Solution – Problem 1

Problem 1 Solution

1. Equilibrium Positions

The potential energy is given by \( U(x) = -\frac{kx}{x^2 + x_0^2} \).
The force acting on the particle is \( F = -\frac{dU}{dx} \). Equilibrium positions occur where the force is zero.

\[ \frac{dU}{dx} = -k \left[ \frac{(x^2 + x_0^2)(1) – x(2x)}{(x^2 + x_0^2)^2} \right] \] \[ \frac{dU}{dx} = -k \left[ \frac{x_0^2 – x^2}{(x^2 + x_0^2)^2} \right] = k \frac{x^2 – x_0^2}{(x^2 + x_0^2)^2} \]

Setting \( \frac{dU}{dx} = 0 \): \[ x^2 – x_0^2 = 0 \implies x = \pm x_0 \]

2. Stability Analysis

To determine stability, we evaluate the second derivative \( \frac{d^2U}{dx^2} \) at the equilibrium points.

At \( x = -x_0 \) (Top of the hill): Potential is Maximum \(\rightarrow\) Unstable.
At \( x = +x_0 \) (Bottom of the valley): Potential is Minimum \(\rightarrow\) Stable.

Calculating \( k_{eff} \) at \( x = x_0 \): \[ k_{eff} = \left. \frac{d^2U}{dx^2} \right|_{x=x_0} = \frac{k}{2x_0^3} \] Since \( k_{eff} > 0 \), the position \( x = x_0 \) is indeed the stable equilibrium.

3. Period of Oscillation

For small oscillations about the stable equilibrium, the system behaves like a simple harmonic oscillator with effective spring constant \( k_{eff} \). \[ T = 2\pi \sqrt{\frac{m}{k_{eff}}} \] Substituting \( k_{eff} = \frac{k}{2x_0^3} \): \[ T = 2\pi \sqrt{\frac{m}{k / 2x_0^3}} = 2\pi \sqrt{\frac{2m x_0^3}{k}} \]

Answer: \( x = x_0 \) and \( T = 2\pi \sqrt{\frac{2m x_0^3}{k}} \)