Two Questions (MCQ) on Doppler Effect in Sound

You will find the formulae to be remembered in connection with Doppler effect and some useful multiple choice questions (with solution) on Doppler effect at this location on this site.

[If you want all questions on Doppler effect on this site, you may click on the label ‘Doppler effect’ below this post].

Today I give you two more multiple choice questions on Doppler effect:

(1) A whistle producing sound waves of frequencies 9500 Hz and above is approaching a stationary person with speed v ms^–1. The velocity of sound in air is 300 ms^–1. If the person can hear frequencies up to 10000 Hz, the maximum value of v up to which he can hear the whistle is

(1) 30 ms^–1

(2) 15√2 ms^–1

(3) 15/√2 ms^–1

(4) 15 ms^–1

The apparent frequency (n’) in terms of all possible variables is given by

n’ = n(v+w–v_L)/(v+w–v_S) where ‘n’ is the real frequency of sound, ‘v’ is the velocity of sound, ‘w’ is the velocity of wind, ‘v_L’ is the velocity of listener and ‘v_S’ is the velocity of the source of sound. Note that in the above expression all velocities are in the same direction and the source is behind the listener and is therefore approaching the listener.

Since the wind velocity is zero and the listener is stationary in the first case, the above expression reduces to

n’ = nv/(v–v_S)

[The apparent frequency is therefore greater than the real frequency].

As the person can hear frequencies up to 10000 Hz only, the maximum value of v_S upto which he can hear the whistle is given by

10000 = 9500×300/(300 – v_S)

Therefore, 300 – v_S = 285 so that v_S = 15 ms^–1

[The above question appeared in AIEEE 2006 question paper]

(2) A person moves away with constant velocity ‘v₀’ from a stationary train which blows its whistle. The ratio of the real frequency of the whistle to the apparent frequency as measured by the person is 1.25. If the person is stationary and the whistle is moving away from the person with the same velocity ‘v₀’, the ratio of the real frequency of the whistle to the apparent frequency as measured by the person will be

(a) 1.2

(b) 1.25

(c) 1.4

(d) 1.45

(e) 1.5

The apparent frequency (n₁) as measured by the person moving away from the whistle with velocity v_L is given by

n₁ = n(v–v_L)/v where ‘n’ is the real frequency of the whistle and ‘v’ is the velocity of sound.

Therefore, n/n₁ = v/(v–v_L) = v/(v–v₀).

Since n/n₁ = 1.25,we have v/(v–v₀) =1.25 so that 1– (v₀/v) = 0.8 from which v₀ = 0.2v.

If the person is stationary and the whistle is moving away from the person with the same velocity ‘v₀’, the apparent frequency (n₂) as measured by the person is given by

n₂ = nv/(v +v_S) where v_S = v₀

Therefore, n/n₂ = (v+v₀)/v.

Substituting for v₀ (= 0.2v) we obtain n/n₂ = (v+0.2v)/v = 1.2