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JEE Advanced 2010Paper 1
Exam
JEE Advanced
Year
2010
Paper
Paper 1
Subject
Physics
Chapter
Oscillations

JEE Advanced 2010 · Paper 1 · Physics · Oscillations

Paragraph: When a particle of mass m moves on the x -axis in a...

Paragraph: When a particle of mass mm moves on the xx-axis in a potential of the form V(x)=kx2V(x)=k x^2, it performs simple harmonic motion.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
The corresponding time period is proportional to mk\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=0x=0 in a way different from kx2k x^2 and its total energy is such that the particle does not escape to infinity. Consider a particle of mass m\mathrm{m} moving on the xx-axis. Its potential energy is V(x)=αx4(α>0)V(x)=\alpha x^4(\alpha>0) for ∣x∣|x| near the origin and becomes a constant equal to V0V_0 for ∣x∣≥X0|x| \geq X_0 (see figure).Question: For periodic motion of small amplitude AA, the time period TT of this particle is proportional to

Options

  1. AAmαA \sqrt{\frac{m}{\alpha}}
  2. B1Amα\frac{1}{A} \sqrt{\frac{m}{\alpha}}
  3. CAαmA \sqrt{\frac{\alpha}{m}}
  4. D1Aαm\frac{1}{A} \sqrt{\frac{\alpha}{m}}

Correct answer

B. 1Amα\frac{1}{A} \sqrt{\frac{m}{\alpha}}

Step-by-step solution

[α]=[PEx4]=[ML2T−2L4]=[ML−2T−2][\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]∴[mα]=[L2T2]∴[1Amα]=[T]\begin{array}{ll} \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}]} \end{array}AsdimensionsofamplitudeAs dimensions of amplitude Ais[L].Hence,thecorrectoptionis(b).is [L]. Hence, the correct option is (b).
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