A body of mass m hangs by an inextensible string that passes over a smooth massless pulley that is fitted with a light spring of stiffness k as shown in the figure, such that the spring is initially at natural length. If the body is released from rest and the spring is released, calculate the maximum elongation of the spring.
Let the spring be elongated by x.That means the pulley falls through a distance x.
Consequently the string of the pulley is slackened to the same extend.Therefore the total distance moved by the body = 2x.
Since the spring is elongated by x,△PEspring=12kx2
As the body falls through a distance 2x.(△PE)gr=−mg(2x)
Since the body x initially at rest and comes to rest just at the instant of maximum elongation of the spring ⇒△PE=0
Applying the principle of conservation of energy.