Solutions to the differential equation $my'' + ky = F \sin \omega t$ show resonance when the driving frequency $\omega$ equals the natural frequency $\sqrt{k/m}$. That is, solutions are unbounded when $\omega = \sqrt{k/m}$ and periodic for all other frequencies. It seems that when the sine function is replaced by a sawtooth function, there are more resonant frequencies. Numerical experiments here with $m = k = F = 1$ seem to be resonant when $\omega = 1/n$ for any integer $n$. Are there any theoretical results that prove this conjecture?
$\begingroup$
$\endgroup$
4

8$\begingroup$ Doesn't this follow from the Fourier decomposition of the sawtooth function? $\endgroup$– gmvhCommented Sep 19, 2020 at 18:42

2$\begingroup$ It is easy (for a physicist/engineer!) to see why physically. The discontinuity in the sawtooth function gives the oscillator a "kick". Since there is no damping in the system, it resonates just as well if you "kick" it once every $n$ cycles of its natural frequency of vibration.as if you "kick" it at every cycle. $\endgroup$– alephzeroCommented Sep 20, 2020 at 18:53

2$\begingroup$ This question shows why mathematicians should take more engineering classes ;) $\endgroup$– Massimo OrtolanoCommented Sep 20, 2020 at 19:42

$\begingroup$ The kick argument makes sense. However, it would imply that you always get resonance, regardless of the driving frequency, but you only get resonance for special frequencies. $\endgroup$– John D. CookCommented Sep 20, 2020 at 22:22
Add a comment

1 Answer
$\begingroup$
$\endgroup$
The sawtooth function $f$ has Fourier decomposition $$ f(t) = \frac{1}{2}\frac{1}{\pi}\sum_{n=1}^\infty \frac{1}{n} \sin(n\omega t) $$ Therefore, if $\omega=\frac{\omega_0}{n}$, the $n$th harmonic of $f$ will have angular frequency $n\omega=\omega_0$, resulting in resonance.