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PGRE Oscilloscope question

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  1. Feb 9, 2010 #1

    Pengwuino

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    I have a question about one of the questions on the 1992 Physics GRE. Some fellow students and I and a professor are going over questions to prepare for the PGRE and we came across one question that we can't seem to get a hang of.

    http://grephysics.net/ans/9277/17

    This is the question with a supposed answer which we don't buy and there seems to be a lot of argument over the actual solutions. The solution on the site and what we suppose is ETS's claim (since they say the answer is A as well) is that the oscilloscope is graphing Voltage vs. Time and that the function is basically y(t) = Sin(t) + Sin(2t) and that the answer represents one of the large humps in the function. However, me and my fellow students seem to have come to an agreement that the oscilloscope must be in X-Y mode (and not Y vs. t or however one woudl construe the question to be) and what you would get would be what is apparently called a Lissajous Curve. Now (E) is such a curve but with the ratio of frequencies in reverse (The X is twice the frequency as Y). What does everyone think about this? In no way does the question seem to assert that the X-axis is infact, time.
     
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  3. Feb 9, 2010 #2

    berkeman

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    Why are there so many typos on a practice test? Weird.

    It's hard to see that A is the answer because the two waves are in phase. If the relative phase were changed a bit, you would see the figure rotate a bit to be a more obvious saddle Lissajous pattern. And if you follow a point on the figure as it traces out, you would convince yourself that there's two "up-downs" for every "left-right".

    Can you see that A is a saddle Lissajous pattern seen edge-on (zero phase shift)?
     
  4. Feb 9, 2010 #3

    Pengwuino

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    What do you mean by saddle Lissajous pattern? I'm kinda learning about this on the fly.
     
  5. Feb 9, 2010 #4

    berkeman

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  6. Feb 9, 2010 #5

    Pengwuino

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    So they are out of phase by pi/2?
     
  7. Feb 9, 2010 #6

    berkeman

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    BTW, back in undergrad, I had a primitive 'scope (tube based). It wasn't good for much, but it DID have an x-y display mode. Best feature ever. We used to put 60Hz in on the horizontal, and rock music in on the vertical -- woo-hoo!

    Very fun at parties. "Out here in the Fields" by the Who was especially awesome. Turn up the volume!
     
  8. Feb 9, 2010 #7

    berkeman

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    Which what where? Sorry, I was off in the fields someplace. Do you mean in the practice question? I don't think that's right. y is at negative extreme when x is at negative extreme...
     
  9. Feb 9, 2010 #8

    berkeman

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    Well, actually I guess it depends on where you define the zero points -- maybe you're estimate is right. Let me sketch a bit...
     
  10. Feb 9, 2010 #9

    Pengwuino

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    Actually i can't see any detail in the picture you supplied. I saw pi/2 and saw it corresponded to the answer A) for the problem and made the assumption the pi/2 was teh phase shift.

    And on second thought, a pi/2 phase shift would just change from sin to cos... that doesnt help haha. Guess I need more detail on that pic to see what's going on.
     
  11. Feb 9, 2010 #10

    berkeman

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    When the low-frequency x sinusoid is at zero, the high-frequency y waveform is at positive extreme. How do you define a phase shift between two different frequency sinusiods?
     
  12. Feb 9, 2010 #11

    berkeman

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    Sorry the pic didnt' help. I just did a Google Images search on Lissajous patterns. I'll look again later tonight to try to find something better.
     
  13. Feb 9, 2010 #12

    Pengwuino

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    Actually Eureka! I literally defined the shift was Sin[2t + pi/2] for my Y and what was plotted was exactly the answer. So the phase shift wasnt between them, it was simply one of the inputs having a phase shift... which for pi/2 just means switching to Cos[2t] it looks like. What a problem...
     
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