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Simple Harmonic Motion problem

  1. Feb 15, 2008 #1
    "A particle moves along the x-axis. It is initially at the position 0.270 m, moving with velocity 0.140 m/s and acceleration -0.320 m/s^2. Assume it moves with simple harmonic motion for 4.50 s and x=0 is its equilibrium position. Find its position and velocity at the end of this time interval."


    x=Amplitude * cos2(pi)ft

    Found f using T= radical (x/a) * 2pi = 5.769 so f= 0.1734 Hz

    However, even with solving for f I'm still left with two unknowns, i.e. A and t. Please help...
     
    Last edited: Feb 15, 2008
  2. jcsd
  3. Feb 15, 2008 #2

    Doc Al

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    Staff: Mentor

    You are given data for the initial postion, velocity, and acceleration. Set up equations for each. That will allow you to solve for the amplitude and phase.
     
  4. Feb 15, 2008 #3
    This is how I've done it quickly...

    x(t) = Bsin[2pi.f.t] + Acos[2pi.f.t] NB because at t=0, x=0.27 you can ignore the sin part
    x(t) = Acos[2pi.f.t]
    v(t) = -A.2pi.f.sin[2pi.f.t] (Diffrentiated once x(t) with respect to t)
    a(t) = -A.(2pi.f)^2.cos[2pi.f.t] (Diffrentiated twice x(t) with respect to t)

    x(t=0) = Acos[2pi.f.0]
    x(t=0) = A (cos(0) = 0)
    A = 0.27

    a(t=0) = -0.27.(2pi.f)^2.cos[2pi.f.0]
    a(t=0) = -0.27.(2pi.f)^2 (cos(0) = 0)
    -0.320 = -0.27.(2pi.f)^2
    f = sqrt[0.320/(0.27.(2.pi)^2)]
    f = 0.173

    x(t) = 0.27.cos[2pi.0.173.t]
    x(t=4.5) = 0.27.cos[2pi.0.173.4.5]
    x(t=4.5) = 0.050m


    Dont quote me though its a while since I've done SHM
     
  5. Feb 15, 2008 #4

    Doc Al

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    Staff: Mentor

    Sanity check: Do you think that the particle is at maximum displacement at t = 0?

    Better to start with this:
    [tex]x = A\sin(\omega t + \phi)[/tex]
     
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