How Can I Solve the Equation for a Pendulum with Large Amplitude?

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SUMMARY

The discussion focuses on solving the equation for a pendulum with large amplitudes, specifically addressing the absence of a closed-form solution for this scenario. The equation of motion is derived as a separable differential equation, leading to the relationship \(\frac{d^2\theta}{dt^2} = \omega \frac{d\omega}{d\theta} = \frac{g}{l} \sin(\theta)\). The solution involves integrating to obtain \(\frac{1}{2}\omega^2 = -\frac{g}{l} \cos(\theta) + C\), which results in an elliptic integral that cannot be integrated in closed form. The phase plane analysis reveals that for low starting speeds, the trajectories form ovals around specific points, indicating the complexity of the pendulum's motion.

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Homework Statement


I really would like some help. Next month I am starting a project with the title "pendulum with larger amplitudes", where I have to come up with a solution on how to solve the equation for the pendulum with large amplitude.


Homework Equations


This is the equation I have to come up with, but I have no idea how to get this.
http://img88.imageshack.us/img88/3345/latex2png896efbxa5.png
I know how to get the equation for the pendulum with small amplitudes, its just the rest, that kills me.


The Attempt at a Solution


I have searched the internet and this forum for hours now. And the only information I am able to find is the final equation and theory about the pendulum with small amplitudes.

I would be really happy if some of you guys are able to help me or give me some links with the theory behind the equation.

Thanks, Jonas
 
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There is no "closed form" solution to the "large angle" pendulum problem and, solutions are NOT necessarily periodic- it is possible to give the pendulum an initial speed so that it "goes over the top" and just continues around and around- so your "period" equation couldn't hold for that. One thing you can do is use "quadrature" on the non-linear pendulum problem. Let [itex]\omega= d\theta /dt[/itex]. Then [itex]d^2\theta/dt^2= d\omega /dt= d\omega/d\theta d\theta /dt= \omega d\omega /d\theta[/itex]
The equation of motion of the pendulum becomes
[tex]\frac{d^2\theta}{dt^2}=\omega \frac{d\omega}{d\theta)= \frac{g}{l} sin(\theta)[/itex]<br /> a relatively simple separable differential equation. Integrating you get <br /> [tex]\frac{1}{2}\omega^2= -\frac{g}{l} cos(\theta)+ C[/tex]<br /> Solving for [itex]\omega= d\theta /dx[/itex] gives a rather complicated root involving [itex]cos(\theta)[/itex] which cannot be integrated in closed form- it is, in fact, an "elliptic integral". If, instead, you were to graph [itex]\frac{1}{2}\omega^2= -\frac{g}{l} cos(\theta)+ C[/itex] in the [itex]\theta-\omega[/itex] plane (the "phase plane") you will see that, for sufficiently low starting speeds, the graphs are ovals around the points (0,0), ([itex]\pi[/itex],0), etc. The period will be related to the distances around those ovals.[/tex]
 

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