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A bat is flitting about in a cave, navigating via ultrasonic beeps. Assume that the sound emission frequency of the bat is $40 \mathrm{kHz}$. 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?
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Frequency of sound emitted by the bat, $v=40 \mathrm{kHz}$.
Velocity of bat (source), $v=0.03 v_s$
[ $v_s=$ velocity of sound $]$
Apparent frequency striking the wall,
$v^{\prime}=\frac{v}{v-v_s} v=\frac{v}{v-0.03 v} \times 40 \times 10^3$
$=\frac{40 \times 10^3}{0.97}$
This is reflected by the wall and is received by the bat.
$$
\therefore \quad v_s=0, v_B=0.03 \mathrm{v}
$$
$\therefore \quad$ Apparent frequency (bat hear)
$$
\begin{aligned}
&=\frac{v+v_B}{v} v^{\prime}=\frac{(v+0.03 v)}{v} \times \frac{40 \times 10^3}{0.97} \\
&=42.47 \times 10^3 \mathrm{~Hz}=42.47 \mathrm{kHz} .
\end{aligned}
$$
Velocity of bat (source), $v=0.03 v_s$
[ $v_s=$ velocity of sound $]$
Apparent frequency striking the wall,
$v^{\prime}=\frac{v}{v-v_s} v=\frac{v}{v-0.03 v} \times 40 \times 10^3$
$=\frac{40 \times 10^3}{0.97}$
This is reflected by the wall and is received by the bat.
$$
\therefore \quad v_s=0, v_B=0.03 \mathrm{v}
$$
$\therefore \quad$ Apparent frequency (bat hear)
$$
\begin{aligned}
&=\frac{v+v_B}{v} v^{\prime}=\frac{(v+0.03 v)}{v} \times \frac{40 \times 10^3}{0.97} \\
&=42.47 \times 10^3 \mathrm{~Hz}=42.47 \mathrm{kHz} .
\end{aligned}
$$
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