Figure shows wave fronts coming form a source at equal interval of 103 milli second Frequency of wave - Physics - Wave Optics

Question

Figure shows wave fronts coming form a source at equal interval of
10 3 milli second
Frequency of wave emitted by source is 300 Hz XX' is line along
common diameter of wave fronts passing through 'A' Distance between
two consecutive wave fronts along line XX' is 0 9 m and 1 3 m in
the right and left of point A respectively
At some instant source is at point A Detector is placed at point
'Q' AQ makes an angle 60° with line XX' If frequency of wave
received by detector is f Hz, then f 110 (in Hz) is equal
to

Sandeep K. Gupta · Accepted Answer

Dear Student,

It's a case of Doppler effect when vs<v
 Here v is velocity of sound and vs is velocity of source
Because time interval between two wave front is given hence frequency of source will bef =1T =1103×10-3=300 Hz
 (also given in the query)Because distance between two wave front will be λap
 Consider left part,                        λap=v+vsf=1
3 m  ------(1)For right part-                             λap=v-vsf=0
9 m     ----  -(2)For the source at Q apparant frequency will befap=fvv-vscos600  ---------(3) from equation (1) and (2)v+vsv-vs=1
30 9    ⇒vs=4 v25From eq
 (3)fap=fvv-4v25×12=f 2523=300×2523=326 HzSo, fap110=326110≈3 Hz
Regards