MHT CET · Physics · Waves and Sound
The pitch of the whistle of an engine appears to drop by \(20 \%\) of original value when it passes a stationary observer. If speed of sound in air \(350 \mathrm{~m} / \mathrm{s}\), then speed of engine in \(\mathrm{m} / \mathrm{s}\) is
- A \(1050\)
- B \(175\)
- C \(520.5\)
- D \(87.5\)
Answer & Solution
Correct Answer
(D) \(87.5\)
Step-by-step Solution
Detailed explanation
Aparent pitch heard by the observer \(\mathrm{f}^{\prime}=\frac{4}{5} \mathrm{f}\)
where, \(f\) is the original pitch of the engine.
As the apparent frequency is lowered, thus the engine must be moving away from the stationary observer.
From Doppler effect, apparent frequency heard:
\(\begin{aligned} & \mathrm{f}^{\prime}=\mathrm{f}\left[\frac{\mathrm{V}_{\text {sound }}}{\mathrm{V}_{\text {sound }}+\mathrm{V}_{\text {source }}}\right] \\ & \therefore \frac{4 \mathrm{f}}{5}=\left[\frac{350}{350+\mathrm{V}_{\text {Source }}}\right] \\ & \Rightarrow \mathrm{V}_{\text {Source }}=87.5\end{aligned}\)
where, \(f\) is the original pitch of the engine.
As the apparent frequency is lowered, thus the engine must be moving away from the stationary observer.
From Doppler effect, apparent frequency heard:
\(\begin{aligned} & \mathrm{f}^{\prime}=\mathrm{f}\left[\frac{\mathrm{V}_{\text {sound }}}{\mathrm{V}_{\text {sound }}+\mathrm{V}_{\text {source }}}\right] \\ & \therefore \frac{4 \mathrm{f}}{5}=\left[\frac{350}{350+\mathrm{V}_{\text {Source }}}\right] \\ & \Rightarrow \mathrm{V}_{\text {Source }}=87.5\end{aligned}\)
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