Question

The time period of a simple pendulum of length $l$ 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 oscillations in both the cases is small. Then $T_1/T_2$ is

• A. $5\sqrt{3}$
• B. 1
• C. $\sqrt{\frac{3}{2}}$
• D. $15\sqrt{3}$

Answer 1

Answer: (C)

$\sqrt{\frac{3}{2}}$

Time period of simple pendulum is given by $T_1=2\pi \sqrt{\frac{l}{g}}$
Time period of a physical pendulum is given by $T_2=2\pi \sqrt{\frac{I}{Mg{l}_{cm}}}$
Where,
$I=\frac{M{l}^2}{12}+\frac{M{l}^2}{4}=\frac{M{l}^2}{3}$ and $l_{cm}=\frac{l}{2}$
So we have,
$T_2=2\pi \sqrt{\frac{\left(\frac{M{l}^2}{3}\right)}{Mg\left(\frac{l}{2}\right)}}\Rightarrow T_2=2\pi \sqrt{\frac{2l}{3g}}$
$\Rightarrow \frac{T_1}{T_2}=\sqrt{\frac{3}{2}}$