15.19. 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 (N) is a point where the amplitude of vibration is the minimum and pressure is the maximum.
An antinode (A) 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 ultrasonic waves of large frequencies. When these waves are reflected from the obstacles in their path, they give them the idea about the distance, direction, size and nature of the obstacle.

(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) This is because solids have both, the elasticity of volume and elasticity of shape, whereas gases have only the volume elasticity.

(e) A sound pulse is a combination of waves of different wavelength. As waves of different λ travel in a dispersive medium with different velocities, therefore, the shape of the pulse gets distorted.

15.20. 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

(i) (a)Frequency of the whistle, ν = 400 Hz
Speed of the train, v_T= 10 m/s
Speed of sound, v = 340 m/s
The apparent frequency (v’) of the whistle as the train approaches the platform is given by the
relation:

(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.

15.21. 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^–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 m s^–1? The speed of sound in still air can be taken as 340 m s^–1.

Answer

For the stationary observer:
Frequency of the sound produced by the whistle, ν = 400 Hz
Speed of sound = 340 m/s
Velocity of the wind, v = 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, v_e = 340 + 10 = 350 m/s
The wavelength (λ) of the sound heard by the observer is given by the relation:
λ = v_e/v = 350/400 = 0.875 m
For the running observer:
Velocity of the observer, v_o = 10 m/s
The observer is moving toward the source. As a result of the relative motions of the source and the observer, there is a change in frequency (v‘).
This is given by the relation:

Since the air is still, the effective speed of sound = 340 + 0 = 340 m/s
The source is at rest. Hence, the wavelength of the sound will not change, i.e., λ remains 0.875 m.
Hence, the given two situations are not exactly identical.

Additional Excercises

15.22. A travelling harmonic wave on a string is described by

(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.

Answer

(a) The given harmonic wave is

where,
k = 2π/λ
∴ λ = 2π/k
and, ω = 2πv
∴ v = ω/2π
Speed, v = vλ = ω/k
where,
ω = 12rad/s
k = 0.0050 m^-1
∴ v = 12/0.0050 = 2400 cm/s
Hence, the velocity of the wave oscillation at x = 1 cm and t = 1s is not equal to the velocity of the wave propagation.

(b) Propagation constant is related to wavelength as:
k = 2π/λ
∴ λ = 2π/k = 2 × 3.14 / 0.0050
= 1256 cm = 12.56 m
Therefore, all the points at distance nλ (n = ±1, ±2, …. and so on), i.e. ± 12.56 m, ± 25.12 m, … 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 11 s.

15.23. 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 1/20 or 0.05 Hz?

Answer

(a) A short pip be a whistle has neither a definite wavelength nor a definite frequency. However, its speed of propagation is fized, being equal to speed of sound in air.

(b) No, frequency of the note produced by a whistle is not 1/20 = 0.05 Hz. Rather 0.05 Hz is the frequency of repetition of the short pip of the whistle.

15.24. One end of a long string of linear mass density 8.0 × 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 function of x and t that describes the wave on the string.

Answer

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)
Linear mass density, μ = 8.0 × 10-3 kg m-1