A source of sound having natural frequency f₀ = 1800 Hz is moving uniformly along a line separated from a stationary listener by a distance l = 250 m. The velocity of source is n = 0.80 times the velocity of sound. Find the frequency of sound (in Hz) received by the listener at the moment when the source gets closest to him.

As sound waves take finite time in reaching from one position to other position. The sound waves emitted by the source when it is closest to the listener ( at O) does not reach instantaneously. The frequency heard at this instant should be emitted from some previous position of source. Let this sound ( or disturbance) was created at position 'S'.

The direction of velocity vector of source makes an angle $\mathrm{\theta }$ with the line joining the disturbance position (source position) and observer position as shown in figure. The time in which sound reaches the listener along SL the source reaches point O, closest to the listener. From figure, 

$\cos \mathrm{\theta }=\frac{\mathrm{SO}}{\mathrm{SL}}=\frac{{\mathrm{v}}_{\mathrm{S}}\mathrm{t}}{\mathrm{vt}}=\frac{{\mathrm{v}}_{\mathrm{S}}}{\mathrm{v}}=\mathrm{n}$ ……..(i)

Now using the formula for apparent frequency received by observer 

${\mathrm{f}}_{\text{app }}={\mathrm{f}}_{\text{original }}\left(\frac{\mathrm{v}-{\mathrm{v}}_{\mathrm{o}}}{\mathrm{v}-{\mathrm{v}}_{\mathrm{s}}}\right)$

As Doppler effect is applicable only along the line joining disturbance position (source position) and observer position, hence we should substitute the component of velocity of the source along this line 'SL'.
Hence the frequency heard by the listener is ${\mathrm{f}}_{\mathrm{app}}={\mathrm{f}}_{\mathrm{o}}\left(\frac{\mathrm{v}-0}{\mathrm{v}-{\mathrm{v}}_{\mathrm{S}}\cos \mathrm{\theta }}\right)$  ………(ii)

$\Rightarrow {\mathrm{f}}_{\mathrm{app}}={\mathrm{f}}_{\mathrm{o}}\left(\frac{\mathrm{v}}{\mathrm{v}-\left(\mathrm{nv}\right)\mathrm{n}}\right)=\frac{{\mathrm{f}}_{\mathrm{S}}}{1-{\mathrm{n}}^2}$   …….(iii)

O n substituting numerical values, we get ${\mathrm{f}}_{\mathrm{app}}=5000\mathrm{Hz}$