Question:
The apparent frequency observed by a moving observer away from a stationary source is 20% less than the actual frequency. If the velocity of sound in air is 330 ms\(^{-1}\), then the velocity of the observer is
The apparent frequency observed by a moving observer away from a stationary source is 20% less than the actual frequency. If the velocity of sound in air is 330 ms\(^{-1}\), then the velocity of the observer is
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For observer moving away, frequency decreases.
Updated On: May 8, 2026
- 660 ms\(^{-1}\)
- 330 ms\(^{-1}\)
- 66 ms\(^{-1}\)
- 33 ms\(^{-1}\)
- 20 ms\(^{-1}\)
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The Correct Option is C
Solution and Explanation
Concept:
Doppler effect for moving observer:
\[
f' = f \left( \frac{v - v_o}{v} \right)
\]
Step 1: Given condition. \[ f' = 0.8 f \]
Step 2: Substitute into formula. \[ 0.8f = f \left( \frac{v - v_o}{v} \right) \]
Step 3: Cancel \(f\). \[ 0.8 = \frac{v - v_o}{v} \]
Step 4: Solve for \(v_o\). \[ 0.8v = v - v_o \Rightarrow v_o = 0.2v \]
Step 5: Substitute \(v = 330\). \[ v_o = 0.2 \times 330 = 66\,\text{m/s} \]
Step 6: Conclusion. \[ \boxed{66\,\text{m/s}} \]
Step 1: Given condition. \[ f' = 0.8 f \]
Step 2: Substitute into formula. \[ 0.8f = f \left( \frac{v - v_o}{v} \right) \]
Step 3: Cancel \(f\). \[ 0.8 = \frac{v - v_o}{v} \]
Step 4: Solve for \(v_o\). \[ 0.8v = v - v_o \Rightarrow v_o = 0.2v \]
Step 5: Substitute \(v = 330\). \[ v_o = 0.2 \times 330 = 66\,\text{m/s} \]
Step 6: Conclusion. \[ \boxed{66\,\text{m/s}} \]
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