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While making calculations for simple harmonic motion, we take the force as $F=F(x)$. Then we use Taylor's expansion and calculate as follows:

$$\begin{align} F(x) &=F(0+x) \\ & = F(0)+xF'(0)+\frac{x^2}{2}F''(0)+ \ldots \\ & = a_0+a_1x+a_2x^2+ \ldots \end{align}$$

Now we assume that at $x=0$, no force is acting on the particle, so $a_0=0$.

And then we assume that the displacement $x$ is small i.e. we neglect the terms where the degree of $x$ is $\ge 2$.

So we get the force as $F=a_1x$.

QUESTION: The above italicised portion is a vital assumption in this calculation. If we did not make that assumption, then principle of superposition would not be applicable in cases relating to these concepts. So I know it is a very useful assumption.

But what I want to know is, what exactly are the information or points that we are losing out because of making this assumption? In other words, what kind of accuracy am I sacrificing while making this assumption?

EDIT: As an addendum after reading the comments and answers, I would like to add that the oscillator is not purely harmonic. And my question is what diracology has pointed out: what information am I losing out when I make a harmonic approximation?

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  • $\begingroup$ You are loosing accuracy when neglecting those terms. I don't think there is anything else that gets affected. $\endgroup$ – hxri Apr 17 '16 at 17:51
  • $\begingroup$ @HariPrasad Is anything else lost apart from just accuracy? I am asking for that only. $\endgroup$ – SchrodingersCat Apr 17 '16 at 17:52
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    $\begingroup$ 1. This is not "mathematical physics". 2. It's not really clear what you are asking. Simple harmonic motion is when the force is exactly of the form $F=ax$. That you can, to first order, approximate any force with this is what makes the harmonic oscillator so useful, but, in the case of actual simple harmonic motion, you're not making an approximation. $\endgroup$ – ACuriousMind Apr 17 '16 at 17:59
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    $\begingroup$ If the oscillator is harmonic of course we don't lose anything. But the question is very pertinent as long as we rephrase it: What information we lose when we perform an harmonic approximation? $\endgroup$ – Diracology Apr 17 '16 at 19:01
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    $\begingroup$ @Diracology: For all practical purposes ALL information. Almost all Hamiltonian systems are non-integrable. That's a very unpleasant property that gets lost completely by switching over to the second most simple integrable system. $\endgroup$ – CuriousOne Apr 19 '16 at 7:00
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When you ignore higher orders in the force, assuming it is linear in the position, you are assuming the potential energy is quadratic at most (quadratic, parabolic or harmonic approximation). Then you loose global information about the potential. We are not able to say for example whether the movement of the particle is bounded or unbounded as we increase energy. We can also loose some nice features such as the spontaneous symmetry breaking (SSB). Consider for instance a potential $U=-\phi^2+\lambda\phi^4$, $\lambda>0$. This potential could represent a non linear spring law and presents SSB. Once we make the harmonic approximation around one of the minima, there is no longer the SSB. Also interesting mathematical and physical aspects present in non linear theories may be lost when we linearize it. Take the pendulum example. In the harmonic approximation we obtain that the period is independent of the amplitude, which is actually not true, in general.

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Your wording is a little off. SMH doesn't care about the size of displacement. SHM is the lowest order approximation to a general oscillation. A general oscillation does care about the size of the displacement, and when you ignore the higher order effects you will miss various phenomena. Examples of things you will not be able to describe are harmonic generation in optics, and thermal expansion of solids.

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  • $\begingroup$ This. If the oscillator is truly harmonic (as per the title) then nothing is lost but the approximation is unnecessary. $\endgroup$ – dmckee --- ex-moderator kitten Apr 17 '16 at 18:50
  • $\begingroup$ @dmckee Yeah. That's why I said the wording is off. He or she mentions the Taylor series and ignoring the higher order terms, so I presume the title misrepresents the question. $\endgroup$ – garyp Apr 17 '16 at 20:59
  • $\begingroup$ Can you elaborate a bit on the examples? Or perhaps suggest a source to read? $\endgroup$ – SchrodingersCat Apr 18 '16 at 4:19
  • $\begingroup$ I have limited time right now. The word to search is anharmonic, or anharmonicity. Wikipedia has a short article. I just noticed that the Wikipedia article on second harmonic generation has a good description and visualization. If I remember, I'll look for more. $\endgroup$ – garyp Apr 18 '16 at 15:21

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