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Q. An engine sounding a whistle of frequency $2000\, Hz$ is receding from the stationary observer at $72 \,km / h$. What is the apparent frequency of the observer? The velocity of sound in air is $340 \,m / s$

TS EAMCET 2020

Solution:

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Here a source is moving away from a stationary observer.
So, frequency recorded at source will be lower than true frequency.
Apparent frequency at observer,
$f^{\prime}=f\left(\frac{v}{v+v_{s}}\right)$
where, $v=$ speed of sound $=340\, m / s$
and $v_{s}=72\, km / h =72 \times \frac{5}{18}=20\, ms ^{-1}$
$\therefore f'=2000\left(\frac{340}{340+20}\right)$