Classical Mechanics: Finding force, equilibrium points, turning points

[JordanGo]

Classical Mechanics: Finding force, equilibrium points, turning points...

Homework Statement

The potential energy between two atoms in a molecule is
U(x) = −1/x^6 +1/x^12
Assume that one of the atoms is very heavy and remains at the origin at rest, and the
other (m = 1) is much less massive and moves only in the x-direction.
(a) Find the force F(x).
(b) Find the equilibrium point x0 and check stability. Give a numerical value for x0
(c) If the system has a total energy E = −0.2, find the turning points, and the period
of oscillation.
(d) If the total energy was E = +0.2, describe the motion of the system.
(e) Find the period for small oscillations around equilibrium. Can the energy in part
(c) be considered “small” in this context?

Homework Equations

F(x)=-dU/dx (not sure though)
E=K+U

The Attempt at a Solution

a)
dU/dx = -6/x^7 +12/x^13

b) Set dU/dx to zero and I get:
6x^6-12=0
x=6√2 (I said the negative part is ignored since we are only interested in positive x)
Stability: second derivative of U evaluated at equilibrium point.
d2U/dx2=-42/x^8+156/x^14 = 12.699, so stable

c)
E=K+U=1/2mv^2-1/x^6+1/x^12
v^2=(2x^6-2-0.4x^12)/x^12
turning points when v=0,
5x^6-x^12-5=0

Now I'm stuck...

 

[ehild]

c)
E=K+U=1/2mv^2-1/x^6+1/x^12
v^2=(2x^6-2-0.4x^12)/x^12
turning points when v=0,
5x^6-x^12-5=0

Now I'm stuck...

You completely ignored the given energy, E=-0.2.

Edit: The formula is correct, I was wrong.

ehild

 

[JordanGo]

Sorry, I skipped a few steps, but I simplified the expression for v with E=-0.2 and I still get the same answer. I believe the answer is complex, I do not know how to deal with this.

 

[ehild]

Sorry, I skipped a few steps, but I simplified the expression for v with E=-0.2 and I still get the same answer. I believe the answer is complex, I do not know how to deal with this.

I checked it again, and I see now that your equation 5x^6-x^12-5=0 is correct. Using the quadratic formula, x^6=(5±√ 5)/2. The solutions are real. Take the sixth (real) root of each to get the turning points.

ehild

 

[JordanGo]

ok thanks! Can you show me how you simplified the expression?

 

[ehild]

Denote y=x^6. Your equation transforms to the quadratic -y^2-5y-5=0. Solve with the quadratic formula.

ehild

 

[JordanGo]

Easy enough, thanks for all the help!