Question

The period of a simple pendulum of length $l\text{is}{T}_1$ and time period of a uniform rod of the same length $l$ pivoted about one end and oscillating in a vertical plane is $T_2$. Amplitude of oscillation in both the cases is small. Then $\frac{T_1}{T_2}$ is

Answer 1

Time period of simple pendulum is given by
$T_1=2\pi \sqrt{\frac{1}{g}}$
Time period of a physical pendulum is given by
$T_2=2\pi \sqrt{\frac{I}{Mg{l}_{cm}}}$
Moment of inertia of rod,
$I=\frac{M{l}^2}{12}+\frac{M{l}^2}{4}=\frac{M{l}^2}{3}\text{and}{l}_{cm}=\frac{l}{2}$
So, we have,
$T_2=2\pi \sqrt{\frac{\left(\frac{m{l}^2}{3}\right)}{Mg\left(\frac{1}{2}\right)}}$
$T_2=2\pi \sqrt{\frac{2l}{3g}}$
$\frac{T_1}{T_2}=\sqrt{\frac{3}{2}}$
Final answer : (d)