Finding the Period of Vibration

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Mass of the rod M = 3.0 kg
mass of the block m = 0.5 kg
length of the rod l = 1.25 kg
center of gravity of the rod from the end B (lg) = l/2 = 0.625 m

In equilibrium, let the rod makes an angle Qo with the horizontal,
moment about the end B = 0
=> k*(Qo*0.75)*0.75 = M*g*0.625 + m*g*1.25 ......(1)

Let when the end A is pulled down by a distance delta_0 (Do), the rod makes an ...