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So in my classical mechanics book it states:

"For any sufficiently small displacement, any system of this kind behaves like a harmonic oscillator."

When discussing SHO. So I am curious what is considered to be "sufficiently small". Like with a spring I can extend it to a certain length and it will be so far from the equilibrium that deformation occurs and obviously at that point the spring will no longer behave as an oscillator, is this a good interpretation? I feel like it's not what is meant. I understand that Hooks law is derived from the maclourine series of potential energy and that because of that it has to be sufficiently close to equilibrium but what is considered "close"?

sammy gerbil
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chris360
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2 Answers2

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You left out an important part of the quote :

Notice that this discussion applies to the motion of a particle near a stable equilibrium point of any potential energy function. For sufficiently small displacements, any system of this kind behaves like a harmonic oscillator.

Source: Classical Mechanics by Tom Kibble & Frank Berkshire, p 23

There is no universal definition of how small is "sufficiently small". It depends on the criterion you decide to apply. There is (in general) no definite point at which the behaviour of any stable oscillator suddenly becomes harmonic. It is only an approximation which gradually gets better the closer you get to the exact point of stable equilibrium. It is up to you to decide at what point deviations from ideal harmonic behaviour are insignificant.

What is being expressed here is the idea of a mathematical limit : that any potential energy function $U(x)$ can be approximated by a parabola $U(x)=\frac12 k(x-a)^2$ as accurately as you care to stipulate, provided that you limit yourself to a small enough range of $x$ around a point of stability $x=a$. Stability means that $U'(a)=0$ and $U''(a)>0$ - ie $x=a$ must be a local minimum of the function $U(x)$.

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sammy gerbil
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Well, usually the idea of "sufficiently close" is relative to the dimensions of the other relevant quantities of the problem. For example, in the case of the spring you mentioned, one must take in consideration how its displacement compares to the length of the spring in its equilibrium state. When angles are involved, you can get a good estimate on the order of magnitude of the error by comparing the value obtained by second-order truncation of the Taylor Series of trigonometric functions with the "exact" value. This could be done with a simple calculator, but you can also look it up (the graphic on this Wikipedia, for example, contains information of this kind: https://en.wikipedia.org/wiki/Small-angle_approximation#Graphic).

That said, you're studying classical mechanics, so at this point I believe the most important thing is to get a physical understanding of oscillatory systems, and to this end the small-angle approximation is very good. Even when the error isn't negligible, the first non-zero term of a series expansion "dominates" the behavior of the function and, in particular, oscillatory systems that aren't really harmonic oscillators will behave like one in a small neighborhood of the equilibrium point.

Othin
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