Motion on a Circle

Department of Mathematical and Statistical Sciences

Mathematics Section

Geometry in Motion

Simple Pendulum

The simple pendulum consists of a mass attached to a string of length which is fixed at a point 0 as shown. If the mass is pulled away from the vertical and released from rest it will execute oscillatory motion. With the assumption that the string has negligible mass, it can be shown that the equation of motion of the mass is

$\displaystyle \ddot{\theta} =-\frac{g}{L} \sin \theta$

Here $\theta$ measures the angle the string makes with respect to the vertical, is the length of the string and the acceleration due to gravity.

The above equation is a second order non-linear equation. For small values of $\theta$ we can make the approximation $\sin \theta \approx \theta$ and obtain the linear approximation

$\displaystyle \ddot{\theta} =-\frac{g}{L} \theta.$

This equation is often written as

$\displaystyle \ddot{\theta} =-\omega^2 \theta$

where

$\displaystyle \omega =\sqrt{\frac{g}{L}}.$

It represents simple harmonic motion and the general solution is

$\displaystyle \theta =A \cos \omega t +B\sin \omega t.$

If initially the bob starts at $\theta =\theta_M$ with $\dot{\theta} =0$ , then

$\displaystyle \theta =\theta_M \cos \omega t.$

Since $\vert \sin \theta \vert \leq \vert \theta \vert$ with equality at $\theta =0$ it follows that for the true pendulum equation, the magnitude of acceleration for a given $\theta$ is smaller than for the linearized pendulum. Hence the true pendulum runs "slower" than the linearized pendulum.

Consider our original equation

$\displaystyle \ddot{\theta} =-\frac{g}{L} \sin \theta.$

We will transform this differential equation into standard form.

Note that

$\displaystyle \frac{d}{dt} \left(\frac{1}{2} \dot{\theta}^2 \right) =\ddot{\theta} \dot{\theta}.$

Hence

$\displaystyle \frac{d}{dt} \left(\frac{1}{2} \dot{\theta}^2 \right) =-\frac{g}{L} \sin \theta \dot{\theta}.$

$\displaystyle \int^t_0 \frac{d}{dt} \left(\frac{1}{2} \dot{\theta}^2 \right) dt$	$\displaystyle = \int^t_0 -\frac{g}{L} \sin \theta \dot{\theta} dt$
$\displaystyle \frac{1}{2} \dot{\theta}^2$	$\displaystyle = \int^\theta_{\theta_M} -\frac{g}{L} \sin \theta d \theta$
	$\displaystyle (\theta =\theta_M \,$ and $\displaystyle \, \dot{\theta} =0 \;$ at $\displaystyle t=0)$
$% latex2html id marker 52 $\displaystyle \therefore \quad \frac{1}{2} \dot{\theta}^2$$	$\displaystyle = \frac{g}{L} (\cos \theta -\cos \theta_M)$

$\displaystyle \dot{\theta}^2 =\frac{4g}{L} \left( \sin^2\left(\frac{\theta_M}{2}\right) -\sin^2\left(\frac{\theta}{2}\right)\right) \quad ($ using $\displaystyle \cos \theta =1-2\sin^2\left(\frac{\theta}{2}\right)).$

Next, let

$\displaystyle k=\sin\left(\frac{\theta_M}{2}\right)$ and set $\displaystyle y=\frac{\sin\left(\frac{\theta}{2}\right)}{k}\,.$

Note $-1\leq y \leq 1.$

is now our new dependent variable.

$\displaystyle \frac{dy}{dt}$	$\displaystyle =\frac{\frac{1}{2} \cos (\frac{1}{2}\theta) \dot{\theta}}{k}$
$% latex2html id marker 66 $\displaystyle \therefore \quad \left(\frac{dy}{dt} \right)^2$$	$\displaystyle =\frac{\frac{1}{4} \cos^2 \left(\frac{1}{2} \theta \right) \dot{\theta}^2}{k^2}$
	$\displaystyle =\frac{1}{4} \frac{(1-k^2y^2)}{k^2} \frac{4g}{L} (k^2-k^2y)$
	$\displaystyle =\frac{g}{L} (1-y^2)(1-k^2y^2)$