# Finding Singular points

1. Apr 19, 2008

### 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..

Last edited: Apr 19, 2008
2. Apr 19, 2008

### HallsofIvy

Staff Emeritus
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 $\sin(\theta)= 0$.

3. Apr 19, 2008

### 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.