A bat is flitting about in a cave,navigating | vedclass

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(B) Let $v$ be the speed of sound in air and $f = 40\; kHz$ be the frequency emitted by the bat.The bat moves toward the wall with speed $v_b = 0.03v$

Vedclass · Accepted Answer

$42.5$

Vedclass · Answer

(B) Let $v$ be the speed of sound in air and $f = 40\; kHz$ be the frequency emitted by the bat.The bat moves toward the wall with speed $v_b = 0.03v$.First,the wall acts as a stationary observer receiving the sound. The frequency $f'$ heard by the wall is:$f' = f \left( \frac{v}{v - v_b} ight) = 40 \left( \frac{v}{v - 0.03v} ight) = \frac{40}{0.97}\; kHz$.Next,the wall reflects this sound,acting as a stationary source,and the bat acts as an observer moving toward the source with speed $v_b = 0.03v$. The frequency $f''$ heard by the bat is:$f'' = f' \left( \frac{v + v_b}{v} ight) = \left( \frac{40}{0.97} ight) \left( \frac{v + 0.03v}{v} ight) = \frac{40 	imes 1.03}{0.97} \approx 42.47\; kHz$.Rounding to one decimal place,we get $42.5\; kHz$.

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