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SHM Equations of Motion

  1. Jun 18, 2011 #1
    Hi :smile:

    I am a bit lost with the equations for velocity:

    I don't know Calculus yet, so my teacher just gave me the equation:

    -wx0cos(wt) (w being omega)

    He then said: v0 = wx0

    and therefore, concluded: -v0cos(wt)

    and then for when the displacement is maximum at time = 0: v0cos(wt)

    Is this correct? I mean, I am obviously not doubting him but I am a bit confused plus my notes were not very organized on this day...

    Thanks in advance
     
  2. jcsd
  3. Jun 18, 2011 #2

    rl.bhat

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    when t =0, ωt = 0 and cos(ωt) = 1

    So vo = -xo*ω
     
  4. Jun 18, 2011 #3
    So the equation would be:

    v0cos(wt)

    and for sin, how would it work?
     
    Last edited: Jun 18, 2011
  5. Jun 18, 2011 #4

    mukundpa

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    The sign and the function sin or cos depends on the instant you are taking t=0. If t is taken zero when the particle is at the equilibrium position (x=0) than the equation for displacement will be x= A sin wt and that for velocity will be v = Aw cos wt

    thus at extreme position wt = 90 deg, gives x = A and v = 0.
    (A is amplitude and Aw = Vo, the maximum velocity)
     
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