Time period of a simple pendulum of length ' $L$ ' is $T_1$. Time period of a uniform rod of same length '…
Time period of a simple pendulum of length ' $L$ ' is $T_1$. Time period of a uniform rod of same length ' $L$ ' suspended from one end and oscillating in a vertical plane is $T_2$. Amplitude of oscillation is small in both the cases. Then $\frac{T_1}{T_2}$ is
$\sqrt{\frac{2}{3}}$
$\sqrt{\frac{3}{2}}$
$\sqrt{\frac{4}{3}}$
1
Solution
Length of pendulum $=L$
Time period of simple pendulum
$
T_1=2 \pi \sqrt{\frac{L}{g}}
$
Length of $\operatorname{rod}=L$
Time period of rod;
$
\begin{aligned}
& T_2=2 \pi \sqrt{\frac{2 L}{3 g}}\left[\because T=2 \pi \sqrt{\frac{I}{m g r_{c m}}} \text { and } r_{c m}=\ell / 2\right] \\
& \frac{T_1}{T_2}=\frac{2 \pi \sqrt{L / g}}{2 \pi \sqrt{2 L / 3 g}}=\sqrt{\frac{3}{2}}
\end{aligned}
$