An engine sounding a whistle of frequency $1152 \mathrm{~Hz}$ is receding from a stationary observer at $72…

An engine sounding a whistle of frequency $1152 \mathrm{~Hz}$ is receding from a stationary observer at $72 \mathrm{~km} /$ hour. If velocity of sound in air is $340 \mathrm{~m} / \mathrm{s}$, then the frequency of note heard by the observer is
  1. 612 Hz
  2. 1088 Hz
  3. 1224 Hz
  4. 544 Hz

Solution

Concept Source is moving away from a stationary observer $\therefore \mathrm{f}^{\prime}=\mathrm{f}\left(\frac{\mathrm{v}}{\mathrm{v}+\mathrm{v}_{\mathrm{s}}}\right)$ Where, $f$ is the true frequency $v$ is the speed of sound and $v_s$ is the speed of source. Given, $\mathrm{f}=1152, \mathrm{v}=340 \mathrm{~m} / \mathrm{s}$ $\text { and } \mathrm{v}_{\mathrm{s}}=\frac{72 \times 10^3}{3600} \mathrm{~m} / \mathrm{s}=20 \mathrm{~m} / \mathrm{s}$ $\therefore$ the apparent frequency $\mathrm{f}^{\prime}=1152 \frac{\times(340)}{(340+20)} \mathrm{Hz}=1088 \mathrm{~Hz}$

Asked in: MHT CET 2022 (08 Aug Shift 2)

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