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 , 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) = αx4 (α > 0) for | x | near the origin and becomes a constant equal to V0 for | x

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 $\sqrt{\frac{m}{k}}$ , 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 = 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) = αx⁴ (α > 0) for | x | near the origin and becomes a constant equal to V₀ for | x | ≥ X₀ (see figure).

Linked Question 1

If the total energy of the particle is E, it will perform periodic motion only if

E = 0

V₀ > E > 0

E < 0

E > V₀

E > 0 as V(x) is always +ve

V₀ > E as it should not react X₀ as at that point F = 0 on it may go to infinity.

Linked Question 2

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

proportional to $\sqrt{\frac{V_{0}}{{mX}_{0}}}$

proportional to $\frac{V_{0}}{{mX}_{0}}$

zero

proportional to V₀

U = constant

F = 0

Linked Question 3

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

$A \sqrt{\frac{m}{α}}$

$\frac{1}{A} \sqrt{\frac{α}{m}}$

$\frac{1}{A} \sqrt{\frac{m}{α}}$

$A \sqrt{\frac{α}{m}}$

$[α] = \frac{[W]}{[L^{4}]} = \frac{[F]}{[L^{3}]}$

$[k] = \frac{[F]}{[L]}$

$[α] = \frac{[k]}{[L^{2}]}$

$[T] = [{(\frac{m}{k})}^{1 / 2}] = [{(\frac{m}{{αL}^{2}})}^{1 / 2}] = \frac{1}{A} \sqrt{\frac{m}{α}}$