Question

A source of sound $A$ emit waves of frequency $1800\ \text{Hz}$ is falling towards ground with a terminal speed $v$. The observer $B$ on the ground directly beneath the source receives waves of frequency $2150\ \text{Hz}$. The source $A$ receive waves, reflected from ground, of frequency nearly:
(Speed of sound $=343\ \text{m}\cdot {{\text{s}}^{-1}}$)
A. $2150\ \text{Hz}$
B. $2500\ \text{Hz}$
C. $1800\ \text{Hz}$
D. $2400\ \text{Hz}$

Answer 1

Hint: Treat the problem as two separate problems. Determine the velocity of the source using the Doppler effect of sound. Then assume the observer as source for the second part and determine the frequency of sound heard by $A$.
Formula Used:
$f={{f}_{0}}\left( \dfrac{v\pm {{v}_{0}}}{v\mp {{v}_{s}}} \right)$

Complete answer:
Doppler effect is the phenomenon observed whenever there is a relative motion between the source emitting waves and an observer. The general equation for the Doppler Effect can be seen as,
$f={{f}_{0}}\left( \dfrac{v\pm {{v}_{0}}}{v\mp {{v}_{s}}} \right)$
Here, $f$ is the frequency received by the observer, ${{f}_{0}}$ is the frequency emitted by the source, $v$ is the velocity of sound waves, ${{v}_{0}}$ is the velocity of observer and ${{v}_{s}}$ is the velocity of source.
Now, model the equation for the observer at rest exactly beneath the source. The equation for source approaching the observer can be written as,
$f={{f}_{0}}\left( \dfrac{v}{v-{{v}_{s}}} \right)$
Substitute the values as $f=2150\ \text{Hz}$ and ${{f}_{0}}=1800\ \text{Hz}$. The source velocity is calculated as,
$\begin{align}
  & 2150=1800\left( \dfrac{343}{343-v} \right) \\
 & 343-v=\dfrac{1800\times 343}{2150} \\
 & v=56\ \text{m}\cdot {{\text{s}}^{-1}}
\end{align}$
Now, since the waves reflect from the ground, we can treat the ground as a new source that emits the waves of frequency $2150\ \text{Hz}$. Thus, the source A can now be treated as an observer. The frequency of waves received by it can be calculated as,
$\begin{align}
  & f={{f}_{0}}\left( \dfrac{v+{{v}_{s}}}{v} \right) \\
 & =2150\left( \dfrac{343+56}{343} \right) \\
 & =2500\ \text{Hz}
\end{align}$
Thus, the frequency observed by source A is $2500\ \text{Hz}$.

Thus, the correct option is (A).

Note:
Use the general formula for doppler effect carefully. Take care of the signs in the formula for source velocity and observer velocity. Use the speed of sound as provided in the question.