doppler

[Raiden2012]

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The source of a sound wave is stationary. The observer is moving towards the source. There is a steady wind blowing from the observer to the source. How does the wind change the observed frequency?

Increase frequency
Decrease frequency
No change


My answer was decrease. Based on TPR, my (flawed?) reasoning was that the wind was helping to move the observer closer to the sound, increasing frequency and moving the source away, decreasing frequency. I took some random numbers
(speed of sound + speed of observer)/(speed of sound) = 350/340
Adding speed of wind as 10 m/s
(speed of sound + speed of observer + speed of wind) / (speed of sound + speed of wind) = 360/350

So the latter decreased, which resulted in me choosing decreased observed frequency


Can anyone explain why the observed frequency increases?

 

[shaboobly]

probably because the wind is increasing the observer's velocity causing the frequency to appear higher.

 

[Raiden2012]

Thanks but the wind is also blowing from the observer to the detector so the detector also feels the wind and is "blown away from the oncoming observer"

Contrast with a similar question
If both source and observer are stationary and the wind is blowing from the observer to the source, there will be NO doppler effect

 

[Raiden2012]

Or does the wind only have an effect on a moving observer/detector?

 

[milski]

Your formula is slightly incorrect - for wind blowing towards the source, you have to subtract the wind velocity, not add it.

The full formula is f/f0=(c+Vr-Vw)/(c-Vw)=1+Vr/(c-Vw) (for no wind you can substitute 0 for Vw)

Since c-Vw < c, Vr/(c-Vw)>Vr/c and the frequency increases as the speed of wind increases.

Just to clarify a bit:
The velocity of the observer, with respect to the media is Vr-Vw (since the wind is blowing in the same direction as its movement).
The velocity of the transmitter is Vs+Vw=Vw (it's stationary and the wind is blowing towards it).

The general formula is f=(c+Vr)/(c-Vs)f0

 

[Raiden2012]

Your formula is slightly incorrect - for wind blowing towards the source, you have to subtract the wind velocity, not add it.

The full formula is f/f0=(c+Vr-Vw)/(c-Vw)=1+Vr/(c-Vw) (for no wind you can substitute 0 for Vw)

Since c-Vw < c, Vr/(c-Vw)>Vr/c and the frequency increases as the speed of wind increases.

Just to clarify a bit:
The velocity of the observer, with respect to the media is Vr-Vw (since the wind is blowing in the same direction as its movement).
The velocity of the transmitter is Vs+Vw=Vw (it's stationary and the wind is blowing towards it).

The general formula is f=(c+Vr)/(c-Vs)f0

Thanks milski, you're my physics guru. Anyway, I understand the math but I still have difficulty understanding why we have to subtract wind velocity from both numerator and denominator.

I reasoned that the wind velocity would increase the detector/observer's velocity, hence making it move faster towards the detector. Hence, my addition to the numerator that would increase the perceived frequency.

Likewise the wind would push the source/transmitter away from the detector/observer so I added velocity to the source in the denominator. This would decrease the perceived frequency.

I can deal with velocities of observers and source but when velocity of something else is introduced, I'm 😕

 

[milski]

We need to agree on nomenclature first. For me, observer and detector are the same things. Let's talk about source and receiver, so that there is no confusion.

You will have to subtract the wind speed from the speed of the receiver - the receiver is moving in the same direction as the media, so its speed with respect to the media itself is lower.That gets you Vr-Vw in the enumerator. That should make sense, since if the receiver started moving almost as fast as the media, he will need a lot of time to go between each two waves, resulting in a very low frequency.

You have to subtract the wind speed from the source speed because what's happening is equivalent to the transmitter moving towards the receiver through the media. If the speed of the wind was really high, you'll have each wave peak almost blown back to the next one, resulting in a very high frequency. (And if the wind is faster than the speed of sound, you'll get a sonic boom! 😎👍)

 

[milski]

I can deal with velocities of observers and detectors but when velocity of something else is introduced, I'm 😕

The most important part: in f=(c+Vr)/(c-Vs) Vr and Vs are velocities in reference to the media. If the media is moving and you have the velocities in reference to something else (like ground), you have to change the reference frame to the media.

 

[Raiden2012]

Awesome, I finally get it! Milski, you're the man! (or woman)

 

[milski]

Awesome, I finally get it! Milski, you're the man! (or woman)

For most physics related purposes you can treat me as an ideal sphere. 😉