Question

A uniform rod of length $1.8m$ 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, $l=1.8m$
Time period of the oscillation of uniform rod with length l pivoted at one end is given by
$T=2\pi \sqrt{\frac{I}{mgR}}...(i)$
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),
$T=2\pi \sqrt{\frac{\frac{m{l}^2}{3}}{mg\frac{l}{2}}}=2\pi \sqrt{\frac{2}{3}\cdot \frac{l}{g}}=2\pi \sqrt{\frac{2}{3}\times \frac{1.8}{g}}$
$\therefore T=2\pi \sqrt{\frac{1.2}{g}}$
Time period of the simple pendulum of length $l^{{}^{\prime }}$ is given by
$\Rightarrow T^{{}^{\prime }}=2\pi \sqrt{\frac{l^{{}^{\prime }}}{g}}$ (Given, $T^{{}^{\prime }}=T)$
$2\pi \sqrt{\frac{l^{{}^{\prime }}}{g}}=2\pi \sqrt{\frac{1.2}{g}}$
$l^{{}^{\prime }}=1.2m$
Hence, the length of the simple pendulum is $1.2m$.