Approximate 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 a small angle $\alpha$ (less than about $10 \degrees$ or $15 \degrees$) from the vertical and then released.

Let $T$ be the 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:

$T = 2 \pi \sqrt {\dfrac a g}$

where $g$ is the Acceleration Due to Gravity.


Proof

At a time $t$, let:

the rod be at an angle $\theta$ to the 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

From Motion of Pendulum, the equation of motion of the bob is given by:

$\dfrac {a^2} 2 \paren {\dfrac {\d \theta} {\d t} }^2 = g a \paren {a \cos \theta - a \cos \alpha}$

Differentiating with respect to $t$ and simplifying yields:

$a \dfrac {\d^2 \theta} {\d t^2} = -g \sin \theta$

For small angles, $\sin \theta \approx \theta$ and so:

$\dfrac {\d^2 \theta} {\d t^2} + \dfrac g a \theta = 0$

This is an instance of a $2$nd order linear ODE.

From Second Order ODE: $y'' + k^2 y = 0$, this has the solution:

$\theta = c_1 \sin \sqrt {\dfrac g a} t + c_1 \cos \sqrt {\dfrac g a} t$

We have at $t = 0$ that $\theta = \alpha$ and $\dfrac {\d \theta} {\d t} = 0$.

This yields:

$\theta = \alpha \cos \sqrt {\dfrac g a} t$

and so the period of this pendulum is:

$T = 2 \pi \sqrt {\dfrac a g}$

$\blacksquare$


Sources