Classical Mechanics: Finding force, equilibrium points, turning points

In summary, we solved a problem involving finding force, equilibrium points, turning points, and period of oscillation for a system with a heavy atom at rest and a light atom moving in the x-direction. We found the force F(x) to be -6/x^7 +12/x^13, the equilibrium point x0 to be 6√2 with stability, and the turning points to be 5x^6-x^12-5=0 with real solutions. The period of oscillation and motion of the system were also determined for a given total energy.
  • #1
JordanGo
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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...
 
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  • #2


JordanGo said:
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
 
Last edited:
  • #3


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.
 
  • #4


JordanGo said:
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
 
  • #5


ok thanks! Can you show me how you simplified the expression?
 
  • #6


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

ehild
 
  • #7


Easy enough, thanks for all the help!
 

1. What is classical mechanics?

Classical mechanics is a branch of physics that deals with the motion of objects under the influence of forces. It is based on Isaac Newton's laws of motion and is used to describe the behavior of macroscopic objects.

2. How do you find the force in classical mechanics?

In classical mechanics, the force acting on an object can be calculated using Newton's second law of motion, which states that force is equal to mass multiplied by acceleration (F=ma). This means that to find the force, you need to know the mass of the object and its acceleration.

3. What are equilibrium points in classical mechanics?

Equilibrium points in classical mechanics refer to the state of an object where the net force acting on it is zero. This means that the object is either at rest or moving with a constant velocity. In other words, the object's acceleration is zero at equilibrium points.

4. How do you calculate turning points in classical mechanics?

Turning points in classical mechanics are points where the direction of an object's motion changes. These points can be calculated by setting the object's acceleration equal to zero and solving for the position of the object. This position marks where the object changes direction.

5. How is classical mechanics used in real life?

Classical mechanics is used in many practical applications, such as engineering, architecture, and sports. It is used to design structures and machines, predict the motion of projectiles, and understand the mechanics of human movement. It is also the basis for many other branches of physics, including thermodynamics and electromagnetism.

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