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The task is to find the period of small oscillations in the potential

$$U=U_0\tan^2{\Big(\frac{x^2}{a^2}\Big)}.$$

I started with finding the stable equilibrium points:

$\frac{dU}{dx}=0$

$2U_{0}\tan{\Big(\frac{x^2}{a^2}\Big)}\frac{1}{\cos^{2}{\Big(\frac{x^2}{a^2}\Big)}}\frac{2x}{a^{2}}=0$

The solutions of this equation are ( $k$ is an arbitrary non-negative integer number)

$x_{k}=\pm a\sqrt{\pi k}$

If I expand the potential near an extremum point, I get the quadratic form. Hence, the period of oscillations can be found like this:

$T = \frac{2 \pi}{w_{0}}=2\pi \sqrt{\frac{m}{k_{eff}}}= a\sqrt{\frac{m \pi}{2U_{0} k}}$

$k_{eff}=\frac{dU^{2}}{dx^{2}} (x_{k})=\frac{8U_{0}\pi k}{a^{2}}$

I don't see any mistakes in my reasoning. However, my solution does not work for the case in which $k=0$. When $x=0$, the potential function, as well as its first and second derivatives, is zero, which is confusing. Could someone explain me what to do in such situation?

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    $\begingroup$ You've correctly noticed that the harmonic oscillator approximation fails when the second derivative is zero, and so the "period of oscillations" doesn't make sense. So when you ask "what to do in such situation?", what exactly do you want to know? $\endgroup$ – ACuriousMind Mar 3 at 14:22
  • $\begingroup$ Escuse me for my unclear phrasing. I wanted to know why it fails. What makes this particular point so special? If you look at the graph, it looks just like any other minima, but for some reason approximation does not make any sense. $\endgroup$ – palkovna Mar 3 at 14:29
  • $\begingroup$ @ACuriousMind Can't you still have a period of the oscillations? $\endgroup$ – Aaron Stevens Mar 3 at 14:41
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    $\begingroup$ @AaronStevens yes but it’s not harmonic oscillation, with frequency independent of initial conditions. $\endgroup$ – ZeroTheHero Mar 3 at 14:48
  • $\begingroup$ @ZeroTheHero Yes I agree with that, but I wouldn't say "the period of oscillations doesn't make sense". $\endgroup$ – Aaron Stevens Mar 3 at 14:50
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Near $0$, the potential is minimal but $U(x)={{U}_{0}}\frac{{{x}^{4}}}{{{a}^{4}}}+O\left( \frac{{{x}^{6}}}{{{a}^{6}}} \right)$ is not an harmonic potential (second derivative is $0$)

You can calculate the period by separation of variables from the energy equation $\frac{1}{2}m{{\overset{\cdot }{\mathop{x}}\,}^{2}}+U(x)=cst$ .

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  • $\begingroup$ You meant $\frac{x}{a} = \sqrt{\pi n}$? If so, I don't see it. I've calculated the second derivative and it does not equal zero unless $x=0$. Thanks for the idea, I'll try this. $\endgroup$ – palkovna Mar 3 at 14:41
  • $\begingroup$ Sorry ! it was a tipo ! $\endgroup$ – Vincent Fraticelli Mar 3 at 14:49
  • $\begingroup$ Moreover this potential bounds the motion to a single region so there ought to be a single physically relevant minimum. $\endgroup$ – ZeroTheHero Mar 3 at 14:51
  • $\begingroup$ @ZeroTheHero What do you mean by this? I thought that the region where a particle is moving depends on where we "place" a particle. $\endgroup$ – palkovna Mar 3 at 14:56
  • $\begingroup$ sure... but there is a single minimum in this region. Maybe I misunderstand why you have a “k” index... That would be fixed by the region where you place your particle. $\endgroup$ – ZeroTheHero Mar 3 at 14:57

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