Kapitza's pendulum

Example

Let $g > 0$ be the acceleration of gravity, and let $\ell > 0$, $m > 0$, $a > 0$ and $\nu > 0$. A rigid pendulum of length $\ell$ with a point mass $m$ hangs from a pivot that moves up and down with height $a \cos \nu t$. Let $\theta$ be the angle of the rod from the downward vertical. In the frame of the pivot the effective gravity is $g - a \nu^2 \cos \nu t$, so the motion satisfies $$\theta'' + \frac{g - a \nu^2 \cos \nu t}{\ell} \sin \theta = 0.$$ This is the pendulum equation with a periodically varying coefficient, so it is a nonlinear parametric oscillator.

Assume fast and small vibrations: $a \ll \ell$ and $\nu \gg \sqrt{g / \ell}$. Averaging over the fast vibration, the slow part of $\theta$ moves in the effective potential $$U_{\mathrm{eff}}(\theta) = - m g \ell \cos \theta + \frac{m a^2 \nu^2}{4} \sin^2 \theta.$$ Check at the upright position: put $\theta = \pi + \psi$. For small $\psi$, $U_{\mathrm{eff}} = m g \ell + \left( \frac{m a^2 \nu^2}{4} - \frac{m g \ell}{2} \right) \psi^2 + O(\psi^4)$. So $\theta = \pi$ is a minimum of $U_{\mathrm{eff}}$, and the inverted pendulum is stable, when $a^2 \nu^2 > 2 g \ell$.