Question:

An observer moves towards a source at rest with a speed of 25% of the speed of sound in air. If the frequency of the sound emitted by the source is 200 Hz, then the frequency of sound heard by the observer is:

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For moving observer and stationary source, use $f' = f (v\pm v_o)/v$ (plus if moving towards source).
Ensure $v_o$ is a fraction of $v$ if given that way — convert if needed.
Sign conventions matter: approaching increases frequency, receding decreases it.
Updated On: Oct 27, 2025
  • $275\ \mathrm{Hz}$
  • $300\ \mathrm{Hz}$
  • $250\ \mathrm{Hz}$
  • $325\ \mathrm{Hz}$
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The Correct Option is C

Solution and Explanation

• Doppler effect for an observer moving towards a stationary source: \[ f' = f\left(\frac{v+v_o}{v}\right) \] where $v$ is speed of sound and $v_o$ observer speed.
• Given $v_o = 0.25\,v$, $f = 200\ \mathrm{Hz}$. Substituting: \[ f' = 200\left(1 + 0.25\right) = 200 \times 1.25 = 250\ \mathrm{Hz}. \] • Hence the observer hears 250 Hz.
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