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

A train standing in a station-yard, blows a whistle of frequency 400Hz

1.	A train standing in a station-yard, blows a whistle of frequency 400Hz in still air. The wind starts blowing in the direction from the yard to the station with a speed of 10 ms^(-1). What are the frequency wavelength and speed of sound for an observer standing on the station's platform? Is the situation exactly identical 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 air can be taken as 340 ms^(-1)
Answer» SOLUTION : According to formula `(f _(L))/(V.v _(L)) = (f _(S))/(v .+ v _(S))` Here v.= resultant velocity of sound, `v +v_(W) = 340 + 10=350 m//s` `therefore (f _(L))/(350 +0) = (f_(S))/(350 +0) implies f _(L) =f _(S) = 400 Hz` (This is so because here in this cae, there is no relative motion between source and listener). Now, we have the formua, `v. =f _(S) lamda . implies lamda. = (v.)/(f_(S)) = (350)/(400) =0.875m` According to formula, `(f._(L))/(v +v _(L)) = (f _(S))/(v + v _(S))` `therefore (f._(L))/(340 + 10) = (400)/(340 + 0) implies f._(L) = (350 XX 400)/(340)` `therefore f._(L) =411.76 Hz` Now, here in this case, `c = f _(S) lamda implies lamda = (v)/(f _(S)) = (340)/(400) = 0.85 m` THUS, SITUATIONS (i) and (ii) are different. They are not equivalent.