Question

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. $\frac{1}{\sqrt{3}}$
• B. $1$
• C. $\sqrt{\frac{4}{3}}$
• D. $\sqrt{\frac{3}{2}}$

Answer 1

Answer: (D)

Time period of simple pendulum is given by
$T_1=2\pi \sqrt{\frac{l}{g}}$
and time period of uniform rod in give position is given by
$T_2=2\pi \sqrt{\frac{\text{ inertia factor }}{\text{ spring factor }}}$

Here, inertia factor $=$ moment of inertia of rod at one end
$=\frac{m{l}^2}{12}+\frac{m{l}^2}{4}$
$=\frac{m{l}^2}{3}$
Spring factor $=$ restoring torque per unit angular displacement
$=mg\times \frac{l}{2}\frac{\sin \theta }{\theta }$
$=mg\frac{l}{2}[$ if $\theta$ is small $]$
$\therefore$ $T_2=2\pi \sqrt{\frac{m{l}^2/3}{mgl/2}}=2\pi \sqrt{\frac{2}{3}\frac{l}{g}}$
Hence, $\frac{T_1}{T_2}=\sqrt{\frac{3}{2}}$