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

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(D) The apparent frequency $n'$ heard by the observer from the train approaching with speed $v_s$ is given by:$n' = n \left[ \frac{v}{v - v_s} \right]

Vedclass · Accepted Answer

$6$

Vedclass · Answer

(D) The apparent frequency $n'$ heard by the observer from the train approaching with speed $v_s$ is given by:$n' = n \left[ \frac{v}{v - v_s} ight]$The apparent frequency $n''$ heard by the observer from the train moving away with speed $v_s$ is given by:$n'' = n \left[ \frac{v}{v + v_s} ight]$The number of beats per second is the difference between these two frequencies:$\Delta n = n' - n''$$\Delta n = nv \left[ \frac{1}{v - v_s} - \frac{1}{v + v_s} ight] = nv \left[ \frac{(v + v_s) - (v - v_s)}{v^2 - v_s^2} ight] = nv \left[ \frac{2v_s}{v^2 - v_s^2} ight]$Since $v_s^2$ is very small compared to $v^2$ ($4^2 = 16$ vs $320^2 = 102400$),we can approximate $v^2 - v_s^2 \approx v^2$:$\Delta n \approx nv \left[ \frac{2v_s}{v^2} ight] = \frac{2nv_s}{v}$Substituting the given values ($n = 240\,Hz$,$v_s = 4\,m/s$,$v = 320\,m/s$):$\Delta n = \frac{2 	imes 240 	imes 4}{320} = \frac{1920}{320} = 6\,Hz$Thus,the number of beats per second heard by the man is $6$.

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

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