frequency source (a tuning
fork Of frequency 340 Hz) when the tube length is 25.5 cm or 79.3 cm. Estimate the speed of sound In air at the temperature of the experiment.
The edge effects may be neglected

Answer. Frequency Of the turning fork, v = 340 Hz
Since the given pipe is attached with a piston at one end, it will behave as a pipe with one end closed and the other end open, as shown in the
given gure

Such a system produces odd harmonics The fundamental note in a closed pipe is given
by the relation.
λ
l1 =
4

Where,

Length of the pipe, l1 = 25.5cm = 0.255m

∴ λ = 4l1 = 4 × 0.255 = 1.02m

The speed of sound is given by the relation:

v = vλ = 340 × 1.02 = 346.8m/s

Page : 392 , Block Name : Exercise

Q15.16 A steel rod 100 cm long is clamped at its middle. The fundamental frequency of longitudinal vibrations of the rod is given to be 2.53
kHz. What is the speed of sound in steel ?

Length of the steel rod, I = 100cm = 1m

3
Fundamental frequency of vibration, v = 2.53kHz = 2.53 × 10 Hz

When the rod is plucked at its middle, an antinode (A) is formed at its centre, and nodes

(N) are formed at its two ends, as shown in the given figure.

λ
The distance between two successive nodes is
2

λ
∴ l =
2

V = vλ

3
= 2.53 × 10 × 2

3
= 5.06 × 10 m/s

= 5.06km/s

Page : 392 , Block Name : Exercise

Q15.17 A pipe 20 cm long is closed at one end. Which harmonic mode of the pipe is resonantly excited by a 430 Hz source? Will the same
source be in resonance with the pipe if both ends are open? (Speed of sound in air IS 340 m ss ).
−1

First (Fundamental); No

Length of the pipe, / = 20cm = 0.2m

th
Source frequency = n normal mode of frequency, vn = 430Hz

Speed of sound, v = 340m/s

In a closed pipe, the normal mode Of frequency is given by the relation:
v
vn = (2n − 1)
4l

340
430 = (2n − 1)
4×0.2

430×4×0.2
2n − 1 = = 1.01
340

2n = 2.01

n ∼ 1

Hence, the rst mode Of vibration frequency is resonantly excited by the given source.
In a pipe open at both ends, the nth mode of vibration frequency is given by the relation:
nv
vn =
2l

2lvn
n =
v

2×0.2×430
= = 0.5
340

Since the number of the mode of vibration (n) has to be an integer, the given source does not produce a resonant vibration in an open pipe.

Page : 392 , Block Name : Exercise

Q15.18 Two sitar strings A and B playing the note 'Ga' are slightly out Of tune and produce beats of frequency 6 Hz. The tension in the string A
is slightly reduced and the beat frequency is found to reduce to 3 Hz. If the original frequency of A is 324 Hz, what is the frequency of B ?

Frequency of string A, fA = 324Hz

Frequency of string B = f8

Beat's frequency, n = 6Hz

Beat's frequency is given as:

n = |fn ± fB |

6 = 324 ± fB

fB = 330Hz or 318Hz

Frequency decreases with a decrease in the tension in a string. This is because frequency is directly proportional to the square root of tension.
It is given as:
v ∝ √T

Hence, the beat frequency cannot be 330Hz

∴ fB = 318Hz

Page : 392 , Block Name : Exercise

Q15.19 Explain why (or how):
(a) In a sound wave, a displacement node is a pressure antinode and vice versa,
(b) Bats can ascertain distances, directions, nature, and sizes of the obstacles without any "eyes",
(c) A violin note and sitar note may have the same frequency, yet we can distinguish between the two notes,
(d) Solids can support both longitudinal and transverse waves, but only longitudinal waves can propagate in gases, and
(e) The shape of a pulse gets distorted during propagation in a dispersive medium.

Answer. (a) A node is a point where the amplitude of vibration is the minimum and pressure is the maximum. On the other hand, an antinode
is a point where the amplitude of vibration is the maximum and pressure is the minimum Therefore, a displacement node is nothing but a
pressure antinode, and vice versa.

(b) Bats emit very high-frequency ultrasonic sound waves. These waves get re ected back toward them by obstacles. A bat receives a re ected
nave (frequency) and estimates the distance, direction, nature, and size Of an obstacle with the help Of its brain senses.

(c) The overtones produced by a sitar and a violin, and the strengths of these overtones, are different. Hence, one can distinguish between the
notes produced by a
sitar and a violin even if they have the same frequency Of vibration.

(d) Solids have shear modulus. They can sustain shearing stress. Since uids do not have any de nite shape, they yield to shearing stress. The
propagation of a transverse wave is such that it produces shearing stress in a medium. The propagation Of such a wave is possible only in
solids, and not in gases. Both solids and uids have their respective bulk moduli. They can sustain compressive stress. Hence, longitudinal
waves can propagate through solids and uids.

(e) A pulse is actually is a combination Of waves having different wavelengths. These waves travel in a dispersive medium with different
velocities, depending on the nature of the medium. This results in the distortion of the shape of a wave pulse.

Page : 393 , Block Name : Exercise

Q15.20 A train, standing at the outer signal of a railway station blows a whistle of frequency 400 Hz in still air. (i) What is the frequency of the
whistle for a platform observer when the 10 ms (b) recedes from train (a) approaches the platform with a speed of 10 m s , platform with a
−1 −1

speed of 10 m s (ii) What is the speed of sound in each case? The speed of sound in still air can be taken as 340 m s
−1 −1

(i) (a) (a)Frequency of the whistle, v = 400Hz

Speed of the train, vτ = 10m/s

speed of sound, v = 340m/s
′
The apparent frequency (v ) of the whistle as the train approaches the platform is given

by the relation:

′ v
v = ( )v
v−vT

340
= ( ) × 400 = 412.12Hz
340−10

′
(b) The apparent frequency (v )
of the whistle as the train recedes from the platform is

given by the relation:

′′ v
v = ( )v
v+vT

340
= ( ) × 400 = 388.57Hz
340+10

(ii) The apparent change in the frequency of sound is caused by the relative motions of the source and the observer. These relative motions
produce no effect on the speed of
sound. Therefore, the speed of sound in air in both the cases remains the same, i.e., 340 rn/s.

Page : 393 , Block Name : Exercise

Q15.21 A train, standing in a station-yard, blows a whistle Of frequency 400 Hz in still air. The wind starts blowing in the direction from the
yard to the station with at a speed of 10 m s . What are the frequency, wavelength, and speed of sound for an observer standing on the
−1

station's platform? Is the situation exactly identical to the case when the air is still and the observer runs towards the yard at a speed of 10 m
s
−1
? The speed of sound in still air can be taken as 340 m s −1

For the stationary observer: 400Hz; 0.875m; 350m/s

For the running observer: Not exactly identical

For the stationary observer:

Frequency of the sound produced by the whistle, v = 400Hz

Speed of sound = 340m/s

velocity of the wind, v = 10m/s

As there is no relative motion between the source and the observer, the frequency Of the sound heard The wind is increases by by the observer
will be the same as that produced by the source, i.e., 400 blowing toward the observer. Hence, the effective speed of the sound 10 units,
Effective speed of the sound v = 340 + 10 = 350m/s
∘

Page : 393 , Block Name : Exercise

Q15.22 A travelling harmonic wave on a string is described by y(x, t) = 7.5 sin(0.0050x + 12t +
π
)
4

(a) What are the displacement and velocity of oscillation of a point at x — 1 cm, and t — 1 s? Is this velocity equal to the velocity of wave
propagation?
(b) Locate the points of the string which have the same transverse displacements and velocity as the x — 1 cm point at t = 2 s, 5 s and 11 s.

(a) The given harmonic wave is:

π
y(x, t) = 7.5 sin(0.0050x + 12t + )
4

For x = 1cm and t = 1s
π
y = (1, 1) = 7.5 sin(0.0050 + 12 + )
4
π
= 7.5 sin(12.0050 + )
4

= 7.5 sin θ

π 3.14
= 12.0050 + = 12.0050 + = 12.79rad
4 4
180
∘
= × 12.79 = 732.81
3.14
∘ ∘
∴ y = (1, 1) = 7.5 sin(732.81 ) = 7.5 sin 12.81

= 7.5 × 0.2217

= 1.6629 ≈ 1.663cm

d d π
v = y(x, t) = [7.5 sin(0.0050x + 12t + )]
dt dt 4

π
= 7.5 × 12 cos(0.0050x + 12t + )
4

At x = 1cm and t = 1s .

π
v = y(1, 1) = 90 cos(12.005 + )
4

∘ ∘
= 90 cos(732.81 ) = 90 cos(90 × 8 + 12.81 )
∘
= 90 cos(12.81 )

= 90 × 0.975 = 87.75cm/s

Now, the equation of a propagating wave is given by:

y(x, t) = a sin(kx + wt + ϕ)

Where

2π
k =
λ

2π
∴ λ =
k
ω
∴ v =
2π

ω
Speed, v = vλ =
k

Where,

ω = 12rad/S

−1
k = 0.0050m
12
v = = 2400cm/s
0.0050

Hence, the velocity of the wave oscillation at x = 1cm and t = 1s is not equal to the

velocity of the wave propagation.

(b) Propagation constant is related to wavelength as:

2π
k =
λ
2π 2 × 3.14
∴ λ = =
k 0.0050

= 1256cm = 12.56m

Page : 393 , Block Name : Additional Exercise

Q15.23 A narrow sound pulse (for example, a short pip by a whistle) is sent across a medium. (a) Does the pulse have a de nite (i) frequency,
(ii) wavelength, (iii) speed of propagation? (b) If the pulse rate is 1 after every 20 s, (that is the whistle is blown for a split of second after every
20 s), s the frequency of the note produced by the whistle equal to 20 or 0.05 HZ?

Answer. (a) (i) No
(ii) No
(iii) Yes

(b) No
Explanation:
(a) The narrow sound pulse does not have a xed wavelength or frequency. However, the speed of the sound pulse remains the same, which is
equal to the speed of sound in that medium.
(b) The short pip produced after every 20 s does not mean that the frequency of the
Whistle is or 0.05 Hz It means that 0.05 Hz is the frequency of the repetition of the pip of the whistle.
1

20

Substituting the values from (ii) , (iii) and (iv) in (i) we get the displacement equation :
3
y(x, t) = 0.05 sin(1.6 × 10 t − 4.84x)m

Page : 393 , Block Name : Additional Exercise

Q15.24 One end of a long string of linear mass density 8.0 × 10–3 kg m is connected to an electrically driven tuning fork of frequency 256 Hz.
The other end passes over a pulley and is tied to a pan containing a mass of 90 kg. The pulley end absorbs all the incoming energy so that
re ected waves at this end have negligible amplitude. At t = 0, the left end (fork end) of the string x = 0 has zero transverse displacement (y = 0)
and is moving along positive y-direction. The amplitude of the wave is 5.0 cm. Write down the transverse displacement y as function of x and t
that describes the wave on the string

The equation of a travelling wave propagating along the positive y -direction is given by

the displacement equation:

y(x, t) = a sin(wt − kx) … (i)
−3 −1
Linear mass density, μ = 8.0 × 10 kgm

Frequency of the tuning fork, v = 256Hz

Amplitude of the wave, a = 5.0cm = 0.05m … (ii)

Mass of the pan, m = 90kg

Tension in the string, T = mg = 90 × 9.8 = 882N

The velocity of the transverse wave v, is given by the relation:

T
v = √
μ

882
= √ = 332m/s
−3
8.0 × 10

Angular frequency, ω = 2πv
3
= 1608.5 = 1.6 × 10 rad/s

v 332
Wavelength, λ = = m
v 256

2π
∴ Propagation constant, k =
λ

2 × 3.14
−1
= = 4.84m
332

256

Page : 393 , Block Name : Additional Exercise

Q15.25 A SONAR system xed in a submarine operates at a frequency 40.0 kHz. An enemy submarine moves towards the SONAR with a speed
of 360 km h What IS the frequency of sound re ected by the submarine? Take the speed of sound in water to be 1450 m s
−1 −1

Operating frequency of the SONAR system, v = 40kHz

Speed of the enemy submarine, ve = 360km/h = 100m/s

Speed of sound in water, v = 1450m/s

The source is at rest and the observer (enemy submarine) is moving toward it. Hence

′
the apparent frequency (v ) received and reflected by the submarine is given by the

relation:

v + vc
′
v = ( )v
v

1450 + 100
= ( ) × 40 = 42.76kHz
1450
′′
The frequency (v ) received by the enemy submarine is given by the relation:

′′ v ′
v = ( )v
v+vs

Where, vs = 100m/s

′′ 1450
∴ v = ( ) × 42.76 = 45.93kHz
1450−100

Page : 393 , Block Name : Additional Exercise

Q15.26 Earthquakes generate sound waves inside the earth. Unlike a gas, the earth can experience both transverse (S) and longitudinal (P)
sound waves. Typically the speed of S wave is about 4.0 km s , and that of P wave is 8.0 km s . A seismograph records P and S waves from an
−1 −1

earthquake. The rst P wave arrives 4 min before the rst S wave. Assuming the waves travel in straight line, at what distance does the
earthquake occur ?

Let vs and vp be the velocities of S and P waves respectively.

Let L be the distance between the epicentre and the seismograph.

We have:

L = vs ts (i)

L = vP tp (ii)

Where,

ts and tp are the respective times taken by the S and P waves to reach the seismograph

from the epicentre

It is given that:

vP = 8km/s

vs = 4km/s

v s ts = v P tp

4ts = 8tp

ts = 2tp (iii)

It is also given that:

ts − tp = 4 min = 240s

2tp − tp = 240

te = 240

And ts = 2 × 240 = 480s

From equation (iil), we get:

L = 8 × 240

= 1920km

Hence, the earthquake occurs at a distance of 1920 km from the seismograph.

Page : 394 , Block Name : Additional Exercise

Q15.27 A bat is itting about in a cave, navigating via ultrasonic beeps. Assume that the sound emission frequency of the bat is 40 kHz. During
one fast swoop directly toward a at wall surface, the bat is moving at 0.03 times the speed of sound in air. What frequency does the bat hear
re ected off the wall?

Ultrasonic beep frequency emitted by the bat, v = 40kHz

velocity of the bat, vb = 0.03v

Where, v = velocity of sound in air

The apparent frequency of the sound striking the wall is given as:

v
′
v = ( )v
v − vb

v
= ( ) × 40
v − 0.03v

40
= kHz
0.97

This frequency is reflected by the stationary wall (vs = 0) toward the bat.

′′
The frequency (v ) of the received sound is given by the relation:

v+vb
∗ ′
v = ( )v
v

v+0.03v 40
= ( ) ×
v 0.97

1.03×40
= = 42.47kHz
0.97

Page : 394 , Block Name : Additional Exercise