Let a particular solution of the differential equation

$$\ddot{x}+2\gamma \dot{x}+\omega_0^2 x=\sum_{n=-\infty}^{\infty}c_n e^{i n \omega t}$$

be given by

$$x=\sum_{n=-\infty}^{\infty}\frac{c_n}{2\gamma i n\omega+(\omega_0^2-n^2 \omega^2)}e^{i n \omega t}.$$

Now this system is a damped driven oscillator, so the complementary solution is transient, leaving the particular solution. However, I would like to know what happens as $\gamma \to 0 $. I know that the system then becomes underdamped, and resonance occurs when $ \omega_0 \approx n\omega $, but what will be the resultant frequency of the motion (i.e. which of the undamped angular frequency $\omega_0$ or the driving frequency $\omega$ will prevail?) Also, as $\gamma \to 0$, the complementary function will subsist much longer, so what will be the consequence of this on the motion? I guess I am a bit confused about "frequencies of sums" in general.

  • 1
    $\begingroup$ For general initial conditions, the transient doesn't decay when the damping is zero and so the motion has both the natural frequency and the driving frequency (or frequencies) present. If a driving frequency matches the natural frequency, that component will increase linearly with time. Is this all that you're asking or is there something more that I'm not seeing? $\endgroup$ – Alfred Centauri Nov 10 '17 at 1:03
  • $\begingroup$ I think your question falls under the category of singular perturbation: en.wikipedia.org/wiki/Singular_perturbation . It is much like the behavior of polynomial roots of $\epsilon z^n + a_1 z^{n-1} + . .. +a_0 = 0$ as $\epsilon \rightarrow 0$. $\endgroup$ – hyportnex Nov 10 '17 at 2:42
  • $\begingroup$ I'm not too confused about the case when the damping is truly zero (although I'm not sure what you mean by the motion having both frequencies. What does that mean, practically? Isn't there a resultant overall frequency?). But what happens when the damping gets less and less? Thanks for that link about singular perturbations. I wonder at what level of study one would expect to encounter this technique. $\endgroup$ – Nerd Nov 10 '17 at 11:35

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