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Simple Harmonic Motion

  1. Jun 2, 2008 #1
    1. The problem statement, all variables and given/known data
    A particle moves with simple harmonic motion of period [tex] \frac{\pi}{2} [/tex]. Initially it is 8cm from the centre of motion and moving away from the centre with a speed of [tex] 4 \sqrt{2} [/tex] cm/s.
    Find an equation for the position of the particle in time t second.


    2. Relevant equations
    [tex] x = A \cos{ \omega t + \epsilon} [/tex]
    [tex] v^2 = \omega^2 (A^2 - x^2) [/tex]
    [tex] T = \frac{2 \pi}{\omega} [/tex]

    3. The attempt at a solution
    [tex] T = \frac{2 \pi}{\omega} [/tex]
    [tex] \omega = 4 rad s^-1 [/tex]
    [tex] v^2 = \omega^2 (A^2 - x^2) [/tex]
    [tex] 32 = 16(A^2 - 64) [/tex]
    [tex] A = \sqrt{66} [/tex]
    [tex] x = A \cos( \omega t + \epsilon) [/tex]
    [tex] x = \sqrt{66}\cos(4t + \epsilon) [/tex]

    The answer in the book is:
    [tex] x = \sqrt{66}\cos(4t + .175) [/tex]

    I don't understand where the .175 comes from.
    Thanks for any help.
     
  2. jcsd
  3. Jun 2, 2008 #2

    Doc Al

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    Staff: Mentor

    Make use of the initial conditions at t = 0 to solve for the phase.
     
  4. Jun 2, 2008 #3
    Thanks for helping.
    So at t=0 x=8

    [tex] 8 = \sqrt{66}\cos(\epsilon) [/tex]
    [tex] \cos(\epsilon) = \frac{8}{\sqrt{66}} [/tex]
    [tex] \epsilon = \arccos\frac{8}{\sqrt{66}} [/tex]
    [tex] \epsilon = 10.02498786 [/tex]

    Have I made a mistake somewhere?
     
    Last edited: Jun 2, 2008
  5. Jun 2, 2008 #4

    Doc Al

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    Staff: Mentor

    Use radians, not degrees. :wink:
     
  6. Jun 2, 2008 #5
    God, that is embarrassing ha.
    Thanks very much.
     
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