Question:easy
An observer moves towards a stationary source of sound with a velocity equal to one-fourth of the velocity of sound. Then the percentage increase in the apparent frequency observed by the observer is
An observer moves towards a stationary source of sound with a velocity equal to one-fourth of the velocity of sound. Then the percentage increase in the apparent frequency observed by the observer is
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For a stationary source and moving observer,
\[
f'=f\left(1+\frac{v_o}{v}\right).
\]
Hence the percentage increase in frequency is simply
\[
\frac{v_o}{v}\times100.
\]
If \(v_o=\frac{v}{4}\), the increase is \(25\%\).
Updated On: Jul 29, 2026
- \(25\%\)
- \(20\%\)
- \(30\%\)
- \(50\%\)
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The Correct Option is A
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