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

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To determine the stability of equilibrium points for the potential energy function U = Ax² - Bx³, the first derivative f'(x) = 2Ax - 3Bx² is set to zero, yielding equilibrium points at x = 0 and x = 2A/(3B). The stability of these points is assessed by evaluating the second derivative; if it is positive at a point, it indicates a local minimum (stable), while a negative value indicates a local maximum (unstable). The discussion highlights the importance of identifying these derivatives to classify the equilibrium points correctly. Understanding these concepts is crucial for analyzing the stability of forces in potential energy contexts.
Blockade
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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?
 
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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?
 
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