Question:
If a source of sound initially at rest is moving away from a stationery observer with an acceleration of $11~ms^{-2}$, then the time taken for the frequency of sound heard by the observer to become 10% less than the frequency of source is (Speed of sound in air $=330~ms^{-1}$)}
If a source of sound initially at rest is moving away from a stationery observer with an acceleration of $11~ms^{-2}$, then the time taken for the frequency of sound heard by the observer to become 10% less than the frequency of source is (Speed of sound in air $=330~ms^{-1}$)}
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Always convert Doppler shift into velocity first, then apply kinematics.
Updated On: Jun 22, 2026
- 4.4 s
- 1.1 s
- 2.2 s
- 3.3 s \bigskip
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The Correct Option is D
Solution and Explanation
Concept:
For a source moving away from observer, Doppler effect gives:
\[
f' = f \cdot \frac{v}{v+v_s}
\]
where:
• $v$ = speed of sound
• $v_s$ = source velocity ---
Step 1: Apply frequency condition.
Given: \[ f' = 0.9f \] So: \[ 0.9 = \frac{v}{v+v_s} \] ---
Step 2: Solve for source velocity.
\[ 0.9(v+v_s)=v \] \[ 0.9v + 0.9v_s = v \] \[ 0.9v_s = 0.1v \Rightarrow v_s = \frac{v}{9} \] \[ v_s = \frac{330}{9} = 36.67~m/s \] ---
Step 3: Use kinematics (acceleration motion).
\[ v_s = at \] \[ t = \frac{36.67}{11} \approx 3.33~s \] --- Final Answer: \[ (D)\ 3.3~s \]
• $v$ = speed of sound
• $v_s$ = source velocity ---
Step 1: Apply frequency condition.
Given: \[ f' = 0.9f \] So: \[ 0.9 = \frac{v}{v+v_s} \] ---
Step 2: Solve for source velocity.
\[ 0.9(v+v_s)=v \] \[ 0.9v + 0.9v_s = v \] \[ 0.9v_s = 0.1v \Rightarrow v_s = \frac{v}{9} \] \[ v_s = \frac{330}{9} = 36.67~m/s \] ---
Step 3: Use kinematics (acceleration motion).
\[ v_s = at \] \[ t = \frac{36.67}{11} \approx 3.33~s \] --- Final Answer: \[ (D)\ 3.3~s \]
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