Understanding Doppler Shift in Radio-Equipped Plumbing Vans

[StephenD420]

Problem:
Three radio-equipped plumbing vans are broadcasting on the same frequency,f0. Van 1 is moving north of van 2 with speed v, van 2 is fixed, and van 3 is moving west of van 2 with speed v. what change in frequency (f-f0) does van 3 hear from van 2? From van 1?

Attempt at solution:

f= f0 * \sqrt{(1-\beta)/(1+\beta)}

is the frequency rate that van 1 and van 2 will receive signals from one another and at which van 2 and van 3 will receive signals from each other.
However, I am confused as to how to find the rate at which van 3 hears from van 1?

I know f=1/t
but I cannot see how to make this work.



Any help would be greatly appreciated.
Stephen

Pictorial representation attached so the placement of the vans is more clear.

 

[Onamor]

Think about sitting in van3. You see van2 moving away from you in a straight line, so you can slap in the formula and get some answer.
Same is possible for van1. If you are sitting in van3, since van2 and 3 are traveling at the same speed, and at right angles, the distance between them is the hypotenuse of a triangle - that is extending.

Try some geometry/trig and components of vectors to find the change in their distance "relative" to one another - ie the speed of their separation.

Any help?

 

[StephenD420]

ok...using some trig...

the increasing distance that the frequency has to travel would be 2*V*t*cos(45degrees).
But how does this help in solving the problem? would f= f0 *1/(2*V*t*cos(45degrees))? If so what would the time, t, represent? I used unit analysis to figure the formula out, but I do not know the logic behind it.

sorry...I need a little bit more probing.

Thanks so much for your quick reply. I really appreciate your help.
Stephen

 

[Onamor]

An increasing ("changing") distance is a velocity. Do you see?

Its important to understand frames of reference.
If you are sitting inside one of the vans (1 or 3, doesn't matter which - the problem is symmetric) then, since you are not accelerating, if there was nothing else in the universe, you would not be able to tell if you were moving. All you would see is the other van moving away from you.
That is, you can "pretend" that one of the vans is not moving (this is only because of the special geometry of this specific problem). Hence it is just like the first problem.

I hesitate to fill in the equation for you as the answer is meaningless in comparison to the understanding - also there are many (equivalent) equations for the doppler shift.

 

[StephenD420]

Let me see if I am understanding so far...

so since the, let's say, y distance is changing from time zero, t0, the time it takes for a wave to get from van 1 to van 3 when van 1,2,3 are perfectly still in the set up shown, then a time t1 later when another signal from van 1 reaches van 3 the y distance that the radio waves have to travel has increased. So now the waves have to travel a distance of 2*V*t0*cos(45).(I am not sure whether to use t0 or t1 in this distance equation..I think t0 since the increase in distance is a fraction of the distance the waves had to travel at the beginning because van 1 is traveling away from van 3 at 2V, so the time to transverse the increased distance is a fraction of t0...right?)

since radio waves are electromagnetic they travel at the speed of light,c. so the time to transverse the increased distance is t1=(2*V*t0*cos(45))/c.

so..
since f=1/t
and the total time that the wave travels is t= t0 + t1
f = 1/(t0+((2*V*t0*cos(45))/c))How am I doing so far? Am I getting the right ideas and concepts?
Thanks again for all of your help.
Stephen

 

[Onamor]

Doppler shift doesn't just depend on one wave. Its a bit more abstract than that.
The observed frequency (ie. once it has been "deformed") is

where [URL]http://upload.wikimedia.org/math/1/0/0/1008ebafed8bcbdcbdb4db1dfe5cc579.png[/URL] is the relative velocity of v_s, the source velocity, and v_r the reciever velocity. (Its positive when they are being separated)

But you want the change in frequency, so you you need to take away f0, the initial broadcasting frequency. As stated in the original post.

That gives you this (ignore the third part, its true, but we don't need it)
[URL]http://upload.wikimedia.org/math/7/8/0/780b605dfde90013fcb5999cb347ac28.png[/URL]

So your know everything but the relative velocity of the source and reciever.
But you can work that out from the geometry.

Draw the velocity vectors of van 1 and 3. They are directed at right angles, their length is their magnitudes, both v. Remeber these are velocity vectors, so they will not change in size (in this problem) like the position vectors would.
Now you need to know how fast the vans are separating. This is the hypotenuse of the velocity triangle. Do you see? Say if not.

 

[StephenD420]

I have looked at what you posted and worked with the equations...but I am still not there...

here is what the book says the answer is...
f = f0/(1+(sqrt(2)*v)/c)
how would you get that from your formulas?

Thanks.
Stephen

 

[Onamor]

This is the "changed" frequency. In the original post it asked for (f-f0), but doesn't really matter, one leads to the other.

Anyway, the sqrt(2)*v is the hypotenuse of the velocity triangle, from pythagoras.
This is V_s,r in the first equation I posted.
The minus or plus is just convetion, or equivalently, whether you "sit" in van 1 or 3.

You can get the sqrt(2) from trigonometry of the triangle - its 2*cos45. Like you had, you just got a bit confused with position and velocity.

 

[StephenD420]

if you plug into your formula you get

f=f0*(1+ (sqrt(2)v/c))

but the book says that f0 is divided by (1+ (sqrt(2)v/c)) not multiplied by to make the answer
f=f0/(1+ (sqrt(2)v/c)).

What am I missing?

Thanks.
Stephen