Analyze the movement after displacing the mass xo horizontally

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Homework Help Overview

The discussion revolves around analyzing the movement of a mass attached to two springs after a small horizontal displacement, denoted as xo. The problem involves concepts from mechanics, particularly relating to oscillatory motion and potential energy in spring systems.

Discussion Character

  • Exploratory, Assumption checking, Mathematical reasoning

Approaches and Questions Raised

  • Participants explore the potential energy in the springs using Hooke's Law and discuss deriving equations of motion. Questions arise regarding the nature of the displacement (horizontal vs. vertical) and its implications on the analysis. Some suggest the complexity of the motion may require advanced methods like Lagrangian mechanics.

Discussion Status

The discussion is active, with various interpretations of the problem being explored. Some participants have proposed differential equations to describe the motion, while others are questioning the assumptions about the displacement and its effects on the analysis. No consensus has been reached, but several productive lines of inquiry are underway.

Contextual Notes

Participants note the potential complications introduced by gravity and the need for clarity on the displacement's orientation. There is also mention of the limitations of simple harmonic motion in this context.

Feynmanfan
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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 length 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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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.
 
Feynmanfan said:
As you can see in the picture we have this mass attached to two springs of proper length l and I'm asked to analyze the movement after displacing the mass xo horizontally (being xo very small).
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
 
Andrew Mason said:
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

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 :

[tex]\ddot{x} = - \frac{k}{ml^2}(x^3)[/tex]

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.
 
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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.
 
Curious3141 said:
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 :

[tex]\ddot{x} = - \frac{k}{ml^2}(x^3)[/tex]

which I don't know how to solve.
Right. I get the same equation:

The restoring force is:

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

Since:

[tex]L^2 + x^2 = (L + dl)^2 = L^2 + 2Ldl + dl^2[/tex]

ignoring the dl^2 term for small x,

[tex]dl = x^2/2L[/tex]

substituting in (1),

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

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/

AM
 
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