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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

  1. A proportional to \(V_0\)
  2. B proportional to \(\frac{V_0}{m X_0}\)
  3. C proportional to \(\sqrt{\frac{V_0}{m X_0}}\)
  4. D Zero
Verified Solution

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).
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