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Annoying Spring Problem

  1. Apr 21, 2005 #1
    A mass of 1.64 kg stretches a vertical spring 0.300 m. If the spring is stretched an additional 0.123 m and released, how long does it take to reach the (new) equilibrium position again?

    What I tried:
    Found k using F=kx (mg=kx)
    using that k, I calculated the force in the extended spring (F=k(0.123))
    using that force, I calculated the acceleration of the mass (F=ma)
    finally, using x=(1/2)at^2 I calculated t.

    I must be missing something obvious b/c this doesnt seem like a difficult problem... :confused:
  2. jcsd
  3. Apr 21, 2005 #2


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    Nope,it's incorrect.The motion is not with constant acceleration.U need to find the period of vibration.

  4. Apr 21, 2005 #3
    Yeah thats what I thought....

    So I calculated k in the first step above.
    Then I used omega = sqrt(k/m) to get omega.
    Then I used f = omega/2pi
    Then period = 1/f and it's still wrong. Any other mistakes I made?

    Thanks for the help :smile:
  5. Apr 21, 2005 #4


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    Yes,the time required is only 1/4-th of the period...Can u see why?

  6. Apr 21, 2005 #5
    Wow hey thanks a lot. I guess brain just isn't working today. :smile:
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