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Simple Harmonic Motion and time calculation

  1. Oct 29, 2005 #1

    An object is undergoing simple harmonic motion with period 0.305 s and amplitude 6.05 cm. At [tex]t=0[/tex] the object is instantaneously at rest at [tex]x=6.05[/tex] cm.

    Calculate the time it takes the object to go from [tex]x=6.05[/tex] cm to x = -1.49 cm.

    I'm thinking [tex]\omega = \frac{2\pi}{T} = 20.6[/tex] rad/s. If we assume left to be the negative direction, then the [tex]\Delta x = -0.0754[/tex] m.

    So, I can use the equation [tex]x = A\cos(\omega t + \phi)[/tex]. [tex]\phi = 0[/tex].

    [tex]-0.0754 = .0605cos(20.6t)[/tex].

    But, wait a minute here, this can't possibly be right. If I divide both sides by .0605, then I have cos equaling a magnitude greater than one.

    I'm stuck.
    Last edited: Oct 29, 2005
  2. jcsd
  3. Oct 29, 2005 #2

    Tom Mattson

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    Your problem is that you set [itex]\Delta x[/itex] equal to [itex]x[/itex]. [itex]\Delta x[/itex] is the displacement between the initial and final positions, and [itex]x[/itex] is the [itex]x[/itex]-coordinate of the particle, which they tell you is [itex]-1.49cm[/itex].
  4. Oct 29, 2005 #3
    Thanks. I guess the problem wasn't as hard as I thought it was.
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