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Small oscillations criteria

by fluidistic
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fluidistic
#1
Mar6-12, 09:32 AM
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I'm not sure where to post this question. In classical mechanics many problems are simplified in the approximation of "small angles" or "small oscillations".
Wikipedia gives the following criteria or approximations:
[itex]\sin \theta \approx \theta[/itex].
[itex]\cos \theta \approx 1 - \frac{\theta ^2}{2}[/itex]
[itex]\tan \theta \approx \theta[/itex].
But in some books I find the relation:
[itex]\cos \theta \approx 1[/itex].
In other words they discard any terms of second degree and higher, keeping terms of degree 0 and 1 only.
Now when I tackle a problem of small oscillations I do not know what criteria to use. Of course keeping terms of second order gives a more accurate result... but I am not sure this is the standard.
What is your experience with this?
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f95toli
#2
Mar6-12, 09:44 AM
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You normally keep as many terms as you can (unless you are doing a "back of the envelope calculation", that is), but if you try to keep higher-order terms you usually end up with an equation you can not solve analytically.
On the other hand you might also miss something if you throw away terms that are actually important. This is something you'll learn from experience and there are no exact rules.

That said, if I am even a little bit unsure I always make sure that my solution is valid (="good enough")in the region I am interested in but solving the full equation numerically.
fluidistic
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Mar6-12, 10:24 AM
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Quote Quote by f95toli View Post
You normally keep as many terms as you can (unless you are doing a "back of the envelope calculation", that is), but if you try to keep higher-order terms you usually end up with an equation you can not solve analytically.
On the other hand you might also miss something if you throw away terms that are actually important. This is something you'll learn from experience and there are no exact rules.

That said, if I am even a little bit unsure I always make sure that my solution is valid (="good enough")in the region I am interested in but solving the full equation numerically.
I see, thanks a lot. Tomorrow I have an exam of classical mechanics. If I have to make such approximation I'll consider the more accurate first and if I see the equation of motions are way too complicated I'm going to try with the lesser accurate albeit easier to solve analytically expression.

chrisbaird
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Mar6-12, 11:04 AM
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Small oscillations criteria

More specfically, the "small oscillations approximation" usually refers to the fact when most oscillations are small enough, the oscillation becomes harmonic, and the mathematics of harmonic oscillations is well-developed and very robust. So, for any particular problem, the "small oscillations approximation" would be to make whatever expansion and throw away whatever terms are necessary to end up with simple harmonic motion.
fluidistic
#5
Mar6-12, 11:22 AM
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Quote Quote by chrisbaird View Post
More specfically, the "small oscillations approximation" usually refers to the fact when most oscillations are small enough, the oscillation becomes harmonic, and the mathematics of harmonic oscillations is well-developed and very robust. So, for any particular problem, the "small oscillations approximation" would be to make whatever expansion and throw away whatever terms are necessary to end up with simple harmonic motion.
Yes. Usually if I have a complicated potential and I want to use the approximation for small oscillations, I approximate it via a Taylor expansion of order 2 and then the Lagrangian of the system is simple enough to be solved in the same way than any harmonic problem.
However sometimes getting the correct (non approximated) expression for V, the potential function, is quite complicated and using the relations I wrote in the first post of this thread can simplify the problem from start. So that I'd get the approximated potential function V without passing from the true potential. My main problem is that I wasn't sure what approximation to make for small angles. But since I'm usually/always taking a second order Taylor expansion of the potential, I think it's extremely important to keep the [itex]\frac{\theta ^2 }{2}[/itex] term of [itex]\cos \theta[/itex].


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