# Potential energy and equilibrium

1. Oct 28, 2007

### gills

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

2. Relevant equations

dU/dx = 0

d^2U/dx^2 > 0 = minimum (stable equilibrium)

d^2U/dx^2 < 0 = maximum (unstable equilibrium)

3. The attempt at a solution

I think i've found the solution. I just want someone to see if i'm correct or not:

dU/dx = Uo [(3/Xo^3)*X^2 + (2a/Xo^2)*X + (4/Xo)] = 0 --->

I then used the quadratic formula to find the roots using --->
a = (3/Xo^3)
b = (2a/Xo^2)
c = (4/Xo) ----> and ended up with two solutions for when X=0 --->

X1 = Xo(-2a + $$\sqrt{4a^2 - 48}$$ / 6) ---> and --->

X2 = Xo(-2a - $$\sqrt{4a^2 - 48}$$ / 6)

I then found d^2U/dx^2 = --->

Uo [(6/Xo^3)*X + (2a/Xo^2)] ---> I then plugged in the different X values -->

For X1 =
Uo * [(6/Xo^3) * Xo(-2a + $$\sqrt{4a^2 - 48}$$ / 6] + 2a/Xo^2 --->

Uo * [(-2a + $$\sqrt{4a^2 - 48}$$) / Xo^2 + 2a/Xo^2] --->

Uo * ($$\sqrt{4a^2 - 48}$$/Xo^2)

For X2 =

The opposite sign which =

Uo * (-$$\sqrt{4a^2 - 48}$$/Xo^2)

Therefore -->

-$$\sqrt{12}$$ > a > $$\sqrt{12}$$

where X1 computed in d^2U/dx^2 > 0

and X2 computer in d^2U/dx^2 < 0

Therefore, for the two equilibria, X1 is in stable equilibrium

and X2 is in unstable equilibrium

2. Oct 28, 2007

### gills

hmmm, not sure if this is correct

3. Oct 28, 2007

### gills

nothing but crickets

4. Oct 31, 2007

### Staff: Mentor

Oops. I thought I had responded, but obviously only in my mind.

Your solution is perfect. The only thing I would change is this:

I'd write it as:
$$a^2 > 12$$

Or:
$$a \in (-\infty, -\sqrt{12}) \cup (\sqrt{12}, +\infty)$$