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Simple Harmonic Motion of a horizontal spring

  1. Jan 11, 2005 #1

    Bri

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    A 1.00-kg object is attached to a horizontal spring. The spring is initially stretched by 0.100 m, and the object is released from rest there. It proceeds to move without friction. The next time the speed of the object is zero is 0.500 s later. What is the maximum speed of the object?

    I know the equation v(max) = wA = A(k/m)^.5
    But I can't figure out how to solve for w. I figured the amplitude is .1, since it's released from that position and won't go any further... is that right?
    The only thing I could think of is using the equation v = -wAsin(wt + phi) and using v=0, A=.1, t=.5, and phi=0, but that just comes out to w=0.
     
  2. jcsd
  3. Jan 12, 2005 #2
    The displacement x = A cos(wt + phi)
    Try using the boundary condition that when x = -0.1 , t = 0.5 s knowing that phi = 0.
     
  4. Jan 12, 2005 #3

    dextercioby

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    HINT:Use the law of conservation of mechanical energy.You have the Amplitude ,the mass,the period of oscillation (so u can figure out the elestic constant) and u can easily find the maximum velocity.

    Daniel.
     
  5. Jan 14, 2005 #4
    "The next time the speed of the object is zero is 0.500 s later"

    Then the period is T=2*0.5 s= 1 s. And now w=2*pi/T.
     
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