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Fractional energy in a damped harmonic oscillator

  1. Apr 14, 2007 #1
    1. The problem statement, all variables and given/known data
    Show that the fractional energy lost per period is
    [tex]\frac{\Delta E}{E} = \frac{2\pi b}{m\omega_0} = \frac{2\pi}{Q}[/tex]
    where [tex]\omega_0 = \srqt{k/m}[/tex] and [tex]Q = m\omega_0 / b[/tex]

    2. Relevant equations
    [tex]E = 1/2 k A^2 e^{-(b/m)t} = E_0 e^{-(b/m)t}[/tex]

    3. The attempt at a solution
    [tex]\Delta E = 1/2 k A^2 e^{-(b/m)(t + T)} - 1/2 k A^2 e^{-(b/m)t}[/tex] where [tex] T = 2\pi / \omega_0[/tex]
    [tex]\frac{\Delta E}{E} = e^{-(b/m)T} - 1[/tex]
    [tex]\frac{\Delta E}{E} = e^{-\frac{2\pi b}{m\omega_0}} - 1[/tex]
    What should I do now?
    Last edited: Apr 14, 2007
  2. jcsd
  3. Apr 14, 2007 #2
    Using the Taylor series of e^x,
    [tex]e^{-(2\pi b)/(m\omega_0)} = 1 + -\frac{2\pi b}{m\omega_0}[/tex]
    Can you explain why I would drop the rest of the terms?
    [tex]\frac{\Delta E}{E} = -\frac{2\pi b}{m\omega_0}[/tex]
    first of all, is this right?
    second, how do i account for the negative sign?
  4. Apr 14, 2007 #3


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    Homework Helper

    All this assumes the damping is small, i.e. b << m \omega_0, so you can drop the higher terms of the Taylor series.

    Re the negative sign, the question asks for the fractional energy lost.

    Energy lost in 1 cycle (a positive number) = Initial energy - final energy.

    Your equation found the change in energy as

    Change in energy in 1 cycle (a negative number) = final energy - initial energy.
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