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

where:
 * $k = \sin \left({\dfrac \alpha 2}\right)$
 * $K \left({k}\right)$ 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 $s$.


 * MotionOfPendulum.png

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

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

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

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

So:

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

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

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


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

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


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

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

Differentiating with respect to $\phi$ we have:


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

Thus:

Then:

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

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.