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Conservative overdamped harmonic oscillator?

  1. Jun 12, 2013 #1
    This isn't homework. I'm reviewing calculus and basic physics after many years of neglect.

    I want to show that a damped harmonic oscillator in one dimension is nonconservative. Given F = -kx - [itex]\small\mu[/itex]v, if F were conservative then there would exist P(x) such that [itex]\small -\frac{dP}{dx} = F[/itex]. I want to show that no such function, P(x), exists.

    The easy way would be to find a closed curve around which the integral of Fdx would be zero, but since Fdx is a 1-dimensional 1-form, this doesn't seem to be a meaningful way to do it.

    So I think brute force has to prevail. It should be true that:
    [tex]\small W=\int_{x_1}^{x_2}Fdx = \int_{x_1}^{x_2}(-kx-\mu v)dx = \frac{1}{2}kx_1^2-\frac{1}{2}kx_2^2-\mu\int_{x_1}^{x_2}\frac{dx}{dt}dx = \frac{1}{2}kx_1^2-\frac{1}{2}kx_2^2-\mu\int_{t_1}^{t_2}\left(\frac{dx}{dt}\right)^2 dt[/tex]
    So let [itex]\small\omega_{\circ}=\sqrt{k/m}\mbox{ , }\zeta=\frac{\mu}{2\sqrt{mk}}\mbox{ , }\omega_1=\left\{\begin{matrix}\omega_{\circ}\sqrt{\zeta^2-1},&\zeta>1\\\omega_{\circ}\sqrt{1-\zeta^2},&\zeta<1\end{matrix}\right.[/itex]
    For underdamped [itex]\small\zeta<1\Rightarrow x=e^{-\zeta\omega_{\circ}t}(C_1 cos\omega_1 t + C_2 sin\omega_1 t)[/itex]
    [tex]\small\Rightarrow W=\frac{1}{2}kx_1^2-\frac{1}{2}kx_2^2-\mu\int_{t_1}^{t_2}e^{-2\zeta\omega_{\circ}t}[(-\zeta\omega_{\circ}C_1+\omega_1 C_2) cos\omega_1 t + (-\omega_1 C_1-\zeta\omega_{\circ} C_2) sin\omega_1 t]^2 dt[/tex]
    Therefore x(t) is not 1-1 [itex]\small\Rightarrow \int_{x_1}^{x_2}vdx[/itex] is multivalued implies W is not a function implies p(x) doesn't exist (since W=-[itex]\small\Delta[/itex]P) implies F is not conservative. Similarly for [itex]\small\zeta=1[/itex].

    But in the overdamped case, [itex]\small\zeta[/itex]>1, x(t) is a non-oscillating decaying exponential which never crosses equilibrium, implying x(t) is 1-1, implying W is a function, implying F is conservative. But how can this be? How can a frictional damping force, which dissipates energy as heat, ever be conservative?
     
  2. jcsd
  3. Jun 14, 2013 #2
    Hi there,

    as you glance at, your conclusion is wrong.
    I guess the culprit is the line of argument, "<...>implying x(t) is 1-1, implying W is a function, implying F is conservative<..>".
    Even for a non-conservative system one can write the work in moving from a to b as a functional of the motion, as you do. Proving it that such functional is actually a function of the final position only is very different.
    To can prove what you are after by simply showing that for two different functions x1(t) and x2(t), with the condition x1(0)=x2(0) and x1(t_given)=x2(t_given), the computed work is different.
    Hope it somehow helps.
     
  4. Jun 15, 2013 #3
    I guess I'm trying to convince myself that it's not always possible to find a function, f(x), such that [itex]\small f_{x}(x(t)) = (\frac{dx}{dt})^2[/itex]. I wish I could think of a simple straightforward way of showing that.
     
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