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 (PKS\thetaZSK)

Let 'F ' be the force applied on the rod, then spring elongated by \(\displaystyle{\left(\triangle{x}\right)}={\frac{{{L}\theta}}{{{2}}}}\)

Since , the moment of force \(\displaystyle{\left(\tau\right)}=-{\frac{{{F}{L}}}{{{2}}}}\)

Or \(\displaystyle{\left(\tau=-{\frac{{{F}{L}}}{{{2}}}}=-{\frac{{{k}{L}\theta}}{{{2}}}}\cdot{\frac{{{L}}}{{{2}}}}\right.}\)

\(\displaystyle=-{\frac{{{k}{L}^{{2}}}}{{{4}}}}\cdot\theta\)

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, \(\displaystyle{T}={2}\pi\cdot\sqrt{{{\frac{{{m}}}{{{3}{k}}}}}}\)

Here m=0.6 kg and k=1850 N/m

T=0.06533s

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 (PKS\thetaZSK)

Let 'F ' be the force applied on the rod, then spring elongated by \(\displaystyle{\left(\triangle{x}\right)}={\frac{{{L}\theta}}{{{2}}}}\)

Since , the moment of force \(\displaystyle{\left(\tau\right)}=-{\frac{{{F}{L}}}{{{2}}}}\)

Or \(\displaystyle{\left(\tau=-{\frac{{{F}{L}}}{{{2}}}}=-{\frac{{{k}{L}\theta}}{{{2}}}}\cdot{\frac{{{L}}}{{{2}}}}\right.}\)

\(\displaystyle=-{\frac{{{k}{L}^{{2}}}}{{{4}}}}\cdot\theta\)

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, \(\displaystyle{T}={2}\pi\cdot\sqrt{{{\frac{{{m}}}{{{3}{k}}}}}}\)

Here m=0.6 kg and k=1850 N/m

T=0.06533s