2 question is: how to move out the particle from the potential well? I am stuck on a problem that I have a simple harmonic oscillation for which I want to draw a phase space. I can do this inside the potential well, nevertheless can not do it outside. In other words I can draw only spirals(due to the fact that my equation has a damping part) How I try to solve my problem. For instance I have x0 = [0,1;velocity]. Which velocity I should give to make my diff eq. get out the bonds of potential well? I suppose. $$H = p^2/2m + U(q)$$ $$H = x'^2/2 - w^2 x$$ $$T = x'^2/2$$ $$U(q) = -w^2 x$$ random point $w = 0.1$ $x = 0.1 =>$ $$U = - (0.1)^2*0.1 = - 2 *10^-3$$ peak point $w = 0.1$ $x = 2pi =>$ $$U = - (0.1)^2*1 = -2 *10^-2$$ diff (delta = peak-random) points is: $$\delta = -2 *10^-2 - - 2 *10^-3$$ energy to go over peak is: $$x'^2/2 = \delta +\text{some_extra_amount_to_be_precise_that_particle_will_go_over_the_well}$$ let this be $\delta' = 0.1;$ $$x'^2 = 2(delta+\delta')$$ $$x' = (2(delta+\delta'))^(1/2)$$ However this does not help me.
Here is the picture for such initial conditions
x01=[1,vel];
x02=[1,1];
x03=[1,10];
x04=[1,100];
x05=[1,1000];
The full code below =) :
time = 700;
tspan = 0:0.01:time;
w=0.1;
g=0.01;
k=1;
delta = -2 *10^-2 - - 2 *10^-3;
newdelta = 0.1;
vel = (2*(delta+newdelta))^(1/2);
x01=[1,vel];
x02=[1,1];
x03=[1,10];
x04=[1,100];
x05=[1,1000];
x06=[1,10000];
options = odeset('RelTol',1e-4,'AbsTol',[1e-4 1e-4]);
[t,x1]=ode45('test_cos',tspan,x01, options,w, g);
[t,x2]=ode45('test_cos',tspan,x02, options,w, g);
[t,x3]=ode45('test_cos',tspan,x03, options,w, g);
[t,x4]=ode45('test_cos',tspan,x04, options,w, g);
[t,x5]=ode45('test_cos',tspan,x05, options,w, g);
[t,x6]=ode45('test_cos',tspan,x06, options,w, g);
figure(1)
plot(x1(:,1),x1(:,2),'--','LineWidth' ,1,'Color','r');
hold on;
plot(x2(:,1),x2(:,2),'--','LineWidth' ,1,'Color', 'g');
plot(x3(:,1),x3(:,2),'--','LineWidth' ,1,'Color', 'b');
plot(x4(:,1),x4(:,2),'--','LineWidth' ,1,'Color', 'k');
plot(x5(:,1),x5(:,2),'--','LineWidth' ,1,'Color', [1.0,0.4,0.0] );
plot(x6(:,1),x6(:,2),'--','LineWidth' ,1,'Color', [1 0 1] );
hold off;
function xDot = test_cos(t,x,~,w,g)
xDot = [x(2); ...
- (g*x(2)+w^2*x(1) ) ];
end
3 Small question - does matlab has a way to make extra condition( for instance, I want to say that 100% guarantee nothing can go out from this segment $[0;3\pi]$ ( for example the potential barrier is infinity after $3\pi$ and before $0$.
I want to find such energy(such velocity that some body can get out of the bonds, here is the picture of it.
Hello everybody, a small wisdom: there is actually a benefit to relax from time to time. After a walk, I have discovered that my question: is nonsense =),
if I use the potential energy like this $U(x)=−w^2x.$
I assert that the particle can not make any movement as I showed on draw by hand picture.
Because, again, the potential energy here is the parabola, and it is parabola that is defined at all the straight line x.
To make a movement as I draw on a picture - potential energy should be $U(x)=−w^2\sin(x)$ (or any other periodic function), where $sin$ - creates holes and humps which I wanted.
After I change the potential energy: $$U(x)=−w^2x ==> U(x)=−w^2\sin(x) $$ I get what I wanted:
After huge zoom, I can see the red and green curves, they show how the trajectories slow down because of the drag coefficient g=0.01 and get attracted by the potential holes:
