# 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.