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Position of glider on an air track

  1. May 30, 2009 #1
    1. The problem statement, all variables and given/known data
    An air-track glider is attached to a spring. The glider is pulled to the right and released from rest at t=0.00 s. It then oscillates with a period of 11.1 s and a maximum speed of 44.7 cm/s. What is the amplitude of the oscillation? (answer A= 7.90e-01 m)
    What is the glider's position at t=0.555 s?
    2. Relevant equations

    x(t)= Acos ([tex]\omega[/tex]t+[tex]\phi[/tex])

    3. The attempt at a solution

    so i solved this question already and got it right by assuming that the phase constant to be zero and solving for x. I was wondering if there was another way to solve for the position??
     
  2. jcsd
  3. May 30, 2009 #2

    nrqed

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    No, that's the only way to do it.
    Do you see why it is correct to set the phase constant equal to zero here?
     
  4. May 30, 2009 #3
    i'm assuming it had to with the fact that when the glider was released it waas not at it's equilibrium point..is that a correct assumption??
     
  5. May 30, 2009 #4

    nrqed

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    It's part of it but that's not sufficient.
    We have to know that it was released from the right of its equilibrim position *and* that it was released with no initial velocity (it was not kicked one way or another). That tells us that it starts with x equal to the maximum amplitude, so x(t) is a pure cosine curve, with no phase constant.

    For example, it it had been released from rest but at the left of the equilibrium position, we would have needed to use pi (or -pi) for the phase constant. If it had had an initial velocity, the phase constant wold be some other value.
     
  6. May 30, 2009 #5
    if it was released from rest at the left of the equilibrium how can we determine if it is pi or -pi without any additional information??..and thank you for answering my questions :))
     
  7. May 30, 2009 #6

    nrqed

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    You are welcome.

    I said pi or -pi because it makes no difference (since a difference of 2pi in the phase constant does not change anything to a cosine function). Any calculation done with pi or -pi would give the same answer.
     
    Last edited: May 30, 2009
  8. May 30, 2009 #7
    ok thanx :))
     
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