Question

A bat is flitting about in a cave, navigating via ultrasonic beeps. Assume that the sound emission frequency of the bat is 40kHz. During one fast swoop directly toward a flat wall surface. The bat is moving at 0.03 times the speed of sound in air. What frequency does the bat hear reflected off the wall ?

Answer 1

At first we need to find the apparent frequency when the wall is the listener and the bat is acting as a source, in this case we would find the apparent frequency which would help is in finding the apparent frequency when the bat is acting as a listener and the wall is acting like a source.
Formula used: ${n}'=n\left( \dfrac{v\pm {{v}_{o}}}{v\pm {{v}_{s}}} \right)$

Complete step-by-step answer:
We know that apparent frequency is,
${n}'=n\left( \dfrac{v\pm {{v}_{o}}}{v\pm {{v}_{s}}} \right)$ here n is the actual frequency , v is velocity of sound in air,${{v}_{o}}$ is the velocity of the observer, ${{v}_{s}}$is the velocity of the source.
This is because of the Doppler effect,
We are considering the wall as the observer, let us consider that the frequency of that wall is
${{n}_{1}}=40\left( \dfrac{v\pm 0}{v-0.3v} \right)$ , as the observer is stationary the velocity of the observer is zero.
Now as the bat is approaching the wall the sign would be a minus sign as apparent frequency is more than the actual frequency.
${{n}_{1}}=\dfrac{40}{0.97}$,
Now in the second case the bat is acting as the listener and the wall is acting as the source so,
Now let the apparent frequency be,
${{n}_{2}}={{n}_{1}}\left( \dfrac{v+0.3v}{v\pm 0} \right)$,
${{n}_{2}}={{n}_{1}}\left( 1.03 \right)$
So ${{n}_{2}}=42.47KHz$.

Note: We don’t need the velocity of sound in air as, both the velocity will get canceled, we know that whenever the apparent frequency is higher than the actual frequency, the –ve or +ve sign comes. when the frequency of the observer is greater the +ve sign comes when the frequency of the listener is greater the –ve sign comes.