Equilibrium and Potential Energy

1. The problem statement, all variables and given/known data
"The potential energy as a function of position for a certain particle is given by
[tex]U(x) = U_0 \left( \frac {x^3}{(x_0)^3} + a \frac{x^2}{(x_0)^2} + 4 \frac{x}{x_0} \right)[/tex]
where [itex]U_0[/itex], [itex]x_0[/itex], and [itex]a[/itex] are constants. For what values of [itex]a[/itex] will there be two static equilibria? Comment on the stability of these equilibria."

2. Relevant equations
[tex]U(x) = U_0 \left( \frac {x^3}{(x_0)^3} + a \frac{x^2}{(x_0)^2} + 4 \frac{x}{x_0} \right)[/tex]

3. The attempt at a solution

I know several things - I know what the graph should basically look like(can change direction, degree of curvature, etc) to meet the conditions of the problem (and it is attached. i assumed positive constants, and i assumed constants except for [itex]a[/itex] to be 1), and i know (from trial and error) the approximate value of [itex]a[/itex] to begin to create two points of equilibrium (which will just be maxima and minima on the graph; when [itex]\frac{dU}{dx} = 0[/itex]). That value is between 3.4 and 3.5.

i took the first derivative of the formula and set it equal to zero, giving me
[tex]0 = U_0 \left( \frac {3x^2}{(x_0)^3} + a \frac{2x}{(x_0)^2} + \frac{4}{x_0} \right)[/tex]

I solved for a, and it gave me
[tex]a = \frac{-3x}{2x_0} + \frac{-2x_0}{x}[/tex]

I solved for x, and that yielded something more cryptic:
[tex]x = \frac{a \pm \sqrt{a^2 -12} }{3/x_0}[/tex]

I took the second derivative and a became
[tex]a = \frac{-3x}{x_0}[/tex]

x became
[tex]x = \frac{-ax_0}{3}[/tex]

My problem is, I don't know how to interpret this with numbers as constants. Furthermore, i checked with a friend, and he said he saw someone with an answer in terms of [itex]a^2[/itex], which sounds completely ridiculous... but I have no idea at this point. I think if i get past this first portion i can solve the second portion because i know relative maxima yield unstable equilibrium, and relative minima yield metastability. Could someone help me out?

 

:)

heh... not yet. i figured it would be simple because i'm in introductory engineering physics. i'll try the calculus forum and the advanced physics forum.

 

Awesome... thank you.

If i set the piece under the radical to zero, i get
[tex]a = \pm \sqrt{12}[/tex]

Both radicals(yours and mine) work the same... i just factored out a [itex]4{x_0}^2[/itex] and square rooted that to get it out of the radical... i like them to be as simple as possible.

When i graph it it seems to work out how i want to now. Thank you very much:)