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:

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

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

Given

${ℒ}\{f(t)\}=F(s)$

$\frac{{\mathrm{d}}^2\theta }{\mathrm{d}{t}^2}=f^{{\text{'}}^{\prime }}(t)$

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

We have

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