A man is watching two trains,one leaving and | vedclass

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(A) The frequency of sound heard by the man from the approaching train is given by:${n_a} = n\left( {\frac{v}{{v - {v_s}}}} \right) = 240\left( {\frac

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(A) The frequency of sound heard by the man from the approaching train is given by:${n_a} = n\left( {\frac{v}{{v - {v_s}}}} ight) = 240\left( {\frac{{320}}{{320 - 4}}} ight) = 240\left( {\frac{{320}}{{316}}} ight) \approx 243.04\, Hz$The frequency of sound heard by the man from the receding train is given by:${n_r} = n\left( {\frac{v}{{v + {v_s}}}} ight) = 240\left( {\frac{{320}}{{320 + 4}}} ight) = 240\left( {\frac{{320}}{{324}}} ight) \approx 237.04\, Hz$The number of beats heard by the man per second is the difference in frequencies:$Beat\ frequency = {n_a} - {n_r} = 243.04 - 237.04 = 6\, Hz$Alternatively,using the formula for beat frequency:$Beat\ frequency = \frac{{2nv{v_s}}}{{{v^2} - v_s^2}} = \frac{{2 	imes 240 	imes 320 	imes 4}}{{{320^2} - {4^2}}} = \frac{{614400}}{{102400 - 16}} = \frac{{614400}}{{102384}} \approx 6\, Hz$

$A$ man is watching two trains,one leaving and the other coming in with equal speeds 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 $= 320\, m/s$) will be equal to

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