Question

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 $340 \,m / s$

• A. $1889 \,Hz$
• B. $2889 \,Hz$
• C. $3889 \,Hz$
• D. $4889 \,Hz$

Answer 1

Answer: (A)

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 $v_{s}=72\, km / h =72 \times \frac{5}{18}=20\, ms ^{-1}$
$\therefore f'=2000\left(\frac{340}{340+20}\right)$