Second derivative of energy as frequency of oscillations

Question

Is there a way to algebraically see why when I take the second derivative of a potential energy in a point where it is minimal (force is zero), I generally get the frequency (squared) of the oscillations around this point?

Answer 1

In a harmonic oscillator,

$$F=-kx$$

where $k$ is a constant and $x$ is the displacement from the mean position. Thus,

$$\frac{\mathrm dF}{\mathrm dx}=-k\tag{1}$$

about the equilibrium point. Now,

$$F=-\frac{\mathrm dV}{\mathrm dx}\tag{2}$$

where $V$ is the ootential energy at any point. Now substituting equation $(2)$ in equation $(1)$, we get

$$-\frac{\mathrm d^2V}{\mathrm dx^2}=-k$$

Cancelling the minus signs, and dividing by the mass of the body ($m$), we obtain

\begin{align} \frac 1 m \frac{\mathrm d^2V}{\mathrm dx^2}&=\frac k m\\ &=\omega^2 \end{align}

where $\omega$ is the angular frequency of oscullation.

Answer 2

The other answers have begun by assuming a force linear in displacement. Let us take a step back and see why harmonic oscillators are popular in physics.

Consider perturbing a system slightly off a stable equilibrium. This means that it is perturbed from a minima (by a small amount $\epsilon$). Taylor expanding the potential energy (as ideally, kinetic energy is zero at minima) we get:

$$U(a+\epsilon) \approx U(a) + \epsilon\frac{\partial U(x)}{\partial x}\bigg|_a + \frac{\epsilon^2}{2}\frac{\partial^2U(x)}{\partial x^2}\bigg|_a$$

Now the first term in RHS is just a constant thus can be zeroed out, the second term is nothing but the force which is zero (minima). What we’re left with is the following: $$U(a+\epsilon) \approx \frac{\epsilon^2}{2}\frac{\partial^2U(x)}{\partial x^2}\bigg|_a$$

Now, the second derivative evaluated at the minima is a constant, say $k$. In other words, the potential curve looks parabolic near the minima: $$U(x)=\frac{1}{2}kx^2$$

Obtaining frequency from here has already been demonstrated in the other answers so I shall omit it.

Answer 3

Let's consider the one-dimensional case for the sake of simplicity. Further, assume the oscillations are simple-harmonic and that the force is conservative. Then by definition for such oscillations, $$F_{\text{rest}} = -m \omega^2 x$$ where $F_{\text{rest}}$ is the restoring force. But we know that $F_{\text{rest}}$ can be written as the negative-gradient of a potential, say $V$, like so: $$F_{\text{rest}} = -\frac{dV}{dx}$$ which implies $$ \frac{dV}{dx} = m \omega^2 x \implies \frac{d^2V}{dx^2} = m \omega^2.$$ Thus, the second-order derivative of the potential is proportional to the square of the frequency of oscillations, $\omega$.