# Potential energy problem

• Aug 10th 2010, 08:13 AM
FGT12
Potential energy problem
The potential energy function of a particle of mass m is $V=\frac{-1}{2}c(x^2-a^2)^2$, where a,c > 0. Sketch V as a function of x and describe the possible types of motion in the three cases (a) E > 0, (b) $E < \frac{-1}{2}ca^4$ and (c) $\frac{-1}{2}ca^4< E < 0$

I can sketch the graph which gives you an upside-down parabola, but i cannot see how to determine the motion, i think it have something to do with T+V=E (conservation of energy)
• Aug 10th 2010, 08:25 AM
Ackbeet
Think of the potential energy function as rising and falling ground (with gravitational potential energy equal to mgh, this analogy is actually fairly precise!). Then, the energy E you can plot on the same graph as the potential energy, only this time it's just a horizontal line. The intersections of V with the horizontal line E represent the turning points. For example: suppose you had a hill V=x, and you had a ball rolling in from the negative x direction. If the energy E were equal to 0, it would intersect V right at the origin, right? Well, that's where the ball would approach, slowly, and then when it hit the origin, it would start rolling back out to negative infinity, and never come back. That's the sort of analysis required in this problem. Make sense?

Another example: the simple harmonic oscillator (pendulum, spring, RLC circuit, etc.). This time, the potential looks like V=x^2. It's a valley. Suppose your energy E = 1. (Note that E must always be greater than V_min; that's a law of physics.) Then the intersection of E with V occurs at +1 and -1. Therefore, the ball is going to go to +1, reverse direction, go to -1, reverse direction, go to +1, etc. It will oscillate between the two turning points. Make sense?

Final example: V = e^(-x^2). This is the bell curve. The peak of it is at the origin. If 0<E<1, then the ball could come in from either positive or negative infinity, hit the turning point (intersection) on the corresponding side of the origin, and then bounce right back out to infinity. But, if the ball has energy greater than 1, it will be able to get over the hill. Then it will go over the hill, lose some kinetic energy in the process, and regain it as it goes back downhill and out to the opposite infinity from which it came.

Those are some ideas. Does this help?