frequency of the sound emitted by the the bat, f0 = 35 kHz
speed of sound, v = 343 m/s
speed of the bat , vb = 6 m/s
Here the detector is the bat ( as it hears the echo) and the source is the wall from where the sound gets reflected.
To find the frequency of the reflected sound from the wall, we have to consider the bat (moving towards the wall) as the source and the stationary wall as the observer.
frequency of the reflected sound from the wall, f1 = f0 * v / ( v + vs)
[ Please note that the observer is stationary, the source is moving towards the observer]
here, vs = speed of the source = - vb (as the source is moving towards the observer) = - 6.0 m/s
f1 = (35 kHz) * ( 343 / (343 - 6)) = 35.623 kHz
Frequency of the sound detected by the bat, f2 = f1 * (( v+ vo)/ v) [ Please note that the source is stationary, the observer is moving towards the source]
here, v0 = speed of the observer = vb (as the observer is moving towards the source) = 6.0 m/s
The frequency of the echo that the bat hears , f2 = (35.623) * (( 343 + 6 ) / 343) =36.2461 kHz (Ans)
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