# Thread: oscillation of a ring on parabola

2. ## no idea with (ii)

no idea with ii

3. Originally Posted by szpengchao
one ring is hang on a parabola with equation y= x^2/4a
ignore friction. the ring is left at a height h from rest.

find its period of oscilation.
Start by getting the equations of motion using Lagranges equation. Here is a reference that saves me the trouble of doing it: (Example 3.2)

Schaum's Outline of Theory and ... - Google Book Search

4. Originally Posted by szpengchao
one ring is hang on a parabola with equation y= x^2/4a
ignore friction. the ring is left at a height h from rest.

find its period of oscilation.
Or you could attack the thing head-on.

Newton's Law in the two coordinate directions says
$\displaystyle m \frac{d^2x}{dt^2} = -\frac{x}{\sqrt{x^2 + a^2}}$

$\displaystyle m \frac{d^2y}{dt^2} = -mg - \frac{a}{\sqrt{x^2 + a^2}}$

That system looks horrible to solve, but the thing is that I can't come up with a good generalized coordinate for this problem. The best I can offer is to put an angular variable at the focus of the parabola.

However, things aren't all bleak. All we need is the coefficients of the expansion:
$\displaystyle \frac{d^2x}{dt^2} = A + Bx + Cx^2 + ~...$

Presumably if h is small then A = 0, the coefficients of the powers of x higher than 1 are small, and B is negative. In that case
$\displaystyle \frac{d^2x}{dt^2} \approx Bx$
which is the harmonic oscillator equation. I leave it to you to check that these conditions come out. (I haven't checked.)

-Dan

Edit: After seeing the original posting of this problem it is clear that the problem is requiring an exact, not approximate solution. The only way I can think of to get this is to use the Lagrangian technique. But again, I can't think of a good generalized coordinate here.