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A simple gravity pendulum is the simplified model for a pendulum: a point mass suspended from a massless cord suspended from a pivot in a vacuum subject to gravity.

In my class, I studied very simplified spring mass systems where the solution to the second order linear homogeneous equation $y''+C_1y'+C_2y=0$ was the position function of the point mass attached to the spring (where $C_1$ and $C_2$ are the damping constant and the spring constant respectively).

I was told that a simple gravity pendulum could be modeled with a similar second order differential equation, and I would appreciate if someone could derive this equation in a didactic and systematic manner, so I could fully appreciate and understand this model (which, I know, is not very realistic, but interesting to me nonetheless). Thank you.

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up vote 3 down vote accepted

Unfortunately I don't have enough reputation to post images, so you'll have to look at this picture instead.

Consider a pendulum of length $L$ and mass $m$. We denote the angle the pendulum makes with the vertical $\theta$, so that the pendulum is at rest when $\theta = 0$.

The dynamics of the pendulum are governed by Newton's second law, $\tau = I \alpha$. Here $\tau$ is the torque around the pivot, $I$ is the rotational inertia of the pendulum and $\alpha$ is its angular acceleration.

Make the following substitutions:

  1. Replace $\alpha$ with $\theta''$. Acceleration is the second derivative of position.
  2. Replace $I$ with $mL^{2}$ . This is the rotational inertia of a point mass $m$ a distance $L$ from the pivot.
  3. Replace $\tau$ with $-L mg \sin(\theta)$. The torque generated by a force depends on the distance between the pivot point and where the force acts, $L$, and the component of the force perpendicular to the line joining the pivot and the mass, $mg\sin(\theta)$. The negative sign is because gravity acts as a restoring force: when $\theta$ is positive the torque acts in the negative direction and vice versa.

These substitutions give: $$-mg\sin(\theta)L = mL^{2}\theta''.$$

We cancel $mL$ on both sides and rearrange: $$\theta'' + \frac{g}{L}\sin(\theta) = 0.$$

This equation has solutions in terms of elliptic functions, which are complicated, but we can avoid all that unpleasantness by making the small-angle approximation: $\sin(\theta) \sim \theta.$ So provided that the pendulum only makes small oscillations its motion is described by $$\theta'' + \frac{g}{L}\theta = 0.$$

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