Motion of a 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:

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

where:

• $\displaystyle k = \sin \left({\frac \alpha 2}\right)$

• $K \left({k}\right)$ 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 $s$

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:

$\displaystyle \frac 1 2 m v^2 = m g \left({a \cos \theta - a \cos \alpha}\right)$

We have that:

• $s = a \theta$
• $\displaystyle v = \frac{\mathrm d s}{\mathrm d t} = a \frac{\mathrm d \theta}{\mathrm d t}$

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

So:

	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle v\)	\(=\)	\(\displaystyle \frac{\mathrm d s}{\mathrm d t} = a \frac{\mathrm d \theta}{\mathrm d t}\)	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)
	\(\displaystyle \)	\(\displaystyle \implies\)	\(\displaystyle \)	\(\displaystyle \frac {a^2} 2 \left({\frac{\mathrm d \theta}{\mathrm d t} }\right)^2\)	\(=\)	\(\displaystyle g a \left({a \cos \theta - a \cos \alpha}\right)\)	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)
	\(\displaystyle \)	\(\displaystyle \implies\)	\(\displaystyle \)	\(\displaystyle \frac{\mathrm d \theta}{\mathrm d t}\)	\(=\)	\(\displaystyle -\sqrt {\frac {2 g \left({\cos \theta - \cos \alpha}\right)} a}\)	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)
	\(\displaystyle \)	\(\displaystyle \implies\)	\(\displaystyle \)	\(\displaystyle \frac{\mathrm d t}{\mathrm d \theta}\)	\(=\)	\(\displaystyle -\sqrt {\frac a {2 g \left({\cos \theta - \cos \alpha}\right)} }\)	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)
	\(\displaystyle \)	\(\displaystyle \implies\)	\(\displaystyle \)	\(\displaystyle \frac T 4\)	\(=\)	\(\displaystyle -\sqrt {\frac a {2 g} } \int_\alpha^0 \frac {\mathrm d \theta} {\sqrt {\cos \theta - \cos \alpha} }\)	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)
	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)	\(=\)	\(\displaystyle \sqrt{\frac a {2 g} } \int_0^\alpha \frac {\mathrm d \theta} {\sqrt {\cos \theta - \cos \alpha} }\)	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)

Substituting:

$\displaystyle \cos \theta = 1 - 2 \sin^2 \frac \theta 2, \cos \alpha = 1 - 2 \sin^2 \frac \alpha 2$

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 $\displaystyle k = \sin \frac \alpha 2$:

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

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

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

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

Differentiating with respect to $\phi$ we have:

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

Thus:

	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \mathrm d \theta\)	\(=\)	\(\displaystyle \frac {2 k \cos \phi} {\cos \frac \theta 2} \mathrm d \phi\)	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)
	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)	\(=\)	\(\displaystyle \frac {2k \sqrt {1 - \sin^2 \phi} } {\sqrt{1 - \sin^2 \frac \theta 2} } \mathrm d \phi\)	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)
	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)	\(=\)	\(\displaystyle \frac {2 \sqrt {k^2 - \sin^2 \frac \theta 2} } {\sqrt{1 - k^2 \sin^2 \phi} } \mathrm d \phi\)	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)

Then:

	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle T\)	\(=\)	\(\displaystyle 2 \sqrt {\frac a g} \int_0^{\pi / 2} \frac {2 \sqrt {k^2 - \sin^2 \frac \theta 2} } {\sqrt{1 - k^2 \sin^2 \phi} } \frac {\mathrm d \phi} {\sqrt {k^2 - \sin^2 \frac \theta 2} }\)	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)
	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)	\(=\)	\(\displaystyle 4 \sqrt {\frac a g} \int_0^{\pi / 2} \frac {\mathrm d \phi} {\sqrt{1 - k^2 \sin^2 \phi} }\)	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)

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.

$\blacksquare$

Sources

• George F. Simmons: Differential Equations (1972): $\S 5$