Where Can I Find the Singular Points for a Pendulum System?

[ash4sigh]

hi,

given the system ml^{2}\theta''+b\theta'+mglsin(\theta)

how do I find the singular points??

or any system for that matter - trying the isocline method just not working! tedious..

 

[HallsofIvy]

First that isn't a system, it is a single equation (actually what you wrote isn't even an equation but I assume that was supposed to be "= 0").

Start by writing it as a system of equations: let \omega= \theta' so that \theta"= \omega' and your one equation becomes two first order equations:
ml^2\omega'+ b\omega+ mglsin(\theta)= 0 and \theta'= \omega or
ml^2\omega'= -b\omega- mglsin(\theta) and \theta'= \omega.

Now "singular points" (or "equilibrium points), points that are single point solutions to the system, are those (\theta, \omega) points where the right hand sides of those equations are 0. (I'm very surprised you didn't know that.)

In other words, you must solve the pair of equations b\omega+ mgl sin(\theta)= 0 and \omega= 0. And that, obviously, reduces to solving \sin(\theta)= 0.

 

[ash4sigh]

it is a pendulum system - not sure where the second theta came from in the first term though..

so it'll be (\theta= 0+k\Pi,w=0) where k is an integer

thank you very much - I have a million and one questions to ask

the hard part with this one is that I am trying to see it from a phase plane perspective - and visualising where the isoclines converge when you can only draw a couple by hand is tough for a newbie..

thank you for your time, I will be sure to be back.