Obtain the equation for the frequency observed by a stationary observer when the source is moving.

(N/A) For this,we use the following sign convention: the direction from the observer to the source is taken as the positive direction of velocity.
Consider a source $S$ moving with velocity $v_{s}$ and an observer who is stationary in a frame where the medium is also at rest.
Let the speed of a wave of angular frequency $\omega$ and period $T_{0}$,both measured by an observer at rest with respect to the medium,be $v$.
As shown in the figure,at time $t=0$,the source is at point $S_{1}$,located at a distance $L$ from the observer,and emits a crest. This reaches the observer at time $t_{1} = \frac{L}{v} \quad \dots (1)$
At time $t=T_{0}$,the source has moved a distance $v_{s}T_{0}$ and is at point $S_{2}$,located at a distance $L + v_{s}T_{0}$ from the observer. At $S_{2}$,the source emits a second crest. This reaches the observer at time $t_{2} = T_{0} + \frac{L + v_{s}T_{0}}{v}$.
At time $n T_{0}$,the source emits its $(n+1)^{th}$ crest,which reaches the observer at time $t_{n+1} = n T_{0} + \frac{L + n v_{s}T_{0}}{v} \quad \dots (2)$.
The time interval between the arrival of the $(n+1)^{th}$ crest and the first crest at the observer is $\Delta t = t_{n+1} - t_{1} = n T_{0} + \frac{L + n v_{s}T_{0}}{v} - \frac{L}{v} = n T_{0} + \frac{n v_{s}T_{0}}{v} = n T_{0} \left( 1 + \frac{v_{s}}{v} \right)$.
The observed period $T$ is the time interval between consecutive crests,so $T = \frac{\Delta t}{n} = T_{0} \left( 1 + \frac{v_{s}}{v} \right) = T_{0} \left( \frac{v + v_{s}}{v} \right)$.
Since frequency $\nu = \frac{1}{T}$ and $\nu_{0} = \frac{1}{T_{0}}$,the observed frequency is $\nu = \nu_{0} \left( \frac{v}{v + v_{s}} \right)$.