Pendulum equation

Example

Let $g > 0$ be the acceleration of gravity and $\ell > 0$ the length of a rigid massless rod, with a point mass at its end. Let $\theta(t)$ be the angle of the rod from the downward vertical at time $t$. Without friction, the motion satisfies the pendulum equation $$\theta''(t) + \frac{g}{\ell} \sin \theta(t) = 0.$$ As a first-order system, it is $\theta' = \omega$, $\omega' = -\frac{g}{\ell} \sin \theta$. The function $E(\theta, \omega) = \frac{1}{2} \omega^2 - \frac{g}{\ell} \cos \theta$ is a first integral.

Remarks

For small angles, $\sin \theta \approx \theta$. This gives the harmonic oscillator $\theta'' + \frac{g}{\ell} \theta = 0$, with the solution $\theta(t) = \theta_0 \cos\left(\sqrt{g/\ell}\, t\right)$ for $\theta(0) = \theta_0$, $\theta'(0) = 0$, and the period $2 \pi \sqrt{\ell / g}$. For the nonlinear equation, the period depends on the amplitude. It is given by an elliptic integral and it is longer than $2 \pi \sqrt{\ell / g}$. The equilibrium $(0, 0)$, hanging down, is stable. The equilibrium $(\pi, 0)$, standing up, is a hyperbolic saddle, so it is unstable.

Explanations

L1 human

Gravity pulls the mass back toward the lowest point, with a force proportional to $\sin \theta$. For small swings, this is nearly proportional to $\theta$, and the pendulum behaves like a spring. For large swings, the motion is slower, and with enough energy the pendulum turns over the top.