% vim: tw=50 % 25/01/2022 10AM \subsubsection*{Equilibrium Points} A particle placed at an \emph{equilibrium point} $x_0$ will stay there for all time. \[ m \ddot{x} = -\dfrac{V}{x} \] implies equilibrium points obey \[ \left. \dfrac{V}{x} \right|_{x_0} \] i.e. critical points of $V$. We can look at motion near equilibrium point. Taylor expanding, \[ V(x) \approx V(x_0) + \half (x - x_0)^2 V''(x) + \cdots \] \begin{itemize} \item $V''(x) > 0 \implies$ minimum of $V$, potential of harmonic oscillator \[ m \ddot{x} \approx V''(x_0)(x - x_0) \] This point is \emph{stable}. Particle oscillates with frequency $\omega = \sqrt{\frac{V''(x_0)}{m}}$. \item $V''(x_)) \implies$ maximum of $V \implies$ point is \emph{unstable}. \[ x - x_0 \approx A e^{\alpha t} + B e^{-\alpha t} \] with \[ \alpha = \sqrt{-\frac{V''(x_0)}{m}} .\] \item $v''(x_0) = 0 \implies$ more work needed. \end{itemize} \begin{example*}[the pendulum] The equation of motion is \[ \ddot{\theta} = -\frac{g}{x} \sin \theta \] The energy is \[ E = \half ml^2 \dot{\theta}^2 - mgl \cos \theta .\] \begin{itemize} \item $E > mgl \implies \dot{\theta} \neq 0$ for all $t$. \item $E < mgl \implies \dot{\theta} = 0$ at some point $\theta_0$. $\implies$ oscillates back and forth and \[ E = -mgl \cos \theta_0 .\] \end{itemize} Using the general solution for 1-dimensional system, \[ T = \int_0^{\theta_0} \frac{\dd \theta}{\sqrt{\frac{2E}{ml^2} + \left( \frac{2g}{l} \right) \cos\theta}} = 4\sqrt{\frac{l}{g}} \int_0^{\theta_0} \frac{\dd \theta}{\sqrt{2\cos\theta - 2\cos\theta_0}} \] but, for small $\theta$, $\cos\theta \approx 1 - \frac{\theta^2}{2}$: \[ T \approx 4 \sqrt{\frac{l}{g}} \int_0^{\theta_0} \frac{\dd \theta}{\sqrt{\theta_0^2 - \theta^2}} = 2\pi \sqrt{\frac{l}{g}} \] (Note: independent of $\theta_0$). \\ This is the result for the harmonic oscillator, of course. \end{example*}