Understanding Simple Pendulum Stability: Linearization and Derivatives

[xago]

Homework Statement

[PLAIN]http://img825.imageshack.us/img825/2673/58894277.png

Homework Equations

sin(theta) ~ theta for small angles.

The Attempt at a Solution

I'm having a hard time understanding what the question is asking since I don't have a lot of experience in stability. My best attempt is to take sin(theta) ~ theta as the linearized equation, but it says to linearize about an arbitrary angle so what if the angle was not small? The only other thing I know about linearizing about a value is to take the derivative of the function ( the solution to the EOM: theta(t) = Acos(wt) ) which is -Awsin(wt) and solve for t = arbitrary angle. But the derivative is a sin function which is not linear...

 

[Metaleer]

Er, is there an error in the picture? It says third time derivative for the angle, but it's supposed to be second time derivative...

In any case, truncating the Taylor series expansion of the sine,

\sin \theta \approx \sin \theta_0 + \theta \cos \theta_0​

Plugging this into the equation of motion of the pendulum,

\theta '' + \frac{g}{l} \theta \cos \theta_0 = -\frac{g}{l} \sin \theta_0.​

As it is, this is a second order linear equation, but if you want a differential equation with the reference angle set such that there's no constant nonhomogeneous term, you need to look for the angle such that the system is in equilibrium. Setting the angular acceleration to 0, we find

\theta_\text{eq} = - \tan \theta_0.​

We now do a change of variable, \theta^* = \theta - \theta_\text{eq}, so that you're left with the differential equation, now in \theta^*,

\theta^* '' + \frac{g}{l}\theta^* \cos \theta_0 = 0.​

From this equation, you can use the different values of \theta_0 to see if the system is stable or not. Hint: What are the solutions in each case? Are they bounded?

Hope this helps.