A source of sound attached to the bob of a simple pendulum execute $$SHM.$$  The difference between the apparent frequency of sound as received by an observer during its approach and recession at the mean position of the $$SHM$$  motion is $$2\% $$  of the natural frequency of the source. The velocity of the source at the mean position is (velocity of sound in the air is $$340\,m/s$$  )
[Assume velocity of sound source << velocity of sound in air]

Solution :
$$\eqalign{ & {f_1} = {f_0}\left( {\frac{{{V_0}}}{{{V_0} - V}}} \right){f_2} = {f_0}\left( {\frac{{{V_0}}}{{{V_0} + V}}} \right) \cr & {f_1} - {f_2} = {f_0}{V_0}\left( {\frac{1}{{{V_0} - V}} - \frac{1}{{{V_0} + V}}} \right) \cr & = {f_0}{V_0}\left( {\frac{{{V_0} + V - {V_0} + V}}{{V_0^2 - {V^2}}}} \right) \cr & = {f_0}{V_0} \times \frac{{2V}}{{V_0^2}} = {f_0}\frac{{2V}}{{{V_0}}} \cr & {\text{Given}}\,\,\frac{{2V{f_0}}}{{{V_0}}} = 0.02 \times {f_0} \Rightarrow V = 0.01\,{V_0} = 3.4\,m/s \cr} $$