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Average kinetic energy of a damped oscillator

  1. Feb 13, 2014 #1
    For a damped mechanical oscillator, the energy of the system is given by $$E = \frac{1}{2}m \dot{x}^2 + \frac{1}{2}k x^2$$ where ##k## is the spring constant. From there, I've seen it dictated that the average kinetic energy ##\langle T \rangle ## is half of the total energy of the system. This makes sense, since the energy sort of "sloshes" back and forth between kinetic and potential energy, but is there a more formal way to show this is true?
     
  2. jcsd
  3. Feb 13, 2014 #2

    Simon Bridge

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    if you have ##y=f(x)## then the average value of ##y## in ##a<x<b## is the height of a rectangle with the same area as the graph of ##y## vs ##x## in that region. $$y_{avg}=\frac{1}{b-a}\int_a^b f(x)\;\text{d}x$$
     
    Last edited: Feb 13, 2014
  4. Feb 13, 2014 #3
    Oh, that's simple! Thank you!
     
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