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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 √(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 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) = ax4 (α > 0) for |x| near the origin and becomes a constant equal to V0 for |x| ≥ X0 (see figure).For periodic motion of small amplitude A, the time period T of this particle is proportional to(a) A √(m/α)(b) 1/A √(m/α)(c) A √(α/m)(d) 1/A √(α/m) |
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Answer» Correct Answer is: (b) 1/A √(m/α) V = αx4, ∴ [α] = ML-2 T-2. The only expression which has the dimension M0 L0 T is (b). |
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