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Simple Harmonic Motion question

2010 · Shift 1 · Q78
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Simple Harmonic Motion question

2010 · Shift 1 · Q78

JEE AdvancedPhysicsSimple Harmonic MotionMCQ+3 / −1
When a particle of mass m moves on the x-axis in a potential of the form V(x) = kx2, it performs simple harmonic motion. The corresponding time period is proportional to mk\sqrt {{m \over k}}km​​, 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 kx2 and its total energy is such that the particle does not escape to infinity. Consider a particle of mass m moving on the x-axis. Its potential energy is V(x) = α\alphaα x4 (α\alphaα > 0) for | x | near the origin and becomes a constant equal to V0 for (see figure). IIT-JEE 2010 Paper 1 Offline Physics - Simple Harmonic Motion Question 13 English ComprehensionFor periodic motion of small amplitude A, the time period T of this particle is proportional to
  1. A
    Am/αA\sqrt {m/\alpha }Am/α​
  2. B
    1Am/α{1 \over A}\sqrt {m/\alpha }A1​m/α​
  3. C
    Aα/mA\sqrt {\alpha /m}Aα/m​
  4. D
    1Aα/m{1 \over A}\sqrt {\alpha /m}A1​α/m​
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