Step 1: Equilibrium Tension

In equilibrium, the three particles lie on a straight line. Let the charge be $q$, mass $m$, and string length $l$. Let $k_e = \frac{1}{4\pi\epsilon_0}$.

The tension $T$ in the string balances the electrostatic repulsion on the outer particles.

• Force from center particle: $F_1 = \frac{k_e q^2}{l^2}$
• Force from far particle (distance $2l$): $F_2 = \frac{k_e q^2}{(2l)^2} = \frac{k_e q^2}{4l^2}$

$$ T = F_1 + F_2 = \frac{k_e q^2}{l^2} \left( 1 + \frac{1}{4} \right) = \frac{5 k_e q^2}{4 l^2} $$

Step 2: Analysis of Displacement (Center of Mass Frame)

Let the central particle be displaced transversely by $y_c$. Since the center of mass remains stationary (no external transverse forces), the outer particles must move in the opposite direction.

$$ m y_c + m y_{outer} + m y_{outer} = 0 $$ $$ y_c + 2 y_{outer} = 0 \implies y_{outer} = -y_c / 2 $$

The relative transverse displacement between the center and an outer particle is:

$$ \Delta y = y_c – y_{outer} = y_c – (-y_c/2) = \frac{3}{2} y_c $$

The angle $\theta$ made by the string with the horizontal is:

$$ \sin \theta \approx \tan \theta = \frac{\Delta y}{l} = \frac{3 y_c}{2 l} $$

Step 3: Net Restoring Force on Central Particle

The forces acting on the central particle in the transverse direction are components of the tension (pulling in) and the electrostatic repulsion from neighbors (pushing out).

• Vertical component of Tension: $2 T \sin \theta$ (Restoring)
• Vertical component of Electrostatic Repulsion from neighbors: $2 F_1 \sin \theta$ (Destabilizing)
• (Note: Force direction is along the string line).

Net Force $F_{net} = -2 T \sin \theta + 2 F_1 \sin \theta = -2 (T – F_1) \sin \theta$.

From Step 1, we know $T = F_1 + F_2$, so $T – F_1 = F_2 = \frac{k_e q^2}{4l^2}$.

$$ F_{net} = -2 \left( \frac{k_e q^2}{4l^2} \right) \sin \theta = – \frac{k_e q^2}{2l^2} \left( \frac{3 y_c}{2 l} \right) $$ $$ F_{net} = – \frac{3 k_e q^2}{4 l^3} y_c $$

Step 4: Calculate Period

The equation of motion is $m \ddot{y}_c = -K_{eff} y_c$, with $K_{eff} = \frac{3 k_e q^2}{4 l^3}$.

Angular frequency $\omega = \sqrt{\frac{K_{eff}}{m}} = \sqrt{\frac{3 k_e q^2}{4 m l^3}}$.

Substitute $k_e = \frac{1}{4\pi\epsilon_0}$:

$$ \omega = \sqrt{\frac{3 q^2}{16 \pi \epsilon_0 m l^3}} $$

Time Period $T_{period} = \frac{2\pi}{\omega}$:

$$ T_{period} = 2\pi \sqrt{\frac{16 \pi \epsilon_0 m l^3}{3 q^2}} = 8\pi \sqrt{\frac{\pi \epsilon_0 m l^3}{3 q^2}} $$