What is meant by small oscillations for this potential energy function?

[zenterix]

Homework Statement

A particle of mass $m$ moves on the $x$-axis with potential energy

$$V(x)=\frac{E_0}{a^4}x^4+4ax^3-8a^2x^2$$

Relevant Equations

a) Find the positions at which the particle is in stable equilibrium.

b) Find the angular frequency of small oscillations about each stable equilibrium position.

c) What do you mean by small oscillations? Be quantitative and give a separate answer for each point of stable equilibrium.

The force on the particle is

$$F(x)=-V'(x)$$

By equating this to zero we find points of stable equilibrium. There are three such points

$$x=0$$

$$x=a^3\left ( \frac{-3a^2\pm\sqrt{9a^4+16E_0}}{2E_0} \right )$$

Next we use a Taylor approximation to $F$.

$$F(x+\Delta x) \approx F(x)+F'(x)\Delta x+\frac{F''(x)}{2}(\Delta x)^2$$

$$=-V'(x)-V''(x)\Delta x-\frac{V'''(x)}{2}(\Delta x)^2$$

We can then evaluate this Taylor approximations near each of the three points of equilibrium.

Near $x=0$ we get

$$F(\Delta x)\approx 16a^2\Delta x-24a(\Delta x)^2$$

Near the other two points of equilibrium we get

$$F(x_0+\Delta x)\approx -\frac{18a^6\pm6a^4\sqrt{9a^4+16E_0}+32E_0a^2}{E_0}\Delta x-\frac{-12a^2\pm 12\sqrt{9a^4+16E_0}}{a}(\Delta x)^2$$

For each of these three cases, if we only consider the linear term in the Taylor approximation, then the functional form of $F$ in the approximation is that of Hooke's law.

In the first case, with $k=0$ and in the other two cases with $k=\frac{18a^6\pm6a^4\sqrt{9a^4+16E_0}+32E_0a^2}{E_0}$.

The angular frequencies of oscillation near each of these equilibrium points are then given by $\omega=\sqrt{\frac{k}{m}}$.

This approximation is, apparently, only good for small oscillations. My question is about item (c) in the original problem: what do we mean by small oscillations?

It seems that we need the quadratic term in the Taylor approximation and all subsequent terms to be small compared to the first order term.

In the book I am reading, Georgi's "The Physics of Waves", it seems we must have

However, I don't quite totally understand the condition 1.25 above (and have asked about it before).

How do we apply this condition in the current problem. For example, for the equilibrium point $x=0$ we have $V'''(0)=24a$ and so $|xV'''(0)|=|24xa|$. We also have $V''(0)=-16a^2$.

Thus it seems that for $x$ near 0, $|xV'''(0|$ approaches zero and so is much smaller than $|V''(0)|$.

Similarly, $V^{(4)}(x)=\frac{24E_0}{a^4}$ and $|xV^{(4)}(0)|=\left | \frac{24E_0x}{a^4}\right | << |24a|=|V'''(0)|$ for small $x$.

All subsequent terms in the Taylor approximation are zero.

 

[kuruman]

By equating this to zero we find points of stable equilibrium.

The equilibrium is not necessarily stable when the first derivative of the potential function is zero. Setting the derivative eequal to zero finds the extrema, local maxima or minima. You can (a) evaluate the second derivative at the points where the first vanishes or (b) plot the function and see for yourself.

 

[zenterix]

$$V''(0)=-16a^2$$

Thus, $0$ is a local max of the potential and a local min of the force on the particle.

If we move away from $0$ into a positive $x$ then the force tends to take the particle farther away from $0$. If we move into a negative $x$ then it seems the force restores the particle to $0$. Thus, $0$ is a semi-stable equilibrium?

For the other two points with zero derivative, the second derivative at those points are parameterized expressions as can be seen below