Physical pendulum: A uniform meter stick is freely pivoted about the 0.20-m mark. If it is allowed to swing in a vertical plane with a small amplitude and friction, what is the frequency of its oscillations?
A) 0.55 Hz
B) 0.66 Hz
C) 0.92 Hz
D) 1.1 Hz
E) 1.3 Hz

Question

Quizplus · Accepted Answer

The frequency of oscillation of a physical pendulum can be calculated using the formula $f = \frac{1}{2\pi} \sqrt{\frac{g}{L}}$, where $g$ is the acceleration due to gravity (approximately $9.8 \, \text{m/s}^2$) and $L$ is the effective length from the pivot point to the center of mass of the pendulum. For a uniform meter stick pivoted at the 0.20-m mark, the center of mass is at the 0.50-m mark from one end, making the distance from the pivot to the center of mass $0.50 - 0.20 = 0.30 \, \text{m}$. Plugging these values into the formula gives $f = \frac{1}{2\pi} \sqrt{\frac{9.8}{0.30}}$, which approximates to 0.66 Hz.

Physical Pendulum: a Uniform Meter Stick Is Freely Pivoted About