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

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(D) Let $n = 240\,Hz$ be the frequency of the whistles,$v_s = 4\,m/s$ be the speed of the trains,and $v = 320\,m/s$ be the speed of sound.The apparent

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$6$

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(D) Let $n = 240\,Hz$ be the frequency of the whistles,$v_s = 4\,m/s$ be the speed of the trains,and $v = 320\,m/s$ be the speed of sound.The apparent frequency $n'$ heard from the train approaching the observer is given by the Doppler effect formula:$n' = n \left[ \frac{v}{v - v_s} ight]$The apparent frequency $n''$ heard from the train moving away from the observer 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'' = n \left[ \frac{v}{v - v_s} - \frac{v}{v + v_s} ight]$$\Delta n = 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 $v_s$ is very small compared to $v$,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 values:$\Delta n = \frac{2 	imes 240 	imes 4}{320} = \frac{1920}{320} = 6\,Hz$Thus,the number of beats per second heard by the man is $6$.

Process	Condition
$(I)$ Adiabatic	$(A)$ $\Delta W = 0$
$(II)$ Isothermal	$(B)$ $\Delta Q = 0$
$(III)$ Isochoric	$(C)$ $\Delta U \neq 0, \Delta W \neq 0, \Delta Q \neq 0$
$(IV)$ Isobaric	$(D)$ $\Delta U = 0$

$A$ man is watching two trains,one leaving and the other coming with 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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