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Question

(B) Let $v$ be the speed of sound and $V$ be the speed of the train/observer. The natural frequency is $n$.In the first case,the source (train) moves

Vedclass · Accepted Answer

$1.25$

Vedclass · Answer

(B) Let $v$ be the speed of sound and $V$ be the speed of the train/observer. The natural frequency is $n$.In the first case,the source (train) moves away from the stationary observer:$n' = n \left( \frac{v}{v + V} \right)$Given $\frac{n}{n'} = 1.2$,so $\frac{v + V}{v} = 1.2 \implies 1 + \frac{V}{v} = 1.2 \implies \frac{V}{v} = 0.2$.In the second case,the source is at rest and the observer moves away from it:$n'' = n \left( \frac{v - V}{v} \right) = n \left( 1 - \frac{V}{v} \right)$The ratio of the natural frequency to the apparent frequency is $\frac{n}{n''} = \frac{1}{1 - \frac{V}{v}}$.Substituting $\frac{V}{v} = 0.2$:$\frac{n}{n''} = \frac{1}{1 - 0.2} = \frac{1}{0.8} = 1.25$.Thus,the ratio is $1.25: 1$.

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