Ans.

The time period of the simple pendulum equivalent to a compound pendulum is given by,

\( \displaystyle{T=2\pi\sqrt{\frac{K^2+l_1^2}{l_1g}}} \)
[ to know the derivation of the above equation (CLICK HERE) ]

or, \( \displaystyle{T=2\pi\sqrt{\frac{L}{g}}} \)

where, \( \displaystyle{L=\frac{K^2+l_1^2}{l_1}} \)

where, K is the radius of gyration, g is the acceleration due to gravity, \( L \) is the length of the equivalent simple pendulum and \( l_1 \) is the distance of centre of suspension O from the centre of gravity G of the metal bar of length l.

Let us consider \( M \) be the mass of this metal bar of length \( l \), then the moment of inertia of the rod is

\( \displaystyle{I=\frac{1}{12}Ml^2 }\)
[ to know the derivation (CLICK HERE) ]

or, \( \displaystyle{MK^2=\frac{1}{12}Ml^2} \)

or, \( \displaystyle{K^2=\frac{l^2}{12}} \)

and from Fig. 1, \( l_1=\frac{l}{2} \)

Therefore, \( \displaystyle{L=\frac{\frac{l^2}{12}+{\left(\frac{l}{2}\right)}^2}{\frac{l}{2}}} \)

or, \( \displaystyle{L=\frac{\frac{l^2}{4}\left(\frac{1}{3}+1\right)}{\frac{l}{2}}} \)

or, \( \displaystyle{L=\frac{l}{2}\left(\frac{4}{3}\right)} \)

or, \( \displaystyle{L=\frac{2l}{3}} \)