Question:
A siren emitting sound of frequency \(800\text{ Hz}\) is going away from a static listener with a speed of \(30\text{ m/s}\). Frequency of sound heard by the listener is \((\text{Velocity of sound in air}=340\text{ m/s})\)
A siren emitting sound of frequency \(800\text{ Hz}\) is going away from a static listener with a speed of \(30\text{ m/s}\). Frequency of sound heard by the listener is \((\text{Velocity of sound in air}=340\text{ m/s})\)
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When the source moves away from the observer, apparent frequency decreases. Use \(f'=f\frac{v}{v+v_s}\).
Updated On: May 7, 2026
- \(286.5\text{ Hz}\)
- \(418.2\text{ Hz}\)
- \(733.3\text{ Hz}\)
- \(644.5\text{ Hz}\)
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The Correct Option is C
Solution and Explanation
This is a Doppler effect problem.
The source is moving away from a stationary listener.
For a source moving away from a stationary observer, apparent frequency is:
\[
f' = f\left(\frac{v}{v+v_s}\right).
\]
Here:
\[
f=800\text{ Hz},
\]
\[
v=340\text{ m/s},
\]
\[
v_s=30\text{ m/s}.
\]
Substitute the values:
\[
f'=800\left(\frac{340}{340+30}\right).
\]
\[
f'=800\left(\frac{340}{370}\right).
\]
\[
f'=800(0.9189).
\]
\[
f'\approx 735.1\text{ Hz}.
\]
The nearest given option is:
\[
733.3\text{ Hz}.
\]
Hence, the frequency heard by the listener is:
\[
733.3\text{ Hz}.
\]
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