Question

The time period of oscillation of a simple pendulum of length $l$ is $T_{1}$ and the time period of oscillation of a uniform rod of the same length $l$ pivoted at 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 given position is given by
$T_{2}=2\pi \sqrt{\frac{I n e r t i a \, f a c t o r}{s p r i n g \, f a c t o r}}$



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
$=mg\times \frac{l}{2} \, \frac{sin \theta }{\theta }$
$=mg\times \frac{l}{2}$ (if $\theta $ is small)
$\therefore $ $T_{2}=2\pi \sqrt{\frac{m l^{2} / 3}{m g l / 2}}=2\pi \sqrt{\frac{2}{3} \, \, \frac{l}{g}} \, $
Hence, $\frac{T_{1}}{T_{2}}=\sqrt{\frac{3}{2}}$