A bat flies at a steady speed of 4 ,ms^{-1} e | vedclass

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(C) The bat acts as both the source and the observer moving towards a stationary wall.First, the frequency $f'$ received by the wall is given by the D

Vedclass · Accepted Answer

$92.1 	imes 10^{3} \,Hz$

Vedclass · Answer

(C) The bat acts as both the source and the observer moving towards a stationary wall.First, the frequency $f'$ received by the wall is given by the Doppler effect formula for a moving source and stationary observer:$f' = f \left( \frac{v}{v - v_b} ight)$where $v = 330 \,ms^{-1}$ is the speed of sound and $v_b = 4 \,ms^{-1}$ is the speed of the bat.Next, the wall reflects this sound, and the bat acts as a moving observer receiving the reflected frequency $f''$ from a stationary source (the wall):$f'' = f' \left( \frac{v + v_b}{v} ight)$Substituting $f'$ into the equation for $f''$:$f'' = f \left( \frac{v}{v - v_b} ight) \left( \frac{v + v_b}{v} ight) = f \left( \frac{v + v_b}{v - v_b} ight)$$f'' = 90 	imes 10^{3} \left( \frac{330 + 4}{330 - 4} ight) = 90 	imes 10^{3} \left( \frac{334}{326} ight)$$f'' \approx 90 	imes 10^{3} 	imes 1.0245 = 92.21 	imes 10^{3} \,Hz$.Rounding to the nearest provided option, the frequency is $92.1 	imes 10^{3} \,Hz$.

$A$ bat flies at a steady speed of $4 \,ms^{-1}$ emitting a sound of $f = 90 \times 10^{3} \,Hz$. It is flying horizontally towards a vertical wall. The frequency of the reflected sound as detected by the bat will be (take velocity of sound in air is $330 \,ms^{-1}$)

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