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A bead is oscillating in horizontal direction as shown in the figure, our aim is to find the angular frequency of the oscillating bead

First, we can write the potential as: $$V(l)=\frac{1}{2}k\cdot l^2 $$ here $\quad l=\sqrt {x^2+l^2_0}\ \ - l_0 \quad$ then: $$V(x) =\frac{1}{2}k\cdot \left(\sqrt {x^2+l^2_0}\ \ - l_0 \right)^2 $$ Taking the second derivative of this: $$V''(x)= \dfrac{k\left(\sqrt{x^2+l_0^2}\left(-x^4-l_0^2x^2\right)+\left(x^2+l_0^2\right)^\frac{3}{2}\left(2x^2+l_0^2\right)-l_0^3x^2-l_0^5\right)}{\left(x^2+l_0^2\right)^\frac{5}{2}}$$ our equilibrium position on $x$ direction is $x_{eq}=0$ plugging this into $$ \omega = \sqrt\frac{V''(x_{eq})}{m}$$ gives: $$ \omega = \sqrt\frac{V''(0)}{m} = 0$$ It seems a bit silly because it looks obvious that it should oscillate with a nonzero angular frequency. Is there any way to find this angular frequency?

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  • $\begingroup$ @AaronStevens Thank you for commenting, I have checked the derivative on the paper and also checked it on Wolfram to see if I miss something but couldn't find any errors. Working with $l$ seems a bit complicated but it's worth giving a try. $\endgroup$ – Tolga A. Oct 28 '18 at 11:02
  • $\begingroup$ Tolga, I may be mistaken about this but I believe that, contrary to the usual prescription of looking at small displacements to linearize an oscillator, this one requires looking at large displacements. I've edited my answer to address this point. $\endgroup$ – Alfred Centauri Oct 28 '18 at 15:46
  • $\begingroup$ Possible duplicate of Understanding transverse oscillation in 1 mass, 2 spring systems $\endgroup$ – sammy gerbil Oct 29 '18 at 14:12
  • $\begingroup$ Your conclusion is correct : the angular frequency for small amplitude oscillations ($A\to 0$) is zero. It follows from your assumption that the tension in the spring is zero at the equilibrium position. If the spring has non-zero tension in the equilibrium position, you will get a non-zero oscillation frequency for small amplitudes. $\endgroup$ – sammy gerbil Oct 29 '18 at 14:17
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Is there any way to find this angular frequency?

The force due to this potential is

$$F = -kx\left(1 - \frac{l_0}{\sqrt{l^2_0 + x^2}}\right)\approx -\frac{k}{2\,l^2_0}x^3,\quad x\ll l_0$$

and so this non-linear system is not approximately linear in the small displacement regime like, e.g., a pendulum. It doesn't seem likely that the motion will be described by a pure sinusoid of angular frequency $\omega$


After having some more coffee, it occurred to me that in the large displacement regime, the force is approximately

$$F \approx -kx\left(1 - \frac{l_0}{\sqrt{x^2}}\right)\approx -kx,\quad x\gg l_0$$

Thus, for a large enough amplitude, I would expect that the motion is approximately sinusoidal with angular frequency $\omega \approx \sqrt{k/m}$. There will be, of course, additional frequency components present though I would expect these to become relatively smaller as the amplitude gets larger.

I'm reminded of a BJT class B push-pull circuit where there is so-called crossover distortion as one transistor 'turns off' and the other transistor 'turns on'. The voltage transfer curve is quite non-linear in this region but quite linear elsewhere.

enter image description here

Image credit

Thus, small amplitude sinusoidal input signals are highly distorted by this circuit but the distortion content of the output rapidly decreases as the input amplitude increases.

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    $\begingroup$ Yeah I noticed that it would work out for large displacements as well, but I didn't know how to handle it. Very cool. $\endgroup$ – BioPhysicist Oct 28 '18 at 17:08
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The issue is that near equilibrium, your force is not linear, so we can't approximate the system as a simple harmonic oscillator.

Your horizonal force is $$F_x=\frac{-kx\left(\sqrt{x^2+l_0^2}-l_0\right)}{\sqrt{x^2+l_0^2}}=-kx+\frac{kxl_0}{\sqrt{x^2+l_0^2}}$$

Unfortunately this does not get rid of your issue, since $\frac{dF}{dx}=0$ when $x=0$, so near $x=0$ (or $x\ll l_0$) the force is not linear (i.e. does not look like $F=-kx$ for $k\neq0$). Therefore, your conclusion is correct. It just means we cannot find a quantity that represents an angular frequency for SHM.

Note that in general just because the force itself is not of the form $F=-kx$ does not mean we cannot approximate the force to have this form near equilibrium. Unfortunately, for this specific case the force cannot be approximated in this way.

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Note that the formula you are using for $\omega$ assumes simple harmonic motion, and the potential you are using does not lead to simple harmonic motion.

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