Equilibrium and Potential Energy

In summary, the potential energy 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. To find two static equilibria, the values of a must be such that the expression inside the square root in the quadratic formula is greater than zero. The second derivative can be used to determine the stability of the equilibria.
  • #1
38
0

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?
 

Attachments

  • untitled.GIF
    untitled.GIF
    1.5 KB · Views: 414
Last edited:
Physics news on Phys.org
  • #2
have you tried posting on advanced physics.
 
  • #3
:)

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
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 [tex]\frac{dU}{dx}[/tex], 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:
[tex]
x = \frac{-2ax_0 \pm \sqrt{4a^2 x_0^2 - 48x_0^2}}{6}
[/tex]
From this, it follows that [tex]4a^2 x_0^2 - 48x_0^2[/tex] must be greater than zero. It should not be equal to zero, because then we will only have one extremum.
 
Last edited:
  • #5
I solved for x, and that yielded something more cryptic:
[tex]x = \frac{a \pm \sqrt{a^2 -12} }{3/x_0}[/tex]

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.

I took the second derivative and a became

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
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:)
 

What is equilibrium?

Equilibrium is a state in which the forces acting on an object are balanced, resulting in a stable and unchanging system.

How is potential energy related to equilibrium?

Potential energy is the energy an object possesses due to its position or configuration within a system. In equilibrium, potential energy is at a minimum, meaning there is no tendency for the object to move or change its position.

What factors affect equilibrium and potential energy?

The factors that affect equilibrium and potential energy include the mass of the object, the distance between objects, and the strength of the forces acting on the object.

What are examples of systems in equilibrium?

Examples of systems in equilibrium include a pendulum at rest, a book sitting on a table, and a ball rolling at constant speed on a flat surface.

How is equilibrium important in everyday life?

Equilibrium is important in everyday life because it allows for stability and balance in systems. For example, our bodies maintain equilibrium to keep our temperature and blood pressure at consistent levels. Equilibrium also plays a role in engineering, economics, and other fields.

Suggested for: Equilibrium and Potential Energy

Replies
10
Views
529
Replies
4
Views
383
Replies
21
Views
523
Replies
6
Views
1K
Replies
3
Views
782
Back
Top