# Analyze the movement after displacing the mass xo horizontally

1. Feb 20, 2005

### Feynmanfan

Dear friends,

I need some help with this problem. As you can see in the picture we have this mass attached to two springs of proper lenght l and I'm asked to analyze the movement after displacing the mass xo horizontally (being xo very small).

I don't want you to solve the problem for me. Just give me a hint how I should begin. I know the oscillator is anharmonic and I believe the best thing to do is analyze its potential and draw the phase plane. But I don't know how to do that.

Thanks for your help

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2. Feb 20, 2005

### s_a

What I'd do is determine the potential energy (U) stored in the springs for any value of q (using Hooke's Law), and then use the fact that the force F = -dU/dq. From this you can determine the acceleration in terms of the position q and hence form a simple 2nd order DE to solve for the equation of the motion of the mass.

3. Feb 20, 2005

### Andrew Mason

Are you sure xo is a horizontal displacement? If so, wouldn't we have to know the distance q when the mass displaced horizontally?

AM

4. Feb 20, 2005

### Curious3141

Perhaps it's a plan view ? Anyway, the problem would be even more complicated if we had to consider gravity.

Assuming that that is a plan view and that all the displacements are in one horizontal plane, the simplest approximate differential equation I could come up with to describe the motion was :

$$\ddot{x} = - \frac{k}{ml^2}(x^3)$$

which I don't know how to solve. It certainly isn't simple harmonic motion.

The exact d.e. is a lot more complicated and almost certainly unsolvable exactly.

k, BTW, is the spring constant of one of the identical springs.

Last edited: Feb 20, 2005
5. Feb 20, 2005

### Curious3141

But if q is indeed a vertical displacement and x is a horizontal one, then it becomes an oscillation in two planes. I think that one would call for solution with a Lagrangian. It's been a long time since I did anything like that, and it isn't elementary.

6. Feb 20, 2005

### Andrew Mason

Right. I get the same equation:

The restoring force is:

(1)$$F = -2k dL sin(\theta) \approx -2kdL(x/L)$$

Since:

$$L^2 + x^2 = (L + dl)^2 = L^2 + 2Ldl + dl^2$$

ignoring the dl^2 term for small x,

$$dl = x^2/2L$$

substituting in (1),

$$F = -2k(x^2/2L)(x/L) = -kx^3/L^2 = m\ddot x$$

Perhaps someone with access to Maple or Mathematica will be able to tackle this. I found this general solution to third order differential equations.

http://virtual.cvut.cz/dyn/examples/examples/equations/eqs7/ [Broken]

AM

Last edited by a moderator: May 1, 2017
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