# Stability Criterion

1. Feb 21, 2009

### nmatthies

Hi! I'm a maths student taking a maths by computer course and this week's homework deals with the kapitsa pendulum (reversed pendulum).

1. The problem statement, all variables and given/known data

Kapitsa Pendulum is a rigid-rod pendulum the suspension point of which is vibrated. It turns out that vibration can stabilize the upper equilibrium point which is unstable in absence of vibration. Here, we will study this effect using a simplified linear non-autonomous equation $$\ddot{x} = x(1+Asin(wt))$$ (1), where constants A >> 1 and w >> 1 model the amplitude and cyclic frequence of the suspension point vibrations. $$x_{0}=0$$ represents the upper equilibrium of the pendulum.

2. Relevant equations

Theory for Kapitsa pendulum predicts that the motion consists of 2 parts, $$x(t)=X(t)+ \tilde{x}(t)$$ (2), with fast oscillations $$\tilde{x}(t)= \frac {-AX(t)} {w^2} sin(wt)$$ (3), and a slowly varying motion X(t) which satisfies the following equation, $$\ddot{X}=(1-\frac {A^2} {2w^2}) X$$ (4).

3. The attempt at a solution

Now one of the questions is to find the analytical solution of the equation (4) and derive the stability of the point X=0 in terms of the parameters A and w.

The analytical solution I have found is $$X(t)=X(0)cos(\sqrt{\frac {A^2} {2w^2} -1}t)$$ (the t is outside of the square root). but I honestly do not know where to start for the stability criterion. What defines stability at a point? and how to express it in terms of A and w? Help will be greatly appreciated.

edit:

does it have something to do with $$\frac {A^2} {2w^2} -1 < 0 or \frac {A^2} {2w^2} -1 > 0$$?

Last edited: Feb 21, 2009
2. Feb 22, 2009

### D H

Staff Emeritus
Bingo. You were given that the slow solution follows

$$\ddot X = \left(1-\frac{A^2}{2\omega^2}\right)X$$

To make it easier to see what is going on, denote $b\equiv 1-A^2/(2\omega^2)$. With this,

$$\ddot X = bX$$

The solutions to this equation behave very differently if b is positive or negative.