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
$=m g \times \frac{l}{2} \frac{\sin \theta}{\theta} $
$=m g \frac{l}{2}[$ 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, $\frac{T_{1}}{T_{2}}=\sqrt{\frac{3}{2}}$