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Ratio of Kinetic to Potential Energy of Simple Harmonic Oscillator

  1. Jan 14, 2010 #1
    1. The problem statement, all variables and given/known data

    Calculate the ratio of the kinetic energy to the potential energy of a simple harmonic oscillator when its displacement is half its amplitude.

    2. Relevant equations

    KE=1/2mv2 = 1/2kA2sin2(wt)
    U=1/2kx2 = 1/2kA2cos2(wt)
    KEmax=1/2kA2
    Umax=1/2KA2
    3. The attempt at a solution

    What I'm thinking is that when the displacement is half the amplitude, the simple harmonic oscillator is halfway to the equilibrium point. When the displacement is half of the amplitude, is the potential energy half of its maximum value? At this point, wouldn't the kinetic energy be half of its maximum value as well? I assumed this is true and attempted to answer the question, which yielded an answer of 1, which is incorrect. Is there something wrong with my logic?
     
  2. jcsd
  3. Jan 14, 2010 #2

    Matterwave

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    Yes, since the potential energy term goes as the square of the displacement, and not linearly with displacement, half the amplitude does not correspond to half the maximum potential energy.

    Just plug x=1/2A into your equation for U=1/2kx^2 and then use conservation of energy to find the kinetic energy at that point.
     
  4. Jan 14, 2010 #3
    Oh, I see what you're saying...but I'm a little lost at the conservation of energy part. I don't understand how to use that here. I know the potential energy is equal to 1/2k(A/2)^2 but isn't the kinetic energy just equal to 1/2mv^2? I understand that kinetic energy is related to the amplitude of the motion because velocity is related to amplitude...but I'm still not able to see how they are connected.
     
  5. Jan 14, 2010 #4

    Matterwave

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    The total energy of the system E=T+U where T is kinetic, and U is potential energy. If you have U and E you can find T. In a SHO E is a constant and does not change. You just found U. Do you know what E is?
     
  6. Jan 14, 2010 #5
    I took a closer look at my notes. Is the total energy equal to the square of the amplitude?
     
  7. Jan 14, 2010 #6
    Oh woops, I don't think that makes sense.
     
  8. Jan 14, 2010 #7

    Matterwave

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    Ok, at maximum amplitude, v=0, correct? That means at maximum amplitude, T=0. Now if E=T+U (at ALL times), and if T=0 then E=?
     
  9. Jan 14, 2010 #8
    Yup, that makes sense...so E = U = 1/2kA^2, so at this point U is at its maximum value. As you said, for a simple harmonic oscillator, E is a constant...so this would mean that E is always equal to 1/2kA^2, right?

    E=1/2kA^2
    U=1/8kA^2
    T=?

    Is the above correct? Now I'm starting to understand what you're saying. We did some problems like this in class...if you don't know E, you can always choose the point where either T or U is equal to 0 to figure it out...right? I mean, you'd only have to do that if you didn't know one or the either, like in this question.
     
  10. Jan 14, 2010 #9

    Matterwave

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    Yes, you are on the right track. Now just figure out T from the above, and take a ratio.

    The important thing to note here is that in a SHO, there is no damping. As such there is no dissipation of the energy, so E remains constant.
     
  11. Jan 14, 2010 #10
    I got T=3, which is the correct answer. Thanks so much for all your help! :)
     
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