# Non-linear spring

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).

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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.

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).

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.

a particle undergoes periodic motion under the influence of potential function:
$$V=kx^n$$

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.

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

hold on--I obviously did not do the physics right as I said it would decrease the period.
No, I think your physics is excellent.
Sorry 5 years ago I could've helped with the math.

AlephZero