# Period of oscillation of magnet levitated over another magnet

The situation is similar to what we used to do as kids, take a vertical wood dowel, with a ring magnet placed at the bottom, and another ring magnet opposing it, floating on top.

More precisely, it would be a round or square magnet, levitated over a similar magnet fixed in place on the bottom. The levitated magnet would have horizontal and rotational motion constrained, so that only vertical motion is allowed.

The Levitated magnet is then hit so that it oscillates vertically. The levitated magnet also can have a weight placed on it.

I'm 30 years past my last college physics class and lack the background to approach this problem analytically. I've tried several experiments to get a handle on it from a practical standpoint, but can't explain what I'm seeing.

I guessed that adding more weight on top of the levitating magnet would decrease the period. But it appears to have little or no effect. (I suspect what's happening here is that the magnets get pushed closer together, and that somehow offsets the increase in weight)

• Could you show what work you have done? – Asher Apr 13 '16 at 1:20

You are kind of on the right path here. The forces between magnetic poles are not linear but are increasing roughly with a Coulomb kind of law: https://en.wikipedia.org/wiki/Force_between_magnets.

If we linearize such a force of the form

$F(r) = {{F_0r_0^2}\over {r^2}}$

around an equilibrium position $r_0$, then we get

$F(r_0+dr) \approx F_0({{r_0^2} \over {r_0^2}} - 2{{r_0^2dr}\over {r_0^3}})=F_0(1 - 2{{dr}\over {r_0}}$),

i.e. the first order restoring force term $-2F_0{{dr}\over {r_0}}$ gets stronger as the equilibrium distance decreased due to adding the weight. The increase in "spring constant" compensates for the increase in mass and the frequency of small oscillations stays roughly constant.

• @faradayent: I am still thinking about that one. Intuitively it seems because the stronger magnet on the bottom pushes the top one into a stronger but "softer" field, but I haven't calculated that, yet, and I am not sure that it actually drops out of the $1/r^2$ type potential. – CuriousOne Apr 13 '16 at 2:35
• @faradayent: I am still thinking about that one. Intuitively it seems because the stronger magnet on the bottom pushes the top one into a stronger but "softer" field, but I haven't calculated that, yet, and I am not sure that it actually drops out of the $1/r^2$ type potential. The other problem is that this Coulomb type potential only works for the far field of individual poles, i.e. the magnets have to be long relative to the distance, and the distance has to be large relative to the thickness of the magnet, but neither condition is well met. – CuriousOne Apr 13 '16 at 3:53
• Thanks CuriousOne. You helped to substantiate what I was seeing. I've figured out another soln to what I was trying to accomplish. I'd like to mark your solution as the correct one, but I don't immediately see how on this board. – faradayent Apr 13 '16 at 18:48