Find the Time period using First Integral

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SUMMARY

The discussion focuses on determining the time period of a particle confined to a potential energy described by \( V(x) = -\frac{V_{0}}{\cosh^2(\alpha{x})} \). The participant's approach involves setting the total energy \( E \) to zero, leading to an integral that diverges, suggesting an infinite time period. However, the consensus is that for a periodic solution to exist, the energy must be negative (\( E < 0 \)), indicating a bound state is necessary for periodic motion.

PREREQUISITES
  • Understanding of classical mechanics principles
  • Familiarity with potential energy functions
  • Knowledge of integral calculus, particularly improper integrals
  • Experience with energy conservation in mechanical systems
NEXT STEPS
  • Study the concept of bound states in quantum mechanics
  • Learn about classical turning points and their significance in motion
  • Explore the implications of negative potential energy on particle dynamics
  • Investigate the mathematical techniques for evaluating improper integrals
USEFUL FOR

This discussion is beneficial for physics students, particularly those studying classical mechanics and potential energy, as well as educators seeking to clarify concepts related to energy states and periodic motion.

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?
 

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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.
 

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