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Phase portrait for a conservative negative potential

  1. Oct 14, 2012 #1
    Hi,
    i was studying Classical Mechanics course introduced by the brilliant Prof. V. Balakrishnan

    he was studying the phase portrait of a simple harmonic oscillator in one dimension , where the total enegry of the system is E=[itex]\frac{1}{2}[/itex]m[itex]\dot{q}[/itex][itex]^{2}[/itex]+[itex]\frac{1}{2}[/itex]mω[itex]^{2}[/itex]q[itex]^{2}[/itex] where E is constant.
    in minute 54 he suggested to change the sign of the potential to be negative:
    E=[itex]\frac{1}{2}[/itex]m[itex]\dot{q}[/itex][itex]^{2}[/itex]-[itex]\frac{1}{2}[/itex]mω[itex]^{2}[/itex]q[itex]^{2}[/itex]
    the motion is not harmonic any more and the potential well is inverted to be opened in the negative y-direction, he asked to complete the phase portrait for this new motion,
    i tried to think of it physically and i was half way to get it, but i failed,
    here is the solution http://www-physics.ucsd.edu/students/courses/fall2010/physics200a/LECTURES/200_COURSE.pdf page 32

    the reason that i failed is that i couldn't able to know why the parabola was rotated to be symmetric around the x-axis?
     
    Last edited by a moderator: Sep 25, 2014
  2. jcsd
  3. Oct 18, 2012 #2
    I found one thing that convinced me somehow, the reason that the family of hyperbolas changed to be symmetric about the x-axis is that, if they will continue to be symmetric around the y-axis, phase trajectories will soon come to intersect each other, which can't be true.
     
  4. Oct 18, 2012 #3
    it was like drawing the hyperbola as y=f(x) so it becomes symmetric around the y-axis for the first group of hyperbolas, then drawing it as x=f(y) so it is symmetric around the x-axis for the second group of hyperbolas.
     
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