Fall 2009/JonathanS

Problem

A simple pendulum with a length L = 0.5m is pulled back and released from an initial angle $\theta _{0}=12^{o}$. Find it's location at t = 3s.

Solution

Assuming no damping and a small angle($\theta <15^{o}$), the equation for the motion of a simple pendulum can be written as:

${\mathrm {d} ^{2}\theta \over \mathrm {d} t^{2}}+{g \over \ell }\theta =0.$

We can then use the Laplace Transform to convert from the time(t) domain into the s domain.

Given

${\mathcal {L}}\{f(t)\}=F(s)$

${\mathrm {d} ^{2}\theta \over \mathrm {d} t^{2}}=f^{''}(t)$

${\mathcal {L}}\{f^{''}(t)\}=s^{2}F(s)-sf(0)-f^{'}(0)$

We have

${\mathcal {L}}\{{\mathrm {d} ^{2}\theta \over \mathrm {d} t^{2}}+{g \over \ell }\theta \}=s^{2}F(s)-sf(0)-f^{'}(0)+{g \over \ell }\theta$