The system consists of two masses $m$ connected by a rigid rod. For small amplitude oscillations, the horizontal displacement $x$ for both balls is approximately the same (as per the hint in the problem). The rod ensures they move in unison horizontally.

When displaced by a small horizontal distance $x$, the cords make angles $\theta_1$ and $\theta_2$ with the vertical.
For small angles: $$ \sin \theta_1 \approx \frac{x}{l_1} \quad \text{and} \quad \sin \theta_2 \approx \frac{x}{l_2} $$ The restoring force is the horizontal component of tension (which balances gravity vertically, so $T \approx mg$). $$ F_1 = -mg \sin \theta_1 \approx -mg \frac{x}{l_1} $$ $$ F_2 = -mg \sin \theta_2 \approx -mg \frac{x}{l_2} $$

Total mass of the system is $M = 2m$. The net restoring force is the sum of forces from both strings: $$ F_{net} = F_1 + F_2 = -mgx \left( \frac{1}{l_1} + \frac{1}{l_2} \right) $$ Using Newton’s Second Law ($F = Ma$): $$ 2m \cdot a = -mgx \left( \frac{l_1 + l_2}{l_1 l_2} \right) $$ $$ a = – \left[ \frac{g(l_1 + l_2)}{2 l_1 l_2} \right] x $$

Comparing this with the standard SHM equation $a = -\omega^2 x$: $$ \omega^2 = \frac{g(l_1 + l_2)}{2 l_1 l_2} $$ The time period $T = \frac{2\pi}{\omega}$ becomes: $$ T = 2\pi \sqrt{\frac{2 l_1 l_2}{g(l_1 + l_2)}} $$