Posted by Elisa on Monday, August 4, 2008 at 11:04pm.
The answer depends upon whether the bat is flying towards the wall or away from it (or parallel to it). The problem should have stated which it is.
If flying away or towards the wall, there are TWO Doppler shifts involved. The wall "receives" and reflects a Doppler-shifted sound, and when the bat hears it, it is shifted again.
Do you need a numerical answer? If the bat is flying towards the wall,
f' = f*[V+Vs) / V]
f' = unknown, f = 35 kHz, V = 343 m/s (in dry air, 20 deg C), Vs = (5.0*2 = 10 m/s) <--Based on drwls</> suggestion, the source moves with a velocity of 5.0 m/s toward the wall, but then the bat continues to move at 5.0 m/s toward the reflected waves, doubling the speed between bat the source</> and bat the observer to 10 m/s </>.
If the bat is flying away from the wall,
f' = f*[V-Vs) / V]. Use the same values as above.
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