# oscillating particle problem

• Aug 18th 2010, 10:50 AM
FGT12
oscillating particle problem
A particle has mass m and moves in the region x > 0 under the force $F = -m\omega^2(x - \frac{a^4}{x^3})$, where $\omega$ and $a$ 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
$-m\omega^2(x - \frac{a^4}{x^3})= 0$
$x = \frac{a^4}{x^3}$
$x = a$ as x>0.

*for the period
$k = V''(a) = -F'(a)$
$V''(x) = m\omega^2 + 3m\omega^2a^4x^-4$
$\frac{V''(a)}{m} = \frac{k}{m} = \omega^2 + 3\omega^2$
$\tau = \frac{2\pi}{\omega}= \frac{2\pi}{\sqrt(4\omega^2}$
$\tau = \frac{\pi}{\omega}$

I tried to use $T + V = E$
at x=a V=0 so
$\frac{1}{2}mv^2 + 0 = E$
$E = \frac{1}{2}mv^2$

as max, min x T=0
so $0 + \int -F(x) dx = \frac{1}{2}mv^2$

however i get lost with constants of integration
• Aug 18th 2010, 12:21 PM
Ackbeet
I don't think your method works for finding the turning points. Finding where the force is zero is not the same as finding where the velocity is zero, which is where your turning points are. I would use turning point analysis to find the limiting values of x in the motion. That is, integrate the force to get the potential energy. Then set the potential energy equal to the total energy (define equilibrium to be zero potential energy; then the total energy is all kinetic at the equilibrium position) to find where the turning points are. Make sense?
• Aug 19th 2010, 11:35 PM
chisigma
Quote:

Originally Posted by FGT12
A particle has mass m and moves in the region x > 0 under the force $F = -m\omega^2(x - \frac{a^4}{x^3})$, where $\omega$ and $a$ 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
$-m\omega^2(x - \frac{a^4}{x^3})= 0$
$x = \frac{a^4}{x^3}$
$x = a$ as x>0.

*for the period
$k = V''(a) = -F'(a)$
$V''(x) = m\omega^2 + 3m\omega^2a^4x^-4$
$\frac{V''(a)}{m} = \frac{k}{m} = \omega^2 + 3\omega^2$
$\tau = \frac{2\pi}{\omega}= \frac{2\pi}{\sqrt(4\omega^2}$
$\tau = \frac{\pi}{\omega}$

I tried to use $T + V = E$
at x=a V=0 so
$\frac{1}{2}mv^2 + 0 = E$
$E = \frac{1}{2}mv^2$

as max, min x T=0
so $0 + \int -F(x) dx = \frac{1}{2}mv^2$

however i get lost with constants of integration

The global problem is a little complex and the best is to divide it in 'slots'. The second order DE describing the motion of the particle is...

$\displaystyle x^{''} = - \omega^{2}\ (x-\frac{a^{4}}{x^{3}})$ , $x(0)= a$ , $x^{'} (0)= v$ (1)

A DE like (1) requires the substitution...

$\displaystyle x^{''} = \frac{d x^{'}}{d t} = \frac{d x^{'}}{d x}\ \frac{d x}{d t} = x^{'}\ \frac{d x^{'}}{d x}$ (2)

... so that (1) becomes...

$\displaystyle x^{'} \ \frac{d x^{'}}{d x} = - \omega^{2}\ (x-\frac{a^{4}}{x^{3}})$, $x^{'} (a) = v$ (3)

The (3) is a first oder DE with separable variables and its solution procedure is 'standard'. Applying that procedure You find...

$\displaystyle x^{'} = \sqrt {v^{2} + \omega^{2}\ (2 a^{2} - x^{2} - \frac{a^{4}}{x^{2}})}$ (4)

The 'first slot' ends here. Now You have (4) that is also a first order DE with separable variables and its 'initial condition' is $x(0)= a$. Successive slots are left to You...

Kind regards

$\chi$ $\sigma$