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

AI Thread Summary
The discussion focuses on a non-linear oscillator with a specific potential energy function, emphasizing the conditions for oscillatory motion based on conservation of energy. It is established that for oscillation to occur, the initial position must satisfy 0 < x_0 < k/α, and the initial velocity must be less than v_0 < k/α√(k/m). The conservation of energy principle indicates that the motion will be oscillatory if the total energy remains constant, with velocity reaching zero at two distinct points. The relationship between initial velocity and oscillatory motion is explored, highlighting that exceeding a certain velocity threshold prevents the system from being bound. The discussion concludes by suggesting that further analysis of energy limits and potential energy behavior is necessary to fully understand the implications of initial velocity on oscillation.
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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