Question

A physical pendulum consists of a meter stick that is pivoted at a small hole drilled through the stick a distance d from the 50 cm mark. The period of oscillation is 2.54 s. Find d.

Answer 1

equivalent length
Leq = I / md
where I is the moment of inertia about the pivot
and m is the mass
and d is the distance from the pivot to the center of mass

By the parallel axis theorem,
I = mL²/12 + md², so
Leq = (mL²/12 + md²) / md = L²/12d + d

period T = 2π√(Leq / g) → square everything
T² = 4π²(L²/12d + d) / g
gT² / 4π² = L²/12d + d → plug in g and T and drop units for ease (d is in meters):
9.8*(2.54)² / 4π² = 1/12d + d
1.6 = 1/12d + d → multiply by d
1.6d = 1/12 + d²
0 = d² - 1.6d + 1/12

This quadratic has roots at
d = .0539 cm = 5.39 cm◄ solution
and d =1.546 m ← greater than 1 m; disregard