The acceleration of this particle for |x| > X0 is

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).

The acceleration of this particle for |x| > X₀ is

(a) proportional to V₀

(b) proportional to V₀ / mX₀

(c) proportional to √(V₀ / mX₀)

(d) zero

Answer 1

Correct Answer is: (d) zero

For x > X₀, V is constant. Hence, force = 0 and therefore acceleration = 0.