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Q.

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

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a

6

b

2

c

0

d

12

answer is A.

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Detailed Solution

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Frequency of sound heard by the man from approaching train
$n_{a} = n (\frac{v}{v - v_{s}}) = 240 (\frac{320}{320 - 4}) = 243 Hz$
Frequency of sound heard by the man from receding train
$n_{r} = n (\frac{v}{v + v_{s}}) = 240 (\frac{320}{320 + 4}) = 237 Hz$
Hence, number of beats heard by man per sec $= n_{a} - n_{r} = 243 - 237 = 6$

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