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# 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? # 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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Other asked 2021-05-10
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? ## Answers (1) 2021-05-12
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

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