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 :
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 & Solution
Correct Answer
(D) Zero
Step-by-step Solution
Detailed explanation
For \(|x|>X_0\), potential energy is constant. Hence, kinetic energy, speed or velocity will also remain constant.
\(\therefore\) Acceleration will be zero.
Hence, the correct option is (d).
\(\therefore\) Acceleration will be zero.
Hence, the correct option is (d).
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