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 - [latex]\small\mu[/latex]v, if F were conservative then there would exist P(x) such that [latex]\small -\frac{dP}{dx} = F[/latex]. 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:

[latex]\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[/latex]
So let [latex]\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.[/latex]

For underdamped [latex]\small\zeta<1\Rightarrow x=e^{-\zeta\omega_{\circ}t}(C_1 cos\omega_1 t + C_2 sin\omega_1 t)[/latex]

[latex]\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[/latex]

Therefore x(t) is not 1-1 [latex]\small\Rightarrow \int_{x_1}^{x_2}vdx[/latex] is multivalued implies W is not a function implies p(x) doesn't exist (since W=-[latex]\small\Delta[/latex]P) implies F is not conservative. Similarly for [latex]\small\zeta=1[/latex].

 

But in the overdamped case, [latex]\small\zeta[/latex]>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?

 

Shouldn't it be enough to show that [latex]\int_{x_1}^{x_2} Fdx[/latex] ≠ [latex]-\int_{x_2}^{x_1} Fdx[/latex]

Shouldn't it be enough to show that [latex]\int_{x_1}^{x_2} Fdx[/latex] ≠ [latex]-\int_{x_2}^{x_1} Fdx[/latex]

Thank you for the reply. I think you're right. Obviously, different paths will give different amounts of work, and so the force is nonconservative.

I just want to be able to see clearly 2 things:

1) why does the 1-1 nature of x(t) in the overdamped case seem to imply path independence?

2) why, exactly, is it not always possible to find a function, call it [latex]\phi(x)\mbox{, such that }\phi_{x}(x(t)) = (\frac{dx}{dt})^2[/latex] ?

Edited June 14, 201312 yr by inkliing