A train standing in a station yard, blows a whistle of frequency 400 Hz in still air. The wind starts blowing in the direction from the yard to the station with a s speed of `10ms^(-1)` . What are the frequency , wavelength and speed of sound for an observer standing on the station's platform? Is the situation exactly identical to the case when the air is still and the observer runs towards the yard at a speed of `10 ms^(-1)` ? The speed of sound in still can be taken as `340 ms^(-1)`

Explanation

Given `f = 400 Hz, v_(w ) = 10ms^(-1) , v = 340 ms^(-1)`
speed of sound `= v + v_(w) = 340 + 10 = 350 ms^(-1) , lambda = ( v )/( f ) = ( 340)/( 400) = 0.85m`
(i) Apparent frequency of sound `f' = ( fv)/((v + v_(w)))`,
Here`v_(0) = 0` and `v_(a) = 0`
i.e., `f ' = ( 400 xx 340)/( 350) = 388.6 Hz`
i.e.,` f' ~~ = 389 Hz`
New wavelength of sound `lambda'`
`lambda' = ( v + w ) /( f ) = ( 350)/( 400)`
i.e., `lambda' = 0.875m`
Speed of sound as heard by an observer will be `350ms^(-1)` and
`Delta lambda = 0.875 - 0.875 = 0.020 m`.
(ii) When the air is still `v_(w) = 0 `
apparent frequency `f' = f ((v+ v_(0))/(v))`
i.e., `f' = ( 400 xx 350)/( 340) = 411.8 Hz`
i.e., ` f ' ~~ 412 Hz`
New wavelength of sound will be `( v)/( f ) = lambda `
i.e., `lambda ' = lambda `
There will be no change in the wavelength of sound w.r.t. the observer in motion. Relative velocity of sound w.r.t. the observer `= 350 ms^(-1)`