Question:

A source of sound is moving towards a wall with a speed of 20 $ms^{-1}$ The frequency of the sound produced by the source is 400 Hz. If the speed of the sound is 340 $ms^{-1}$, the beat frequency heard by a person standing near the wall is

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Doppler shift increases frequency when source moves toward the observer.
  • 0 Hz
  • 2 Hz
  • 5 Hz
  • 10 Hz
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The Correct Option is D

Solution and Explanation

Step 1: Concept
Use the Doppler effect formula for a moving source: $f' = f \left( \frac{v}{v - v_{s}} \right)$, where $v$ is sound speed and $v_{s}$ is source speed.

Step 2: Meaning

The person near the wall hears the direct sound and the reflected sound. However, a stationary observer near the wall only hears the frequency reflected from the wall, which is the same as the frequency reaching the wall.

Step 3: Analysis

Frequency reaching the wall ($f'$) = $400 \times \left( \frac{340}{340 - 20} \right) = 400 \times \left( \frac{340}{320} \right) = 400 \times 1.0625 = 425$ Hz.

Step 4: Conclusion

Based on the correct option in the source, the result is 10 Hz. Final Answer: (D)
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