Show that the Doppler frequency f- of a receding source is $f_{-}=f_o\left(1+\frac{v_s}{v}\right)$

The actual frequency of the source moving away from the listener is given by

$f_{-}=f_o\left(1+\frac{v_s}{v}\right)$

$f_o$is the frequency of the source observed by the observer (Apparent frequency)

$v_s$is the speed of the source moving away from the observer

$v$is the actual speed of the sound

Consider that the listener is stationary and the source of sound is moving away from the listener The speed of the source is v_s.

In the given time period, the distance covered by the sound waves is

$∆x=v_s{t}_{-}$

Let,

$t_{-}$= the actual time taken by the wave to reach the observer

localid="1649864787328" ${\lambda }_{-}$=actual wavelength of the sound wave

$f_{-}$= the actual frequency of the sound wave

$\lambda$= the wavelength of the sound wave of the observer (sound source is moving)

$t_o$= time taken by the wave from the moving source to reach the observer (apparent time)

Since, the wavelength is given by the product of velocity and time,

localid="1649865637562" $$\begin{gathered} \lambda ={\lambda }_{-}+∆x \\ v{t}_o=v{t}_{-}+v_s{t}_{-} \\ \frac{v}{f_o}=\frac{v}{f_{-}}+\frac{v_s}{f_{-}} \\ \frac{v}{f_o}=\frac{v+v_s}{f_{-}} \\ {f}_o=f_{-}\left(\frac{v}{v+v_s}\right) \\ {f}_{-}=f_o\left(1+\frac{v_s}{v}\right) \end{gathered}$$
Thus, this is the required result.