Question

A bat emitting an ultrasonic wave of frequency \[4.5 \times {10^4}\,{\text{Hz}}\] at speed of \[6\,{\text{m}}\,{{\text{s}}^{ - 1}}\] between two parallel walls. The frequencies heard by the bat will be:
A. \[4.67 \times {10^4}\,{\text{Hz}}\], \[4.34 \times {10^4}\,{\text{Hz}}\]
B. \[4.34 \times {10^4}\,{\text{Hz}}\], \[4.67 \times {10^4}\,{\text{Hz}}\]
C. \[4.5 \times {10^4}\,{\text{Hz}}\], \[5.4 \times {10^4}\,{\text{Hz}}\]
D. \[4.67 \times {10^3}\,{\text{Hz}}\], \[4.34 \times {10^4}\,{\text{Hz}}\]

Answer 1

Hint: First of all, we will use the doppler’s effect for the two different situations. One is for the source moving towards the wall and the other one is for the source moving away from the wall. We will substitute the required values and manipulate accordingly to obtain the answer.

Formula used:
The apparent frequency is given by the formula, when the we apply the doppler’s effect ahead of the source:
\[{f_1} = f \times \left( {\dfrac{{v + u}}{{v - u}}} \right)\] …… (1)
The apparent frequency is given by the formula, when the we apply the doppler’s effect ahead of the source:
\[{f_2} = f \times \left( {\dfrac{{v - u}}{{v + u}}} \right)\] …… (2)
Where,
\[{f_2}\] and \[{f_1}\] are the apparent frequencies of the ultrasonic sound after reflection from the second and the first wall.
\[v\] indicates the velocity of the sound in air.
\[u\] indicates the velocity of the ultrasonic wave.
\[f\] indicates the original frequency of the ultrasonic wave.

Complete step by step answer:
In the given problem, we are supplied with the following data:
A bat emits an ultrasonic wavelength whose frequency is \[4.5 \times {10^4}\,{\text{Hz}}\].
The speed of the wave is given as \[6\,{\text{m}}\,{{\text{s}}^{ - 1}}\] between the two walls.
We are required to find the frequencies as heard by the bat.
To begin with, we will need to apply doppler’s effect on the above problem.
In this case as we can see, there is a wall in the front of the bat as well as a wall behind the bat.
We will have to apply the doppler’s effect for behind and ahead of the source.

The apparent frequency is given by the formula, when the we apply the doppler’s effect ahead of the source:
${f_1} = f \times \left( {\dfrac{{v + u}}{{v - u}}} \right) \\
\Rightarrow {f_1} = 4.5 \times {10^4} \times \left( {\dfrac{{330 + 6}}{{330 - 6}}} \right) \\
\Rightarrow {f_1} = 4.5 \times {10^4} \times \dfrac{{28}}{{27}} \\
\Rightarrow {f_1} = 4.67 \times {10^4}\,{\text{Hz}} \\$

The apparent frequency is given by the formula, when the we apply the doppler’s effect ahead of the source:
${f_2} = f \times \left( {\dfrac{{v - u}}{{v + u}}} \right) \\
\Rightarrow {f_2} = 4.5 \times {10^4} \times \left( {\dfrac{{330 - 6}}{{330 + 6}}} \right) \\
\Rightarrow {f_2} = 4.5 \times {10^4} \times \dfrac{{27}}{{28}} \\
\Rightarrow {f_2} = 4.34 \times {10^4}\,{\text{Hz}} \\$
Hence, the two frequencies that will be heard are \[4.67 \times {10^4}\,{\text{Hz}}\] and \[4.34 \times {10^4}\,{\text{Hz}}\].

So, the correct answer is “Option A”.

Note:
While solving the problem, most of the students seem to have some confusion regarding the formulas needed to be used for the two different situations. It is important to note that frequency of the wave ahead of the source is associated with the situation when the source approaches the wall at rest and the frequency gets increased. The frequency gets increased when the source moves away from the wall.