Question

A bat flies at a constant speed of 0.04V towards a large tree trunk (V denotes the speed of sound). The bat emits an ultrasonic pulse. The pulse is reflected off the tree and returns to the bat, which can detect and analyse the returning signal. If the returning signal has a frequency of 61kHz, find out the frequency of the original ultrasonic pulse.
A.56 kHz
B. 62 kHz
C. 68 kHz
D. 74 kHz

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.
To this end, first obtain an expression for the apparent frequency of the pulse as heard by the trunk. Then obtain a second expression for the apparent frequency of the reflected pulse as heard by the bat. Substitute the first relation in the second and solve it to arrive at the frequency of the original ultrasonic pulse.

Formula used: Doppler effect for source moving towards stationary observer: $f^{\prime} = f\left(\dfrac{V}{V-v_s}\right)$
Doppler effect for observer moving towards stationary source: $f^{\prime} = f\left(\dfrac{V+v_s}{V}\right)$

Complete step by step answer:
We can easily deduce the relation between the change in the frequency of the sound heard by the observer 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 observer are in relative motion. It suggests that whenever the source approaches a stationary observer, the apparent frequency (as heard by the observer) 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.
We have the bat emitting the ultrasonic pulse as the source and the large tree trunk acts like an obstacle and an observer since it receives the ultrasonic pulse and reflects it off in terms of the relative magnitude of frequency it receives. Thus, there is a relative uniform motion between the bat and the tree trunk, and the bat, which is the source, is approaching the tree trunk.
Let us assume that the frequency of the ultrasonic pulse emitted by the bat be $f\;kHz$
In such as case, the frequency as observed by the tree trunk, accounting for the Doppler effect can be given as:
$f^{\prime} = f\left(\dfrac{V}{V-v_s}\right)$, where V is the speed of light and $v_s$ is the speed of source.
$\Rightarrow f^{\prime} = f\left(\dfrac{V}{V-0.04V}\right) = f\left(\dfrac{V}{0.96V}\right) = \dfrac{f}{0.96}$
Now, when the trunk reflects this pulse with a frequency $61\;kHz$, the trunk acts like a source at rest and the bat acts like an observer in motion, in which case, the frequency as heard by the bat is given as:
$61\;kHz = f^{\prime}\left(\dfrac{V+v_s}{V}\right) = f^{\prime}\left(\dfrac{V+0.04V}{V}\right) = f^{\prime}\left(\dfrac{1.04V}{V}\right) = 1.04f^{\prime}$
Substituting $f^{\prime} = \dfrac{f}{0.96}$, we get:
$61\;kHz = 1.04. \dfrac{f}{0.96} \Rightarrow f = \dfrac{61 \times 0.96}{1.04} = 56.3\;kHz$

So, the correct answer is “Option A”.

Note: Note that the above problem can be solved heuristically devoid of any mathematical evaluation. Since the bat flies towards the tree, the frequency of waves as they hit the tree is higher than the frequency with which they were emitted by the bat. As the waves get reflected off the tree and are detected by a bat which is still flying towards the tree trunk, the frequency shifts higher again. Thus, if the returning signal as detected by the bat is $61\;kHz$, the original pulse must have been emitted at a frequency lower than $61\;kHz$. This also produces the same result since the only option having frequency lower than this frequency is A. $56\;kHz$.