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

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

Substituting values we get

$\frac{{\mathrm{d}}^2\theta }{\mathrm{d}{t}^2}+\frac{9.81}{0.5}\theta =0.$

$\Rightarrow \frac{{\mathrm{d}}^2\theta }{\mathrm{d}{t}^2}+19.62\theta =0.$

Remember the identities

${ℒ}\{f(t)\}=F(s)=\underset{0}{\overset{\mathrm{\infty }}{\int }}{e}^{-st}f(t)dt.$

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

Now we can take the Laplace Transform to change the second order differential equation, from the t domain, into a simple linear equation, from the s domain, that's much easier to work with

${ℒ}\{\frac{{\mathrm{d}}^2\theta }{\mathrm{d}{t}^2}+19.62\theta \}=s^{2}F(s)-sf(0)-f^{\text{'}}(0)+19.62\theta =0$

$\Rightarrow$ $s^2\mathit{\theta }-s\theta (0)-{\theta }^{\text{'}}(0)+19.62\mathit{\theta }=0$

Since we know that $\theta (0)=12^o$ and the initial velocity ${\theta }^{\text{'}}(0)=0$ we get

$s^2\mathit{\theta }-12s+19.62\mathit{\theta }=0$

$\Rightarrow$ $\mathit{\theta }(s^2+19.62)=12s$

$\Rightarrow$ $\mathit{\theta }=\frac{12s}{s^2+19.62}$

$\Rightarrow$ $\mathit{\theta }=\frac{12s}{s^2+(4.429{)}^2}$

Now we can take the inverse Laplace Transform to convert our equation back into the time domain using the identity

${{ℒ}}^{-1}\left\{\frac{s}{s^2+{\omega }^2}\right\}=cos(\omega t)$

We get

${{ℒ}}^{-1}\left\{\frac{12s}{s^2+(4.429{)}^2}\right\}=12cos(4.429t)$

This will give us the angle (in degrees) of the pendulum at any given time t.