Q: A whistle revolves in a circle with angular velocity \omega = 20 rad/s using a string of length 50 cm. If the actual frequency of sound from the whistle is 385 Hz, then the minimum frequency heard by the observer far away from the centre is (velocity of sound v = 340 m/s)

374 Hz. Velocity of whistle: v_s = \omega r = 20 \times 0.5 = 10 m/s. Minimum frequency occurs when source moves away: f_{min} = f \frac{v}{v + v_s} = 385 \frac{340}{340 + 10} = 385 \frac{340}{350} = 374 Hz

Question 1

Two sources are at a finite distance apart. They emit sounds of wavelength \lambda. An observer situated between them on line joining approaches one source with speed u. Then, the number of beat heard/second by observer will be

Accepted Answer

\frac{2u}{\lambda}. Original frequency of each source: n = \frac{v}{\lambda}. When observer approaches source A: n'_A = n(1 + \frac{u}{v}). When observer recedes from source B: n'_B = n(1 - \frac{u}{v}). Beats = n'_A - n'_B = n \cdot \frac{2u}{v} = \frac{v}{\lambda} \cdot \frac{2u}{v} = \frac{2u}{\lambda}

Question 2

Two cars moving in opposite directions approach each other with speed of 22 m/s and 16.5 m/s respectively. The driver of the first car blows a horn having a frequency 400 Hz. The frequency heard by the driver of the second car is (velocity of sound 340 m/s)

Accepted Answer

448 Hz. Both source and observer moving towards each other: f' = f \frac{v + v_o}{v - v_s} = 400 \frac{340 + 16.5}{340 - 22} = 400 \frac{356.5}{318} = 448 Hz

Question 3

A siren emitting a sound of frequency 800 Hz moves away from an observer towards a cliff at a speed of 15 m/s. Then, the frequency of sound that the observer hears in the echo reflected from the cliff is (Take, velocity of sound in air = 330 m/s)

Accepted Answer

838 Hz. Echo frequency received by observer: f' = f \frac{v}{v + v_s} = 800 \frac{330}{330 + 15} = 800 \frac{330}{345} = 838 Hz

Question 4

A source of sound S emitting waves of frequency 100 Hz and an observer O are located at some distance from each other. The source is moving with a speed of 19.4 m/s at an angle of 60° with the source-observer line. The observer is at rest. The apparent frequency observed by the observer (velocity of sound in air is 330 m/s) is

Accepted Answer

103 Hz. Only the component of velocity along the line joining source and observer matters: v_s \cos(60°) = 19.4 	imes 0.5 = 9.7 m/s. f' = f \frac{v}{v - v_s\cos(60°)} = 100 \frac{330}{330 - 9.7} = 100 \frac{330}{320.3} \approx 103 Hz

Question 5

A speeding motorcyclist sees traffic jam ahead of him. He slows down to 36 km/h. He finds that traffic has eased and a car moving ahead of him at 18 km/h is honking at a frequency of 1392 Hz. If the speed of sound is 343 m/s, the frequency of the honk as heard by him will be

Accepted Answer

1412 Hz. Both moving in same direction with observer catching up to source: f' = f \frac{v + v_o}{v + v_s} where v_o = 36 km/h = 10 m/s, v_s = 18 km/h = 5 m/s. f' = 1392 \frac{343 + 10}{343 + 5} = 1392 \frac{353}{348} = 1412 Hz

Question 6

The driver of a car travelling with speed 30 m/s towards a hill sounds a horn of frequency 600 Hz. If the velocity of sound in air is 330 m/s, the frequency of reflected sound as heard by driver is

Accepted Answer

720 Hz. Two Doppler effects: (1) Car approaching hill: f_1 = f \frac{v}{v - v_s} = 600 \frac{330}{330 - 30} = 600 \frac{330}{300} = 660 Hz. (2) Car receding from echo: f' = f_1 \frac{v - v_o}{v} = 660 \frac{330 - 30}{330} = 660 \frac{300}{330} = 600 Hz. Wait, using double Doppler formula: f' = f \frac{v + v_o}{v - v_s} = 600 \frac{330 + 30}{330 - 30} = 600 \frac{360}{300} = 720 Hz

Question 7

A car is moving towards a high cliff. The car driver sounds a horn of frequency f. The reflected sound heard by the driver has a frequency 2f. If v be the velocity of sound, then the velocity of the car, in the same velocity units, will be

Accepted Answer

\frac{v}{3}. Using double Doppler effect: f' = f \frac{v + v_o}{v - v_s} where v_o = v_s = v_c (car velocity). Given f' = 2f: 2f = f \frac{v + v_c}{v - v_c}, so 2(v - v_c) = v + v_c, giving 2v - 2v_c = v + v_c, thus v = 3v_c, or v_c = \frac{v}{3}

Question 8

An observer moves towards a stationary source of sound with a speed 1/5th of the speed of sound. The wavelength and frequency of the source emitted are \lambda and f respectively. The apparent frequency and wavelength recorded by the observer are respectively

Accepted Answer

1.2f, \lambda. Observer moving towards stationary source: f' = f \frac{v + v_o}{v} = f \frac{v + v/5}{v} = f \frac{6v/5}{v} = 1.2f. Wavelength reaching observer remains \lambda (motion doesn't change wavelength).

Question 9

A whistle revolves in a circle with angular velocity \omega = 20 rad/s using a string of length 50 cm. If the actual frequency of sound from the whistle is 385 Hz, then the minimum frequency heard by the observer far away from the centre is (velocity of sound v = 340 m/s)

Accepted Answer

374 Hz. Velocity of whistle: v_s = \omega r = 20 	imes 0.5 = 10 m/s. Minimum frequency occurs when source moves away: f_{min} = f \frac{v}{v + v_s} = 385 \frac{340}{340 + 10} = 385 \frac{340}{350} = 374 Hz

Question 10

A vehicle, with a horn of frequency n is moving with a velocity of 30 m/s in a direction perpendicular to the straight line joining the observer and the vehicle. The observer perceives the sound to have a frequency n + n_1. Then (If the sound velocity in air is 300 m/s)

Accepted Answer

n_1 = 0. When velocity is perpendicular to the line joining source and observer, there is no component of velocity along this line. Therefore, no Doppler effect occurs, and n_1 = 0.

Question 11

A star which is emitting radiation at a wavelength of 5000 Å is approaching the earth with a velocity of 1.50 	imes 10^6 m/s. The change in wavelength of the radiation as received on the earth is

Accepted Answer

25 Å. Doppler shift in light: \Delta\lambda = \lambda \frac{v}{c} where approaching gives negative shift (but magnitude): |\Delta\lambda| = 5000 	imes \frac{1.50 	imes 10^6}{3 	imes 10^8} = 5000 	imes 0.005 = 25 Å

Question 12

Two trains move towards each other with the same speed. The speed of sound is 340 m/s. If the height of the tone of the whistle of one of them heard on the other changes 9/8 times, then the speed of each train should be

Accepted Answer

20 m/s. Two trains approaching with same speed v_t: f' = f \frac{v + v_o}{v - v_s} = f \frac{v + v_t}{v - v_t}. Given f' = \frac{9f}{8}: \frac{9}{8} = \frac{340 + v_t}{340 - v_t}. 9(340 - v_t) = 8(340 + v_t), so 3060 - 9v_t = 2720 + 8v_t, giving v_t = 20 m/s

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