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Homework Help: Oscillation question: dealing with a maximum velocity

  1. Sep 8, 2005 #1
    Hi, I just want to clarify something.
    If there is a particle attached to a spring, in which it's maximum velocity is at t=0 towards THE LEFT.
    Does that mean that when x(t)= Acos(wt+phi) then v(t)= -wAsin(wt+phi)
    therefore v(0)=-wAsin(phi) = - 20
    and Vmax= wa= -20 or positive 20.
    When it is heading towards the left, should i make the twenty negative when plugging it into the equations?
  2. jcsd
  3. Sep 8, 2005 #2


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    Mmmh, yes, conventionally, positive x is to the right, and negative x to the left. Hence, if you are told that the initial speed is to the left, it means that the quantity x is decreasing as time goes by. That is to say, the derivative of x wrt t is negative. But dx/dt = v, so yeah, v<0 at this moment.
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