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Question

(A) Let $n$ be the actual frequency of the whistle,$v$ be the speed of sound,and $v_s$ be the speed of the train.When the train approaches the station

Vedclass · Accepted Answer

$\frac{2 n_1 n_2}{n_1+n_2}$

Vedclass · Answer

(A) Let $n$ be the actual frequency of the whistle,$v$ be the speed of sound,and $v_s$ be the speed of the train.When the train approaches the stationary observer,the apparent frequency is $n_1 = n \left( \frac{v}{v - v_s} \right)$.When the train moves away from the stationary observer,the apparent frequency is $n_2 = n \left( \frac{v}{v + v_s} \right)$.Taking the reciprocals of both equations:$\frac{1}{n_1} = \frac{v - v_s}{nv} = \frac{1}{n} - \frac{v_s}{nv}$$\frac{1}{n_2} = \frac{v + v_s}{nv} = \frac{1}{n} + \frac{v_s}{nv}$Adding these two equations:$\frac{1}{n_1} + \frac{1}{n_2} = \frac{2}{n} \implies \frac{n_1 + n_2}{n_1 n_2} = \frac{2}{n}$Therefore,the actual frequency $n = \frac{2 n_1 n_2}{n_1 + n_2}$.When the observer moves with the train,there is no relative motion between the source and the observer,so the observed frequency is equal to the actual frequency $n$.

When a train is approaching a stationary observer,the apparent frequency of the whistle of the train is $n_1$,and when the train is moving away from the observer,the apparent frequency is $n_2$. The frequency of the whistle noticed by the observer when he moves with the train is . . . . . . .

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