# Oscillation of a particle on a parabolic surface [equation of motion]

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Yeap, @haruspex is great and he has very great patience too :D.

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This problem fooled me at first into thinking that the potential is harmonic because the potential energy is ##U=mgy=mgAx^2##. As such, I reasoned, there must be a clearly defined frequency of oscillations. Then it dawned on me that this is not the case here because the motion is two-dimensional and not restricted only in ##x##. A 2d parabolic cup ##y=Ax^2## with gravity along ##y## does not give rise to an amplitude-independent oscillation frequency. The appropriate shape that does that is the cycloid as is known from solving the brachistochrone problem. Therefore, in this case one needs to apply a small angle approximation for oscillations near the bottom of the parabolic cup.

We can approximate the bottom of the cup to a circle of radius ##R##, and write the vertical height relative to the bottom, $$y=R-\sqrt{R^2-x^2}=\frac{(R)^2-(R-\sqrt{R^2-x^2})^2}{R+\sqrt{R^2-x^2}}=\frac{x^2}{R+\sqrt{R^2-x^2}}=Ax^2.$$All that remains to obtain the answer in post #1 is to find an approximate expression for ##y## when ##x<<R##, solve for ##R## in terms of ##A## and substitute in the expression for the frequency of a pendulum of length ##R##.

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Delta2
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We can approximate the bottom of the cup to a circle of radius R,
Yes, that's ok, but I don't think it is that helpful. A further approximation will be required.
For the actual parabolic shape I got ##(1+4Ax^2)\dot x^2=2gA(X^2-x^2)##, where X is the initial x displacement.
One route from there is to differentiate and approximate for small X, x to obtain the classic SHM ODE. Another is to make the small x approximation immediately to get it into the form ##\dot x^2+\omega^2x^2= constant##, then apply a small X approximation,

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Yes, that's ok, but I don't think it is that helpful. A further approximation will be required.
I don't see where a further approximation is required. The expression $$y=\frac{x^2}{R+\sqrt{R^2-x^2}}$$ is exact for a circular path. The only approximation comes in when one says that for ##x<<R##, one gets ##y\approx\dfrac{x^2}{2R}##, still for a circular path. Now it is always true that, for the actual parabolic path, ##y=Ax^2## for any ##x##. Therefore for small ##x## (in the parabolic path) one can match the parabola to a circle if one sets $$\frac{x^2}{2R}=Ax^2~\Rightarrow~R=\frac{1}{2A}.$$ For a pendulum following a circular path of radius ##R## and in the small angle approximation ##\omega=\sqrt{g/R}##. The answer follows.

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The only approximation comes in when one says that for x<<R,
That's the one.
I am saying I don't see the benefit in approximating it as circular first. There is no great difficulty in writing the exact ODE for the parabola and then doing the small displacement approximation.

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