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Two trains A and B are moving on parallel tracks in opposite direction at same speed of 30 ms^–1. Just when the engines of the two trains are about
to cross, the engine of train A begins to sound a horn. The sound of the horn is composed of components varying in frequency from 900 Hz to 1200 Hz. The speed of sound in air is 330 ms^–1.

Find the frequency spread heard by a passenger in train B.

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750 Hz to 1440 Hz

1080 Hz to 1440 Hz

1250 Hz to 1440 Hz

960 Hz to 1440 Hz

answer is A.

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Detailed Solution

$f_{m a x} = \frac{1200 (1 + \frac{30}{330})}{(1 - \frac{30}{330})} H z = 1440 H z f_{m i n} = \frac{900 (1 - \frac{30}{330})}{(1 + \frac{30}{330})} H z = 750 H z$

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