The equation of motion of a pendulum subjected to gravity reads:

$\displaystyle I\frac{d^2 \theta}{dt^2}=-m g \ell \sin \theta$

By defining

$\displaystyle \omega_0 = \sqrt{\frac{mg\ell}{I}}$

and by using:

$\displaystyle \dot \theta \equiv \frac{d \theta}{dt}, \quad \ddot \theta \equiv \frac{d^2 \theta}{dt^2}$

this becomes:

$\displaystyle \ddot \theta = -\omega_0^2 \sin \theta$

(1)

For small oscillations (i.e. $\sin \theta \approx \theta$ for small $\theta$ ) the equation of motion becomes that of a harmonic oscillator:

$\displaystyle \ddot \theta = -\omega_0^2 \theta$

with solution:

$\displaystyle \theta(t) = A \cos (\omega_0 t + \psi),$

where

and $\psi$ are determined by the initial conditions $\theta(t=0) \equiv \theta_0$ and $\dot \theta(t=0) \equiv \dot \theta_0$ .

The period ( $\tau_0$ ) for small oscillations (small $\theta$ ) is:

$\displaystyle \tau_0 = \frac{2\pi}{\omega_0}$

$\tau_0$ does not depend on $\theta_0$ or $\dot \theta_0$ because of the small angle approximation. Obviously this approximation breaks down at larger angles. An estimate of the deviation from harmonic motion can be obtained as follows. Write $\ddot \theta$ as:

$\displaystyle \ddot \theta=\frac{d \dot \theta}{d\theta}\frac{d\theta}{dt}=\dot... ...c{d\dot\theta}{d\theta}=\frac{d}{d\theta}\left(\frac{\dot \theta^2}{2}\right)$

Integrate (1)

$\displaystyle \int_0^{\dot \theta} d \left( \frac{\dot \theta^{'2}}{2} \right) = -\omega_0^2 \int_{\theta_0}^{\theta}\sin \theta' d\theta'$

The solution is:

$\displaystyle \dot \theta^2=-2\omega_0^2(\cos\theta - \cos \theta_0)$

Taking the square root on both sides and rearranging gives:

$\displaystyle \frac{d\theta}{\sqrt{\cos \theta - \cos \theta_0}}=\pm\sqrt{2}\omega_0dt$

Integrate again:

$\displaystyle \sqrt{2}\omega_0\int_0^t dt' = -\int_{\theta_0}^\theta \frac{d\theta'}{\sqrt{\cos \theta' - \cos \theta_0}}$

We take $\theta_0$ to be positive. Since $\theta<\theta_0$ , the integral on the right hand side is negative. By means of the identity:

$\displaystyle \cos\theta = 1-2\sin^2 \frac{\theta}{2}$

we get the integral:

$\displaystyle 2\omega_0 t = -\int_{\theta_0}^\theta \frac{d\theta'}{\sqrt{\sin^2(\theta_0/2)- \sin^2 (\theta'/2)}}$

Rewrite by introducing the new variable $\beta$ defined by:

$\displaystyle \sin \beta = \frac{\sin(\theta/2)}{\sin(\theta_0/2)}$

The integral becomes:

$\displaystyle \omega_0 t = \int_{\beta}^{\pi/2} \frac{d\beta'}{\sqrt{1-\sin^2(\theta_0/2) \sin^2\beta'}}$

Take $\theta=0$ (so also $\beta=0$ ) and integrate over a quarter of a period:

$\displaystyle \tau= \frac{4}{\omega_0} \int_0^{\pi/2} \frac{d\beta}{\sqrt{1-\sin^2(\theta_0/2) \sin^2\beta}}$

This integral is known as the complete elliptic integral of the first kind. This integral does not have an analytical solution, but it can be approximated using a power series expansion. Note that in the approximation to small angles the period was $\tau_0 = 2\pi/\omega_0$ , giving the ratio:

$\displaystyle \frac{\tau}{\tau_0}= \frac{2}{\pi} \int_0^{\pi/2} \frac{d\beta}{\sqrt{1-\sin^2(\theta_0/2) \sin^2\beta}}$

This integral can be expanded by a power series, so each term in the power series can be integrated separately:

$\displaystyle \frac{\tau}{\tau_0}$	$\displaystyle = \frac{2}{\pi} \int_0^{\pi/2}d\beta\left(1+\frac{1}{2}\sin^2\frac{\theta_0}{2}\sin^2\beta +\ldots \right)$
	$\displaystyle = \frac{2}{\pi}\left[\beta + \frac{1}{4}\sin^2\left(\frac{\theta_0}{2}\right)\left(\beta-\frac{\sin 2\beta}{2}\right) +\ldots \right]_0^{\pi/2}$
	$\displaystyle =\left[1+\frac{1}{4}\sin^2\left(\frac{\theta_0}{2}\right)\right] + \ldots$

Again using the harmonic approximation for $\theta_0$ , approximate $\sin^2(\theta_0/2)\approx \theta_0^2/4$ to get:

$\displaystyle \tau=\tau_0\left(1+\frac{\theta_0^2}{16}+\ldots\right)$

Note that the period increases with $\theta_0$ . The fractional lengthening therefore becomes:

$\displaystyle \frac{\tau-\tau_0}{\tau_0}\approx\frac{\theta_0^2}{16}$

The small angle approximation only differs $1.7\%$ for $\theta_0$