A question about derivation of the potential energy around the stable equilibrium point

Question

I'm learning about harmonic oscillators. In the last lecture my teacher derived the potential energy of a system that has a stable equilibrium point but not a harmonic oscillator.

At the beginning of the lecture he talked about stable, non-stable and neutral equilibrium points. Explained the motion of each system when we exert a force on the object. If we exert a force on a body that is a part of a system that has a non-stable equilibrium point, the body moves and never comes back to the same point at the start. If we exert a force on a body that is a part of a system that has a stable equilibrium point, the body comes back to the starting point (stable equilibrium point) until the body loses It's all energy. That was the main idea.

And then, he reminded us the potential energy of Mass-Spring System and Simple Pendulum.

Potential energy stored in Mass-Spring System is $$U(x)=\frac{1}{2}C{x}^{2}$$ and in Simple Pendulum is $$U(\theta)=MgL=MgL(1-\cos{\theta})=MgL(1-[1-\frac{{\theta}^2}{2!}+\frac{{\theta}^4}{4!}-\dots])=MgL(1-[1-\frac{{\theta}^2}{2!}])=MgL\frac{{\theta}^2}{2}=\frac{MgL{\theta}^2}{2}$$

The purpose of this reminding was to explain that the potential energy stored in a harmonic oscillator system is proportional to the square of displacement of the body. And my teacher said that the $U(q)-q \text{(Potentional energy-displacement)}$ graphic is a parabola.

Then we discussed about a system that has a stable equilibrium point but not a harmonic oscillator.

As you can see in the graph, $q=0$ point is the stable equilibrium point. When the object reaches this point ($q=0$), there is no force applied to the object. Also, this point is a minimum point (so $\frac{\partial U}{\partial q} {|}_{q=0}=0$). Since $F=-\frac{\partial U}{\partial q}$, $F=-\frac{\partial U}{\partial q} {|}_{q=0}=0$.

Everything was fine until here. Then my teacher expanded the potential energy function as a Taylor series :

$$U(q)=U(0)+\frac{q}{1!}\frac{\partial U}{\partial q} {|}_{q=0} + \frac{q^2}{2!}\frac{{\partial}^2 U}{\partial q^2} {|}_{q=0} + \frac{q^3}{3!}\frac{{\partial}^3 U}{\partial q^3} {|}_{q=0} + \dots$$

As the stable equilibrium condition, term $\frac{q}{1!}\frac{\partial U}{\partial q} {|}_{q=0}$ equals to $0$. Also my teacher neglected term $\frac{q^3}{3!}\frac{{\partial}^3 U}{\partial q^3} {|}_{q=0}$ but he didn't neglect the term $\frac{q^2}{2!}\frac{{\partial}^2 U}{\partial q^2} {|}_{q=0}$.

My first question is about this term. Why didn't he neglect this term? What is the reason? (Note that the value of $q$ is really small because the body is around the stable equilibrium point.)

Then he got the following expression for the potential energy ,

$$U(q)\simeq U(0)+\frac{1}{2}{{q}^{2}}\frac{{\partial}^2 U}{\partial q^2} {|}_{q=0}$$

In this expression, my teacher told that the terms $U(0)$ and $\frac{{\partial}^2 U}{\partial q^2} {|}_{q=0}$ are constant. So he wrote that $$U(q)\simeq (const)+(const)q^2$$ but he didn't include the $\frac{1}{2}$.

My second question is : are the terms $U(0)$ and $\frac{{\partial}^2 U}{\partial q^2} {|}_{q=0}$ constant because we are at the stable equilibrium point (as seen in the graph)? Is that correct?

And my third question : why didn't he include $\frac{1}{2}$ in the expression $U(q)=(const)+(const)q^2$?

In the conclusion, he told that for the harmonic oscillators the potential energy stored in the system is proportional to square of the displacement.

Thanks!

Answer 1

Question 1. Since the order $q$ term in the potential vanishes at the equilibrium point, the lowest non-vanishing (and non-constant) term is the $q^2$ term. This term will be small since $q$ is small, but we can't neglect it because it's the "least small" term available.

Question 2. In this context, constant means a number that doesn't depend on $q$. $U(0)$ and $\partial^2 U|_{q=0}$ are constant because they are evaluated at $q=0$. For example, if $U(x) = - \cos x$, $U(0) = - \cos 0 = - 1$ is a constant and so is $\partial^2 U|_{q=0} = \cos 0 = 1$. It has nothing to do with stable equilibrium, we could evaluate these functions at any point and still get constants.

Question 3. Multiplying any constant by any other constant still results in a constant. For example, $1/2 \cdot 1 = 1/2$ is a constant. The point is not to display the specific constants here, just to say that the factors written as $const$ don't depend on $q$.