Why doesn't the speed of the wind have an effect on the apparent frequency?

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?

• It does have an effect when the wind speed is changing. For example, if there is a surge of air from the train toward the boy while the horn is already blaring, then for a time there will be a high frequency shift, until the "buffered" (appalling pun in this context!) waves have cleared from the air, whereupon the frequency will return to normal but with a change in the transfer time in each direction. If the boy and the train were exchanging signals steadily in a round robin fashion, each would observe identical frequencies but an overall slowing of the exchange process like in SR. Jan 25 '18 at 3:54

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 $$f'=f(\frac{v \pm v_{obs}}{v \mp v_{source}})$$ 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: $$f'=f(\frac{v \pm v_{obs}}{v \mp v_{source}})=f(\frac{343m/s + 30m/s}{343+ 30m/s})=f=400Hz$$

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...

• Would a huge and sudden shift in wind speed cause a temporary apparent frequency change then? May 7 '15 at 14:11
• @Mindwin regardless what the theory says about that, I'd recommend not doing the experiment to test it. It's likely very unsafe
– Jim
Apr 7 '17 at 13:17
• @jim Oh, thanks for your advice, i was about to put my firstborn in a wind tunnel together with a speeding train. For science... XD Apr 7 '17 at 13:57

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

• This is only the case in a steady state with a fixed windspeed, however. When the windspeed changes, the frequency is indeed upset, and you'd need to employ a soundlike version of General Relativity (with the air as the inertial frame) to describe the effects. Jan 25 '18 at 4:02