Question

A sound source is falling under gravity. At some time t=0, the detector lies vertically below the source at a depth $H$ as shown. If $v$ is the velocity of sound and $f_0$ is the frequency of the source, then the apparent frequency recorded after $t=2s$ is:

A. $f_0$
B. $\dfrac{f_0(v+2g)}{v}$
C. $\dfrac{f_0(v-2g)}{v}$
D. $\dfrac{f_0 v}{v-2g}$

Answer 1

Try and recall the trend of frequency change with respect to a relative motion between the source and the detector. In other words, we can use the principle of Doppler effect, which states that if a sound wave source is moving towards or away from a detector its frequency as heard by the detector will be greater than or less than the original frequency respectively.
Use this to determine the apparent frequency of the sound wave relative to the detector by first calculating the source velocity at time t=2s and then substituting this in the expression for apparent frequency.

Formula used:
apparent frequency: $f^{\prime} = f_0.\left(\dfrac{v}{v-v_s}\right)$

Complete step-by-step answer:
We can easily deduce the relation between the change in the frequency of the sound as detected by the detector with the motion of the sound source by understanding the Doppler Effect.
The Doppler effect describes the change in the pitch (frequency) of sound waves whenever the source of sound and the detector are in relative motion. It suggests that whenever the source approaches a stationary detector, the apparent frequency (as heard by the detector) will be more than the actual frequency, and whenever the source is receding from a stationary observer, the apparent frequency will be less than the actual frequency.
In this context let us look at the question.
Let the actual frequency of the source be $f_0$ and let the apparent frequency of the sound as heard by the detector be $f^{\prime}$
Let the velocity of the sound wave emitted by the source be $v$
The velocity of the source at any time t can be deduced from the equation of motion:
$v_s = u+at$, where u is the initial velocity, a is the acceleration and t is the time.
In our scenario, the source falls under gravity $\Rightarrow u =0$ and $a=g$
$\Rightarrow v_s=gt$
At $t=2\;s \Rightarrow v_s = 2g$
The apparent frequency as heard by the detector while the source is in motion is given as:
$f^{\prime} = f_0.\left(\dfrac{v}{v-v_s}\right)$
$f^{\prime} = f_0.\left(\dfrac{v}{v-2g}\right)$
Therefore, the correct choice would be D. $\dfrac{f_0 v}{v-2g}$

So, the correct answer is “Option D”.

Note: Remember that the relation for apparent frequency that we have used above is applicable only in situations when only the source is moving towards the detector.
If the source was moving away from the detector, then
$f^{\prime}= f_0.\left(\dfrac{v}{v+v_s}\right)$ since $v_s$ will be taken negative.
If the detector was in motion and the source stationary, then
$f^{\prime} = f_0.\left(\dfrac{v\pm v_s}{v}\right)$, where its ‘+’ for detector moving towards source and ‘-’ for detector moving away from the source.