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**FGT12** A particle has mass m and moves in the region x > 0 under the force $\displaystyle F = -m\omega^2(x - \frac{a^4}{x^3})$, where $\displaystyle \omega$ and $\displaystyle a$ are constants. The particle starts from the position of equilibirum with speed v. Find the limiting values of x in the subsequent motion. Show also that the period of oscillation is independent of v.

so far i have done:

*for the position of equilibrium this is where F=0

$\displaystyle -m\omega^2(x - \frac{a^4}{x^3})= 0$

$\displaystyle x = \frac{a^4}{x^3}$

$\displaystyle x = a$ as x>0.

*for the period

$\displaystyle k = V''(a) = -F'(a)$

$\displaystyle V''(x) = m\omega^2 + 3m\omega^2a^4x^-4$

$\displaystyle \frac{V''(a)}{m} = \frac{k}{m} = \omega^2 + 3\omega^2$

$\displaystyle \tau = \frac{2\pi}{\omega}= \frac{2\pi}{\sqrt(4\omega^2}$

$\displaystyle \tau = \frac{\pi}{\omega}$

I tried to use $\displaystyle T + V = E$

at x=a V=0 so

$\displaystyle \frac{1}{2}mv^2 + 0 = E$

$\displaystyle E = \frac{1}{2}mv^2$

as max, min x T=0

so $\displaystyle 0 + \int -F(x) dx = \frac{1}{2}mv^2$

however i get lost with constants of integration