# Doppler effect with two moving objects including wind

1. Oct 18, 2012

### johnqwertyful

1. The problem statement, all variables and given/known data

An observer is moving to the right. A source is moving to the left. There is a wind moving to the right. What is the observed frequency?

I had a quiz and this came up. I'm not sure if I did it right.

2. Relevant equations

f'=f(v±(vl-vw))/(v±(vs-vw))

3. The attempt at a solution

I just need to know if I add to the top, subtract from the bottom or the other way or what. I added to the bottom, subtracted from the top. Are you supposed to just add or subtract from the top and bottom? In class we said that if the two are stationary, wind doesn't make a bit of difference leading me to believe that you subtract or add from both bottom and top.

Any help? I've looked online and can't find an example with a moving source and observer and wind...

2. Oct 18, 2012

### TSny

Maybe you can get rid of the wind by switching to a frame of reference moving with the wind.

3. Oct 18, 2012

### johnqwertyful

I should have said that the observer and wind aren't at the same velocity. Is that what you meant?

4. Oct 18, 2012

### TSny

Imagine that you analyze the whole situation from a frame of reference moving with the wind. In that frame of reference, the wind velocity would be zero. How could you get the velocity of observer and the velocity of the source relative to this frame of reference?

5. Oct 18, 2012

### johnqwertyful

The numbers were v=700 mph vl=68.1 vs=94.8 vw=53.1 f=1224

I did it that way on the quiz I think.

I did (700+68.1-53.1)/(700-(94.8+53.1))*1224=1585 Hz at first, but then I thought it was too high or wrong or something. So I changed it to something else. I was right the first time. Damn.

The second part I got correct I think, after they pass. (700-68.1+53.1)/(700-94.8+53.1)*1224=1130. Right? I had the first one correct at first, but yeah. Darn second guessing yourself.

Oh well, I have over 100% homework grade so far, and got As on my other quizzes and he drops one. So I'm not too stressed out. I just wanted to know it to actually know it.

6. Oct 18, 2012

### johnqwertyful

Thanks for the help, though!

7. Oct 18, 2012

### TSny

Yes, that would be the correct answer. In the frame of reference with no wind, the speed of the observer would be 68.1 - 53.1 and the speed of the source would be 94.8 +53.1.

I think you need a plus instead of the red minus. The source is now moving away from the observer, so that would tend to decrease the frequency.

f = [700-(68.1-53.1)]/[700+(94.8+53.1)]*1224

Sounds like you're doing very well!