Motion of Pendulum

Theorem
Consider a pendulum consisting of a bob whose mass is $$m$$, at the end of a rod of negligible mass of length $$a$$.

Let the bob be pulled to one side to an angle $$\alpha$$ and then released.

Let $$T$$ be the time period of the pendulum, that is, the time through which the pendulum takes to travel from one end of its path to the other, and back again.

Then:
 * $$T = 2 \sqrt {\frac a g} K \left({k}\right)$$

where:
 * $$k = \sin \left({\frac \alpha 2}\right)$$


 * $$K \left({k}\right)$$ is the Complete Elliptical Integral of the First Kind.

Proof
At a time $$t$$, let:
 * the rod be at an angle $$\theta$$ to the vertical
 * the bob be travelling at a speed $$v$$
 * the displacement of the bob from where it is when the rod is vertical, along its line of travel, be $$s$$


 * Pendulum.png

At its maximum displacement, the velocity of the bob is zero, so its kinetic energy is zero.

By the Principle of Conservation of Energy we have:
 * $$\frac 1 2 m v^2 = m g \left({a \cos \theta - a \cos \alpha}\right)$$

We have that:
 * $$s = a \theta$$
 * $$v = \frac{\mathrm{d}{s}}{\mathrm{d}{t}} = a \frac{\mathrm{d}{\theta}}{\mathrm{d}{t}}$$

The rate of change of $$s$$ at time $$t$$ is the velocity of the bob.

So:

$$ $$ $$ $$ $$ $$

Substituting:
 * $$\cos \theta = 1 - 2 \sin^2 \frac \theta 2, \cos \alpha = 1 - 2 \sin^2 \frac \alpha 2$$

we get:
 * $$T = 2 \sqrt {\frac a g} \int_0^\alpha {\frac {\mathrm{d} \theta} {\sqrt {1 - 2 \sin^2 \frac \theta 2 - 1 + 2 \sin^2 \frac \alpha 2}}} = 2 \sqrt {\frac a g} \int_0^\alpha {\frac {\mathrm{d} \theta} {\sqrt {\sin^2 \frac \alpha 2 - \sin^2 \frac \theta 2}}}$$

We now put $$k = \sin \frac \alpha 2$$:


 * $$T = 2 \sqrt {\frac a g} \int_0^\alpha {\frac {\mathrm{d} \theta} {\sqrt {k^2 - \sin^2 \frac \theta 2}}}$$

Next, let us introduce the variable $$\phi$$, such that:


 * $$\sin \frac \theta 2 = k \sin \phi$$

... and where $$\phi$$ goes from $$0 \to \pi / 2$$ as $$\theta$$ goes from $$0 \to \alpha$$.

Differentiating with respect to $$\phi$$ we have:


 * $$\frac 1 2 \cos \frac \theta 2 \frac{\mathrm{d}{\theta}}{\mathrm{d}{\phi}} = k \cos \phi$$

Thus:

$$ $$ $$

Then:

$$ $$

The integral:
 * $$\int_0^{\pi / 2} {\frac {\mathrm{d} \phi} {\sqrt{1 - k^2 \sin^2 \phi}}}$$

is the Complete Elliptical Integral of the First Kind and is a function of $$k$$, defined on the interval $$0 < k < 1$$.

Hence the result.