An observer moves towards a stationary source of sound with a...

An observer moves towards a stationary source of sound with a velocity one-fifth of the velocity of sound. The percentage change in the apparent frequency is

Options

Azero
B $5 %$
C $10 %$
D $20 %$

Correct answer

D. $20 %$

Step-by-step solution

By Doppler's effect

\dfrac{{ν}^{'}}{ν} = \dfrac{c + {ν}_{0}}{c}

(where

c

is the velocity of sound)

\dfrac{{ν}^{'}}{ν} = \dfrac{v + \left(v / 5\right)}{v} = \dfrac{6}{5}

∴

Fractional increase

= \dfrac{{ν}^{'} - ν}{ν} = \left(\dfrac{{ν}^{'}}{ν} - 1\right) = \left(\dfrac{6}{5} - 1\right) = \dfrac{1}{5}

∴

Percentage increase

= \dfrac{100}{5} = 20 %

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