Elastic pendulum

Example

Let $g > 0$ be the acceleration of gravity, and let $m, k, \ell_0 > 0$. A point mass $m$ hangs from a fixed pivot on a massless spring with stiffness $k$ and rest length $\ell_0$. It moves without friction in a vertical plane. Let $r$ be the length of the spring and $\theta$ its angle from the downward vertical. The Lagrangian is $$L = \frac{m}{2} \left( r'^2 + r^2 \theta'^2 \right) - \frac{k}{2} (r - \ell_0)^2 + m g r \cos \theta.$$ So the elastic pendulum is a Hamiltonian system with two degrees of freedom. Its equilibrium hangs straight down with length $\ell = \ell_0 + m g / k$. Check: at $\theta = 0$ and $r' = 0$, the radial equation $m r'' = m r \theta'^2 - k (r - \ell_0) + m g \cos \theta$ gives $r'' = 0$ exactly when $k (r - \ell_0) = m g$.