# Tricky problem

1. Nov 27, 2006

### Logarythmic

Consider the one dimentional motion of a particle in the potential

$$V(x)=D(e^{-2ax}-2e^{-ax})$$.

I'm supposed to obtain the expressions for x as a function of time separately for the cases that the total energy E is positive, zero or negative.

I have used

$$\frac{1}{2}m\dot{x}^2 + V(x) - E = 0$$

and got the integral

$$\int \sqrt{\frac{m}{2(E-V(x))}}dx$$

to solve.

First, is this a correct method?

Second, how do I solve this integral?

2. Nov 28, 2006

### dextercioby

Yes, it looks okay. The same kinda substitution i advised in the thread on turning points would be useful.

Daniel.

3. Nov 28, 2006

### KoGs

You guess a solution.