Question

A uniform rod of length $1.8 \,m$ suspended by an end is made to undergo small oscillations. Find the length of the simple pendulum having the mass and time period equal to that of the rod.

• A. 3.6 m
• B. 1.2 m
• C. 2.4 m
• D. 4.2 m

Answer 1

Answer: (B)

Given, length of the rod, $l=1.8\, m$
Time period of the oscillation of uniform rod with length l pivoted at one end is given by
$T=2 \pi \sqrt{\frac{I}{m g R}}\,\,\,...(i)$
where, $R$ is the distance of the point of pivot to the centre of the gravity and $I$ is moment of inertia of the rod.
$\therefore R=\frac{l}{2}$ and $I=\frac{m l^{2}}{3}$
$\therefore $ From Eq. (i),
$T=2 \pi \sqrt{\frac{\frac{m l^{2}}{3}}{m g \frac{l}{2}}}=2 \pi \sqrt{\frac{2}{3} \cdot \frac{l}{g}}=2 \pi \sqrt{\frac{2}{3} \times \frac{1.8}{g}}$
$\therefore T=2 \pi \sqrt{\frac{1.2}{g}}$
Time period of the simple pendulum of length $l^{'}$ is given by
$\Rightarrow T^{'} =2 \pi \sqrt{\frac{l^{'}}{g}}\,\,$ (Given, $T^{'}=T) $
$2 \pi \sqrt{\frac{l^{'}}{g}} =2 \pi \sqrt{\frac{1.2}{g}} $
$l^{'} =1.2\,m$
Hence, the length of the simple pendulum is $1.2\, m$.