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The frequency changes by 10% as the source approaches a stationary observer with constant speed vs. What would be the percentage change in frequency as the source recedes the observer with the same speed ? Given, that vs « v ( v = speed of sound in air)
  • a)
    16.7%
  • b)
    8.5%
  • c)
    14.3%
  • d)
    20%
Correct answer is option 'B'. Can you explain this answer?
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Given that vs is the speed of the source, the percentage change in frequency as the source recedes from the observer with the same speed can be calculated using the formula:

Percentage change in frequency = (-v/vs) x 100%

Where v is the speed of the observer.

Since the observer and source have the same speed, v = vs. Therefore, the formula becomes:

Percentage change in frequency = (-v/v) x 100%

Percentage change in frequency = -100%

So, the percentage change in frequency as the source recedes the observer with the same speed is -100%.
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