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
Diffcult
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When train approaches,
$\mathrm{n}^{\prime}=\mathrm{n}\left(\frac{\mathrm{v}}{\mathrm{v}-\mathrm{v}_{\mathrm{t}}}\right)=\frac{1}{\mathrm{n}^{\prime}}=\frac{1}{\mathrm{n}}\left(1-\frac{\mathrm{v}_{\mathrm{t}}}{\mathrm{v}}\right) \Rightarrow \frac{\mathrm{v}_{\mathrm{t}}}{\mathrm{v}}=1-\frac{\mathrm{n}}{\mathrm{n}^{\prime}}$
When train recedes away,
$\mathrm{n}^{\prime \prime}=\mathrm{n}\left(\frac{\mathrm{v}}{\mathrm{v}+\mathrm{v}_{\mathrm{t}}}\right) \quad \Rightarrow \frac{1}{\mathrm{n}^{\prime \prime}}=\frac{1}{\mathrm{n}}\left(1+\frac{\mathrm{v}_{\mathrm{t}}}{\mathrm{v}}\right)$
$ \Rightarrow \frac{1}{{{n^\prime }^\prime }} = \frac{1}{n}\left( {1 + 1 - \frac{n}{{{n^\prime }}}} \right)\quad \Rightarrow \frac{1}{{{n^\prime }^\prime }} = \frac{1}{n}\left( {2 - \frac{n}{{{n^\prime }}}} \right)$
$\Rightarrow \frac{\mathrm{n}}{\mathrm{n}^{\prime \prime}}+\frac{\mathrm{n}}{\mathrm{n}^{\prime}}=2 \quad \Rightarrow \mathrm{n}=\frac{2 \mathrm{n}^{\prime} \mathrm{n}^{\prime \prime}}{\mathrm{n}^{\prime}+\mathrm{n}^{\prime \prime}}$
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