A uniform square plate of edge length a is suspended from one of...

The time period of a physical pendulum is given by $T = 2\pi \sqrt{\dfrac{I}{mgd}}$, where $I$ is the moment of inertia about the axis of suspension and $d$ is the distance from the point of suspension to the center of mass.

For a square plate of edge length $a$ suspended from a corner, the distance to the center of mass is half the diagonal:

$d = \dfrac{a}{\sqrt{2}}$

The moment of inertia of the square plate about an axis passing through its center of mass and perpendicular to its plane is:

$I_{cm} = \dfrac{m(a^2 + a^2)}{12} = \dfrac{ma^2}{6}$

Using the parallel axis theorem, the moment of inertia about the corner is:

$I = I_{cm} + md^2 = \dfrac{ma^2}{6} + m\left(\dfrac{a}{\sqrt{2}}\right)^2 = \dfrac{ma^2}{6} + \dfrac{ma^2}{2} = \dfrac{2ma^2}{3}$

Substituting $I$ and $d$ into the time period formula, we get:

$T = 2\pi \sqrt{\dfrac{\dfrac{2ma^2}{3}}{mg\left(\dfrac{a}{\sqrt{2}}\right)}} = 2\pi \sqrt{\dfrac{2\sqrt{2}a}{3g}} = 2\pi \sqrt{\dfrac{\sqrt{8}a}{3g}}$