# Equation of motion via Euler-Lagrange equation

1. Feb 13, 2012

### fluidistic

1. The problem statement, all variables and given/known data
A bead of mass m slides without friction along a wire which has the shape of a parabola y=Ax² with axis vertical in the Earth's gravitational field g.
a)Find the Lagrangian, taking as generalized coordinate the horizontal displacement x.
b)Write down the Lagrange's equation of motion.

2. Relevant equations
Solved the problem.

3. The attempt at a solution
a)$L=\frac{m\dot x ^2}{2}(1+4A^2 x^2)-mgAx^2$.
b)$\ddot x (1+4A^2x^2 )-8A^2 \dot x ^2 x +2gAx=0$.
For the fun of it, I wanted to get some information about the motion of the particle, by looking at either small oscilations or by solving the equation of motion.
However:
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1)For small oscillations I'd need to approximate the potential energy function by a quadratic function but it's already done so I'm "lucky" on that part. This is supposed to make the equation of motion more friendly.
2)I'm currently self studying mathematical methods in physics and I must say that the equation of motion equation looks really terrible! It's non linear and of degree two.
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So I was wondering if I could either solve this differential equation via some method or at least get some info about the behaviour of the motion. Using intuition the motion must be periodic, but I find nothing more than this. Any help is appreciated.

Last edited: Feb 13, 2012
2. Feb 13, 2012

### ehild

You can find the frequency of small oscillations assuming that x and dx/dt are small, so ignoring the squared terms.

The system is conservative, you can find the relation between x (or y) and dx/dt for a given energy, and determine the maximum displacement from equilibrium or the maximum speed.

ehild

3. Feb 13, 2012

### fluidistic

Wow thanks a lot.
The differential equation reduces to $\ddot x +2gAx=0 \Rightarrow x(t) \approx A \cos (\sqrt{2gA}x)+B\sin (\sqrt{2gA}x)$ that is, if A is greater than 0.
P.S.:thanks for spotting my sign error.

4. Feb 14, 2012

### ehild

If A is negative, x=0 is unstable equilibrium. There will not be oscillations, the bead will just slide down along the parabola if displaced from x=0.

ehild

5. Feb 14, 2012

### fluidistic

I understand this with intuition, however how could I "prove" it mathematically or at least see it?
If I still consider $\ddot x +2gAx=0$, this would give $x(t)=Ce^{\sqrt {-2gA}x }+De^{-\sqrt {-2gA}x }$. I know that this function diverges when x tends to either positive or negative infinite. However this DE was built under the assumption that $x\approx 0$, therefore I do not think it is well suited to study the behaviour of the particle for great |x|.

6. Feb 15, 2012

### ehild

x=0 is a solution of the original differential equation. If both x and xdot are small, the differential equation is like Hook's Law mx'' = -2mgAx, and the force is opposite to the displacement when A>0. The displaced particle oscillates around the equilibrium point.
If A<0, mx" = 2mg|A|x, the force is in the same direction as the displacement, the displaced particle will move away from equilibrium.

ehild

7. Feb 15, 2012

### vela

Staff Emeritus
The x in the argument of the sine and cosine and in the exponents should be t.

8. Feb 15, 2012

### ehild

Of course, you are right!!! I did not notice .

ehild