Chapter 15 Waves Solutions
Question - 11 : - The transverse displacement of a string (clamped at its both ends) is given by
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 represent 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.
Answer - 11 : -
(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:
Hence, the given function represents a stationary wave.
(b) A wave travelling along the positive x-direction is given as:
The wave travelling along the negative x-direction is given as:
The superposition of these two waves yields:
The transverse displacement of the string is given as:
Comparing equations (i) and (ii), we have:
∴Wavelength, λ = 3 m
It is given that:
120π = 2πν
Frequency, ν = 60 Hz
Wave speed, v = νλ
= 60 × 3 = 180 m/s
(c) The velocity of a transverse wave travelling in a string is given by the relation:
Where,
Velocity of the transverse wave, v =180 m/s
Mass of the string, m = 3.0× 10–2 kg
Length of the string, l =1.5 m
Mass per unit length of the string, 
Tension in the string = T
From equation (i), tension can beobtained as:
T = v2μ
= (180)2 × 2 × 10–2
= 648 N
Answer - 12 : -
(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:
For x = 0.375 m and t = 0
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 traveling wave, (ii) a stationary wave or (iii) none at all:
(a) y = 2 cos (3x) sin (10t)
(b)
(c) y = 3 sin (5x – 0.5t) + 4 cos (5x – 0.5t)
(d) y = cos x sin t + cos 2x sin 2t
Answer - 13 : -
(a) The given equation represents a stationary wavebecause the harmonic terms kx and ωt appearseparately in the equation.
(b) The given equation does not contain any harmonicterm. Therefore, it does not represent either a travelling wave or a stationarywave.
(c) The given equation represents a travelling wave asthe harmonic terms kx and ωt are in thecombination of kx – ωt.
(d) The given equation represents a stationary wavebecause the harmonic terms kx and ωt appearseparately in the equation. This equation actually represents the superpositionof 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 × 10–2 kg and its linear mass density is 4.0 × 10–2 kg m–1. What is (a) the speed of a transverse wave on the string, and (b) the tension in the string?
Answer - 14 : -
(a) Mass of the wire, m = 3.5 × 10–2 kg
Linear mass density,
Frequency of vibration, ν = 45 Hz
∴Length of the wire,
The wavelength of the stationary wave (λ)is related to the length of the wire by the relation:
For fundamental node, n =1:
λ = 2l
λ = 2 × 0.875 = 1.75 m
The speed of the transverse wave in the string is givenas:
v = νλ=45 × 1.75 = 78.75 m/s
(b) The tension produced in the string is given by therelation:
T = v2µ
= (78.75)2 × 4.0 × 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.
Answer - 15 : -
Frequency of the turning fork, ν = 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:
Where,
Length of the pipe, 
The speed of sound is given by the relation:
= 340 ×1.02 = 346.8 m/s
Answer - 16 : -
Length of the steel rod, l = 100 cm = 1 m
Fundamental frequency of vibration, ν = 2.53 kHz = 2.53 × 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
.
The speed of sound in steel is given by the relation:
v = νλ
= 2.53 × 103 ×2
= 5.06 × 103 m/s
= 5.06 km/s
Answer - 17 : -
Answer: First (Fundamental); No
Length of the pipe, l = 20cm = 0.2 m
Source frequency = nth normalmode of frequency, νn = 430 Hz
Speed of sound, v = 340 m/s
In a closed pipe, the nth normalmode of frequency is given by the relation:
Hence, the first mode of vibration frequency is resonantlyexcited by the given source.
In a pipe open at both ends, the nth modeof vibration frequency is given by the relation:
Since the number of the mode of vibration (n)has to be an integer, the given source does not produce a resonant vibration inan 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 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?
Answer - 18 : -
Frequency of string A, fA =324 Hz
Frequency of string B = fB
Beat’s frequency, n = 6 Hz
Frequency decreases with a decrease in the tension in astring. This is because frequency is directly proportional to the square rootof tension. It is given as:
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.
Answer - 19 : -
(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 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.
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 m s–1, (b) recedes from the platform with a speed of 10 m s–1? (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.
Answer - 20 : -
(i) (a)Frequency of the whistle, ν = 400 Hz
Speed of the train, vT= 10m/s
Speed of sound, v = 340 m/s
The apparent frequency 
of the whistle as the trainapproaches the platform is given by the relation:
(b) The apparent frequency 
of the whistle as the trainrecedes from the platform is given by the relation:
(ii) The apparent change in the frequency of sound is caused bythe relative motions of the source and the observer. These relative motionsproduce no effect on the speed of sound. Therefore, the speed of sound in airin both the cases remains the same, i.e., 340 m/s.