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Question: A bat moving at \(10\,m{s^{ - 1}}\) towards a wall sends a sound signal of \(8000\,Hz\) towards it. ...

A bat moving at 10 ms−110\,m{s^{ - 1}} towards a wall sends a sound signal of 8000 Hz8000\,Hz towards it. On reflection it hears a sound of frequency ff . The value of ff in HzHz is close to (speed of sound =320 ms−1 = 320\,m{s^{ - 1}} )
A. 82588258
B. 80008000
C. 85168516
D. 84248424

Explanation

Solution

Here we have to use the principle of Doppler Effect of sound.Doppler Effect of sound: The Doppler effect is the difference in a wave’s frequency in response to an observer that shifts relative to the source of the wave. The change of pitch heard as a vehicle with a horn approaches and retreats from an observer is a typical example of Doppler Effect.

Complete step by step answer:
Given, Speed of the bat vs=10ms−1{v _s} = 10m{s^{ - 1}}
We know that,
Speed of sound in air v=320ms−1v = 320m{s^{ - 1}}
Let f∘{f_ \circ } be the frequency of the sound the bat sends towards the wall.
f∘=8000 Hz{f_ \circ } = 8000\,Hz
After reflection the frequency of the sound is ff .
f=f∘(v+vsv−vs) ⇒f=(320+10320−10)×8000 ⇒f=330310×800 ∴f=8516 Hz  f = {f_ \circ }\left( {\dfrac{{v + {v_s}}}{{v - {v_s}}}} \right) \\\ \Rightarrow f = \left( {\dfrac{{320 + 10}}{{320 - 10}}} \right) \times 8000 \\\ \Rightarrow f = \dfrac{{330}}{{310}} \times 800 \\\ \therefore f = 8516\,Hz \\\
Thus, the value of ff in HzHz is close to 85168516. Hence, option C is correct.

Additional information:
The Doppler Effect is caused when waves are transmitted out at a normal rate or frequency by the source of a waveform, such as sound or light, but there is continuous relative motion between the source and the observer, allowing frequency measured to change.
The explanation for Doppler Effect is that each subsequent wave crest is produced from a position closer to the observer than the crest of the previous wave as the wave source travels towards the observer. Thus, each wave takes significantly less time than the previous wave to enter the observer.

Note: Here we have to see whether the sound wave of the bat is decreasing from the wall or not. Then only we can apply the Doppler effect. Also we have to correctly remember the formula of the Doppler Effect otherwise the answer will be wrong.