# Why can some oscillations be modeled by Simple Harmonic Motion, while others cannot?

For some oscillators an increase in the driving amplitude changes the period (frequency) of the oscillation, but the simple harmonic oscillator does not predict this type of behavior. Why?

• – rob Mar 30 '16 at 17:40
• That doesn't really answer my question. – paul Mar 30 '16 at 17:46
• At increased amplitude higher-order terms in the potential (quartic, etc.) become non-negligible. – rob Mar 30 '16 at 17:47
• Well, that would depend on the higher order terms now, wouldn't it? – Jon Custer Mar 30 '16 at 18:18
• Imagine a pendulum. A pendulum is not a simple harmonic oscillator. I'm not a physicist, and I don't know what equation would accurately describe the motion of a pendulum, but whatever it is, it's going to have some terms that become more significant as amplitude increases, and it's going to have some that don't, and just by an amazing coincidence, the limit as amplitude approaches zero is going to look like an SHO. Same goes for a mass and spring system where the spring does not actually obey Hooke's Law, but rather, some law that approaches Hooke's Law as displacement approaches zero. – Solomon Slow Mar 30 '16 at 19:27

Consider the example of a plane pendulum with bob mass $m$ and length $\ell$, where the potential energy of the pendulum bob is $$U = -mg\ell \cos\theta$$ up to a constant. For small angles $\theta\ll 1\rm\,radian$ this is approximately simple harmonic, because neglecting higher-order terms in the Taylor expansion $$\cos\theta = 1 - \frac{\theta^2}{2!} + \frac{\theta^4}{4!} - \cdots$$ gives a potential proportional to $\theta^2$, just like a spring. However the full equation of motion for a plane pendulum, $$m\ell\ddot\theta = -mg\sin\theta$$ is a differential version of a transcendental equation. The exact solution seems to involve the Jacobian elliptic functions.