Here $L$ is the length of th whole massless rod and $3$ balls have been attached to the rod having lengths of $\frac{L}{3}$. We have to find the time period of the pendulum as a whole.
Here is my approach. Since we are used to calculating pendulum related problems for one bob,i tried to reduce the system to an equivalent one ball system.
Let us assume mass of each ball is $m$. Now when the pendulum is displaced with a slight angular displacement,the restoring force acting on each of the ball is $mg\sin \theta=mg\theta$ when $\theta$ is small. Since balls experience $mg\theta$ force in the same direction, in the equivalent one ball system,the net force on that corresponding ball will be $3mg\theta$. Now from hooks law, $F=kx$ where $k$ is the SHM constant and $x$ is displacement. Here the equivalent one ball will travserse the same angle as these $3$ balls, so $\theta_{equivalent ball}$ will be equal to $\theta$. Therefore $3mg\theta=kL\theta$ so $k=\frac{3mg}{L}$. And we know $\omega^2=\frac{k}{m}=\frac{3g}{L}$. And so $T=\frac{2\pi}{\omega}=2\pi\sqrt{\frac{L}{3g}}$.
Is this correct?If wrong,please correct my thinking process and kindly post the actual solution.