# Unconstrained motion

1. Oct 27, 2008

### Benzoate

1. The problem statement, all variables and given/known data
A particle P F=36/x^3-9/x^2 (X>0)

Show that each motion of P consists of either(i) a periodic oscillation between two extreeme points , or (ii) an unbounded motion with one extreme point, depending upon the value of the total energy . Initially P is projected from the point x=4 with speed 0.5
. Show that P oscillates between two extremes points and find the period of the motion.
[You may make use of the formula

Show that there is a single equilibrium position for P and that it is stable. Find the period of small oscillations about this point.
2. Relevant equations

.5*m*v^2+V(x)=E

-dV/dx=F(x)

3. The attempt at a solution

V(x)=-F*dx=

V(x)=18/x^2-9/x

.5*v^2+18/x^2-9/x=E(x)
plugging v=.5 and x=4.0m
.5*(.5)^2+18/(4)^2-9/4=E

tau=2*$$\int$$ dx/[2*(E-V(x))].5

E=-1.00

therefore

v^2=-1-(18/x^2)+9/x

plugging in v=.5 and x=4 I can now find the extreme points ;

for the particle to be stable d^2V/dx^2 >0

2. Oct 28, 2008

### Benzoate

Maybe I should rephrase the question : Anybody having trouble reading my OP or have trouble understanding what I am trying to conveyed?

Last edited: Oct 28, 2008
3. Oct 28, 2008

### Benzoate

I will redo my solution because I think I calculated E wrong.

1/2mv^2+V(x)=E(x)

(1/2)*(1)*v^2+18/x^2-9/x=E

[(.5)^2/2+18/16+9/4=E

E=3.5

v^2=3.5-18x^2-9/x
.125=3.5-18/x^2-9/x

0=3.375-18/x^2-9/x

at turning points , E=0

using the quadratic formula I find that x=4 and x= 2.6667; In actuality , x=3 and x=6. Where did I go wrong?