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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?

docscience
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paul
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2 Answers2

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In general it's nonlinearity that leads to the type of behavior you describe.

Simple harmonic motion is usually modeled in terms of linear differential equations, but real physical systems tend to be nonlinear to some degree and sometimes the non-linearities can become large as a function of the amount of energy you put into the system.

As physicists or engineers we tend to force the nonlinear behavior into simple non-linear differential equations like the Vander-Waals model, the duffing oscillator, but more often than not there is the saturation nonlinearity (systems can only hold so much energy before they literally break)

For linear systems if you excite the system with a pure frequency you expect to see the same frequency at the system output. If you double the amplitude you would expect to see a doubling of the output. But if you excite a system and see changes in frequency with changes in the amplitude of your excitation, or non-proportional changes in the amplitude, then the likelihood is there is a nonlinear behavior lurking within.

docscience
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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.

There are an infinite number of ways an oscillator can be non-harmonic, so there isn't a general answer to your question as stated.

rob
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