# Period $T$ of oscillation with cubic force function

How would I find the period of an oscillator with the following force equation?

$$F(x)=-cx^3$$

I've already found the potential energy equation by integrating over distance:

$$U(x)={cx^4 \over 4}.$$

Now I have to find a function for the period (in terms of $A$, the amplitude, $m$, and $c$), but I'm stuck on how to approach the problem. I can set up a differential equation:

$$m{d^2x(t) \over dt^2}=-cx^3,$$

$$d^2x(t)=-{cx^3 \over m}dt^2.$$

But I am not sure how to solve this. Wolfram Alpha gives a particularly nasty solution involving the hypergeometric function, so I don't think the solution involves differential equations. But I don't have any other leads.

How would I find the period $T$ of this oscillator?

Since $$\frac1 2mv^2+U(x)=U(A)$$ We have $$dt=\frac{dx}v=\frac{dx}{\sqrt{2(U(A)-U(x))/m}}=\frac{dx}{\sqrt{c(A^4-x^4)/(2m)}}$$ Then $$\frac T4=\int_0^{\frac T4}dt=\int_0^A\frac{dx}{\sqrt{\frac{c}{2m}(A^4-x^4)}}$$ Thus $$T=4\int_0^A\frac{dx}{\sqrt{\frac{c}{2m}(A^4-x^4)}}$$

Starting from $$\frac{1}{2} \left( v(x)^2 - v_0^2 \right)= - \frac{c}{m} x^4$$

with initial velocity $v_0$ when $x=0$, the time relationship is

$$t = \int_0^x \frac{1}{v(x)}\,{\rm d} x$$

I use in intermediate variable $\xi$ for distance $x = \sqrt{\frac{2 m v_0^2}{c}}\, \xi$ .

I integrate the energy relationship to get

$$t = \int_0^x \frac{1}{\sqrt{v_0^2 - \frac{c x^4}{2 m} }} \,{\rm d} x=\sqrt{ \frac{2 m}{c v_0^2} } \int_0^\xi \frac{1}{\sqrt{1-\xi^4}}\,{\rm d} \xi$$

$$t = \sqrt{ \frac{2 m}{c v_0^2} }\; {\rm EllipticF}( \sin^{-1}\xi, -1)$$

Note that the elliptic integral has a taylor expansion of

$${\rm EllipticF}(x,m) \approx x + \frac{m}{6} x^3 - \frac{m}{30} x^5 + \ldots$$

which makes the above solution approximately (for small displacements)

$$t = \frac{ \tau}{2\pi} \; \sin^{-1}\xi$$

with period $\tau = 2 \pi \sqrt{ \frac{2 m}{c v_0^2} }$ and final solution

$$x = \sqrt{\frac{2 m v_0^2}{c}}\,\sin \left( \frac{2 \pi t}{\tau} \right)$$

• +1, But I am not able to reproduce the step from the original integral $\int_0^\xi \frac{1}{\sqrt{1-\xi^4}}\,{\rm d} \xi$ to some definition of the elliptic integral of the first kind. Could you add some information ? – Trimok Sep 25 '13 at 9:45
• Use Alpha to confirm it. – ja72 Sep 28 '13 at 17:06