Obtain an expression for the apparent frequency of sound for a source moving away and towards a stationary observer.

Explanation

Let `S_(1)` be the position of the source at t= 0. Let L be the distance `b//w` the observed and the source at S,
Let v be the velocity of sound in air.
Time taken for sound ( say crest ) to reach the observer
`t_(1) =( L)/( v)` ......(1)
Let the source travel further from` S_(1) `,and occupy the position `S_(2)` away from a stationary observer.
distance `S_(1)S_(2) = v_(s) T_(0)`
For a crest at `S_(2)` to reach the observer, it takes the time `t_(2) = T_(0) + (L+v_(s) T_(0))/( v)` ....(2)
At time nT, the source emits its `(n+1)^(th)` crest and this reaches the observer at time.
`t_(n+1) = n T_(0) + ( L+ v_(s) ( nT_(0)))/(v_(0))`
Time of observation of n crests by an observer will be nT.
`nT = ( nT_(0) + ( L + nv_(s) T_(0))/(v))-((L)/( v))`
Time for I crest = `( nT_(0) + ( Lnv_(s) T_(0))/( v) - ( L)/( v)) + n `
`T = ( nvT_(0)+L+nv_(s) T_(0)-L)/(nv)`
i.e, `T= T_(0)+ (v_(s) T_(0))/( v)`

or `T = T_(0) ((v+ v_(s))/(v))`
Apparent frequency of sound heard by a stationary observer for a source of sound moving away from him.
`f = f_(0)((v)/(v+v_(s)))`
and apparent wavelength `lambda =( v+ v_(s))/(f_(0))` , change ( increase ) in wavelength.
`Deltalambda = ( v+ v_(s))/( f_(0))- ( v)/(f_(0)) = ((1)/( f_(0))) ( v+ v_(s)-v)= (v_(s)) /(f_(0))`
Percentage increase in `lambda = ( Delta lambda)/( lambda ) xx 100 = ( v_(s))/( f_(0) lambda)xx100`
For a source moving towards the observer , change `v_(s)` to `- v_(s)`
Then `f= f_(0) ((v)/(v-v_(s)))`
Apparent wavelength `=( v- v_(s))/( f_(0))`, here `lambda lt lambda_(0)`