Question

A source of sound $S$ is moving with a velocity of \[\text{50m/s}\] towards a stationary observer. The observer measures the frequency of the source as \[1000Hz\]. What will be the apparent frequency of the source when it is moving away from the observer after crossing him? (Take velocity of sound in air is $350m/s$)
\[\begin{align}
  & A.857Hz \\
 & B.807Hz \\
 & C.750Hz \\
 & D.1143Hz \\
\end{align}\]

Answer 1

Hint: Doppler Effect is the change in frequency when the position of the observer changes with respect to the source. However, here we are assuming that the velocity of the wave is constant during the interaction. Also, the wave is either approaching the observer or moving away from the observer, only.

Formula used:
$f=\left(\dfrac{c\pm v_{o}}{c\pm v_{s}}\right) f_{0}$, where, $f$ is the apparent frequency of the sound, $f_{0}$ is the actual or real frequency of the sound, $c$ is the speed of the sound wave and $v_{s},v_{o}$ is the speed of the moving source and observer respectively.

Complete step-by-step answer:
Let us consider the given sound wave to travel at a speed $c=350m/s$. Given that the source is moving towards stationary observer at a velocity of \[\text{50m/s}\], then the speed of the observer is $v_{o}=0m/s$, and the speed of the source with respect to the observer is $v_{s}=+50m/s$, as the source is moving towards observer.
Also given that the apparent or observed frequency of the sound wave is $f=1000Hz.$
From Doppler’s law we know $f=\left(\dfrac{c\pm v_{o}}{c\pm v_{s}}\right) f_{0}$
Here, the Doppler’s law will become $f=\left(\dfrac{c}{c-v_{s}}\right) f_{0}$, as the source is moving towards the stationary observer .
Now substituting the values, we get
$1000=\left(\dfrac{350}{350-50}\right)f_{0}$
$\Rightarrow f_{0}=1000\left(\dfrac{300}{350}\right)$
$\Rightarrow f_{0}=875Hz$
Now the if the source moves past the observer, with $f_{0}=875Hz$, then the frequency of the new wave is given as $f\prime =\left(\dfrac{c}{c+v_{s}}\right)f_{0}$
Substituting, we get,
$f\prime=\left(\dfrac{350}{350+50}\right)\times 875=\left(\dfrac{350}{400}\right)\times 875=750Hz$

So, the correct answer is “Option c”.

Note: Doppler law is used only when the speed of the source and the observer are both less than the speed of the sound wave in the medium. Then clearly the relative frequency of a moving source or observer or both, is lesser than the frequency of the sound wave. This law can be extended to any waves.