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## GR0877 Problem 45

PROBLEM STATEMENT: This problem is still being typed.

SOLUTION: (C) According to the general Doppler formula $\displaystyle f_o = \frac{c+v_o}{c+v_s}f$, where $f_o$ is observed frequency, $c$ is the speed of sound waves in the medium, $v_s$ and $v_o$ are the speed of the source and the observer relative to the medium, respectively. Sign convention: the speed $v_s$ is taken to be positive if the source is moving away from the observer at speed $v_s$, while $v_o$ is taken to be positive if the observer is moving toward the source at speed $v_o$. In our case, the source and observer are moving in such a way that $v_o = v_s$ (and equal to the speed of the wind $w$). Thus, we have no Doppler effect at all: $f_o = f$.

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1. 11.10.2011 at 02:13

This is basically a trick question. It’s set up like a Doppler Effect problem, but note that the source never moves. The sound waves travel faster due to the wind, but they arrive at the listener at the same rate, so the frequency is unchanged.

• 11.10.2011 at 12:59

Yes, that’s a tricky one. Only 15% of all test takers answered this question correctly!

2. 10.11.2011 at 00:23

Damn. I knew there was no Doppler shift, but I was racking my brain to find the relation between frequency and speed, imagining there was some other shift going on. The only choice I eliminated was the right one! Boo to you, GRE.

3. 03.10.2012 at 15:43

in your solution i think it should be (c-w)/(c-w), as relative to the medium the observer is moving away from the source (negative sign in numerator) and relative to the medium the source is moving toward the observer (negative sign in denominator). then you still get frequency unchanged. does this make sense?

• 05.10.2012 at 17:44

Well, the formula is correct if the sign convention given above is implied. Think about it in this way: 1) if $v_o = 0$ and $v_s=0$ then there is no change in frequency: $f_o = f$. 2) If the source doesn’t move relative to the medium, then, if observer moves toward the source, its frequency must increase: $\displaystyle f_o = \frac{c + v_o}{c} f = \left( 1 + \frac{v_0}{c} \right) f$, so the plus sign in the numerator is correct. 3) the same goes for the denominator.

• 06.10.2012 at 00:05

Thanks for the reply. In this problem, though, the source is moving toward the observer with the wind speed, and the observer is moving away from the source with the wind speed. With the sign convention above, source moving toward observer = negative sign in denominator, and observer moving away from source = negative sign in numerator. It’s the same result in the end, but do you see the issue I’m having?

4. 06.10.2012 at 19:45

Thanks for the reply. In this problem, though, the source is moving toward the observer with the wind speed, and the observer is moving away from the source with the wind speed. With the sign convention above, source moving toward observer = negative sign in denominator, and observer moving away from source = negative sign in numerator. It’s the same result in the end, but do you see the issue I’m having?

Oh, I see where the misunderstanding is. There is no relative speed between the source and the observer. They both move with the same speed in the same direction relative to the medium. Thus, $v_o = v_s$ and we actually don’t need the sign convention here at all (it doesn’t matter whether the sign is plus or minus BOTH in the numerator and the denominator). I remember when I was typing this solution I was trying to be as more general in the formula, as possible for you to use it in similar problems where the sign convention is crucial.

• 23.10.2016 at 17:05

I know this is real old by now..but for anyone studying for the PGRE by looking through these solutions:

I believe crunchy fouton is correct. If we define the source at the origin and take one dimensional motion:

S ——> O
———————————————–>x+

Where the S is the source, O is the observer, and the small arrow is the direction of the medium;s velocity. Notice that both the source and observer have negative velocity (equal in magnitude to the medium’s velocity) relative to the medium since the medium travels in the +x direction while the source and observer are still. So, you do get the -V_m in both the top and bottom but as crunchy fouton points out, this leave you with the same answer.

• 23.10.2016 at 23:29

That’s exactly what I said in the above comment.

5. 28.08.2016 at 11:39

As a quick guide to knowing the answer is (C), notice that if you assume there is some sort of Dopler shift calculation needed, there is no simple cancellation with the numbers involved. This is almost NEVER the case in a GRE problem. Annoying numbers almost always cancel out nicely.

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