In the overhead view of, a long uniform rod of mass m0.6 Kg is free to rotate in a horizontal planeabout a vertical axis through its center .A spring

In theoverhead view of , a long uniform rodof mass ( m = 0.6 Kg) is freeto rotate in a horizontal plane about avertical axis through its center.
A spring with force constant( k = 1850 N/m ) is connected horizontally between one end of the rod and a fixed wall.
When the rod is inequilibrium, it is parallel to the wall.
Let ' T ' be the time-period of the small oscillations that result when the rod is rotated slightly and released
Let rod ( length L) rotates by angle ($\theta$)
Let 'F ' be the force applied on the rod, then spring elongated by ${\displaystyle \left(\mathrm{△}x\right)=\frac{L\theta }{2}}$
Since , the moment of force ${\displaystyle \left(\tau \right)=-\frac{FL}{2}}$
Or ${\displaystyle (\tau =-\frac{FL}{2}=-\frac{kL\theta }{2}\cdot \frac{L}{2}}$
${\displaystyle =-\frac{k{L}^2}{4}\cdot \theta }$
So, the time period ${\displaystyle \left(T\right)=2\pi \cdot \sqrt{\frac{l}{C}}}$
where ${\displaystyle I=\frac{m{L}^2}{12}}$
${\displaystyle C=\frac{k{L}^2}{4}}$
So, ${\displaystyle T=2\pi \cdot \sqrt{\frac{m}{3k}}}$
Here m=0.6 kg and k=1850 N/m
T=0.06533s