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S Sep 27, 2020 at 15:24 history suggested gmvh
added "mp.mathematical-physics" as alternative top-level tag
Sep 27, 2020 at 13:55 review Suggested edits
S Sep 27, 2020 at 15:24
Sep 20, 2020 at 22:22 comment added John D. Cook 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, 2020 at 19:42 comment added Massimo Ortolano This question shows why mathematicians should take more engineering classes ;-)
Sep 20, 2020 at 18:53 comment added alephzero 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, 2020 at 15:51 history edited Rodrigo de Azevedo CC BY-SA 4.0
edited title
Sep 20, 2020 at 13:58 history edited Rodrigo de Azevedo CC BY-SA 4.0
edited title
Sep 20, 2020 at 2:34 history became hot network question
Sep 19, 2020 at 18:58 vote accept John D. Cook
Sep 19, 2020 at 18:49 answer added gmvh timeline score: 18
Sep 19, 2020 at 18:42 comment added gmvh Doesn't this follow from the Fourier decomposition of the sawtooth function?
Sep 19, 2020 at 18:41 history edited user44143 CC BY-SA 4.0
top-level tag and more descriptive title
Sep 19, 2020 at 18:33 history asked John D. Cook CC BY-SA 4.0