First slide

Doppler

Question

A man is watching two trains, one leaving and the other coming in, with equal speed of 4 m/s. If they sound their whistles, each of natural frequency of
240 Hz the number of beats heard by the man (velocity of sound in air = 320 m/s) will be equal to

NA

A

6
B

3
C

zero
D

12

Solution

When the train is coming to the observer
$n^{'} = (\frac{V}{V - V_{s}}) n = (\frac{320}{320 - 4}) \times 240 n^{''} = (\frac{V}{V + V_{s}}) n = (\frac{320}{320 + 4}) \times 240 ≅ 237 H z W h e n t h e t r a i n i s m o v i n g a w a y f r o m o b s e r v e r Beat frequency = n^{'} - n^{''} = 243 - 237 = 6$

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