An observer moves towards a stationary source of sound, with a velocity one fifth of the velocity of sound. What is the percentage increase in the app - Physics MCQ AIEEE 2026 | ExamTest.live | ExamTest.live

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An observer moves towards a stationary source of sound, with a velocity one fifth of the velocity of sound. What is the percentage increase in the apparent frequency?

A $zero$
B $0.5%$
C $5%$
D $20%$

Solution & Step-by-step Explanation

Apparent frequency f^{'} = f (\frac{v + v _{o}}{v}), where v is speed of sound and v_{o} is speed of observer.Given v_{o} = v /5 . f^{'} = f (\frac{v + v /5}{v}) = f (\frac{6/5 v}{v}) = 1.2 f Change in frequency Δ f = 1.2 f - f = 0.2 f .Percentage increase = \frac{0.2 f}{f} \times 100 = 20% .

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An observer moves towards a stationary source of sound, with a velocity one fifth of the velocity of sound. What is the percentage increase in the apparent frequency?

A

zero

B

0.5%

C

5%

D

20%

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