# Small oscillations on a constraint curve

1. Apr 2, 2013

### tmode

1. The problem statement, all variables and given/known data

From Goldstein Classical Mechanics, 6.16:

A mass particle moves in a constant vertical gravitational field along the curve defined by y=ax4 , where y is the vertical direction. Find the equation of motion for small oscillations about the position of equilibrium.

3. The attempt at a solution

When taking a quadratic approximation as in the standard way to solve small oscillations, the eigenfrequency vanishes, giving a result without oscillation, which is clearly not what I'm looking for. I'm honestly not sure how to proceed with this.

2. Apr 2, 2013

### ehild

Hi tmode, welcome to PF. Show what you tried in detail, please. How did you take the constraint into account?

ehild

3. Apr 3, 2013

### tmode

Hello, thank you.

I took the constraint into account by substituting $y=ax^4$ and $\dot{y}=4ax^3$ into the Lagrangian: $$L = T-V = \frac{1}{2}m(\dot{x}^2 + \dot{y}^2) - mgy$$
To yield, without any small oscillation approximation: $$L = \frac{1}{2}m\dot{x}^2 (1+16 a^2 x^6) -mgax^4$$
However, when you try to Taylor expand the $m(1+16a^2 x^6)$ and $mgax^4$ about the equilibrium point x=0 (ie. small x), at quadratic order you get back m in the former and 0 in the latter, resulting in ω=0 for this treatment as a small oscillation approximation, and therefore no oscillation. For the potential, the first non-zero Taylor expansion approximation is the original value $mgax^4$, which would be very small in any case, on the order of x^4.

Taking the Euler-Lagrange equation to get equation of motion, we end up with $$0=m\ddot{x}(1+16a^2 x^6) + 48m\dot{x}^2 a^2 x^5 + 4mga x^3$$
I don't think this is all the question wants answered, since it's not taking any sort of small oscillation approximation, nor does it give an explicit equation of motion.

Last edited: Apr 3, 2013
4. Apr 3, 2013

### ehild

The equation you derived is correct. I do not see anything you can do more. Even ignoring the terms with x^5 and x^6 it is not an equation for SHM.

ehild

5. Apr 3, 2013

### Staff: Mentor

The motion is not harmonic, why do you expect to find a equation for harmonic motion?

In first non-vanishing order, we get $\ddot{x} = -4gax^3$, which is just the derivative of the potential => okay.

6. Apr 5, 2013

### tmode

Alright, thanks both of you. In the context of the level of difficulty of the rest of the problems from that chapter, it seemed far too simple leaving it there, seeming like an incomplete solution. Thanks again.