Published July 29, 2020｜2 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 $x > 0$.
^{[1]}

- [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 $U(0) = 0$.↩