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 infinitely. 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).
If the total energy of the particle is E, it will perform periodic motion only if.

• A. $E<0$
• B. $E>0$
• C. $V_0>E>0$
• D. $E>V_0$

Answer 1

The correct option is C $V_0>E>0$

If $E>V_o\to$ Particle will escape

If $E>0\to$ Particle will not be in SHM

$\therefore$ Periodic motion is preferred only if $0<E<V_o$