See image — AITS & Test Series Chemistry Question

💡 Solution & Explanation

Suppose by falling down through a height h, the mass m compresses the spring balance by a length x. The P.E. lost by the mass = mg (h+ x) This is stored up as energy of the spring by compression = 2 1k 2 x  mg(h + x) = 2 1k 2 x or 2 1k 2 x – mgx – mgh = 0 m k h or 2 2mg 2mgh 0 k k    x x Solving this quadratic equation, we get 2 2mg 2mg 8mgh k k k 2                x = mg mg 2kh 1 k k mg   In the equilibrium position, the spring will be compressed through the distance mg/k and hence the amplitude of oscillation is A = mg 2kh 1 k mg  Energy of oscillation = 2 2 1 1 mg 2kh kA k 1 2 2 k mg              =  2 (mg) mgh 2k = 11 J