Obtain the equation of frequency observed by an observer for a moving source and a moving observer at different velocities.

(N/A) Let us take the direction from the observer to the source as the positive direction.
Let the source and the observer be moving with velocities $v_{s}$ and $v_{0}$ respectively as shown in the figure.
Suppose at time $t=0$,the observer is at $O_{1}$ and the source is at $S_{1}$,with $O_{1}$ being to the left of $S_{1}$.
The source emits a wave of velocity $v$,frequency $\nu$,and period $T_{0}$,all measured by an observer at rest with respect to the medium.
Let $L$ be the distance between $O_{1}$ and $S_{1}$ at $t=0$,when the source emits the first crest. Since the observer is moving,the velocity of the wave relative to the observer is $v+v_{0}$. Therefore,the first crest reaches the observer at time $t_{1}=\frac{L}{v+v_{0}}$.
At time $t=T_{0}$,both the observer and the source have moved to their new positions $O_{2}$ and $S_{2}$ respectively. The new distance between the observer and the source $O_{2}S_{2}$ would be $L+(v_{s}-v_{0})T_{0}$.
At $S_{2}$,the source emits a second crest. This reaches the observer at time $t_{2}=T_{0}+\frac{L+(v_{s}-v_{0})T_{0}}{v+v_{0}}$.
At time $nT_{0}$,the source emits its $(n+1)^{th}$ crest,and this reaches the observer at time $t_{n+1}=nT_{0}+\frac{L+n(v_{s}-v_{0})T_{0}}{v+v_{0}}$.
The time interval between the arrival of the first and $(n+1)^{th}$ crest is $\Delta t = t_{n+1} - t_{1} = nT_{0} + \frac{n(v_{s}-v_{0})T_{0}}{v+v_{0}} = nT_{0} \left(1 + \frac{v_{s}-v_{0}}{v+v_{0}}\right) = nT_{0} \left(\frac{v+v_{0}+v_{s}-v_{0}}{v+v_{0}}\right) = nT_{0} \left(\frac{v+v_{s}}{v+v_{0}}\right)$.
The observed period is $T' = \frac{\Delta t}{n} = T_{0} \left(\frac{v+v_{s}}{v+v_{0}}\right)$.
Thus,the observed frequency is $\nu' = \frac{1}{T'} = \nu \left(\frac{v+v_{0}}{v+v_{s}}\right)$.