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Homework Statement
"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."
Homework 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]
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?
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