When a train approaches a stationary observer | vedclass

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Question

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

Vedclass · Accepted Answer

$n = \frac{2n' n''}{n' + n''}$

Vedclass · Answer

(C) Let $n$ be the actual frequency of the whistle,$v$ be the speed of sound,and $v_t$ be the speed of the train.When the train approaches the stationary observer:$n' = n \left( \frac{v}{v - v_t} \right) \Rightarrow \frac{n}{n'} = \frac{v - v_t}{v} = 1 - \frac{v_t}{v} \Rightarrow \frac{v_t}{v} = 1 - \frac{n}{n'}$When the train recedes away from the stationary observer:$n'' = n \left( \frac{v}{v + v_t} \right) \Rightarrow \frac{n}{n''} = \frac{v + v_t}{v} = 1 + \frac{v_t}{v}$Substituting the value of $\frac{v_t}{v}$ from the first equation into the second:$\frac{n}{n''} = 1 + \left( 1 - \frac{n}{n'} \right) = 2 - \frac{n}{n'}$Rearranging the terms:$\frac{n}{n''} + \frac{n}{n'} = 2 \Rightarrow n \left( \frac{n' + n''}{n' n''} \right) = 2$$n = \frac{2n' n''}{n' + n''}$

When a train approaches a stationary observer,the apparent frequency of the whistle is $n'$ and when the same train recedes away from the observer,the apparent frequency is $n''$. Then,the apparent frequency $n$ when the observer moves with the train is

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