Consider an undamped harmonic oscillator. It may be driven at it's natural frequency, $\omega_0^2 = \frac{k}{m}$.

According to Feynman, and other sources, were this to happen, the amplitude of the oscillations would be infinite. I have tried solving the differential equation myself, and the result I got seemed slightly different.

In my solution, the amplitude of the oscillations is directly proportional to time, meaning that the oscillations will get very big quite quickly, and will only become infinite as $t \to \infty$. This page seems to confirm my result:

In contrast, the second case, $\omega = \omega_0$, will have some serious issues at t increases. The addition of the t in the particular solution will mean that we are going to see an oscillation that grows in amplitude as t increases. This case is called resonance and we would generally like to avoid this at all costs.

Am I correct in this? I haven't found any other material which supports this idea, but perhaps I haven't looked hard enough.

  • $\begingroup$ It's $\omega_0^2=k/m$ or $\omega_0=\sqrt{k/m}$. I've proposed an edit to that affect. $\endgroup$ – eepperly16 Jun 11 '15 at 17:06

Absolutely. What Feynman means when he says the amplitude is infinite is indeed that the amplitude grows limitless as $t \to \infty$.

When we talk about the amplitude of a harmonic oscillator, we typically refer to the oscillator in its steady state. However, the undamped system at resonance never reaches a steady state and grows to have an infinite amplitude. This is why we say the amplitude is infinite.

  • $\begingroup$ "However, the undamped system at resonance never reaches a steady state and grows to have an infinite amplitude." That would explain it. Thanks very much $\endgroup$ – user44024 Jun 11 '15 at 17:09
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    $\begingroup$ @jdp407: the point is that if you are driving the oscillator you are doing work on it so you are increasing its energy. To make the amplitude infinite you need to make the energy infinite, and to do that you need to do an infinite amount of work on it. Obviously this takes an infinite time. $\endgroup$ – John Rennie Jun 11 '15 at 17:12
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    $\begingroup$ Careful. I think this is only true exactly at resonance. At other frequencies, the average power transfer is zero. $\endgroup$ – Rob Jeffries Jun 11 '15 at 20:53

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