Question

 An observer travelling with a constant velocity of 20ms^-1, passes close to a stationary source of sound and notices that there is a change of frequency of 50 Hz as he passes the source. What is the frequency of the source? 

Answer 1

The source is stationary so $v_s =0$ while the observer/listener's velocity is $v_L=20\ m/s$ .

What is given is the change in frequency which is 50 Hz, this means $f_L=f_s+50\ Hz.$ Thus, from

$f_L={f_s(v+v_L) \over (v + v_s)}$ we obtain

$f_S + 50Hz = \dfrac{v+v_L}{v}f_S\ \ \ \Rightarrow\ \ \ 50Hz= (\dfrac{v+v_L}{v}-1)f_S,\\f_S =(50Hz)(\dfrac{v+v_L}{v}-1)^{-1}=\ \ \ (50Hz)(\dfrac{340m/s+20m/s}{340m/s}-1)^{-1}=850\ Hz$

Therefore, the frequency of the sound emitted by the source is 850 Hz.