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Velocity vs time graph simple harmonic motion phase constant

  1. Feb 1, 2015 #1
    1. The problem statement, all variables and given/known data
    http://i.imgur.com/u8vUv5a.jpg
    Find the phase constant.

    2. Relevant equations
    x(t)=Acos(ωt + Φ)
    v(t)=-Aωsin(ωt + Φ)
    Vmax = ωA
    ω=2π/T
    3. The attempt at a solution
    ω = 2π/12 = 0.5236
    A = 60/0.5236 = 114.59 cm
    v(0) = -30 = -114.59(0.5236)sinΦ
    0.5 = sinΦ
    Φ = π/6 and 5π/6. Which angle is correct?
     
  2. jcsd
  3. Feb 1, 2015 #2

    TSny

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    T ≠ 12
    There are different ways to see which is correct. For example, you could sketch a graph of V(t) for Φ = π/6 and for Φ = 5π/6 and compare your sketches with the given graph. Or, you could evaluate V(t) for some particular value of t (not equal to zero), say t = 1, for Φ = π/6 and for Φ = 5π/6 and compare to the given graph.
     
  4. Feb 1, 2015 #3
    What's the period and how to find it?
    So my period and amplitude and angles are incorrect?
     
    Last edited: Feb 1, 2015
  5. Feb 1, 2015 #4

    haruspex

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    Looks like 12 to me.
     
  6. Feb 1, 2015 #5
    thanks, 5π/6 was correct. Confirmed by mastering physics.
     
  7. Feb 1, 2015 #6

    TSny

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    Must be my eyesoo) (or my brain o0)) But it doesn't look like the graph of V(t) goes through exactly one cycle in 12 seconds.

    [Oh wait! Now that I look at it again for the 7th time, I see that the graph doesn't meet the t axis at t = 12. Very sorry about that. I agree that T = 12.]
     
  8. Feb 3, 2015 #7

    ehild

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    The other way to choose the correct phase angle:
    The v(t) graph is increasing at t=0. The slope of the v(t) graph is the acceleration, a(t) = dv/dt = -Aω2cos(ωt + Φ). It has to be positive at t=0. Which phase angle to choose?
     
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