Motion of Pendulum

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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 $s$.
MotionOfPendulum.png

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

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