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Position vs time graph simple harmonic motion phase constant

  1. Dec 9, 2016 #1
    1. The problem statement, all variables and given/known data
    http://imgur.com/a/FDfAp
    BJKwTzy.png
    What is the phase constant?
    2. Relevant equations
    x(t) = A*cos(ωt+Φ)

    3. The attempt at a solution
    If I'm not mistaken at t = 0 the graph starts at half the amplitude or 5. Also the amplitude of this graph is 10, and at t = 0 angular velocity is also 0.
    5 = 10*cos(0 + Φ)
    Φ = arccos(5/10)
    Φ = π/3
    The answer is supposed to be -2π/3, I'm not sure what I did wrong or if I'm even doing the question right.
     
    Last edited by a moderator: Dec 10, 2016
  2. jcsd
  3. Dec 10, 2016 #2

    ehild

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    The phase constant is different if you suppose x(t) as x(t) = A*cos(ωt+Φ) or x(t) = A*cos(ωt-Φ) or x(t) = A*sin(ωt+Φ), x(t) = A*sin(ωt-Φ).
    You also have to take into account if the function increases or decreases at t=0. The function in question increases, so the phase can not be pi/3. But it does not mean that -2pi/3 is correct.
     
  4. Dec 10, 2016 #3

    haruspex

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    The arccos function is defined to return a value in a certain range, but there are infinitely many solutions to cos(Φ)=0.5.
    What other solutions are there in, say, (-π, π), or (0, 2π) if you prefer?
    These all give the right value at t=0, but what about an instant later?
     
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