1. Not finding help here? Sign up for a free 30min tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Pendulum with larger amplitudes

  1. Nov 18, 2007 #1
    1. The problem statement, all variables and given/known data
    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.

    2. Relevant equations
    This is the equation I have to come up with, but I have no idea how to get this.
    I know how to get the equation for the pendulum with small amplitudes, its just the rest, that kills me.

    3. 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
  2. jcsd
  3. Nov 18, 2007 #2


    User Avatar
    Staff Emeritus
    Science Advisor

    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]
    a relatively simple separable differential equation. Integrating you get
    [tex]\frac{1}{2}\omega^2= -\frac{g}{l} cos(\theta)+ C[/tex]
    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.
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Have something to add?

Similar Discussions: Pendulum with larger amplitudes
  1. A Pendulum (Replies: 8)

  2. Quantum amplitude (Replies: 1)

  3. Scattering amplitude (Replies: 1)