# 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 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 with force constant k = 1850 N/m is connected horizontally betweenone end of the rod and a fixed wall. When the rod is in equilibrium, it is parallel to the wall. What isthe period of the small oscillations thatresult when the rod is rotated slightly and released?
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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 $\left(\mathrm{△}x\right)=\frac{L\theta }{2}$
Since , the moment of force $\left(\tau \right)=-\frac{FL}{2}$
Or $\left(\tau =-\frac{FL}{2}=-\frac{kL\theta }{2}\cdot \frac{L}{2}$
$=-\frac{k{L}^{2}}{4}\cdot \theta$
So, the time period $\left(T\right)=2\pi \cdot \sqrt{\frac{l}{C}}$
where $I=\frac{m{L}^{2}}{12}$
$C=\frac{k{L}^{2}}{4}$
So, $T=2\pi \cdot \sqrt{\frac{m}{3k}}$
Here m=0.6 kg and k=1850 N/m
T=0.06533s