Consider a source moving towards an observer at the speed of `upsilon_(s)=0.95upsilon` . Deduce the observed frequency if original frequency is 500Hz. What would happen if `upsilon_(s)gtupsilon?` (Jet planes moving faster than sound are common). Here , `upsilon` is velocity of sound in air.

Explanation

As source moves towards observer,
`v^(')=(upsilonv)/(upsilon-upsilon_(s))=(upsilonxx500)/(upsilon-0.95upsilon)=(500upsilon)/(0.05upsilon)=10000Hz`
When `upsilongtupsilon,v^(')` becomes negative which has no physicall meaning. Doppler's formula is not applicable in this case.