Question

A simple pendulum of length $l$ is displaced so that its taught string is horizontal and then released. A uniform bar pivoted at one end is simultaneously released from its horizontal position. If their motions are synchronous, what is the length of the bar?

• A. $\frac{3l}{2}$
• B. $l$
• C. $2l$
• D. $\frac{2l}{3}$

Answer 1

The correct option is A $\frac{3l}{2}$

For a simple pendulum
$T=2\pi \sqrt{\frac{l}{g}}$
For a compound pendulum, let $L^{\prime }$ be the length, $I$ be the moment of inertia about the pivot.
$T^{\prime }=2\pi \sqrt{\frac{I}{Mgd}}=2\pi \sqrt{\frac{M{L}^{\prime 2}}{3Mg\frac{L^{\prime }}{2}}}$
$=2\pi \sqrt{\frac{2L^{\prime }}{3g}}$

$T=T^{\prime }$ as the motion is synchronous.
$\Rightarrow \sqrt{\frac{l}{g}}=\sqrt{\frac{2L^{\prime }}{3g}}\Rightarrow L^{\prime }=\frac{3l}{2}$