Let the pivot be at the origin. The system consists of two rods of length $l$ pivoted at one end, carrying masses $m$ at the other ends. The masses are connected by a spring of stiffness $k$.

Let $\theta$ be the angle each rod makes with the horizontal symmetry axis. The vertical distance of each mass from the axis is $y = l \sin\theta$. The total length of the spring at any instant is: $$ x = 2l \sin\theta $$

Since the floor is frictionless and horizontal, gravity is balanced by the normal force. The only restoring force arises from the spring. For the system to oscillate about a stable equilibrium position $\theta_0$, the net torque at that position must be zero. Since there are no external forces driving the rods to a specific angle other than the spring itself, the equilibrium position corresponds to the relaxed state of the spring.

Therefore, at equilibrium angle $\theta_0$, the spring force is zero. $$ \text{Relaxed length } l_0 = 2l \sin\theta_0 $$

We displace the system by a small angle $\alpha$ from equilibrium, so $\theta = \theta_0 + \alpha$. The linear displacement of each mass perpendicular to the rod is $ds = l \alpha$.

Calculating the Restoring Torque:
The change in the spring’s length ($\Delta x$) for a small change in angle $d\theta$ is given by differentiating $x = 2l \sin\theta$: $$ dx = \frac{d}{d\theta}(2l \sin\theta) \cdot d\theta = 2l \cos\theta \cdot d\theta $$ For a small angular displacement $\alpha$, the extension in the spring is $\Delta x \approx 2l \cos\theta_0 \cdot \alpha$.

The spring force $F_s$ is: $$ F_s = k \Delta x = k (2l \cos\theta_0 \cdot \alpha) $$ This force acts vertically (perpendicular to the symmetry axis). To find the torque $\tau$ on one rod about the pivot, we need the component of this force perpendicular to the rod ($F_{\perp}$) multiplied by the rod length ($l$).

The angle between the vertical spring and the rod is $(90^\circ – \theta)$. Therefore: $$ F_{\perp} = F_s \cos\theta_0 $$ $$ \tau_{\text{restoring}} = – (F_{\perp}) l = – (F_s \cos\theta_0) l $$ Substituting $F_s$: $$ \tau_{\text{restoring}} = – [k (2l \cos\theta_0 \cdot \alpha) \cos\theta_0] l $$ $$ \tau_{\text{restoring}} = – 2 k l^2 \cos^2\theta_0 \cdot \alpha $$

Using Newton’s Second Law for rotation $\tau = I \ddot{\alpha}$ for one mass (where $I = ml^2$): $$ m l^2 \ddot{\alpha} = – 2 k l^2 \cos^2\theta_0 \cdot \alpha $$ $$ m \ddot{\alpha} + 2 k \cos^2\theta_0 \cdot \alpha = 0 $$ $$ \ddot{\alpha} + \left( \frac{2 k \cos^2\theta_0}{m} \right) \alpha = 0 $$

This represents Simple Harmonic Motion (SHM) with angular frequency $\omega$: $$ \omega^2 = \frac{2 k \cos^2\theta_0}{m} $$

The problem states that the period of oscillation is $T = 2\pi\sqrt{m/k}$. This implies the given frequency is: $$ \omega_{\text{given}} = \sqrt{\frac{k}{m}} \implies \omega_{\text{given}}^2 = \frac{k}{m} $$

Equating our derived frequency to the given frequency: $$ \frac{2 k \cos^2\theta_0}{m} = \frac{k}{m} $$ $$ 2 \cos^2\theta_0 = 1 $$ $$ \cos^2\theta_0 = \frac{1}{2} \implies \cos\theta_0 = \frac{1}{\sqrt{2}} $$ $$ \theta_0 = 45^\circ $$

Finally, we substitute $\theta_0$ back into the expression for the relaxed length $l_0$: $$ l_0 = 2l \sin(45^\circ) $$ $$ l_0 = 2l \left( \frac{1}{\sqrt{2}} \right) $$ $$ l_0 = \sqrt{2} l $$