# Quartic Oscillators

• Jun 24th 2009, 09:53 AM
calculusfrk
Quartic Oscillators
Hi, here is a practice question. I'm not sure where the equation of motion comes in to play. Any help would be much appreciated. Thanks!

The equation of motion for a particle of mass 1 in a quartic oscillator V(x)=(1/4)x^4 is ẍ+x^3=0. Suppose that the maximum amplitude of the oscillator is x(max). Find an expression for the time T that it takes to go from x=0 to x=x(max) and show that this time is inversely proportional to x(max).
• Jun 25th 2009, 01:46 AM
the_doc
Using conservation of energy (let maximum of $\displaystyle x$ be $\displaystyle x_0$) you must have:

$\displaystyle V(x_0) = \frac{1}{2}m v^2 + V(x)$

but since $\displaystyle m=1$ we have

$\displaystyle v^2 =2 \left( V(x_0) - V(x)\right)$

$\displaystyle \iff v^2 = \frac{1}{2} \left(x_0^4 - x^4 \right)$

so

$\displaystyle \frac{dx}{dt} = \sqrt{\frac{1}{2} \left(x_0^4 - x^4 \right)}$

which is of separable variable form so

$\displaystyle \int_{0}^{T} \frac{1}{\sqrt{2}} \, \mathrm{d}t = \int_{0}^{x_0} \frac{1}{\sqrt{x_0^4 - x^4}} \, \mathrm{d}x$.

Let $\displaystyle x= x_0 u$ then $\displaystyle \mathrm{d}x = x_0 \, \mathrm{d}u$ so this becomes

$\displaystyle \frac{T}{\sqrt{2}} = \int_{0}^{1} \frac{1}{x_0^2 \sqrt{1 - u^4}} x_0 \, \mathrm{d}u$

$\displaystyle \iff T = \frac{\sqrt{2}}{x_0} \int_{0}^{1} \frac{1}{\sqrt{1 - u^4}} \, \mathrm{d}u$.

At this point we have answered the question as the integral has no dependence on $\displaystyle x_0$ so is simply a numerical constant and so we've shown that the time period is inversely proportional to the amplitude $\displaystyle x_0$.

If you must know what the integral evaluates to:

$\displaystyle \int_{0}^{1} \frac{1}{\sqrt{1 - u^4}} \, \mathrm{d}u = \frac{1}{4} \beta \left(\frac{1}{4},\frac{1}{2} \right) = \frac{4 \left[\left(\frac{1}{4}\right)! \right]^2}{\sqrt{2 \pi}}$,

so not anything nice in terms of elementary functions!
• Jun 25th 2009, 06:01 PM
zorop
then x'' + x^3 = 0 doesn't have any meaning here?
x'' =v*dv/dx = - x ^3 then v dv = - x^3 dx
so we can get the v= dx/dt
is this right?
• Jun 25th 2009, 08:46 PM
zorop
k, maybe it's the same in the end
• Jun 26th 2009, 03:53 AM
mr fantastic
Quote:

Originally Posted by zorop
then x'' + x^3 = 0 doesn't have any meaning here?
x'' =v*dv/dx = - x ^3 then v dv = - x^3 dx
so we can get the v= dx/dt
is this right?

The solution to $\displaystyle \ddot{x} + x^3 = 0$ gives the Conservation of Energy equation that the_doc used as his/her starting point:

$\displaystyle \ddot{x} + x^3 = 0$

$\displaystyle \Rightarrow \frac{d^2 x}{dt^2} + x^3 = 0$

$\displaystyle \Rightarrow \frac{d}{dx} \left[ \frac{v^2}{2}\right] + x^3 = 0$

$\displaystyle \Rightarrow \frac{v^2}{2} + \frac{x^4}{4} = C$.