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Finding the maximum kinetic energy of a harmonic oscillator

  1. Jan 28, 2012 #1
    1. The problem statement, all variables and given/known data
    A harmonic oscillator with a vertical mass on a string has a hanging mass of 2m and a spring constant of K. It oscillates with an amplitude of Z. When its position is at a distance Z/2 of the equilibrium point, its potential energy is Ui. What is the maximum kinetic energy Km?


    2. Relevant equations
    E=1/2mV2 + 1/2mA2


    3. The attempt at a solution
    I realized that the max kinetic energy must be equal to the to the max potential energy because total energy is constant. I just don't know where to put the equations to make it fit.

    Can someone show me the set up? The answer give was Km = 4Ui

    Thanks in advance!
     
  2. jcsd
  3. Jan 28, 2012 #2
    Can you see that ω^2 = k/2m ?
    Also max velocity = ωZ
    So you can get an expression for the max KE
    The velocity at a displacement x is given by v^2 = ω^2(A^2-x^2)
    can you use this to get the KE at a displacement Z/2
    If you combine this lot you should get the answer
    Hope this helps..... I will follow this up if you have trouble
     
  4. Feb 9, 2012 #3
    I don't think I used your method, but I did end up figuring it out. Thanks for your help though.
     
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