Timeline for Intuition - why does the period not depend on the amplitude in a pendulum?
Current License: CC BY-SA 3.0
13 events
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| Jul 20, 2017 at 21:31 | history | edited | JMac | CC BY-SA 3.0 |
Clairified assumptions, added clairification.
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| Jul 20, 2017 at 18:37 | comment | added | JMac | @CarlWitthoft I have elaborated a bit on that in my comment here; but "for a small perturbation" the math does state that they will be the exact same, as long as that assumption holds. In real life, the assumption never holds; but it's close enough that for practical purposes you can say it does. In real life you also never get a massless string with a point mass suspended; so it falls apart for many reasons; that we can choose to ignore depending on how you formulate the problem. | |
| Jul 20, 2017 at 17:47 | comment | added | Carl Witthoft | This is just plain wrong. The difference is tiny, but it's there. | |
| Jul 20, 2017 at 17:40 | comment | added | Eric Lippert | @NoamChai: It becomes more clear if we add a single word to the final sentence. The higher velocity allows the higher pendulum to complete its swing in about the same amount of time as the lower, even though it has a longer path. The reason this works is that, for small perturbations the additional velocity is almost exactly the amount needed to account for the additional distance. For large perturbations, the additional velocity is not almost exactly that amount. | |
| Jul 20, 2017 at 17:06 | comment | added | JMac | @NoamChai As far as I know it technically doesnt work for any perturbation perfectly. When the angle is small the periodic bevaivour appears linear (and approximately is), and this relationship only applies when it is linear. Since it is only ever approximately linear, they are only approximately the same period. | |
| Jul 20, 2017 at 16:43 | comment | added | Noam Chai | And physical intuition for why his answer does not work in big perturbation? | |
| Jul 20, 2017 at 16:40 | comment | added | Joafigue | Because for small perturbations we use an aproximation (taylor series, order 1), but if we want to correctly describe higher amplitudes, we have to extend the aproximation (taylor of higher order) which reflects the dynamics including the amplitude of the oscilation | |
| Jul 20, 2017 at 16:00 | comment | added | Noam Chai | So why in big pertubatuions this consideration doesn't work? @JMac | |
| Jul 20, 2017 at 15:59 | vote | accept | Noam Chai | ||
| Jul 20, 2017 at 15:37 | vote | accept | Noam Chai | ||
| Jul 20, 2017 at 15:59 | |||||
| Jul 20, 2017 at 15:04 | comment | added | JMac | @ZaKh Because it also travels further. I did mention that in my answer. | |
| Jul 20, 2017 at 14:53 | comment | added | user65035 | Why at the same amount of time as the lower , if its moving at a higher velocity than the other pendulum ? | |
| Jul 20, 2017 at 14:08 | history | answered | JMac | CC BY-SA 3.0 |