# How to find if equilibrium points of a force is un/stable?

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1. Apr 7, 2016

1. The problem statement, all variables and given/known data
U = Ax2 - Bx3

2. Relevant equations
du/dx = 2Ax - 3Bx2

3. The attempt at a solution

If I was given a potential energy function U = Ax2 - Bx3 and am asked to find:

1) The expression for the force as a function of x.

2) The equilibrium points and determine if are they stable or unstable?

So, for 1):
Would I differential the function giving like so?

U' = f'(x) = 2Ax - 3Bx2

Now for 2):
Would I set f'(x) = 0 to find the equilibrium points?

f'(x) = 2Ax - 3Bx2 = 0

In return I get the points of x through the quadratic equation:
x = 0 and X = A/B

If this is all correct how can I determine if a equilibrium point is stable or unstable?

2. Apr 7, 2016

### ehild

Recall the force is negative derivative of the potential energy.
X=0 is correct, but you have a mistake in the other equilibrium point.

The equilibrium is stable if the potential energy is minimum in that point and unstable if it is maximum.

3. Apr 7, 2016

Oh sorry, x = 2A/(3B).
How I find the max and min of potential energy?

4. Apr 7, 2016

### ehild

You found the positions of the extremes, at x=0 and at x=2A/3B.
Have you learnt what should be the second derivative at a maximum and at a minimum?