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

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

Substituting values we get

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

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

Remember the identities

${\mathcal {L}}\{f(t)\}=F(s)=\int _{0}^{\infty }e^{-st}f(t)\,dt.$

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

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

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

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

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

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

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

$\Rightarrow$ ${\boldsymbol {\theta }}={12s \over {s^{2}+19.62}}$

$\Rightarrow$ ${\boldsymbol {\theta }}={12s \over {s^{2}+(4.429)^{2}}}$

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

${\mathcal {L}}^{-1}{\bigg \{}{s \over {s^{2}+\omega ^{2}}}{\bigg \}}=cos(\omega t)$

We get

${\mathcal {L}}^{-1}{\bigg \{}{12s \over {s^{2}+(4.429)^{2}}}{\bigg \}}=12cos(4.429t)$

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