Condition on Total Energy for Periodic Motion in Potential Well

To determine the condition on the total energy $E$ for the particle to perform periodic motion, we analyze the potential energy curve $V(x)$:

1. Minimum Potential Energy:
   Near the origin, the potential energy is given by:
   $V(x) = \alpha x^4 \quad (\alpha > 0)$
   The minimum potential energy occurs at $x = 0$:
   $V_{\text{min}} = V(0) = 0$
   The kinetic energy of the particle is given by:
   $K = E - V(x) \ge 0 \implies E \ge V(x)$
   For the particle to move (i.e., not remain permanently at rest at $x = 0$), its total energy must be strictly greater than the minimum potential energy:
   $E > 0$

2. Bound State Condition (Confinement):
   For $|x| \ge X_0$, the potential energy becomes a constant value $V_0$.

   • If $E \ge V_0$, the particle has sufficient energy to overcome the potential well and escape to infinity ($|x| \to \infty$). Such motion is unbounded and therefore non-periodic.
   • If $E < V_0$, the particle encounters classical turning points at $x = \pm x_t$ (where $|x_t| < X_0$ and $V(\pm x_t) = E$). The particle is trapped within the potential well, oscillating back and forth between $-x_t$ and $+x_t$, which constitutes periodic motion.

Combining both conditions:

• For periodic motion to occur, the total energy must satisfy:
  $0 < E < V_0$
  This corresponds directly to Option C.

• Since the question asks "will perform periodic motion only if", it specifies a necessary condition ($A \implies B$). Since $0 < E < V_0$ implies $E > 0$, the condition $E > 0$ is also a strictly necessary condition, making Option B valid as well.

Therefore, the correct options are B and C.