# Phase portrait for a conservative negative potential

1. Oct 14, 2012

### IWantToLearn

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=$\frac{1}{2}$m$\dot{q}$$^{2}$+$\frac{1}{2}$mω$^{2}$q$^{2}$ where E is constant.
in minute 54 he suggested to change the sign of the potential to be negative:
E=$\frac{1}{2}$m$\dot{q}$$^{2}$-$\frac{1}{2}$mω$^{2}$q$^{2}$
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. Oct 18, 2012

### IWantToLearn

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.

3. Oct 18, 2012

### IWantToLearn

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.