# Resonance arising when harmonic oscillator is excited using sawtooth

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?

• Doesn't this follow from the Fourier decomposition of the sawtooth function?
– gmvh
Sep 19 '20 at 18:42
• 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. Sep 20 '20 at 18:53
• This question shows why mathematicians should take more engineering classes ;-) Sep 20 '20 at 19:42
• 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. Sep 20 '20 at 22:22

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.