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 like a meter stick pivoted at a point other than its center of mass can be found using the formula $f = \frac{1}{2\pi}\sqrt{\frac{g}{l_{\text{eff}}}}$, where $g$ is the acceleration due to gravity (approximately $9.8 \, \text{m/s}^2$) and $l_{\text{eff}}$ 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 it 0.30 m from the pivot. Plugging these values into the formula gives a frequency close to 0.66 Hz.

A Uniform Meter Stick Is Freely Pivoted About the 0