JEE Advanced · Physics · 15. Oscillations
Paragraph :
When a particle of mass \(m\) moves on the \(x\)-axis in a potential of the form \(V(x)=k x^2\), 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 \(\mathrm{m}\) moving on the \(x\)-axis. Its potential energy is \(V(x)=\alpha x^4(\alpha>0)\) for \(|x|\) near the origin and becomes a constant equal to \(V_0\) for \(|x| \geq X_0\) (see figure).
Question :
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 & Solution
Correct Answer
(C) \(V_0>E>0\)
Step-by-step Solution
Detailed explanation
If \(E>V_0\), particle will escape. But simultaneously for oscillations, \(E>0\). Hence, the correct answer is \(V_0>E>0\) or the correct option is (c).
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