The pitch of a whistle of an engine appears to drop by $30 \%$ of original value when it passes a stationary…
- 840
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- 150
Solution
If $n$ is the original (actual) frequency of the whistle and $\mathrm{n}^{\prime}$ the frequency as the engine passes the stationary observer, then $\begin{array}{ll} & \frac{\mathrm{n}^{\prime}}{\mathrm{n}}=\left(\frac{\mathrm{v}}{\mathrm{v}+\mathrm{v}_{\text {cngine }}}\right) \\ \therefore \quad & 0.7=\frac{350}{350+\mathrm{v}_{\text {engine }}} \\ & 0.7 \mathrm{y}_{\text {engine }}+245=350 \\ \therefore \quad & \mathrm{v}_{\text {engine }}=150 \mathrm{~m} / \mathrm{s}\end{array}$
Asked in: MHT CET 2024 (03 May Shift 1)