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Homework Help: Unconstrained motion

  1. Oct 27, 2008 #1
    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*[tex]\int[/tex] 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. jcsd
  3. Oct 28, 2008 #2
    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
  4. Oct 28, 2008 #3
    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?
     
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