Small Oscillations on a Parabola problem.

[jameson2]

Homework Statement

Find the frequency of oscillations of a particle (mass m) which is free to move along the parabola y= -ax^2 + 2ax - a, and is attached to an ideal spring whose other end is fixed at (1,l) A force F is required to extend the spring to length l. a can be any real number.

Homework Equations

Lagrangian eqation, which can be used to get equation of motion.
Potential energy U=0.5kx^2
Stable equilibrium is at a point where the first derivative of U is 0 and second derivative is positive.

The Attempt at a Solution

First I need to get an equation for the potential energy to complete the lagrangian:
this requires the extension of the spring, which is what I'm having trouble with. Letting r be the extension: I tried just using Pythagoras' theorem and got the following
(l+r)^2=(x-1)^2 + (l-y)^2 and then neglecting the r^2 part due to the oscillations being small, Iihave this :
r=(1/2l)(x^2 -2x +1 -2ly +y^2)
then I think I need this in terms of x and not y, so I filled in the equation of the parabola for y, to get r in terms of x, and hence potential energy in terms of x.
However this gives me a really messy equation that I'm not going to bother typing out as it's clearly wrong. I don't see where I can change this approach to soving the problem though, any ideas?

 

[Jhero]

Thank you for your question. The approach you have taken to find the potential energy and equation of motion is on the right track, but there are a few issues with your calculations.

Firstly, when using Pythagoras' theorem, you need to consider the entire length of the spring, not just the extension. This means that the right-hand side of your equation should be (l-r)^2 instead of (l+r)^2.

Secondly, when neglecting the r^2 term, you also need to neglect the 2lr term, as it is of the same order of magnitude. This means that your final equation for r should be r = (1/2l)(x^2 - 2x + 1 - 2ly + y^2 - 2lr).

Lastly, when substituting the equation of the parabola for y, you need to be careful with the signs. The equation should be y = -ax^2 + 2ax - a, so when substituting it into your equation for r, you should have r = (1/2l)(x^2 - 2x + 1 + 2alx - al).

Once you have corrected these mistakes, you should be able to simplify the potential energy equation and find the equilibrium point where the first derivative is 0 and the second derivative is positive. From there, you can find the frequency of oscillations using the Lagrangian equation and the equation of motion.

I hope this helps, and please let me know if you have any further questions. Best of luck with your calculations!