A boy is standing in front of stationary train. The train blows a horn of $400Hz$ frequency . If the wind is blowing from train to boy at speed at $30m/s$, the apparent frequency of sound heard by the boy will be?
The answer: The frequency remains the same at $400Hz$
MY QUESTION:
Why doesn't the speed of the wind have an effect on the apparent frequency?
The problem is equivalent with considering stationary air and both train and boy moving 30m/s relative to the ground (and air). As they move in the same direction relative to the air, there is no relative motion between them (the observer and source are neither approaching nor receding). So there is no reason to expect a Doppler shift.
The Doppler Shift formula $f'=f(\frac{v \pm v_\text{obs}}{v \mp v_\text{source}})$ only works if the wind (or the medium that the sound is moving in) is constant. Therefore, if the wind is moving at a constant speed, change the reference frame so that the wind is stationary.
In your case, change the reference frame so that it is moving from the train to the boy at $30m/s$. This way, the wind is stationary in the new reference frame. Then, calculate the speed of both the observer and source in the new reference frame: $f'=f(\frac{v \pm v_\text{obs}}{v \mp v_\text{source}})=f(\frac{343\text{m/s} + 30\text{m/s}}{343\text{m/s}+ 30\text{m/s}})=f=400\text{Hz}$
Note: If the windspeed changes, the frequency will be different as the reference frame will be changing all the time.
Because that is the result when you examine the process in detail.
For example:
The boy and the train are in a static relationship. The train could sound its whistle for as long as the power source held out. If the boy received more waves per second than the train produced, where would the extra waves come from?
Or:
The wind is snatching the waves from the train and speeding them up in the direction of the boy, while at the same time increasing the velocity of the wave relative to the ground, stretching out each wave by the same amount. Since:$$f\times \lambda=v$$increasing $\lambda$ and $v$ by the same factor must leave $f$ unchanged...
The formula for apparent frequency as heard by observer when velocity of sound and wind are in same direction is given by
$$n^\prime=\frac{v+v_w-v_o}{v+v_w-v_s}n$$ Where $n$=original frequency
$n^\prime$=apparent frequency
$v$=velocity of sound
$v_w$=velocity of wind
$v_s$=velocity of source of sound
$v_o$=velocity of observer
Since $v_o=0$ & $v_s=0$
$$n'=\frac{v+v_w-0}{v+v_w-0}n=n$$ So it is the same as the original frequency
When you hear a loud music performance from a distance away, the pitch does often noticeably warble up and down a little. This is caused by changes in wind speed affecting the perceived frequency.
Most of these answers are not accounting for the fact that the wind velocity is relative to the source/receiver, which means the equations, while correct, are missing a sign change.
f′=(c+vw−vr)/(c-vw−vs)f
where c is the speed of sound in air (767 mph), vw is positive for the source and negative for the receiver. There is an initial compression of the sound at the source of the horn which remains constant thereafter. So if the horn is emitting a 400Hz tone (f=400), its is shifted by the wind such that:
f'=(767mph+30mph+0)/(767mph-30mph+0)*400 = 432Hz. So the listener hears a constant 432 Hz tone instant of the initial 400Hz tone. This change is generally too low for a person to notice, so the effect is audibly negligible. Also, since the shift is constant, the observer would not be aware unless they were able to directly compare the sound at the source to the sound at their position.
