Question

A uniform rod of length l is suspended by an end and is made to undergo small oscillations. Find the length of the simple pendulum having the time period equal to that of the rod.

Answer 1

Let $A\to$ suspension point, $B\to$ Centre of Gravity, $l=\frac{1}{2}h=\frac{1}{2}$ Moment of intertia about A is,

$l_{CG}+m{h}^2=\frac{m{l}^2}{l2}+m\frac{l^2}{4}$

$=\frac{m{l}^2}{12}+\frac{m{l}^2}{4}=\frac{m{l}^2}{3}$

$\therefore T=2\pi \sqrt{\frac{1}{mg\left(\frac{l}{2}\right)}}$

$=2\pi \sqrt{\frac{2m{l}^2}{3mgl}}=2\pi \sqrt{\frac{2l}{3g}}$

Let the time period 'T' is equal to the time period of simple pendulum of length 'x',

$\therefore T=2\pi \sqrt{\left(\frac{x}{g}\right)}$

So, $\frac{2l}{3g}=\frac{x}{y}$

$\Rightarrow x=\frac{2l}{3}$

$\therefore$ Length of the simple pendulum

$=\frac{2l}{3}$