- #1
inkliing
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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 - \small\muv, if F were conservative then there would exist P(x) such that \small -\frac{dP}{dx} = F. 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:
\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
So let \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.
For underdamped \small\zeta<1\Rightarrow x=e^{-\zeta\omega_{\circ}t}(C_1 cos\omega_1 t + C_2 sin\omega_1 t)
\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
Therefore x(t) is not 1-1 \small\Rightarrow \int_{x_1}^{x_2}vdx is multivalued implies W is not a function implies p(x) doesn't exist (since W=-\small\DeltaP) implies F is not conservative. Similarly for \small\zeta=1.
But in the overdamped case, \small\zeta>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?
I want to show that a damped harmonic oscillator in one dimension is nonconservative. Given F = -kx - \small\muv, if F were conservative then there would exist P(x) such that \small -\frac{dP}{dx} = F. 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:
\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
So let \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.
For underdamped \small\zeta<1\Rightarrow x=e^{-\zeta\omega_{\circ}t}(C_1 cos\omega_1 t + C_2 sin\omega_1 t)
\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
Therefore x(t) is not 1-1 \small\Rightarrow \int_{x_1}^{x_2}vdx is multivalued implies W is not a function implies p(x) doesn't exist (since W=-\small\DeltaP) implies F is not conservative. Similarly for \small\zeta=1.
But in the overdamped case, \small\zeta>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?