A man is watching two trains, one leaving and the other coming in with equal spe

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 frequency 240 Hz, the number of beats heard by the man (velocity of sound in air is 320 m/s) will be equal to

Options:

A . 6

B . 3

C . 0

D . 12

Answer: Option A
:
A
Apparent frequency due to train which is coming in is

m_{1} = \frac{v}{v - v_{s}} n

Apparent frequency due to train which is leaving is

m_{2} = \frac{v}{v - v_{s}} n

So the number of beats is

n_{1} - n_{2} = (\frac{1}{316} - \frac{1}{324}) 320 \times 240 \Rightarrow n_{1} - n_{2} = 6

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