Q: A block of mass ‘m’ is attached to the light spring of force constant K and released when it is in its natural length. Find amplitude of subsequent oscillations.
Sol: From conservation of energy $\large m g x = \frac{1}{2} K x^2 $
The maximum displacement of the spring in subsequent motion will be 2mg/K
From F = Kx, mg = Kx The equilibrium position of the system will occur at the extension of mg/K
amplitude $\large = \frac{2mg}{K} – \frac{mg}{K} = \frac{mg}{K} $