Answer
The period of oscillation is $~~1.64~s$
Work Step by Step
It was found that the distance from the end of the meter stick down to the point $P$ is $\frac{2L}{3}$
Then the distance between $C$ and $P$ is $h = \frac{L}{6}$
We can find the period of oscillation:
$T = 2\pi~\sqrt{\frac{I}{mgh}}$
$T = 2\pi~\sqrt{\frac{\frac{1}{12}mL^2+m(\frac{L}{6})^2}{mgh}}$
$T = 2\pi~\sqrt{\frac{\frac{L^2}{12}+\frac{L^2}{36}}{(g)(\frac{L}{6})}}$
$T = 2\pi~\sqrt{\frac{\frac{L}{2}+\frac{L}{6}}{g}}$
$T = 2\pi~\sqrt{\frac{2L}{3g}}$
$T = 2\pi~\sqrt{\frac{(2)(1.0~m)}{(3)(9.8~m/s^2)}}$
$T = 1.64~s$
The period of oscillation is $~~1.64~s$
