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← IIT-JEE 2007 · Paper 2
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IIT-JEE 2007 · Paper 2, Physics question 16

Passage

Paragraph for Question Nos. 14 to 16

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 engine of train A blows a long whistle.

Assume that the sound of the whistle is composed of components varying in frequency from f1=800f_1=800 Hz to f2=1120f_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.

(Figure: intensity against frequency, a flat-topped band from f1f_1 to f2f_2.)

(The paragraph is not printed in the official archive copy (which reproduces only the questions with their answers); its text is restored from the JEE 2007 question paper.)

The spread of frequency as observed by the passengers in train B is

Choose one option
Answer and worked solution

Official answer: A

Doppler factor. The source (A, behind) moves towards B at 20 m/s and the observer (B) moves away at 30 m/s:
f′=f 340−30340−20=310320f.f'=f\,\frac{340-30}{340-20}=\frac{310}{320}f.

Spread. Every frequency is scaled by the same factor, so the spread becomes 310320×320=310\frac{310}{320}\times320=310 Hz.

(A)

Topics:Doppler effect
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