# Physical pendulum and meter stick

A physical pendulum consists of a meter stick that is pivoted at a small hole drilled through the stick a distance x from the 50. cm mark. The period of oscillation is observed to be 1.656 s. Find the distance x.

I know I have to find the moment of Inertia and I believe the equation for this is I=1/2mL^2 + mx^2 is this the right equation? Then use the equation T= 2pie(I/mgx)^1/2.

This is what I did but it's not right so if any one can point out any errors will be greatly appreciated:
I=1/2mL^2 + mx^2
I=1/2m(0.5)^2 + mx^2
I=0.125m + mx^2

T=2pie(I/mgx)^1/2
1.656=2pie{(0.125m + mx^2)/(9.8mx)}^1/2
0.263= {(0.125+x)/9.8}^1/2
0.0692(9.8)=0.125+x
0.553=x

Is this right?