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When finding the period of a pendulum beyond the small angle approximation, we have to use integration for small interval of $\theta$ and elliptical integration.

I was trying to apply this situation for the driven pendulum. When using small angle approximation, the equation of motion would be like below. $$I \ddot{x}(t)+b \dot x(t)+m g l x(t)=F\cos(\omega t)$$

  • $I$ : moment of inertia
  • $b$ : damping coefficient
  • $m$ : mass
  • $g$ : gravity acceleration
  • $l$ : distance between CM of the system and the origin of rotation
  • $F$ : amplitude of external force
  • $\omega$ : angular frequency of external force
  • $x$ : angular displacement
  • $t$ : time

When not using small angle approximation, above equation would be changed. $$I \ddot x(t)+b\dot x(t)+m g l\sin(x(t))=F\cos(\omega t)\cos(x(t)).$$ Then, how do I solve this differential equation? I tried to simulate the result with Mathematica's DSolve, but it didn't show the solution. Although I plotted the graph with NDSolve, I want to know the way to solve that differential equation.

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  • $\begingroup$ The reason one uses the small angle approximation is that the equation becomes easy to solve analytically. I suspect the more general case does not have an analytical solution, so in a sense your plots are the closest you can come to a solution. $\endgroup$
    – alarge
    Commented Jan 14, 2015 at 14:10
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    $\begingroup$ Note that not every ODE can be solved analytically; numerical methods are attractive alternatives to the ones without analytic solutions. $\endgroup$
    – Kyle Kanos
    Commented Jan 14, 2015 at 14:11
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    $\begingroup$ See this PDF. Even for the simplest undamped undriven case there is no analytic solution to the equations of motion. For a driven damped non-harmonic oscillator you have no chance! $\endgroup$
    – John Rennie
    Commented Jan 14, 2015 at 14:11
  • $\begingroup$ Thanks for your comments. I should better use the numerical solutions. $\endgroup$
    – Linear Chaos
    Commented Jan 14, 2015 at 14:22
  • $\begingroup$ @YeongWooSong If you used Mathematica, try recreating some of the pictures that pop up when you search "Damped Driven Pendulum" on google images: google.com/search?q=driven+damped+pendulum&tbm=isch Hopefully you'll end up convinced that your teachers were right to leave out the nonlinear term! (Most of those images probably have a forcing of $F\cos(\omega t)$ instead of $F\cos(\omega t)\cos(x(t))$, so you shouldn't expect to reproduce them exactly) $\endgroup$
    – user12029
    Commented Jan 14, 2015 at 21:42

2 Answers 2

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As was mentioned in the comments, the differential equation you give is not solvable analytically. What one can do is one can go beyond the small angle approximation in a controlled fashion by Taylor expanding the sine and cosine function and find (e.g. expanding up to order $x^3$)

$$I \ddot x + b\dot x + mgl\left( x - \frac{x^3}{6}\right) = F \cos(\omega t) \left( 1 - \frac{x^2}{2} \right) $$

This still is a non-linear differential equation and might not be solvable either, but it opens the door for Perturbation Theory, which is the major tool to use when going beyong linear approximations.

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  • $\begingroup$ Aha. I didn't thought about the taylor expanding. Thanks for your idea. $\endgroup$
    – Linear Chaos
    Commented Jan 14, 2015 at 14:23
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    $\begingroup$ @YeongWooSong I find that surprising, since Taylor expansion is how you get the small angle equation in the first place. $\endgroup$
    – Danu
    Commented Jan 14, 2015 at 14:41
  • $\begingroup$ @Danu: Is it? $\endgroup$
    – user541686
    Commented Jan 14, 2015 at 23:39
  • $\begingroup$ @Danu I only thought about the first term or the expansions. Didn't thought about the second, third power of it. $\endgroup$
    – Linear Chaos
    Commented Jan 15, 2015 at 2:03
  • $\begingroup$ @Danu: Did you notice & click on the link in my comment? The linear approximation doesn't depend on the Taylor series, rather than Taylor series combines the linear approximation with higher-order approximation... I feel like it's backwards to put it the other way. $\endgroup$
    – user541686
    Commented Jan 15, 2015 at 7:28
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The solutions for a forced/driven pendulum can be chaotic, in the sense of chaos theory, so the period may not even exist! For a tutorial on see these course notes for example. What you can do with an forced/driven pendulum is to simulate it and compute the various chaos-theory related parameters: Lyapunov exponents, Poincare map, etc. There are numerous resources on the web for that. Here are a few such resources:

  • A Mathematica CDF, which has some [foot]notes on the implementaion.
  • There also various (more or less annoying) applets on the web with a similar simulation. Essentially they are all based on Runge-Kuta numerical solutions. You can find more concise/intelligible Java source code in Tao Pang An Introduction to Computational Physics, pp. 92-93
  • A short paper on a Python scipy implementation.
  • A question here about Lyapunov exponents for the driven pendulum equation.

Etc. If all you want is to compute the period for non-small angles for an unforced pendulum, Wikipedia has some answers, most notably the fairly fast (in practice) Carvalhaes and Suppes approximation.

Addendum for the more mathematically inclined: the perturbation method doesn't always work for the forced pendulum equation. In particular it doesn't work for Hubbard’s pendulum equation, which is an instance of the general forced pendulum equation. There's a relatively recent (2008) computer-assited proof of chaotic behavior in that case.

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