The period of a physical pendulum is given by $T = 2\pi \sqrt{\frac{I}{mgh}}$, where $I$ is the moment of inertia about the pivot, $m$ is the mass of the stick, and $h$ is the distance from the pivot to the center of mass.

For a uniform stick of length L pivoted at one end, the distance to the center of mass is $h = L/2$.

The moment of inertia of the stick about its center of mass is $I_{cm} = \frac{1}{12}mL^2$. Using the parallel-axis theorem, the moment of inertia about the pivot point at the end is:

$$I = I_{cm} + mh^2 = \frac{1}{12}mL^2 + m\left(\frac{L}{2}\right)^2 = \frac{1}{12}mL^2 + \frac{1}{4}mL^2 = \frac{1}{3}mL^2$$

Substituting $I$ and $h$ into the period formula:

$$T = 2\pi \sqrt{\frac{\frac{1}{3}mL^2}{mg(L/2)}} = 2\pi \sqrt{\frac{2L^2}{3gL}}$$ $$T = 2\pi \sqrt{\frac{2L}{3g}}$$