Answer
For a simple pendulum of a mass on a string:
$\omega = \sqrt{\frac{mgd}{I}} = \sqrt{\frac{g}{L}}$
Work Step by Step
For a mass on a string of length $L$, the moment of inertia is $I = mL^2$. The distance $d$ from the center of mass to the pivot is $L$.
We can find an expression for the angular frequency as:
$\omega = \sqrt{\frac{mgd}{I}}$
$\omega = \sqrt{\frac{mgL}{mL^2}}$
$\omega = \sqrt{\frac{g}{L}}$
