# 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 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. – alephzero Sep 20 at 18:53
• This question shows why mathematicians should take more engineering classes ;-) – Massimo Ortolano Sep 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. – John D. Cook Sep 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.