How to find if equilibrium points of a force is un/stable?

Click For Summary

Homework Help Overview

The discussion revolves around determining the stability of equilibrium points derived from a potential energy function given by U = Ax² - Bx³. Participants are exploring the relationship between potential energy, force, and equilibrium conditions.

Discussion Character

  • Exploratory, Conceptual clarification, Mathematical reasoning

Approaches and Questions Raised

  • Participants discuss differentiating the potential energy function to find the force and setting the derivative to zero to identify equilibrium points. There is also a focus on determining the stability of these points based on the nature of the potential energy at those locations.

Discussion Status

Some participants have provided guidance on the relationship between the second derivative of the potential energy and the stability of equilibrium points. Multiple interpretations of the equilibrium points are being explored, with some corrections noted regarding their values.

Contextual Notes

There is a mention of potential confusion regarding the calculation of equilibrium points and their stability, as well as the need to analyze the second derivative to classify these points. Participants are also reflecting on the implications of the potential energy function's behavior at the identified points.

Blockade
Messages
68
Reaction score
0

Homework Statement


U = Ax2 - Bx3

Homework Equations


du/dx = 2Ax - 3Bx2

The Attempt at a Solution



If I was given a potential energy function U = Ax2 - Bx3 and am asked to find:

1) The expression for the force as a function of x.

2) The equilibrium points and determine if are they stable or unstable?

So, for 1):
Would I differential the function giving like so?

U' = f'(x) = 2Ax - 3Bx2

Now for 2):
Would I set f'(x) = 0 to find the equilibrium points?

f'(x) = 2Ax - 3Bx2 = 0

In return I get the points of x through the quadratic equation:
x = 0 and X = A/B

If this is all correct how can I determine if a equilibrium point is stable or unstable?
 
Physics news on Phys.org
Blockade said:

Homework Statement


U = Ax2 - Bx3

Homework Equations


du/dx = 2Ax - 3Bx2

The Attempt at a Solution



If I was given a potential energy function U = Ax2 - Bx3 and am asked to find:

1) The expression for the force as a function of x.

2) The equilibrium points and determine if are they stable or unstable?

So, for 1):
Would I differential the function giving like so?

U' = f'(x) = 2Ax - 3Bx2

Recall the force is negative derivative of the potential energy.
Blockade said:
Now for 2):
Would I set f'(x) = 0 to find the equilibrium points?

f'(x) = 2Ax - 3Bx2 = 0

In return I get the points of x through the quadratic equation:
x = 0 and X = A/B

If this is all correct how can I determine if a equilibrium point is stable or unstable?
X=0 is correct, but you have a mistake in the other equilibrium point.

The equilibrium is stable if the potential energy is minimum in that point and unstable if it is maximum.
 
ehild said:
Recall the force is negative derivative of the potential energy.

X=0 is correct, but you have a mistake in the other equilibrium point.

The equilibrium is stable if the potential energy is minimum in that point and unstable if it is maximum.

Oh sorry, x = 2A/(3B).
How I find the max and min of potential energy?
 
Blockade said:
Oh sorry, x = 2A/(3B).
How I find the max and min of potential energy?
You found the positions of the extremes, at x=0 and at x=2A/3B.
Have you learned what should be the second derivative at a maximum and at a minimum?
 

Similar threads

  • · Replies 2 ·
Replies
2
Views
1K
  • · Replies 1 ·
Replies
1
Views
1K
Replies
3
Views
881
Replies
2
Views
1K
Replies
2
Views
1K
Replies
4
Views
2K
  • · Replies 29 ·
Replies
29
Views
4K
Replies
8
Views
20K
Replies
6
Views
2K
  • · Replies 21 ·
Replies
21
Views
9K