Punjab State Board PSEB 11th Class Physics Book Solutions Chapter 15 Waves Textbook Exercise Questions and Answers.
PSEB Solutions for Class 11 Physics Chapter 15 Waves
PSEB 11th Class Physics Guide Waves Textbook Questions and Answers
Question 1.
A string of mass 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?
Solution:
Mass of the string, M = 2.50 kg
Tension in the string, T = 200 N
Length of the string, l = 20.0 m
Mass per unit length, µ = \(\frac{M}{l}=\frac{2.50}{20} \) = 0.125 kg m-1
The velocity (υ) of the transverse wave in the string is given by the relation:
υ = \(\sqrt{\frac{T}{\mu}}=\sqrt{\frac{200}{0.125}}=\sqrt{1600} \) = 40 m/s
∴ Time taken by the disturbance to reach the other end, t = \(\frac{l}{v}=\frac{20}{40}\) = 0.5 s
Question 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 340ms-1?(g=9.8ms-2)
Solution:
Height of the tower, s = 300 m
The initial velocity of the stone, µ = 0
Acceleration, a = g = 9.8 m/s2
Speed of sound in air = 340 rn/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:
s=ut1+\(\frac{1}{2}\) gt12
300 = 0+ \( \frac{1}{2}\) × 9.8 × t12
∴ t1 =\( \sqrt{\frac{300 \times 2}{9.8}}\) = 7.82 s
Time taken by the sound to reach the top of the tower,
t2 =\(\frac{300}{340}\) = 0.88 s
Therefore, the time after which the splash is heard, t = t1 +t2
= 7.82+0.88= 8.7 s .
Question 3.
A steel wire has a length of 12.0 m and a mass of 2.10 kg. What should be the tension in the wire so that the speed of a transverse wave on the wire equals the speed of sound in dry air at 20°C = 343 ms-1.
Solution:
Length of the steel wire, l = 12 m
Mass of the steel wire, m = 2.10 kg
Velocity of the transverse wave, ν = 343 m/s
Mass per unit length, µ =\(\frac{m}{l}=\frac{2.10}{12}\) = 0.175 kg m-1
For tension T, velocity of the transverse wave can be obtained using the relation:
υ = \(\sqrt{\frac{T}{\mu}}\)
∴ T = υ2µ = (343)2 × 0.175 = 20588.575 ≈ 2.06 ×104 N
Question 4.
Use the formula υ = \(\sqrt{\frac{\gamma \boldsymbol{P}}{\rho}}\) to explain why the speed of sound in
air
(a) is independent of pressure,
(b) increases with temperature,
(c) increases with humidity.
Solution:
(a) Given the relation:
υ = \( \sqrt{\frac{\gamma \boldsymbol{P}}{\rho}} \) ……………………………. (i)
Density, ρ = \(\frac{\text { Mass }}{\text { Volume }}=\frac{M}{V}\)
where, M = Molecular weight of the gas; V = Volume of the gas
Hence, equation ‘(i) reduces to:
υ = \(\sqrt{\frac{\gamma P V}{M}}\) …………………………………………. (ii)
Now, from the ideal gas equation for n = 1 :
PV = RT
For constant T, PV = Constant
Since both M and γ are constants, υ = 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) Given the relation:
υ = \( \sqrt{\frac{\gamma \boldsymbol{P}}{\rho}} \) …………………………………. (i)
For one mole of an ideal gas, the gas equation can be written as:
PV = RT
P = \(\frac{R T}{V}\)
Substituting equation (ii) in equation (i), we get:
υ = \(\sqrt{\frac{\gamma R T}{V \rho}}=\sqrt{\frac{\gamma R T}{M}}\) ……………………………… (iii)
where, M = mass = ρV is a constant; γ and R are also constants We conclude from equation (iii) that v ∝ \(\sqrt{T}\) .
Hence, the speed of sound in a gas is directly proportional to the square root of the temperature of the gaseous medium, i. e., the speed of the sound increases with an increase in the temperature of the gaseous medium and vice versa.
(c) Let υm and υd be the speeds of sound in moist air and dry air respectively.
Let ρm and ρd be the densities of moist air and dry air respectively.
Take the relation:
υ = \(\sqrt{\frac{\gamma P}{\rho}}\)
Hence, the speed of sound in moist air is:
υm = \(\sqrt{\frac{\gamma P}{\rho_{m}}}\) ………………………….. (i)
And the speed of sound in dry air is:
υd = \(\sqrt{\frac{Y P}{P_{d}}}\) ………………………………… (ii)
On dividing equations (i) and (ii), we get:
\(\frac{v_{m}}{v_{d}} \)
On dividing equations (i) and (ii), we get:
\(\frac{v_{m}}{v_{d}}=\sqrt{\frac{\gamma P}{\rho_{m}}} \times \frac{\rho_{d}}{\gamma P}=\sqrt{\frac{\rho_{d}}{\rho_{m}}} \)
However, the presence of water vapour reduces the density of air, i.e.,
ρd>ρm
∴ υm>υd
Hence the speed of sound in moist air greater than it is in dry air.
Thus, in a gaseous medium, the speed of sound increase with humidity.
Question 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-υt or x+υt, i.e., y=f(x±υt). Is the converse true? Examine if the following functions for y can
possibly represent a traveIliig wave:
(a) (x—υt)2 (b)log \(\left[\frac{x+v t}{x_{0}}\right] \) (c) \(\frac{1}{(x+v t)}\)
Solution:
No, 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 finite for all values of x and t.
(a) Does not represent a wave
Explanation :
For x = 0 and t = 0, the function (x – υt)2 becomes 0.
Hence, for x = 0 and t = 0, the function represents a point and not a wave,
(b) Represents a wave Explanation:
For x = 0 and t = 0, the function log \(\left(\frac{x+v t}{x_{0}}\right)\) = log 0 = ∞
Since the function does not converge to a finite value for x = 0 and t = 0, it represents a travelling wave.
(c) Does not represents a wave
Explanation :
For x = 0 and t = 0, the function
\(\frac{1}{x+v t}\) = log \(\frac{1}{0} \) = ∞
Since the function does not converge to a finite value for x = 0 and t = 0, it does not represent a travelling wave.
Question 6.
A bat emits the ultrasonic sound of frequency 1000 kHz in air. If the sound meets a water surface, what is the wavelength of (a) the reflected sound, (b) the transmitted sound? Speed of sound in air is 340 ms-1 and in water 1486 ms-1.
Solution:
(a) Frequency of the ultrasonic sound, υ = 1000 kHz = 106 Hz
Speed of sound in water, υa = 340 m/s
The wavelength (λr)of the transmitted sound is given as:
λr = \(\frac{v}{v}=\frac{340}{10^{6}}\) = 3.4 × 10-4 m
(b) Frequency of the ultrasonic sound, v = 1000 kHz = 106 Hz
Speed of sound in water, υw, =1486 m/s
The wavelength of the transmitted sound is given as:
λt= \(\frac{1486}{10^{6}}\) = 1.49 × 10-3 m
Question 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-1? The operating frequency of the scanner is 4.2 MHz.
Solution:
Speed of sound in the tissue, υ = 1.7 km/s = 1.7 x 10 3 m/s
Operating frequency of the scanner, v = 4.2 MHz = 4.2 x 106 Hz
The wavelength of sound in the tissue is given as:
λ = \(\frac{v}{v}=\frac{1.7 \times 10^{3}}{4.2 \times 10^{6}}\) = 4.1 x 10-4m.
Question 8.
A transverse harmonic wave on a string is described by y(x, t) = 3.0sin(36t+0.018x+\(\frac{\pi}{4}\))
where x and y are in cm and t in s. The positive direction of x is from left to right.
Is this a travelling wave or a stationary wave?
(a) 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?
Solution:
(a) Yes.
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
y(x, t) = 3.0 sin( 36t +0.018x+\(\frac{\pi}{4}\)) …………………………………. (ii)
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:
ω = 36 rad/s and k = 0.018 cm-1
We know that
v = \(\frac{\omega}{2 \pi}\) and λ = \(\frac{2 \pi}{k}\)
Also,
υ = vλ
∴ υ = \(\left(\frac{\omega}{2 \pi}\right) \times\left(\frac{2 \pi}{k}\right)\) = \(\frac{\omega}{k}=\frac{36}{0.018} \) = 2000 cm/s = 20 m/s
Hence, the speed of the given travelling wave is 20 m/s.
(b) Amplitude of the given wave, a =3 cm (Given)
Frequency of the given wave:
v = \(\frac{\omega}{2 \pi}=\frac{36}{2 \times 3.14}\) = 5.73 Hz
(c) On comparing çquations (i) and (ii), we find that the initial phase angle, Φ = \(\frac{\pi}{4}\)
(d) The distance between two successive crests or troughs is equal to the
wavelength of the wave.
Wavelength is given by the relation:
k= \(\frac{2 \pi}{\lambda}\)
∴ λ = \(\frac{2 \pi}{k}=\frac{2 \times 3.14}{0.018}\) = 348.89 cm = 3.49 m.
Question 9.
For the wave described in question 8, plot the displacement (y) versus (t) graphs for s =0,2 and 4 çm. 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?
Solution:
All the waves have different phases. The given transverse harmonic wave is
y(x,t) = 3.0 sin (36t+0.018x + \(\frac{\pi}{4}\))
For x = 0, the equation reduces to
y(0,t) = 3.0 sin (36t+\(\frac{\pi}{4}\)) ………………………….. (i)
Also,
ω = \(\frac{2 \pi}{T}\) = 36 rad/s
∴ T = \(\frac{2 \pi}{\omega}=\frac{2 \pi}{36} \) = \(\frac{\pi}{18}\) s
For different values of t, we calculate y using eq. (i). These values are tabulated below
On plotting y versus t graph, we obtain a sinusoidal curve as shown in figure below.
Similar graphs are obtained for x = 2 cm and x = 4 cm.
The oscillatory motion in travelling wave differs from one point to another only in terms of phase. Amplitude and frequency of oscillatory motion remain the same in all the three cases.
Question 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 difference between oscillatory motion of two points separated by a distance of
(a) 4m
(b) 0.5m
(c) \(\frac{\lambda}{2}\)
(d) \(\frac{3 \lambda}{4}\)
Solution:
Equation for a travelling harmonic wave is given as
y{x,t) = 2.0 cos 2π(10t -0.0080x +0.35)
= = 2.0 cos (20πt – 0.016πx+0.70π)
where, propagation constant, k = 0.0160π
Amplitude, a = 2 cm
Angular frequency, ω = 20 π rad/s
Phase difference is given by the relation:
Φ =kx=\(\frac{2 \pi}{\lambda} \)
(a)
For x=4m=400cm
Φ =0.016 π × 400 =6.4 π rad
(b) For 0.5 m=50cm
Φ = 0.1016 π × 50 = 0.8 π rad
(c) For x= \(\frac{\lambda}{2}\)
Φ = \(\frac{2 \pi}{\lambda} \times \frac{\lambda}{2}\) = π rad
(d) For x= \(\frac{3 \lambda}{4}\)
Φ = \(\frac{2 \pi}{\lambda} \times \frac{3 \lambda}{4} \) = 1.5π rad
Question 11.
The transverse displacement of a string (clamped at its both ends) is given by y(x,t) = 0.06 sin \(\frac{\mathbf{2} \pi}{\mathbf{3}}\) x cos (120πt) where x and y are in m and t in s. The length of the string is 1.5 m and its mass is 3.0 × 10-2 Kg Answer the following
(a) Does the function represents a travelling wave or a stationary wave?
(b) Interpret the wave as a superposition of two waves travelling in opposite directions. What is the wavelength,
frequency, and speed of each wave?
(c) Determine the tension in the string.
Solution:
(a) The general equation representing a stationary wave is given by the displacement function:
y(x,t) = 2asinkxcosωt
This equation is similar to the given equation:
y(x,t)= 0.06 sin\(\left(\frac{2 \pi}{3} x\right)\) cos (120πt)
Hence, the given function represents a stationary wave.
(b) A wave travelling along the positive x -direction is given as
y1 =asin(ωt -kx)
The wave travelling along the negative x -direction is given as:
y2 = -asin(ωt +kx)
The superposition of these two waves yields:
y= y1+y2 = asin(ωt -kx)-asin(ωt +kr)
= asin(ωt)cos(kx) – asin(kx)cos(ωt)- asin(ωt)cos(kx) – asin(kx)cos(ωt)
= -2asin(kx)cos(ωt)
= – 2asin \(\left(\frac{2 \pi}{\lambda} x\right)\)cos (2πcvt) …………………………….. (i)
The transverse displacement of the string is given as y(x,t) = 0.06 sin \(\left(\frac{2 \pi}{3} x\right)\)cos (120πt) ………………………………. (ii)
Comparing equations (i) and (ii), we have
\(\frac{2 \pi}{\lambda}=\frac{2 \pi}{3}\)
∴ Wavelength, λ = 3 m
it is given that
120 π =2πv
Frequency, ν =60 Hz
Wave speed, υ = vλ
=60 × 3=180 m/s
(c) The velocity of a transverse wave travelling in a string is given by the relation
υ = \(\sqrt{\frac{T}{\mu}} \) ………………………… (iii)
where, µ = Mass per unit length of the string = \(\frac{m}{l}=\frac{3.0}{1.5} \times 10^{-2}\)
=2 x 10-2 kgm-1
T = Tension in the string = T
From equation (iii), tension can be obtained as
T =ν2µ=(180)2 x 2 x 10-2 =648 N
Question 12.
(i) For the wave on a string described in question 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?
Solution:
(I) (a) Yes, except at the nodes; All the points on the string oscillate with the same frequency, except at
the nodes which have zero frequency.
(b) Yes, except at the nodes;
All the points in any vibrating loop have the same phase, except at the nodes.
(C) No;
All the points in any vibrating loop have different amplitudes of vibration.
(ii) The given equation is .
y(x,t) = 0.06 sin \(\left(\frac{2 \pi}{3} x\right)\)cos (120πt)
For x = 0.375m and t =0
Amplitude = Displacement 0.06sin \(\left(\frac{2 \pi}{3} x\right) \cos 0^{\circ}\)
= 0.06 sin \(\left(\frac{2 \pi}{3} \times 0.375\right) \times 1\)
= 0.06 sin(0.25π) = 0.06 sin\(\left(\frac{\pi}{4}\right)\)
= 0.06 x \(\frac{1}{\sqrt{2}}\) = 0.042 m
Question 13.
Given below are some functions of x and t to represent the displacement (transverse or longitudinal) of an elastic wave. State which of these represent (i) a travelling wave, (ii) a stationary wave or (iii) none at all:
(a) y = 2cos(3x)sin(10t)
(b) y = 2\(\sqrt{x-v t}\)
(c) y = 3sin(5x – 0.5t) + 4cos(5x – 0.5t)
(d) Y = cos x sin t + cos 2x sin 2t
Solution:
(a) The given equation represents a stationary wave because the harmonic terms kx and cot 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 cot are in the combination of kx – cot.
(d) The given equation represents a stationary wave because the harmonic terms kx and cot appear separately in the equation. This equation actually represents the superposition of two stationary waves.
Question 14.
A wire stretched between two rigid supports vibrates in its fundamental mode with a frequency of 45 Hz. The mass of the wire is 3.5 x10-2 kg and its linear mass density is 40 x 10-2 kgm-1. What is
(a) the speed of a transverse wave on the string, and
(b) the tension in the string?
Solution:
Mass of the wire, m = 3.5×10 -2 kg
Linear mass density, μ = \(\frac{m}{l}\) = 4.0 × 10-2 kg m-1
Frequency of vibration, μ = 45 Hz
∴ Length of the wire, l = \(\frac{m}{\mu}=\frac{3.5 \times 10^{-2}}{4.0 \times 10^{-2}}\) =0.875 m
The wavelength of the stationary wave (λ,) is related to the length of the wire by the relation:
λ = \(\frac{2 l}{n}\)
where, n = Number of nodes in the wire For fundamental node, n = 1:
λ =2l
λ =2 x 0.875 = 1.75 m
(a) The speed of the transverse wave in the string is given as
υ = vλ = 45 x 1.75 = 78.75 m/s
(b) The tension produced in the string is given by the relation:
T =υ2 μ
= (78.75)2 x 4.0 x 10-2 =248.06 N
Question 15.
A metre-long tube open at one end, with a movable piston at the other end, shows resonance with a fixed 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.
Solution:
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 figure.
Such a system produces odd harmonics. The fundamental note in a closed pipe is given by the relation l1 = \(\frac{\lambda}{4}\)
where, length of the pipe, = 25.5 cm = 0.255 m
λ = 4l1 =4 x 0.255 = 1.02 m
The speed of sound is given by the relation:
υ = vλ = 340 x 1.02 = 346.8 m/s
Question 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?
Solution:
Length of the steel rod, l = 100 cm = lm
v Fundamental frequency of vibration, v = 2.53 kHz = 2.53 x 103 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.
The distance between two successive nodes is \(\frac{\lambda}{2}\)
l = \( \frac{\lambda}{2}\)
λ=2l=2 x l=2m
The speed of sound in steel is given by the relation:
v =vλ = 2.53 x 103 x 2
= 5.06 x 10 3 m/s = 5.06 km/s
Question 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 ms-1).
Solution:
First (Fundamental); No
Length of the pipe, l = 20 cm = 0.2 m
Source frequency = nth normal mode of frequency, vn = 430 Hz Speed of sound, ν = 340 m/s
In a closed pipe, nth the rth normal mode of frequency is given by the relation
vn = (2n -1) \(\frac{v}{4 l}\) ; n is an integer = 0,1,2,3 ………………
430 = (2n -1) \(\frac{340}{4 \times 0.2} \)
2n-1 = \(\frac{430 \times 4 \times 0.2}{340}\) = 1.01
2n =1.01+1
2n = 2.01
n ≈ 1
Hence, the first 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:
vn = \(\frac{n v}{2 l}\)
n = \(\frac{2 l v_{n}}{v}=\frac{2 \times 0.2 \times 430}{340}\) = 0.5
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.
Question 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 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?
Solution:
Frequency of string A, fA = 324 Hz
Frequency of string B = fB
Beat’s frequency, n = 6 Hz
Beat’s frequency is given as
n = |fA ±fB|
6 =324 ±fB
fB =330 Hz or 318 Hz
The frequency decreases with a decrease in the tension in a string. This is because the frequency is directly proportional to the square root of the tension. It is given as
v ∝ \(\sqrt{T}\)
Hence, the beat frequency cannot be 330 Hz.
∴ fB= 318 Hz
Question 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.
Solution:
(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 reflected back toward them by obstacles. A bat receives a reflected wave (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 fluids do not have any definite 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 fluids have their respective bulk moduli. They can sustain compressive stress. Hence, longitudinal waves can propagate through solids and fluids.
(e) A pulse is actually 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.
Question 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 train
(a) approaches the platform with a speed of 10 ms-1,
(b) recedes from the platform with a speed of 10 ms-1?
(ii) What is the speed of sound in each case? The speed of sound in still air can be taken as 340 ms -1.
Solution:
(i)
(a) Frequency of the whistle, ν = 400 Hz
Speed of the train, υT = 10 m/s
Speed of sound, υ = 340 m/s
The apparent frequency (v’) of the whisde as the train approaches the platform is given by the relation
υ’ = \(=\left(\frac{v}{v-v_{T}}\right) \mathrm{v}=\left(\frac{340}{340-10}\right) \times 400\) = 412.12 Hz
(b) The apparent frequency (v”) of the whistle as the train recedes from the platform is given by the relation
v” = \(\left(\frac{v}{v+v_{T}}\right) \mathrm{v}=\left(\frac{340}{340+10}\right) \times 400\) = 388.57 Hz
(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 m/s.
Question 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 ms-1. What are the frequency, wavelength, and speed of sound for an observer standing on the 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 ms-1? The speed of sound in still air can be taken as 340 ms-1.
Solution:
For the stationary observer:
Frequency of the sound produced by the whistle, v = 400 Hz
Speed of sound = 340 m/s
Velocity of the wind, ν = 10 m/s
As there is no relative motion between the source and the observer, the frequency of the sound heard .by the observer will be the same as that produced by the source, i. e., 400 Hz.
The wind is blowing toward the observer. Hence, the effective speed of the sound increases by 10 units, i.e.,
Effective speed of the sound, υe = 340 +10 = 350 m/s
The wavelength (λ) of the sound heard by the observer is given by the relation:
λ = \(\frac{v_{e}}{v}=\frac{350}{400}\) = 0.857 m
For the running observer:
Velocity of the observer, υ0 = 10 m/s
The observer is moving toward the source. As a result of the
motions of the source and the observer, there is a change in (v’).
This is given by the relation:
υ’ = \(\left(\frac{v+v_{o}}{v}\right) v=\left(\frac{340+10}{340}\right) \times 400\) = 411.76 Hz
Since the air is still, the effective speed of sound = 340 + 0 = The source is at rest. Hence, the wavelength of the sound will i. e., λ remains 0.875 m
Hence, the given two situations are not exactly identical.
Additional Exercises
Question 22.
A travelling harmonic wave on a string is described by
y(x,t) = 7.5 sin (0.0050x + 12t+\(\frac{\pi}{4}\))
(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.
Solution:
(a) The given harmonic wave is
y(x,t) = 7.5sin(0.0050x + 12t+\(\frac{\pi}{4}\))
For x = 1 cm and t = 1 s,
y(1, 1) = 7.5sin(0.0050x + 12t+\(\frac{\pi}{4}\))
= 7.5 sin (12.0050+\(\frac{\pi}{4}\)) = 7.5sinθ
where, 0 = 12.0050 + \(\frac{\pi}{4}\) = 12.0050 + \( \) = 12.79 rad 4
= \(\frac{180}{3.14} \times 12.79\) = 732.810
∴ y(1,1) = 7.5 sin (732.810) = 7.5sin (90 × 8 +12.81°) = 7.5 sin 12.81°
= 7.5 × 0.2217
= 1.6229 ≈ 1.663 cm
The velocity of the oscillation at a given point and time is given as
At x = 1 cm and t = 1 s
v = y(1, 1) = 90 cos(12.005 +\(\frac{\pi}{4}\))
= 90 cos (732.81 ° ) = 90 cos (90 x 8 +12.81 ° ) = 90cos(12.81°) = 90 x 0.975 = 87.75 cm/s
Now, the equation of a propagating wave is given by
y(x,t) = asin(kx +ωt +Φ)
where, k = \(\frac{2 \pi}{\lambda} \)
∴ λ = \(\frac{2 \pi}{k}\)
And ω = 2πv
∴ v = \(\frac{\omega}{2 \pi}\)
Speed, υ = vλ = \(\frac{\omega}{k}\)
where, ω = 12 rad/s
k = 0.0050 cm-1
∴ v = \(\frac{12}{0.0050}\) = 2400 cm/s
Hence, the velocity of the wave oscillation at x = 1 cm and t = 1 s is not equal to the velocity of the wave propagation.
(b) Propagation constant is related to wavelength as:
k = \(\frac{2 \pi}{\lambda}\)
∴ λ = \(\frac{2 \pi}{k}=\frac{2 \times 3.14}{0.0050}\) = 1256 cm = 12.56 cm
Therefore, all the points at distances nλ{n = ±1,±2… and so on), i.e., ±12.56 m, + 25.12m, … and so on for x =1 cm, will have the same displacement as the x = 1 cm points at t = 2 s, 5 s and 11s.
Question 23.
A narrow sound pulse (for example, a short pip by a whistle) is sent across a medium,
(a) Does the pulse have a definite
(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), is the frequency of the note
produced by the whistle equal to \(\frac{1}{20}\) or 0.05 Hz?
Solution:
(a) (i) No;
(ii) No;
(iii) Yes;
Explanation:
The narrow sound pulse does not have a fixed 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) No;
The short pip produced after every 20 s does not mean that the frequency of the whistle is \(\frac{1}{20}\) or 0.05 Hz. It means that 0.05 Hz is the frequency of the repetition of the pip of the whisde.
Question 24.
One end of a long string of linear mass density 8.0 x 10-3 kg m-1 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 reflected 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 a function of x and t that describes the wave on the string.
Solution:
The equation of a Gravelling wave propagating along the positive y-direction is given by the displacement equation
y(x, t) = a sin (cot – kx) ………………………. (i)
Linear mass density, μ = 8.0 x 10 -3 kg m-1
Frequency of the tuning fork, v = 256 Hz
The amplitude of the wave, a = 5.0 cm = 0.05 m ……………………………. (ii)
Mass of the pan, m = 90 kg
Tension in the string, T = mg = 90 x 9.8 = 882 N
The velocity of the transverse wave υ, is given by the relation:
υ = \(=\sqrt{\frac{T}{\mu}}=\sqrt{\frac{882}{8.0 \times 10^{-3}}}\) = 332 m/s
Angular Frequency, ω = 2πv
= 2 x 3.14 x 256
= 1607.68 = 16 x 103 rad/s ………………………….. (iii)
Wavelength λ = \(\frac{v}{v}=\frac{332}{256}\)m
∴ propagation constant, k = \(\frac{2 \pi}{\lambda}=\frac{2 \times 3.14}{\frac{332}{256}}\) = 4.84 m-1 ……….. (iv)
Substituting the values from equations (ii), (iii), and (iv) in equation (i), we get the displacement equation:
y(x,t) = 0.05sin(1.6 x 103t -4.84 x)
where x and y are in and t in s.
Question 25.
A SONAR system fixed in a submarine operates at a frequency 40.0 kHz. An enemy submarine moves towards the SONAR with a speed of 360 km h-1. What is the frequency of sound reflected by the submarine? Take the speed of sound in water to be 1450 ms-1.
Solution:
Operating frequency of the SONAR system, v = 40 kHz
Speed of the enemy submarine, ve = 360 km/h = 100 m/s
Speed of sound in water, v = 1450 m/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:
The frequency (v”) received by the enemy submarine is given by the relation
v’ = \( =\left(\frac{v+v_{e}}{v}\right) v=\left(\frac{1450+100}{1450}\right) \times 40\) = 42.76 kHz
The frequency (V”) received by the enemy submarine is given by the relation
v” = \(\left(\frac{v}{v-v_{s}}\right) v^{\prime}\)
Where vs = 100 m/s
∴ v” = \(\left(\frac{1450}{1450-100}\right) \times 42.76 \) = 45.93 kHz
Question 28.
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 4.0kms-1 , and that of P wave is 8.0 kms1. A
seismograph records P and S waves from an Earthquake. The first P wave arrives 4 min before the first S wave. Assuming the waves travel in straight line, at what distance does the Earthquake occur?
Solution:
Let νs 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 = νstsub>s ……………………….. (i)
L = νptsub>p …………………………(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
νp =8km/s
νs =4km/s
From equations (i) and (ii), we have
υsts = υptp
4ts = 8tp
ts = 2tp …………………………..(iii)
It is also given that
ts – tp =4 min=240s
2tp-tp= 240
tp = 240
and = 2×240 =840 s
From equation (ii), we get
L =8×240=1920 km
Hence, the Earthquake occurs at a distance of 1920 km from the seismograph.
Question 27.
A bat is flitting 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 flat wall surface, the bat is moving at 0.03 times the speed of sound hi air. What frequency does the bat hear reflected off the wall?
Solution:
Ultrasonic beep frequency emitted by the bat, ν = 40 kHz
The velocity of the bat, νb = 0.03 ν
where, ν = velocity of sound in air
The appartment frequency of the sound strìking the wall is given as
v’ = \(\left(\frac{v}{v-v_{b}}\right) v=\left(\frac{v}{v-0.03 v}\right) \times 40 \) = \(\frac{40}{0.97}\) kHz
This frequency is reflected by the stationary wall (νs = 0) toward the bat.
The frequency (ν”) of the received sound is given by the relation:
ν” = \(\left(\frac{v+v_{b}}{v}\right) \mathrm{v}^{\prime}=\left(\frac{v+0.03 v}{v}\right) \times \frac{40}{0.97}=\frac{1.03 \times 40}{0.97}\) = 42.47 kHz