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 , and make the following assumptions:
- As discussed, we assume .
- For simplicity, we set .
- Since we are doing physics, we assume that is differentiable.
- To avoid the particle getting stuck, we will assume that is strictly increasing on .
In what follows, we assume
. Suppose we launch the particle rightwards at time
with kinetic energy
. By time-symmetry, the particle must come to rest at time
is the period of the oscillator. Thus
is the velocity of the particle. Since
is the kinetic energy of the particle, we have
Making the substitution
, we obtain
The proportionality constant throughout depends only on
and fixed numerical factors. If the oscillator is to have a fixed period, the integral
must be independent of
. Under some mild continuity assumptions, it is a fun analysis exercise to show that
must be constant.
But integrating the equality
the harmonic oscillator potential.