Doppler Effect radio broadcasting

[atarr3]

Homework Statement

Three radio-equipped plumbing vans are broadcasting on the same frequency f_0. Van 1 is moving north of van2 with a speed of v, van 2 is fixed, and van 3 is moving west of van 2 with a speed of v. What change in frequency (f-f_0) does van 3 hear from van 1?

Homework Equations

I'm not really sure what equation to use because the doppler shift function doesn't work here.\

The Attempt at a Solution

I'm not really sure where to start from thise.

Thanks in advance for all of your help, guys.

 

[sylas]

I'm not really sure what equation to use because the doppler shift function doesn't work here.

Actually; yes it does. So for a start, let's see the function. What is it?

Cheers -- sylas

 

[atarr3]

Well for a source and receiver receding from one another:

f=[(1-v/c)^1/2/(1+v/c)^1/2]f_0

I guess I am just having trouble figuring out the velocity then.

 

[sylas]

Well for a source and receiver receding from one another:

f=[(1-v/c)^1/2/(1+v/c)^1/2]f_0

I guess I am just having trouble figuring out the velocity then.

Yes, that is the key. What is the relative velocity of the two vans. How would you approach that? Any ideas?

 

[atarr3]

I actually have no idea. I'm assuming it's v<<c, so addition of velocities wouldn't apply here. I had initially tried to find the velocity using the Pythagorean theorem, but that didn't really work either. Maybe break it into x and y components and find the velocity for that?

 

[sylas]

I actually have no idea. I'm assuming it's v<<c, so addition of velocities wouldn't apply here. I had initially tried to find the velocity using the Pythagorean theorem, but that didn't really work either. Maybe break it into x and y components and find the velocity for that?

Have you done vector arithmetic yet? If so, use that. If not, x and y components should work. Pythagorean theorem will be handy.

 

[atarr3]

Well the velocity would be v times the square root of 2 at an angle of 45 degrees. So if we use the relativistic doppler shift equation f = [(1 + v/c*cos(theta))/(1-v^2/c^2)^(1/2)]*f_0, I end up with [(1 + v/c)/(1-2v^2/c)^(1/2)]*f_0, but I don't feel like that's right.

 

[sylas]

Well the velocity would be v times the square root of 2 at an angle of 45 degrees. So if we use the relativistic doppler shift equation f = [(1 + v/c*cos(theta))/(1-v^2/c^2)^(1/2)]*f_0, I end up with [(1 + v/c)/(1-2v^2/c)^(1/2)]*f_0, but I don't feel like that's right.

You have not yet substituted for all instances of v.

 

[atarr3]

You mean into the doppler shift equation?

 

[sylas]

You mean into the doppler shift equation?

You have the Doppler shift equation, and you need to use the appropriate expression for v in that equation.

 

[atarr3]

So I'm plugging in the v I found from adding the two vectors?

 

[sylas]

So I'm plugging in the v I found from adding the two vectors?

You gave a doppler formula, and it looked correct. You proposed a value for v, and it looked correct. When you put them together, however, it was a bit of a mess.

Try writing it all out step by step. Step 1: Doppler formula. Step 2: The expression to use for v between van 1 and van 3. Step 3: the result of plugging step 2 into step 1.

 

[atarr3]

I tried it again and I am still getting the same thing. 1 + v/c in the numerator. The square root of (1 - 2*beta^2) in the denominator.

 

[sylas]

I tried it again and I am still getting the same thing. 1 + v/c in the numerator. The square root of (1 - 2*beta^2) in the denominator.

Write it out, in the three steps I recommended, in a post here. Then we can se what you are doing.

 

[atarr3]

Ok so according to my book the equation is f = [(1 + beta*cos(theta))*f_0]*gamma.
Step 2, finding the velocity between van 1 and van 3, is found using the Pythagorean theorem. Both are moving at a 90 degree angle from on another, and both are moving at velocity v. That gives us v^2 + v^2 = 2v^2. The square root of this is the velocity, and since both x and y components of velocity are equal, it must be moving at a 45 degree angle, which is our theta.

For step three, cos(theta) is root(2)/2. Plugging v and cos(theta) into the formula gives me:

[1+root(2v^2)/c*root(2)/2)]*f_0/root(1-2v^2/c^2). The numerator simplifies to 1+root(4v^2)/(2*c) = 1+root(v^2)/c = 1+v/c.

Where am I going wrong?

 

[sylas]

Ok so according to my book the equation is f = [(1 + beta*cos(theta))*f_0]*gamma.

Hm... that is a different formula from the simple Doppler formula you gave in the first instance. It is a more general case when the velocity is not directly along the line of sight.

In the particular case you have described, I don't think you need this, unless I misunderstand the problem. But maybe I misunderstand the problem!

Are you proposing that van 1 and van 3 both left van 2 in the same instant, and proceed at speed v to the north and to the west? In that case, I believe you can use the formula you gave in the first instance in message 3.

Can you tell me, are you expected to use the formula with beta and theta? In this case, you will need to describe what theta and beta represent, so that you can apply it to the problem appropriately.

Cheers -- sylas

 

[atarr3]

Ohhhh ok. I thought we had to use that formula because it was moving at an angle, but it's not really moving at an angle, just according to the velocity. But it's still moving away from Van 3 directly. Ok. I got the right answer now that I used that original formula. Thank you so much for your help.

 

[sylas]

Ohhhh ok. I thought we had to use that formula because it was moving at an angle, but it's not really moving at an angle, just according to the velocity. But it's still moving away from Van 3 directly. Ok. I got the right answer now that I used that original formula. Thank you so much for your help.

No problem. And thank you. I have learned something too... I have never actually used the full Doppler effect taking into account movement not along the line of sight, and thinking about that has helped clarify a few matters in my own mind.

Cheers -- sylas