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 so that the rod is at an angle $\alpha$ from the vertical and then released.
Let $T$ be the time period of the pendulum, that is, the time through which the bob takes to travel from one end of its path to the other, and back again.
Then:
- $T = 4 \sqrt {\dfrac a g} \map K k$
where:
- $k = \map \sin {\dfrac \alpha 2}$
- $\map K k$ is the complete elliptic 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 speed of the bob is $0$, so its kinetic energy is $0$.
By the Principle of Conservation of Energy:
- $\dfrac 1 2 m v^2 = m g \paren {a \cos \theta - a \cos \alpha}$
We have that:
- $s = a \theta$
- $v = \dfrac {\d s} {\d t} = a \dfrac {\d \theta} {\d t}$
The rate of change of $s$ at time $t$ is the speed of the bob.
So:
\(\ds v\) | \(=\) | \(\ds \frac {\d s} {\d t} = a \frac {\d \theta} {\d t}\) | ||||||||||||
\(\ds \leadsto \ \ \) | \(\ds \frac {a^2} 2 \paren {\frac {\d \theta} {\d t} }^2\) | \(=\) | \(\ds g a \paren {a \cos \theta - a \cos \alpha}\) | |||||||||||
\(\ds \leadsto \ \ \) | \(\ds \frac {\d \theta} {\d t}\) | \(=\) | \(\ds -\sqrt {\frac {2 g \paren {\cos \theta - \cos \alpha} } a}\) | |||||||||||
\(\ds \leadsto \ \ \) | \(\ds \frac {\d t} {\d \theta}\) | \(=\) | \(\ds -\sqrt {\frac a {2 g \paren {\cos \theta - \cos \alpha} } }\) | |||||||||||
\(\ds \leadsto \ \ \) | \(\ds \frac T 4\) | \(=\) | \(\ds -\sqrt {\frac a {2 g} } \int_\alpha^0 \frac {\d \theta} {\sqrt {\cos \theta - \cos \alpha} }\) | |||||||||||
\(\ds \) | \(=\) | \(\ds \sqrt {\frac a {2 g} } \int_0^\alpha \frac {\d \theta} {\sqrt {\cos \theta - \cos \alpha} }\) |
Substituting:
- $\cos \theta = 1 - 2 \sin^2 \dfrac \theta 2$
- $\cos \alpha = 1 - 2 \sin^2 \dfrac \alpha 2$
we get:
- $\ds T = 2 \sqrt {\frac a g} \int_0^\alpha {\frac {\d \theta} {\sqrt {1 - 2 \sin^2 \frac \theta 2 - 1 + 2 \sin^2 \frac \alpha 2} } } = 2 \sqrt {\frac a g} \int_0^\alpha {\frac {\d \theta} {\sqrt {\sin^2 \frac \alpha 2 - \sin^2 \frac \theta 2} } }$
We now put $k = \sin \dfrac \alpha 2$:
- $\ds T = 2 \sqrt {\frac a g} \int_0^\alpha {\frac {\d \theta} {\sqrt {k^2 - \sin^2 \frac \theta 2} } }$
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:
- $\dfrac 1 2 \cos \dfrac \theta 2 \dfrac {\d \theta} {\d {\phi}} = k \cos \phi$
Thus:
\(\ds \d \theta\) | \(=\) | \(\ds \frac {2 k \cos \phi} {\cos \frac \theta 2} \rd \phi\) | ||||||||||||
\(\ds \) | \(=\) | \(\ds \frac {2 k \sqrt {1 - \sin^2 \phi} } {\sqrt {1 - \sin^2 \frac \theta 2} } \rd \phi\) | ||||||||||||
\(\ds \) | \(=\) | \(\ds \frac {2 \sqrt {k^2 - \sin^2 \frac \theta 2} } {\sqrt {1 - k^2 \sin^2 \phi} } \rd \phi\) |
Then:
\(\ds T\) | \(=\) | \(\ds 2 \sqrt {\frac a g} \int_0^{\pi / 2} \frac {2 \sqrt {k^2 - \sin^2 \frac \theta 2} } {\sqrt {1 - k^2 \sin^2 \phi} } \frac {\d \phi} {\sqrt {k^2 - \sin^2 \frac \theta 2} }\) | ||||||||||||
\(\ds \) | \(=\) | \(\ds 4 \sqrt {\frac a g} \int_0^{\pi / 2} \frac {\d \phi} {\sqrt {1 - k^2 \sin^2 \phi} }\) |
The integral:
- $\ds \int_0^{\pi / 2} {\frac {\d \phi} {\sqrt {1 - k^2 \sin^2 \phi} } }$
is the complete elliptic integral of the first kind and is a function of $k$, defined on the interval $0 < k < 1$.
Hence the result.
$\blacksquare$
Also see
Sources
- 1972: George F. Simmons: Differential Equations ... (previous) ... (next): $1$: The Nature of Differential Equations: $\S 5$: Falling Bodies and Other Rate Problems
- 1989: Ephraim J. Borowski and Jonathan M. Borwein: Dictionary of Mathematics ... (previous) ... (next): elliptic integral
- 2008: Ian Stewart: Taming the Infinite ... (previous) ... (next): Chapter $9$: Patterns in Nature: Differential equations