Energy for Linear 1D Systems - 1D potential

[thecourtholio]

Homework Statement

A particle of mass m moves along the x–axis under the influence of force $F_x=-ax+bx^3$ , where a and b are known positive constants.

(a) Find, and sketch, the particle's potential energy, taking U(0) = 0

(b) Identify and classify all equilibrium points

(c) Find the frequencies of small–amplitude oscillation about all stable equilibrium points, if any

(d) For what range of total mechanical energies will the particle remain bound, executing periodic motion

(e) For what range of total mechanical energies will the particle ultimately escape to infinity

(f) Find the kinetic energy as a function of position in case (e)

Homework Equations

$$U(x)-U(0)=-\int_0^x \vec{F}\cdot d\vec{r}$$
For extrema: $$\frac{dU}{dx}=0$$
Frequency: $$\omega^2=\frac{1}{m}\left . \frac{d^2U}{dx^2}\right |_{x=equilibrium\ point}$$
$$E=T(x)+U(x)$$

The Attempt at a Solution

For (a) I found $$U(x)=\frac{1}{2}ax^2-\frac{1}{4}bx^4$$ and the plot looks kind of like a big M (two peaks, one trough in the middle).

For (b), I found equilibrium points at $x=0,\pm\sqrt{\frac{a}{b}}$, with $x=0$ being a stable equilibrium.
So for (c), $$\omega^2=\frac{1}{m}\left . \frac{d^2U}{dx^2}\right |_{x=0}$$ $$ \omega=\sqrt{\frac{a}{m}}$$
But now, I'm kind of stuck on part (d). I know that the total mechanical energy is obviously $E=T(x)+U(x)$, where $T(x)$ is the KE. I think that the particle will be bound between $-\sqrt{\frac{a}{b}}<x<\sqrt{\frac{a}{b}}$. But I'm a little unsure of how to write this quantitatively. I'm also a little confused as to how to express it as a range of energies.

 

[kuruman]

and the plot looks kind of like a big M (two peaks, one trough in the middle).

You have three equilibrium points one of which at x = 0 is stable. The particle is bound in the valley of the "M". How deep is this valley? What would happen if the energy of the particle is greater than the depth of the valley? Remember, the total energy is constant and you can draw this constant energy in the same plot as the potential energy. Doing this at several values of the total energy might show you what's going on.

 

[thecourtholio]

You have three equilibrium points one of which at x = 0 is stable. The particle is bound in the valley of the "M". How deep is this valley? What would happen if the energy of the particle is greater than the depth of the valley?

What would quantify "how deep" the valley is? And that was another thing that was confusing me was the constant energy. I've seen a lot of the plots of $U(x)$ vs. $x$ in the literature include a horizontal line indicating the total energy but none of them really explained why they chose where they placed it relative to the $U(x)$ curve.

 

[thecourtholio]

What would happen if the energy of the particle is greater than the depth of the valley?

Oh okay. So if the particle's energy is lower than the maximums of $U(x)$ it will oscillate indefinitely but if its energy is greater than that potential maximum then it can escape right? In this case that means that if $E<U(\pm\sqrt{\frac{a}{b}})$ it will oscillate but if $E>U(\pm\sqrt{\frac{a}{b}})$ then it can escape. Is that right?

 

[kuruman]

That is right. In such plots the horizontal line is just a line at whatever value the total energy of the particle is. There is an additional issue to consider: At a given position x, the vertical distance between the constant energy line and the potential line represents the kinetic energy. At the point where the constant energy line intersects the potential energy line the kinetic energy is zero because the total energy is equal to the potential energy. These are called the "turning points" between which the particle oscillates. Different values of total energy give you different turning points.