# Equilibrium and Potential Energy

1. Apr 5, 2007

### xaer04

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
$$U(x) = U_0 \left( \frac {x^3}{(x_0)^3} + a \frac{x^2}{(x_0)^2} + 4 \frac{x}{x_0} \right)$$
where $U_0$, $x_0$, and $a$ are constants. For what values of $a$ will there be two static equilibria? Comment on the stability of these equilibria."

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

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 $a$ to be 1), and i know (from trial and error) the approximate value of $a$ to begin to create two points of equilibrium (which will just be maxima and minima on the graph; when $\frac{dU}{dx} = 0$). 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
$$0 = U_0 \left( \frac {3x^2}{(x_0)^3} + a \frac{2x}{(x_0)^2} + \frac{4}{x_0} \right)$$

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

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

I took the second derivative and a became
$$a = \frac{-3x}{x_0}$$

x became
$$x = \frac{-ax_0}{3}$$

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 $a^2$, 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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Last edited: Apr 5, 2007
2. Apr 5, 2007

### denverdoc

have you tried posting on advanced physics.

3. Apr 5, 2007

### xaer04

:)

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.

4. Apr 5, 2007

### Saketh

In order for there to be two real values where potential energy is at an extremum, the two values of x must be real.

When you solve for $$\frac{dU}{dx}$$, you end up with a quadratic expression in x. If you solve for x with the quadratic formula, you will get an expression with a radical. In order for the values of x to be real, the expression inside the radical must be greater than zero.

When I did it through elementary calculus and algebra, I got something like this:
$$x = \frac{-2ax_0 \pm \sqrt{4a^2 x_0^2 - 48x_0^2}}{6}$$
From this, it follows that $$4a^2 x_0^2 - 48x_0^2$$ must be greater than zero. It should not be equal to zero, because then we will only have one extremum.

Last edited: Apr 5, 2007
5. Apr 5, 2007

### eigenglue

You have an equilibrium point where x has two real values, which would be wherever the term inside the square root is greater than 0.

You don't need to set the second derivative to zero. You compare the second derivative to zero to determine whether the points above were minima or maxima, which tells you whether the equilibrium points were stable or unstable.

6. Apr 5, 2007

### xaer04

Awesome... thank you.

If i set the piece under the radical to zero, i get
$$a = \pm \sqrt{12}$$

Both radicals(yours and mine) work the same... i just factored out a $4{x_0}^2$ 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:)