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:

Right half of a quadratic potential, with a delta function at zero.Right half of a quadratic potential, with a delta function at zero.

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)U(x), and make the following assumptions:

  • As discussed, we assume U(x)=U(x)U(x) = U(-x).
  • For simplicity, we set U(0)=0U(0) = 0.
  • Since we are doing physics, we assume that U(x)U(x) is differentiable.
  • To avoid the particle getting stuck, we will assume that U(x)U(x) is strictly increasing on x>0x > 0.[1]
In what follows, we assume x>0x > 0. Suppose we launch the particle rightwards at time t=0t = 0 with kinetic energy EE. By time-symmetry, the particle must come to rest at time t=T/4t = T/4, where TT is the period of the oscillator. ThusT=40T/4dt=40EdtdxdxdUdU=40EdUvU(x),\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 vv is the velocity of the particle. Since EU=12mv2E - U = \frac{1}{2} m v^2 is the kinetic energy of the particle, we haveT0EdUU(U)EU=0E1UEUUU(U):=f(U)dU.\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=Esin2θU = E \sin^2{\theta}, we obtainT0π/2f(Esin2θ)dθ.T \propto \int_0^{\pi/2} f(E \sin^2{\theta}) \, \dd{\theta}.The proportionality constant throughout depends only on mm and fixed numerical factors. If the oscillator is to have a fixed period, the integral0π/2f(Esin2θ)dθ\int_0^{\pi/2} f(E \sin^2{\theta}) \, \dd{\theta}must be independent of EE. Under some mild continuity assumptions, it is a fun analysis exercise to show that ff must be constant.[2]But integrating the equalitydUU=UUdx=dxf(U)=Cdx\frac{\dd{U}}{\sqrt{U}} = \frac{U'}{\sqrt{U}} \, \dd{x} = \frac{\dd{x}}{f(U)} = C \, \dd{x}yields 2U=Cx2 \sqrt{U} = Cx,[3]the harmonic oscillator potential.


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