Question

A uniform rod of mass $M$ and length $L$ is pivoted about point $O$ as shown in the figure given. It is slightly rotated from its mean position so that it performs angular simple harmonic motion. For this physical pendulum, determine the time period of oscillation.

• A. $2\pi \sqrt{\frac{L}{g}}$
• B. $\pi \sqrt{\frac{7L}{3g}}$
• C. $2\pi \sqrt{\frac{2L}{3g}}$
• D. $noneofthese$

Answer 1

The correct option is B $\pi \sqrt{\frac{7L}{3g}}$

In the present case, $d=\frac{L}{4}$

$I=\frac{M{L}^2}{12}+M{\left(\frac{L}{4}\right)}^2=\frac{7}{48}M{L}^2$

So, time period of the physical pendulum is $T=2\pi \sqrt{\frac{I}{Mgd}}$

$=2\pi \sqrt{\frac{\frac{7}{48}M{L}^2}{Mg\times \frac{L}{4}}}=\pi \sqrt{\frac{7L}{3g}}$