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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 :
For periodic motion of small amplitude \(A\), the time period \(T\) of this particle is proportional to

  1. A \(A \sqrt{\frac{m}{\alpha}}\)
  2. B \(\frac{1}{A} \sqrt{\frac{m}{\alpha}}\)
  3. C \(A \sqrt{\frac{\alpha}{m}}\)
  4. D \(\frac{1}{A} \sqrt{\frac{\alpha}{m}}\)
Verified Solution

Answer & Solution

Correct Answer

(B) \(\frac{1}{A} \sqrt{\frac{m}{\alpha}}\)

Step-by-step Solution

Detailed explanation

\([\alpha]=\left[\frac{\mathrm{PE}}{x^4}\right]=\left[\frac{\mathrm{ML}^2 \mathrm{~T}^{-2}}{\mathrm{~L}^4}\right]=\left[\mathrm{ML}^{-2} \mathrm{~T}^{-2}\right]\)
\(\therefore {\left[\frac{m}{\alpha}\right]=\left[\mathrm{L}^2 \mathrm{~T}^2\right]}\)
\(\therefore {\left[\frac{1}{A} \sqrt{\frac{m}{\alpha}}\right]=[\mathrm{T}]}\)
As dimensions of amplitude \(A\) is [L].
Hence, the correct option is (b).
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