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Simple Harmonic Motion - Finding the Phase

  1. Mar 8, 2010 #1
    1. The problem statement, all variables and given/known data
    Here is a position-time graph for a ball on a spring in SHM.

    What is the phase of the particle at point a on the graph?

    2. Relevant equations
    phase = ωt + ɸ0


    3. The attempt at a solution
    First I found the initial phase, ɸ0. 0.5A / A = 0.5 = cos(ɸ0). Then ɸ0 = -π/3. It's negative because the particle is moving to the right.

    Now for point a, I figured this is at 0.75T, where T is the period.This is kind of hard to see so it could be wrong. But then:

    phase = (2π/T)(0.75T) - (π/3) = (7/6)π or 210 degrees. However, the answer is -120 degrees.

    I also tried with points b and c but they were wrong too. Can anyone help?
     
    Last edited: Mar 9, 2010
  2. jcsd
  3. Mar 8, 2010 #2

    vela

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    Your assumption that ta=0.75 T is wrong. What's the increase in phase going from the very left to the first maximum? Similarly, what's the increase in phase from the first minimum to point a?
     
  4. Mar 8, 2010 #3
    I'm not sure how to figure out the increase in phase directly... does it depend on knowing time still?

    Because I'm confused how my assumption that t at a = 0.75T is wrong. The time from the first crest to the first trough is definitely 0.5T, right? And the time from 0 to a trough or crest is 0.25T. So the time from 0 to a should be 0.5(0.25T) = (1/8)T...
     
    Last edited: Mar 8, 2010
  5. Mar 8, 2010 #4

    vela

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    The time from 0 to the first crest is not 0.25 T. Think about the unit circle. What angles gives you cos x1=0.5 and cos x2=1? What's the difference x2-x1? How much time does that increase correspond to?
     
  6. Mar 8, 2010 #5
    x1 = 60
    x2 = 0

    x2 - x1 = 60

    So... 360 degrees corresponds to one revolution (T), then 60/360 is the amount of time?

    But that can't be right either, because (2pi / T)(60/360)T - (pi/3) = 0
     
  7. Mar 8, 2010 #6

    vela

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    That wasn't meant to give you the answer directly. It was to get you to see that the time from t=0 (-60 degrees) to get to the crest (0 degrees) isn't 0.25 T as you were assuming.
     
  8. Mar 8, 2010 #7
    Ah, I figured it out - I think this is right now:

    So I need to solve the equation cos((2pi/T)t - (pi/3)) = -0.5

    cos(x) = -0.5 when x = 120 and 240, 480 and 600, etc.

    So the answer is 240 degrees (same as -120 degrees).

    But if the question didn't show the diagram, how would you know which angle to use (120 or 240?)
     
  9. Mar 8, 2010 #8

    vela

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    You wouldn't be able to because without additional information, like the diagram, you don't know in which direction the particle is moving.
     
  10. Mar 8, 2010 #9
    Okay, that makes sense :)

    Thanks for your help!
     
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