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

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

$\Rightarrow$ Failed to parse (unknown function "\s"): {\displaystyle \s^2\theta-s\theta(0)-\theta^{ '}(0)+19.62\theta=0}