# Non-linear spring

1. Jan 18, 2007

### Ja4Coltrane

On a physics test today, one of the free response questions asked about a non linear spring where F=-kx^3.
Now I got everything right, but there was one thing that I felt I did not have any concrete mathematical basis for--I just thought about it. I had to say whether increasing the amplitude would increase, decrease, or keep the period the same. I said decrease (which, apparentely is correct) just by considering the exponential-esque relationship between displacement and acceleration and by comparing that to the constant period SHM. Can anyone explain the mathematics better than this? Thanks (sorry--unnecesarily long post).

2. Jan 18, 2007

### Mindscrape

Well, it looks like you figured out the physics. If you stretch it further then the time it takes to complete an oscillation takes longer.

As far as the math goes, you would be best of doing it quantitatively instead of analytically. If you have Matlab or Mathematica you can do the differential equations for the spring and then look at their graphs.

3. Jan 18, 2007

### Ja4Coltrane

you're saying that I should integrate over the amplitude to find the speed that it moves at over each displacement element dx and find time by summing up all of the dx/v(x).

4. Jan 18, 2007

### Mindscrape

So, you have to know how to formulate the differential eqn.

$$F=ma=m \frac {d^2 x}{dt^2}$$

By Newton's second law, we know that the sum of all forces will be the force, so:

$$m \frac{d^2 x}{dt^2} = -kx^3$$
or
$$m \frac{d^2 x}{dt^2} + kx^3 = 0$$

This is the differential equation you need to solve for. Because of the x^3 term the equation is tough to solve analytically (because it is nonlinear), so I would find some kind of program that can show you the quantitative solution for different initial values of displacement. I think I am probably telling you to do something where you have no idea what I am talking about. But if you are seriously interested then I can try to explain.

5. Jan 19, 2007

### tim_lou

a particle undergoes periodic motion under the influence of potential function:
$$V=kx^n$$
can be reduced to quadrature.

consider:
$$\frac{1}{2}m\dot{x}^2=E-kx^n$$
$$\frac{dx}{\sqrt{E-kx^n}}=\sqrt{\frac{2}{m}}dt$$

integrate, get the period:
$$\tau=4\int_0^A\frac{dx}{\sqrt{E-kx^n}}\sqrt{\frac{m}{2}}$$
simplify, and you will get a result similar to:

edit: wrong!!! $$\tau\sim E^{\frac{n}{2}-1}$$

edit: sorry, the last step was incorrect!

when you do the integral carefully, things turn out to be:
$$\tau\sim E^{\frac{1}{n}-\frac{1}{2}}$$

so, when n=2, period is independent of the energy,
when n>2, the exponents is negative, as energy increases, tau decreases.
when n<2, the exponents is positive, as energy increases, tau increases.

Last edited: Jan 19, 2007
6. Jan 19, 2007

### Ja4Coltrane

hold on--I obviously did not do the physics right as I said it would decrease the period.

7. Jan 19, 2007

### curly_ebhc

No, I think your physics is excellent.
Sorry 5 years ago I could've helped with the math.

8. Jan 20, 2007

### AlephZero

Ja4Coltrane is right, the period decreases as the amplitude increases.

Mindscrape was wrong about the period, but the rest of his posts look correct.

Considering the relation between displacement and acceleration was a good way to figure it out.

Google for "Duffing equation" if you want more. There is a huge amount of research that has been done about solutions this equation, especially if you have an external force applied to the spring, not free oscillations. There are no simple "formulas" for the general solutions though.

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