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

Question

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)

Answer 1

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).