Timeline for Approximations for a spring pendulum's equations of motion in 2D
Current License: CC BY-SA 3.0
14 events
| when toggle format | what | by | license | comment | |
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| Mar 21, 2017 at 13:57 | vote | accept | frakbak | ||
| Mar 21, 2017 at 13:56 | comment | added | frakbak | @ZeroTheHero Oh, I see. I was plugging into (1), not (3). Thanks for the references, I see there is plenty to learn here. | |
| Mar 21, 2017 at 12:58 | comment | added | ZeroTheHero | @frakbak I numbered the equations for the purpose of discussion and verified the algebra, which seems correct. I also added some minor instructions, using the numbered equations for clarity. Classical perturbation theory at a level accessible to most physicists is discussed in the texbook of Hand & Find on Analytical Mechanics, and also Tai L. Chow, Classical Mechanics. I'm sure there are others (probably Fetter and Walecka must have this somewhere; as always Goldstein has something to say on this topic.) | |
| Mar 21, 2017 at 12:53 | history | edited | ZeroTheHero | CC BY-SA 3.0 |
added equation numbering for discussion purposes
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| Mar 21, 2017 at 9:55 | comment | added | frakbak | @ZeroTheHero Thanks--I'm not sure about your algebra though in the $\theta$ equation. Shouldn't we get the zeroth-order term $(g-l\omega_\theta^2)A\cos\omega_\theta t + 2AB\omega_\theta\omega_a\sin\omega\theta t\sin\omega_a t - AB\omega_\theta^2\cos\omega_\theta t\cos\omega_a t$ (plus higher order terms)? And $g-l\omega_\theta^2=0$, right? Otherwise I am really missing something. I gather my difficulties on this problem stem from the fact that I know nothing about perturbation theory! | |
| Mar 20, 2017 at 13:05 | history | edited | ZeroTheHero | CC BY-SA 3.0 |
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| Mar 20, 2017 at 12:53 | history | edited | ZeroTheHero | CC BY-SA 3.0 |
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| Mar 20, 2017 at 10:24 | comment | added | J.G. | @frakbak We need a bit more than this. Define $a_0,\,\theta_0$ as the linearised equations' roots, then define $\delta a=a-a_0,\,\delta\theta=\theta-\theta_0$. The $\epsilon^2$ terms should tell you something linearisable about the $\delta$s. | |
| Mar 20, 2017 at 10:03 | comment | added | frakbak | @J.G. Expanding out in powers of $\epsilon$, I get $\epsilon(l\ddot\theta+g\theta) + \epsilon^2(a\ddot\theta+2\dot\theta\dot a) + O(\epsilon^3)=0$ and $\epsilon(M\ddot a + ka) + \epsilon^2(-Ml\dot\theta^2+\frac12 Mg\theta^2) + O(\epsilon^3)=0$. Setting the $\epsilon$ coefficients equal to zero completely determines $\theta$ and $a$ (up to constants of integration from the initial conditions). How then does setting the $\epsilon^2$ coefficients equal to zero make sense? | |
| Mar 20, 2017 at 9:20 | comment | added | J.G. | @frakbak If you equate $\epsilon$ coefficients and equate $\epsilon^2$ coefficients, you get two equations and $\epsilon$ is eliminated from each one. | |
| Mar 20, 2017 at 8:24 | comment | added | frakbak | Thanks that helps to clear up part (b). Still confused on part (c). How am I going to solve the equations "to the next order"? Does that mean I save all terms up to order $\epsilon^2$? For the first equation then, won't I get $l\ddot\theta + g\theta = -\epsilon(a\ddot\theta+2\dot\theta\dot a)$? Similarly for the second equation. Now I still have an $\epsilon$ lying around... | |
| Mar 20, 2017 at 4:05 | history | edited | ZeroTheHero | CC BY-SA 3.0 |
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| Mar 20, 2017 at 2:52 | history | edited | ZeroTheHero | CC BY-SA 3.0 |
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| Mar 20, 2017 at 2:44 | history | answered | ZeroTheHero | CC BY-SA 3.0 |