An object producing a pitch of 400 Hz flies past a stationary person. The object was moving in a straight line with a velocity$200\ ms^{- 1}$. What is the change in frequency noted by the person as the person as the object files past him?

Question

Dr. Abubacker · Accepted Answer

(c) From Doppler’s effect, the perceived frequency (v’) is given by
$v^{'} = \left( \frac{v - v_{o}}{v - v_{s}} ight)v\ $
Where $v_{o}$ is velocity of observer, $v_{s}$ of source, v of sound and v the original frequency.
Given, $v_{o} = 0$(stationary), $v = 300\ ms^{- 1}$
$v_{s} = 200\ ms^{- 1},\ \ \ \ \ v = 400\ Hz$
$	herefore v^{'} = \frac{300 	imes 400}{300 - 200}$
$= \frac{300 	imes 400}{100}$
$\Longrightarrow v^{'} = 1200\ Hz$

Q. An object producing a pitch of 400 Hz flies past a stationary person. The object was moving in a straight line with a velocity$200\ ms^{- 1}$. What is the change in frequency noted by the person as the person as the object files past him?

Explanation

Reviewed by Dr. Abubacker

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