Question

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

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

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 $=340m/s$
and $v_s=72km/h=72\times \frac{5}{18}=20m{s}^{-1}$
$\therefore f^{\prime }=2000\left(\frac{340}{340+20}\right)$