# AAMC 10 Frequency

Discussion in 'MCAT Study Question Q&A' started by MedPR, 03.30.12.

1. ### MedPR

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Doppler Effect isn't a problem for me, but I am a little confused about the wording of the question and the explanation. As with any Doppler question, if the object/source have a net movement away from each other, the observed frequency will be lower. So in this problem, as the projectile gets higher, the observed frequency decreases. Though it is approaching 0, the frequency won't actually ever be 0 right?

From the explanation, "frequency shift" seems like it is the same as "change in frequency" and I guess is also the same as "doppler shift". So when the velocity of the projectile hits 0 (at hmax) the frequency shift is 0 (since it is no longer changing because the object/source aren't moving relative to each other) but the overall frequency is still a non-zero number?

So to break this down into really simple and non-physics terms. When the projectile is first shot its velocity is very high since gravity hasn't started to decelerate it yet. Remembering that overall distance is irrelevant in Doppler problems, we consider only that the velocity is very high initially so the Doppler shift ("frequency shift"??) is the greatest. Then as it reaches its max height, it has been decelerated continuously by gravity so its velocity is 0 so the frequency shift is not changing (=0). Then it accelerates downwards and velocity increases even though the distance is decreasing (distance doesn't affect doppler!).

tl&dr: Just wanted to verify that the term "frequency shift" is exactly the same thing as "doppler shift"

2. ### pfaction

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I got this wrong too, for the same f'n reasons. I said, how can it fall to zero? So I chose D, because I said "at the top of the flight it's lowest".

AT THE TOP OF THE FLIGHT. STRAIGHT UPWARDS.

These were key things I missed, and subsequently got the momentum (last q) wrong.

Also, conceptually speaking, think about a rocket that launches really high in the air. It gets softer, softer until you can't hear it, then it roars back before impact.

But I still don't see how it's 0. V+/-VD / V-/+VS
There has to be 340 somewhere in there.
3. ### MedPR

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Doppler shift doesn't have anything to do with how far away the object is, only how quickly the separation is changing.

It's 0 because the question is talking about frequency shift, not frequency. You can think of frequency shift as change in frequency. When both velocities are zero, the observed frequency is exactly the same as the source frequency whether the objects are 1 inch apart or 1 mile apart. The intensity (loudness) of the sound will obviously be different, but the frequency is exactly the same if source and observer are not moving relative to each other. That is the definition of the doppler shift.

As you move away at a velocity that begins to decrease, the frequency shift (doppler shift, change in frequency, whatever you know it as) will start to decrease (decrease just another way to say approach 0). At hmax, vy = 0, so fobs=fsource, so frequency shift is 0 for just this instant in time. Then as the projectile starts to make its way back down, it is accelerating (gravity), so its velocity is increasing and the frequency shift is therefore increasing.
4. ### Mecarter

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Lesson learned here for me: don't base an answer on data contained in a question. I had this right, and then the question stating that the object explodes made me change it, considering that it won't be producing a sound as its bits and pieces descend. Then, the shift would fall continuously in magnitude and go to zero. And that is another thing I learned -- they didn't stick "and fall to zero" on that answer choice, so that's another reason I should not have picked it. But, time stress, etc. -- I'll get better.