
From the figure, the third side of the right angled triangle is
\(x = \sqrt {{L^2} + {L^2}} \)
x= third side of the right angled triangle.
\(\begin{array}{c}x = \sqrt {2{L^2}} \\ = \sqrt 2 \;L\end{array}\)
The center of gravity balanced by the rod is,
\(\begin{array}{l}d = \frac{L}{{2\sqrt 2 }}\\d = {\rm{balanced}}\;{\rm{distance}}\;{\rm{of}}\;{\rm{the}}\;{\rm{rods}}\end{array}\)
Moment of inertia is,
\(I = \frac{2}{3}m{L^2}\)
Frequency of oscillation of thin two rods is,
\(\begin{array}{c}f = \frac{1}{{2\pi }}\sqrt {\frac{{2mgd}}{I}} \;\;\;\;\;\;\;\;\;\;\left( {d = \frac{L}{{2\sqrt 2 }};\;I = \frac{2}{3}m{L^2}} \right)\\f = \frac{1}{{2\pi }}\sqrt {\frac{{2mg\left( {\frac{L}{{2\sqrt 2 }}} \right)}}{{\frac{2}{3}m{L^2}}}} \\ = \frac{1}{{2\pi }}\sqrt {\frac{{6g}}{{4\sqrt 2 \;L}}} \\ = \frac{1}{{4\pi }}\sqrt {\frac{{6g}}{{\sqrt 2 \;L}}} \end{array}\)
Hence the frequency of oscillation is \(\frac{1}{{4\pi }}\sqrt {\frac{{6g}}{{\sqrt 2 \;L}}} \)