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Oscillations (SHM)

  1. May 22, 2012 #1
    1. The problem statement, all variables and given/known data

    I'm trying to solve this problem:

    http://img824.imageshack.us/img824/3513/prob1r.jpg [Broken]

    3. The attempt at a solution

    I rearranged the equation T=2π√m/k to find the spring constant:

    [itex]k= \frac{m}{\left( \frac{T}{2 \pi} \right)^2} = \frac{70}{\left( \frac{6}{2 \pi} \right)^2}= 76.7628 \ N/m[/itex]

    To find the original unstretched length I solve for y in F=ky and subtract it from 60 m:

    F=-ky

    [itex]y= \frac{-F}{k} = \frac{-mg}{k} = \frac{-70 \times 9.81}{76.7628} = -8.957 \ m[/itex]

    The spring is stretched by 8.94 meters from its equilibrium position so the original length is:

    60-8.94=51.0543 m

    Is this right? And how do I find the damping constant for the cord? What formula do I have to use? Any help is greatly appreciated.
     
    Last edited by a moderator: May 6, 2017
  2. jcsd
  3. May 27, 2012 #2

    Simon Bridge

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    This is not SHM - this is damped harmonic motion... you will have been given the formula for that. You can also look it up.
     
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