Two trains A and B are moving with speeds 20 ~m / s and 30 ~m / s respectively in the same direction on the same straight track, with B ahead of A . The engines are at the front ends. The engines of train A blows a long whistle. Assume that the sound of the whistle is composed of components varying in frequency from f_1=800 ~Hz to f_2=1120 ~Hz , as shown in the figure. The spread in the frequency (highest frequency - lowest frequency) is thus 320 ~Hz . The speed of sound in still air is 340 ~m / s . The spread of frequency as observed by the passengers in train B is

Waves PYQ — JEE Advanced 2007 · Shift 2 · Q24 with Solution

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Correct answer: A

Train $A$ emits the whistle, so it is the source.
Passengers in train $B$ are the observers.
Both trains move in the same direction, with $B$ ahead of $A$ .
Since sound travels from $A$ to $B$ , the sound wave travels forward (same direction as train motion).

So:

Source velocity toward observer: $v_s = 20\,\text{m/s}$
Observer velocity away from source, but along the direction of wave travel: $v_o = 30\,\text{m/s}$
Speed of sound: $v = 340\,\text{m/s}$

For sound, when the observer moves in the direction of wave propagation and the source also moves in that direction, the observed frequency is

f' = f\,\frac{v - v_o}{v - v_s}

Here,

f' = f\,\frac{340-30}{340-20} = f\,\frac{310}{320}

Thus every frequency component is multiplied by the same factor:

\frac{310}{320}

Original spread:

\Delta f = f_2 - f_1 = 1120 - 800 = 320\,\text{Hz}

Observed spread:

\Delta f' = \Delta f\cdot \frac{310}{320}

\Delta f' = 320\cdot \frac{310}{320} = 310\,\text{Hz}

\boxed{310\,\text{Hz}}

So the correct option is A.