Published July 29, 20202 minute read

It’s well-known that the period of a simple harmonic oscillator (SHO) is independent of its oscillation amplitude. But is this the only oscillator for which this holds?

No. A simple counterexample is the SHO + ‘brick wall’ potential:  A bit silly, you might say. What if we restrict ourselves to symmetric (even) potential functions?

As intuition suggests, it turns out that the SHO is indeed the unique symmetric constant-period oscillator. We will denote the potential by $U(x)$, and make the following assumptions:

• As discussed, we assume $U(x) = U(-x)$.
• For simplicity, we set $U(0) = 0$.
• Since we are doing physics, we assume that $U(x)$ is differentiable.
• To avoid the particle getting stuck, we will assume that $U(x)$ is strictly increasing on $x > 0$.
In what follows, we assume $x > 0$. Suppose we launch the particle rightwards at time $t = 0$ with kinetic energy $E$. By time-symmetry, the particle must come to rest at time $t = T/4$, where $T$ is the period of the oscillator. Thus\begin{aligned} T &= 4 \int_0^{T/4} \dd{t} \\ &= 4 \int_0^E \frac{\dd{t}}{\dd{x}} \frac{\dd{x}}{\dd{U}} \, \dd{U} \\ &= 4 \int_0^E \frac{\dd{U}}{v \, U'(x)}, \end{aligned}where $v$ is the velocity of the particle. Since $E - U = \frac{1}{2} m v^2$ is the kinetic energy of the particle, we have\begin{aligned} T &\propto \int_0^E \frac{\dd{U}}{U'(U) \, \sqrt{E - U}} \\ &= \int_0^E \frac{1}{\sqrt{U} \, \sqrt{E - U}} \underbrace{\frac{\sqrt{U}}{U'(U)}}_{\defeq f(U)} \, \dd{U}. \end{aligned}Making the substitution $U = E \sin^2{\theta}$, we obtain$T \propto \int_0^{\pi/2} f(E \sin^2{\theta}) \, \dd{\theta}.$The proportionality constant throughout depends only on $m$ and fixed numerical factors. If the oscillator is to have a fixed period, the integral$\int_0^{\pi/2} f(E \sin^2{\theta}) \, \dd{\theta}$must be independent of $E$. Under some mild continuity assumptions, it is a fun analysis exercise to show that $f$ must be constant.But integrating the equality$\frac{\dd{U}}{\sqrt{U}} = \frac{U'}{\sqrt{U}} \, \dd{x} = \frac{\dd{x}}{f(U)} = C \, \dd{x}$yields $2 \sqrt{U} = Cx$,the harmonic oscillator potential.

## Footnotes

1. This condition implies that $U$ is a parameterization of $x$ for $x > 0$. We will frequently omit pullbacks and write $f(U)$ instead of $f(x(U))$.
2. This makes intuitive sense; the integral is essentially a weighted average of $f$ on the interval $[0, E]$. For all such averages to be equal, the function $f$ must be constant.
3. Recall that $U(0) = 0$.