Nonlinear Vertical Spring

1. Dec 8, 2006

Raging Dragon

Solved the problem.

Last edited: Dec 9, 2006
2. Dec 9, 2006

OlderDan

Where do you get that the restoring force is proportional to z³? There is nothing in the problem statement to suggest that the spring is anything but a linear (force proportional to z) spring.

3. Dec 9, 2006

AlephZero

If the force IS given in the question as proportional to z^3, then you can find d by conservation of energy.

The equation of motion for the nonlinear spring is then a form of the Duffing equation, which has been (and still is being) studied in great detail - but AFAIK there is no analytic expression for the period. That suggests to me there is something wrong with the question, or your understanding of it.

4. Dec 9, 2006

Raging Dragon

Problem solved, don't need anymore help.

Last edited: Dec 9, 2006
5. Dec 9, 2006

arildno

You can find "d" by energy conservation:
The potential energy associated to gravity has wholly been converted into spring potential energy at the extremum d.

6. Dec 9, 2006

AlephZero

At the equlibrium position, the spring is streched a distance b and the tension is mg, so kb^3 = mg.

If we let the PE = 0 at z = 0, the initial energy of the system (KE + PE) = 0.

At displacement z (positive upwards) the PE (gravity + spring) is mgz + kz^4/4

At the extremes of the oscillation the velocity is 0, so the KE = 0, and KE + PE = 0 (conservation of energy from the starting condition), so the PE = 0.

So at the extremes mgd + kd^4/4.
Sp d = 0 (duh - we knew that already) or d^3 = -4mg/k = -4b^3.

You can write down the equation of motion as total energy = constant, i.e. 1/2 m(z-dot)^2 = -mgz + kz^4/4. But how you solve that to get the period, I dunno. Maybe you can express the solution as an an elliptic function or something since there's something that looks like sqrt(a + bz^3) in there. I dunno if that's a sensible suggestion or not.