Question

# 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

## The correct option is D $$20\%$$Let velocity of sound in air be $$v$$.  $$\Rightarrow$$ Velocity of observer, $$v_{observer} = \dfrac{v}{5}$$Let $$\nu$$ be the original frequency of the sound source and $$\nu'$$ be the apparent frequency heard by the moving observer.Using Doppler effect when observer is moving towards the stationary source:$$\nu' = \nu \bigg[\dfrac{v_{sound} + v_{observer}}{v_{sound}} \bigg]$$ $$= \nu \bigg[\dfrac{v +\frac{v}{5}}{v} \bigg] = \dfrac{6}{5}\nu$$Percentage increase in frequency $$= \dfrac{\nu' - \nu}{\nu} \times 100 = \dfrac{\frac{6}{5}\nu - \nu}{\nu} \times 100 = 20$$ %Physics

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