Non-linear Oscillator: Understanding Bound Motion through Conservation of Energy

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SUMMARY

The discussion focuses on the behavior of a non-linear oscillator defined by the potential energy function U(x) = (1/2)kx² - (1/3)αx³, where k and α are positive constants. It establishes that oscillatory motion occurs when the initial position x₀ satisfies 0 < x₀ < k/α and the initial velocity v₀ is less than k/α√(k/m). The conservation of energy principle is applied, demonstrating that the total energy E remains constant, and the critical points of the potential energy function are identified as x = 0 and x = k/α. The discussion also emphasizes that for oscillatory motion, the energy must be less than zero at the extremes of displacement.

PREREQUISITES
  • Understanding of non-linear oscillators
  • Familiarity with conservation of energy principles
  • Knowledge of potential energy functions
  • Basic calculus for finding critical points (derivatives)
NEXT STEPS
  • Study the implications of the potential energy function U(x) for non-linear oscillators
  • Learn about the conditions for bound motion in mechanical systems
  • Explore the concept of critical points and their significance in oscillatory motion
  • Investigate the relationship between kinetic energy and potential energy in oscillatory systems
USEFUL FOR

Students and educators in physics, particularly those focusing on mechanics, as well as researchers studying non-linear dynamics and oscillatory systems.

Adorno
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Homework Statement


A non-linear oscillator consisting of a mass on a spring has a potential energy of the form \frac{1}{2}kx^2 - \frac{1}{3}\alpha x^3, where k and \alpha are positive constants, and x is displacement. Using conservation of energy, show that the motion is oscillatory if the initial position x_0 satisfies 0 &lt; x_0 &lt; \frac{k}{\alpha} and the initial velocity satisfies v_0 &lt; \frac{k}{\alpha}\sqrt{\frac{k}{m}}.

Homework Equations


E = T + U = \mathrm{constant}

F = -\frac{dU}{dx}

The Attempt at a Solution


By conservation of energy, the quantity E = \frac{1}{2}mv^2 + U(x) must be constant. So if the motion is oscillatory then the velocity will be zero at two (and only two) different positions, i.e. we have E = U(x_1) = U(x_2). Since E is the maximum value of the potential energy, this is equivalent to saying that the potential energy must reach the value of its local max/min at two (and only two) positions. In other words, the x values must lie between the two critical points of U. By setting dU/dx = 0, we get x = 0 and x = \frac{k}{\alpha} as the two critical points, and so we must have 0 &lt; x &lt; \frac{k}{\alpha}, as required.

I think the above is OK, but feel free to correct me if you see a problem. What I don't understand is how the initial velocity comes into it. The question is saying that if the initial velocity is greater than \frac{k}{\alpha}\sqrt{\frac{k}{m}}, then the motion won't be oscillatory, but I don't know how to derive this. Presumably this has to do with the conservation of energy as well -- I guess I have to use the kinetic energy term \frac{1}{2}mv^2 somehow. Could anyone help with this part?
 
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Draw the potential, and with that draw the force. For small initial velocity the particle can be trapped by the potential.
 
Adorno said:
What I don't understand is how the initial velocity comes into it. The question is saying that if the initial velocity is greater than \frac{k}{\alpha}\sqrt{\frac{k}{m}}, then the motion won't be oscillatory, but I don't know how to derive this. Presumably this has to do with the conservation of energy as well -- I guess I have to use the kinetic energy term \frac{1}{2}mv^2 somehow. Could anyone help with this part?

Consider also that a system is only bound for E < 0 . When the mass is at the extreme of displacement x = \frac{k}{\alpha}, what would the maximum permissible velocity be?

(I think there is some margin in the given conditions: I am getting some dimensionless multipliers on the order of 1 for the limits on displacement and velocity.)
 

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