Merrily Swinging Around

The net torque on the pendulum about the pivot is given by $\tau = - mgl \sin \theta$ . We will use the rotational analogue of Newton's second law motion: net torque ( $\tau$ ) is equal to $I \alpha$ .

$\begin{aligned} I \alpha & = - mgl \sin \theta \\ I \frac{d^2 \theta}{dt^2} & = - mgl \sin \theta \\ ml^2 \frac{d^2 \theta}{dt^2} &= - mgl \sin \theta \end{aligned}$

We obtain the differential equation

$\frac{d^2 \theta}{dt^2} = - \frac{g}{l} \sin \theta$

We cannot find a closed form solution for this differential equation. However, we can find an infinite series which is sufficient to find the dependence between $T$ and $\theta$ .

$T = 2 \pi \sqrt{\frac{l}{g}} \left( 1 + \frac{1}{16} \theta_0^2 + \frac{11}{3072} \theta_0 ^4 + \cdots \right)$

This is an increasing function since all the coefficients are positive. Thus we see that as $\theta_0$ increases, $T$ also increases, although very slowly.

We know that $\omega ^{2}$ $\theta$ = $\alpha$ .

Hence T= 2 $\pi$ / $\omega$ hence T Varies directly upon $\theta$ or amplitude.

This involves a integral which on applying maclaurins expansion gives a form called as elliptic integrals.fine with the result