Can we apply Doppler effect to a source of sound moving faster than the velocity of sound?

Explanation

The Doppler effect is the change in frequency or wavelength of a wave for an observer who is moving relative to the wave source. 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 figure Suppose at time t=0, the observer is at `O_(1)` annd the source is at `S_(1),O_(1)` being to the left of `S_(1)`. the source emits a wave of velocity v, of frequency v and period `T_(0)` all measured by ann observer at rest with respect of the medium.
Let L be the distance betweenn `O_(1) and S_(1)` at t=0, when the source emits the first crest.
Now, 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)=L//(v+v_(0))`.
at time `t=T_(0)`, both the observer and the source have moved to their new position `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)+[L+(v_(S)-v_(0))T_(0)]//(v+v_(0))`
this reaches the observer at time
`t_(n+1)=nT_(0)+[L+n(v_(S)-v_(0))T_(0)]//(v+v_(0))`
Hence in a time interval `t_(n+1)-t_(n),` i.e.,
`piT_(0)+(L+n(v_(S)-v_(0))T_(0)]//(v+v_(0))-L//(v+v_(0))`.
The observer counts of the wave as equal to T given by `T=T_(0)(1+(v_(S)-v_(0))/((v+v_(0)))=T_(0)((v+v_(0))/(v+V_(0)))`
The frequency v observed by the observer is given by `v=v_(0)((v+v_(0))/(v+v_(S)))`
(i) When both observer and source are approaching each other, the source is moving in negative direction
`therefore v=v_(0)((v+v_(0))/(v-v_(S)))`
(ii) When both source and observer are moving away from each other, the observer is moving in negative direction