Solution to 1992 Problem 61

The angular frequency of a physical pendulum is
\begin{align*}\omega = \sqrt{\frac{M g d}{I}}\end{align*}
where I is the moment of inertia about the pivot point and d is the distance from the pivot to the center of mass.
If we let l denote the length of the pendulum and m denote the mass of the identical masses, then
\begin{align*}I_I &= 2 m l^2 \\I_{II} &= \frac{5 m l^2}{4} \\d_I &= l \\d_{II} &= \frac{3 l}{4}\end{align*}
\begin{align*}\frac{f_{II}}{f_I}= \frac{\omega_{II}}{\omega_I} = \frac{\sqrt{\displaystyle \frac{M g d_{II}}{I_{II}}}}{\sqrt{...
Therefore, answer (A) is correct.

return to the 1992 problem list

return to homepage

Please send questions or comments to where X = physgre.