Answer: The beat frequency detect by the woman is f0 [ 2v / ( v - vW) ]
Step-by-step explanation:
given a Doppler effect,
the frequency heard by the listener in terms of frequency of source is expressed as;
fL = [ (v + vL) / (v + vS) ] fS
vL is the velocity of the listener, vS is the velocity of the source and V is the velocity of sound.
Now What beat frequency does the woman detect (in terms of the given quantities)?
The wall will receive and reflect pulses at a frequency of;
fWALL =( v / (v - v0)) f0
so the woman will hear this wave;
fWOMAN = ((v + vW) / v) ( v / (v - vW)) f0
fWOMAN = (( v + vW) / ( v - vW)) f0
Now the beat frequency will be;
fBEAT = ((v + vW) / (v - vW)) f0 - f0
fBEAT= f0 [ ((v + vW) / (v - vW)) - 1 ]
fBEAT= f0 [ 2v / ( v - vW) ]
The beat frequency detect by the woman is f0 [ 2v / ( v - vW) ]