Question

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

Short answer

The relation has been derived.

Step-by-step solution

Write the given information

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

To derive the relation

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.