Question

When a particle of mass m moves on the x-axis in a potential of the form $V\left(x\right)=k{x}^2$. It performs simple harmonic motion. The corresponding time period is proportional to $\sqrt{\frac{m}{k}}$, as can be seen easily using dimensional analysis. However, the motion of a particle can be periodic even when its potential energy increases on both sides of $x=0$ in a way different from $k{x}^2$ and its total energy is such that the particle does not escape to infinity. Consider a particle of mass m moving on the x-axis. Its potential energy is $V\left(x\right)=\alpha x^4(\alpha >0)$ for |x| near the origin and becomes a constant equal to $V_0$ for $|x|\ge X_0$(see figure).
The acceleration of this particle for $|x|>X_0$ is?

• A. Proportional to $V_0$
• B. Proportional to $\frac{V_0}{m{X}_0}$
• C. Proportional to $\sqrt{\frac{V_0}{m{X}_0}}$
• D. Zero

Answer 1

The correct option is D Zero

For $|x|>X_o$

Potential energy becomes almost constant. Thus, no change of velocity occurs. So, acceleration then is ${}^{\prime }{0}^{\prime }$