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Kinetic Energy from a Simple Harmonic motion and spring

  1. Jun 15, 2016 #1
    1. The problem statement, all variables and given/known data
    A 0.26-kg block on a horizontal frictionless surface is attached to an ideal massless spring whose spring constant is 190 N/m. The block is pulled from its equilibrium position at x = 0.00 m to a displacement x = +0.080 m and is released from rest. The block then executes simple harmonic motion along the horizontal x-axis. When the displacement is x = 1.4×10−2 m, what is the kinetic energy of the block?

    2. Relevant equations


    3. The attempt at a solution
    I have everything for the equation except for phi and and the amplitude A. I have found a=10.23 m/s2 and w=27.0327rad/s but I am stuck on how to find phi and A. Is there a relationship between the initial equilibrium position for finding A or phi?

    Thanks for the help!
  2. jcsd
  3. Jun 15, 2016 #2


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    What's wrong with ##PE = 1/2 k x^2##?
    You have the amplitude, so you know the total energy. (Yes, you do have the amplitude)
  4. Jun 15, 2016 #3


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    You do have A.
    What will be the KE of the block at t=0? Put it in this equation to get phi.
  5. Jun 15, 2016 #4
    To use PE=1/2kx^2 wouldn't I need the x at
    Is A= 0.014m?
  6. Jun 15, 2016 #5
  7. Jun 15, 2016 #6


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    No. When it's stretched initially, prior to letting go, what is the KE? At which point in the cycle of a general harmonic oscillator is the KE equal to this value?

    Then you're looking for the KE at the point x = .014m. So you know x for the PE, as you know at which point you want to look at.
  8. Jun 15, 2016 #7
    Oh I got it now!! Thanks so much !! :)
  9. Jun 15, 2016 #8


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    No problem.
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