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A man is watching 2 trains ,one leaving and the other coming in which equals a speed of 4 m/s.If they sound their whistles,each of frequency 240 Hz ,the numbers of beats heard by the man (velocity of sound in air =320 m/s) will be equal to :

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answer is A.

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Detailed Solution

App. frequency due to train which is coming in $n_{1} = \frac{V}{V - V_{2}} n$

App . frequency due to train which is leaving in $n_{2} = \frac{V}{V + V_{2}} n$

So , the number of beats n₁ - n₂= $(\frac{1}{316} - \frac{1}{324}) 320 \times 240 \Rightarrow n_{1} - n_{2} = 6$

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