A particle of mass $m$ and energy $E<0$ moves in a one-dimensional Morse potential:
$$V(x)=V_0(e^{-2ax}-2e^{-ax}),\qquad V_0,a>0,\qquad E>-V_0.$$
Determine the turning points of the movement and the period of the oscillation of the particle.
I have started learning for my exam and this is one of the exercises in my textbook.
Never dealt with this type of question, so these are my thoughts so far:
To get the turning points I was thinking of solving the equation $E=V_0(e^{-2ax}-2e^{-ax})$. I was doing the arithmetics with the absolute value of $E$. But still I couldn't seem to find the values for $x$. At the end I used Wolfram Alpha to find the values but it gave me results with complex values. Is there a simple way to solve this type of equations for $x$?
Anyway, about the period, I assume it's the time it takes for the particle to get from $x_1$ to $x_2$. But how am I supposed to approach this? How would I get a time value just out of the equation for the potential?
I hope someone can help me out here.
Best Answer
Energy conservation dictates $$ E = \frac{1}{2}m\dot{x}^2 + V(x) = \text{const}$$ With some arithmetic it follows $$ \dot{x} = \frac{dx}{dt} = \sqrt{2m^{-1}(E-V(x))}$$ This ODE can be solved via separation of variables, yielding $$ \int_{t_1}^{t_2}dt = \int_{x_1}^{x_2} \frac{dx}{\sqrt{2m^{-1}(E-V(x))}}$$ The integral on the left hand side can be evaluated immediately, where $t_1$ and $t_2$ are understood as the times when the particle is at $x_1$ or $x_2$ respectively. So it is simply half the period.
Observe that the integral on the right diverges when $x$ approaches the turning point $E=V(x)$.
This method of solving Newtons equations in a 1d potential should be treated in any textbook on mechanics.
For a general potential it is in general hard or impossible to find the turning points in closed form. Here however, you can substitute $y=\exp(-ax)$ and solve the corresponding quadratic equation. I'll leave the explicit calculation to you.