Question

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 at a speed of 10 m s–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 m s–1? The speed of sound in still air can be taken as 340 m s–1.

Answer 1

The frequency of sound heard by the observer will be same as that produced by the source due to there is no relative motion .
Hence, f = 400 Hz
Effective speed of sound ( Ve) = Vsound + Vw = 340 + 10 = 350 m/s

Wavelength = speed of wave/frequency = 350/400 = 0.875 m

When observer starts running in still air , then there is a relative motion between the observer the source with respect to medium ,
Here,
Vm { speed of observer } = 0
Velocity of observer ( Vo) = 10 m/s , Vsource = 0

Now, use Doppler's effect of sound
Apparent frequency = f × {( V + Vo)/V}
= 400 × { (340 + 10)/340 }
= 411.76 Hz
As source is rest , wavelength doesn't change i.e
Wavelength = 0.875 m
Speed of sound = V + Vm
= 340 + 0 = 340 m/s

Obviously the situations in two cases are entirely different .

Answer 2

For the stationery observer:400Hz;0.875m;350m/s

for the running observer:not exactly identical

for the stationery observer:

frequency of the sound produced by the whistle,v=400 Hz

speed of sound=340m/s

velocity of the wind,v=10m/s

as there is no relative motion between the source and observer the frequency of the sound heard by the observer will be the same as that produced by the source,i.e,400Hz

the wind is blowing towards the observer hence the effective speed of sound increases by 10 units.

effective speed of the sound,ve=340+10=350m/s

the wavelength of the sound heard by the observer is given by the relation:

wavelengt=ve/v=350/400=0.875 m

for the running observer:

velocity of the observer,v0=10m/s


the observer is moving toward the source as result of the relative motion of the source and your server there is change in frequency.

This is given by the relation:

v1=(v+v0/v)v

=(340+10/340)×400=411.76 Hz

since the air is still the effective speed of sound =340+0=340m/s

the source is at rest in the wavelength of the sound will not change remains 0.875 m.

hence the given 2 situations are not exactly identical.



hope its helps