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Energy of driven damped oscillator

  1. Feb 13, 2009 #1
    Hi all!

    I was considering the Energy of a driven damped oscillator and came upon the following equation:

    given the equation of motion:

    [tex]m\ddot x+Dx=-b\dot x+F(t)[/tex]

    take the equation multiplied by [tex]\dot x[/tex]

    [tex]m\ddot x\dot x+Dx\dot x=-b\dot x^2+F(t)\dot x[/tex]

    and we rewrite it:

    [tex]\frac{d}{dt}(\frac{m\dot x^2}{2}+\frac{Dx^2}{2})=-b\dot x^2+F(t)\dot x[/tex]

    now the LHS appears to be the total energy of the undamped harmonic oscillator, by energy conservation a constant, so it´s rate of change is 0:

    [tex]0=-b\dot x^2+F(t)\dot x[/tex]

    Ok, so far so good :) But here lies my problem. Now we´ve obtained an equation for \dot x which we could solve for x(t). But the result appears to be different from the standart solution :(

    [tex]0=\dot x(-b\dot x+F(t))[/tex]
    [tex]\dot x=0[/tex]
    [tex]-b\dot x+F(t)=0[/tex]

    => [tex]\dot x_1=const:=C[/tex]

    so the solution would read: (would it? Do we have superposition here, since the DE is not linear any more?)


    So could anyone please help me find the mistake :) I would be thankful :)
  2. jcsd
  3. Feb 13, 2009 #2
    This is where you went wrong. The energy here is not conserved within the system, so you can't put the rate of change to zero. The SHM is a closed system, this is not. Here the energy keeps changing, the external force F(t) is pumping in energy while the damping is siphoning it out. Your equation tells you how this happens :
    \frac{d}{dt}(\frac{m\dot x^2}{2}+\frac{Dx^2}{2})=-b\dot x^2+F(t)\dot x[/tex]

    But you can't put the LHS to zero, because energy is not being conserved. Suppose there was no external force and so the term F(t) was absent. The equation shows you how your energy would then decrease due to the damping term. And you know it must, because eventually it loses all the energy and stops.

    So the moral of the story is : when your system is not closed, but interacting and exchanging energy with other systems (in this case whatever is doing the driving and the damping), you cannot use energy conservation for your system.
  4. Feb 13, 2009 #3
    So it´s not like all the energy pumped in gets lost in damping, cuz this would imply a vanishing LHS?
  5. Feb 13, 2009 #4
    the LHS isn't energy, but rate of change of it. If there is no driving term, eventually all the energy gets lost - and the thing comes to rest. Then as energy is changing no longer, the LHS is also zero. At this pint you can see that the rhs is also zero. Theoretically, this happens after infinite time.

    Point is, the function of time on the LHS is not always zero, but becomes zero at t= infinity.
  6. Feb 13, 2009 #5


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    As already pointed out, the LHS is zero for an undamped oscillator. Let's take a look at the bracketed terms:

    [tex]\frac{m\dot{x}^2}{2} = \frac{1}{2}mv^2 = T[/tex]

    [tex]\frac{Dx^2}{2} = V[/tex]

    The first term represents the kinetic energy, whilst the second term represents the potential energy. So the LHS represents the rate of change of the sum of the potential and kinetic energies, in other words the total energy. In this case of an undamped oscillator the total energy remains constant, however in the case of damped motion the total energy is not conserved.
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