Harmonic oscillator with 3 charged particles

[Jean-C]

Homework Statement

I got an alpha particle (charge 2+) fixed at x=0 and an electron fixed at x=2. I then add a fluor ion (charge 1-) to the right of the electron and we note his position x_eq. The question is to find the constant spring (k) relative to the harmonic oscillation made by the fluor atom for small movements around its equilibrium.

Homework Equations

The electric force between two charges :
F=k_ecqQ/r²
where r is the distance between the two charges.
Note : I used k_ec so It is clear it is different from the spring constant k.

The force in an harmonic oscillator :
F=-kx
where k is the constant I'm looking for.

The Attempt at a Solution

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It is easy to see that the motion is that of an harmonic oscillator but I can't demonstrate it mathematically. I start off by finding the equilibrium position by finding where the resulting force is 0 :
F=k_ec(2*-1)/(x_eq-0)² + k_ec(-1*-1)/(x_eq-2)²
With that I found out the the equilibrium position was at 2(2+sqrt(2)).
But then I have no idea how to find the relative constant k since the equation isn't linear in x like F=-kx but rather like F=k/x².

My first idea is to transform the equation to a linear one and then isolate k, but I can't seem to find how. If anyone can give me a cue as to how to do that or to find k I'd be very grateful!
Thanks!

 

[BvU]

Hello again, Jean-C,

Much better !

isn't linear in x like F=-kx but rather like F=k/x².

Not exactly: in these equations x means the deviation from equilibrium, so (your $x)-x_{\rm eq}\$. You want to find the first order coefficient in that F. Effectively you approximate F(x) as a straight line at $x_{\rm eq}$ to find $k_{\rm\; 'spring'}$

 

[Jean-C]

Thanks for the fast answer I really appreciate it! And sorry for the last post, I was very tired...
I understand that I need to observe small deviations from equilibrium, but I must be doing something wrong : do I replace x_eq by x-x_eq and then use the Taylor approximation for the first order (since F(0) is 0 at equilibrium)?

 

[BvU]

Yes. It's called differentiation, I think . Mind you, your $x$ is already defined. The game is now to rewrite F(x) in terms of $x'= x - x_{\rm eq}$. So you don't want to replace $x_{\rm eq}$ .

 

[Jean-C]

Alright, thanks to you I have figured it out! Thanks a lot!

 

[BvU]

You're welcome. Nice exercise.
warning: make sure you have the units right.