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A nonlinear spring whose restoring force is given by $F=-kx^3$ where $x$ is the displacement from equilibrium , is stretched a distance $A$. Attached to its end is a mass $m$. Calculate....(I can do that) ..suppose the amplitude of oscillation is increased, what happens to the period?

Here's what I think: If the amplitude is increased the spring posses more total energy, at equilibrium the spring is traveling faster than before because it posses more kinetic energy. I think in the spring travels faster when it's at a similar displacement from equilibrium, but it has to travel more distance, so I can't conclude anything.

I was think about solving,

$$mx''=-kx^3$$

But realized this is a very hard job.

Any ideas? Thanks.

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    $\begingroup$ "Here's what I think ... so I can't conclude anything." Yes you can! First think about simple harmonic motion, where $F = -kx$ and the period is independent of the amplitude. Now write the equation for your spring as $F =-k'x$ where $k' = kx^2$. The average stiffness of the nonlinear spring over each cycle of the motion increases with amplitude, so the period reduces. $\endgroup$ – alephzero May 8 '17 at 21:25
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The potential energy is $U\left(x\right) = kx^4/4$ since $-d/dx\left(kx^4/4\right) = -kx^3 = F$, and the energy $$ E = \frac{1}{2}m\left(\frac{dx}{dt}\right)^2 + \frac{1}{4}kx^4 $$ is conserved.

From the above you can show that $$ \begin{eqnarray} dt &=& \pm \ dx \sqrt{\frac{m}{2E}}\left(1-\frac{k}{4E}x^4\right)^{-1/2} \\ &=& \pm \ dx \sqrt{\frac{2m}{k}} \ A^{-2} \left[1-\left(\frac{x}{A}\right)^4\right]^{-1/2} \end{eqnarray} $$ where the amplitude $A = \left(4E / k\right)^{1/4}$ can be found from setting $dx/dt = 0$ in the expression for the energy and solving for $x$.

The period is then $$ \begin{eqnarray} T &=& 4 \sqrt{\frac{2m}{k}} \ A^{-2} \int_0^A dx \left[1-\left(\frac{x}{A}\right)^4\right]^{-1/2} \\ &=& 4 \sqrt{\frac{2m}{k}} \ A^{-1} \int_0^1 du \left(1-u^4\right)^{-1/2} \\ &=& \left(4 \sqrt{\frac{2m}{k}} I\right) A^{-1} \\ &\propto& A^{-1} \end{eqnarray} $$ where $u = x/A$ and $I = \int_0^1 du \left(1-u^4\right)^{-1/2} \approx 1.31$ (see this).

You can repeat the above for a more general potential energy $U\left(x\right) = \alpha \left|x\right|^n$, where you should find that

$$ dt = \pm \ dx \sqrt{\frac{m}{2\alpha}} \ A^{-n/2} \left[1-\left(\frac{\left|x\right|}{A}\right)^n\right]^{-1/2} $$

and

$$ \begin{eqnarray} T_n &=& \left(4 \sqrt{\frac{m}{2\alpha}} I_n\right) A^{1-n/2} \\ &\propto& A^{1-n/2} \end{eqnarray} $$

where

$$ I_n = \int_0^1 du \left(1-u^n\right)^{-1/2} $$

can be evaluated in terms of gamma functions (see this).

This is in agreement with the above for $\alpha = k/4$ and $n=4$, and with Landau and Lifshitz's Mechanics problem 2a of section 12 (page 27), where they find that $T_n \propto E^{1/n-1/2} \propto A^{1-n/2}$.

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