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Question regarding simple harmonic motion

  1. Jan 22, 2005 #1
    I have my solution and my doubt in the attachment that followed.
     

    Attached Files:

  2. jcsd
  3. Jan 22, 2005 #2

    Andrew Mason

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    Your solution is right. But the equation of motion might be easier to use if you chose the sin instead of cos:

    [tex] y = 3sin(\pi t - \pi/2)[/tex]

    This is because at t = 0, the mass is at maximum negative amplitude ([itex]\theta = -\pi/2[/itex])

    So:

    [tex]d = y-y_0 = 3sin(\pi t - \pi/2) - 3sin(-\pi/2) = 3sin(\pi t - \pi/2) + 3[/tex]

    For 1), where t = 1 sec., d = 3 +3 = 6
    For 2), where t = .75 sec,[itex]d = 3sin(\pi/4) + 3 = 5.12[/itex]

    AM
     
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