MHT CET · Physics · Waves and Sound
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
- A 612 Hz
- B 1088 Hz
- C 1224 Hz
- D 544 Hz
Answer & Solution
Correct Answer
(B) 1088 Hz
Step-by-step Solution
Detailed explanation
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 \(\mathrm{f}^{\prime}=1152 \frac{\times(340)}{(340+20)} \mathrm{Hz}=1088 \mathrm{~Hz}\)
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