A particle of mass m is attached to one end A of a model spring OA of natural le

Question

A particle of mass m is attached to one end A of a model spring OA of natural length lo and stiffness k. The other end of the spring is attached to a fixed point O. and the spring hangs vertically downwards as shown in Figure Ql. The mass is displaced downwards from its equilibrium position by a distance lo/2 and released from rest. Assume air resistance can be neglected. Figure Q1 By using the law of conservation of mechanical energy, show that the speed of the particle as it passes through the equilibrium position is.

Explanation / Answer

The mass is displaced down by a length of l/2

So the energy stored in the spring = 1/2 k(l/2)2  

Now when the mass passes through the mean position the energy of the spring reduces to 0. So the enrgy of the system = Kinetic energy of the body = 1/2 mv2

=> 1/2 mv2 = 1/2 k(l/2)2  

=> v = l/2(k/m)

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