Find the Time period using First Integral

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Muthumanimaran
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Recently I lend the Classical mechanics book written by Goldstein from the library, In the last page, someone scribbled this problem without any solution, I am just curious and want to give a try the problem mentioned below. I just want to know whether my approach and my solution is correct or not.
1. Homework Statement

Find the Time period of the particle confined to the Potential energy of the form $V(x)=-\frac{V_{0}}{\cosh^2(\alpha{x})}$.

Homework Equations


$$\int_{0}^{t_0}dt=\int_{x_0}^{x}\frac{dx}{\sqrt{\frac{2}{m}(E-V(x))}}$$
Where $'E'$ is the Total Energy
and $'m'$ is the mass of the particle

The Attempt at a Solution


Since the potential energy is negative, the total Energy required to escape from the potential is zero, i.e, $E=0$,
The Potential diagram is attached with this thread, as we can see, the potential goes to 0 as $x$ goes to $\infty$
or $-\infty$, If I take one half of the Potential say from $0$ to $\infty$ and substituting in the above integral becomes,

$$\int_{0}^{\frac{T}{2}}dt=\sqrt{\frac{m}{2V_0}}\int_{0}^{\infty}cosh(\alpha{x})dx$$

Apparently the right hand side goes to infinity. And the time period of the particle confined also goes to infinity. Am I right? I solved the problem but I don't know whether my solution is correct or not. Is there any mistake in my solution?
 
Attachments
  • Graph of a potential well: a negative, bell-shaped curve with minimum of -1 at x = 0, approaching 0 as x goes toward -5 and 5, axes from -10 to 10 horizontally and -1 to 0 vertically, legend reads Potential well.
    potential well.png
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If you take E = 0 there is no classical turning point and the motion is not periodic. In order to have a periodic solution you need to assume a bound state with E < 0.