A body of mass m is attached to one end of a massless spring which is suspended vertically from a fixed point. The mass is held in hand, so that the spring is neighter stretched nor comprssed. Suddenly the support of the hand is removed. The lawest position attained by the mass during oscillation is 4 cm below the point, where it was held in hand.

A body of mass m is attached to one end of a massless spring which is

1.	A body of mass m is attached to one end of a massless spring which is suspended vertically from a fixed point. The mass is held in hand, so that the spring is neighter stretched nor comprssed. Suddenly the support of the hand is removed. The lawest position attained by the mass during oscillation is 4 cm below the point, where it was held in hand.
Answer» Solution :When SUPPORT of the hand is REMOVED, the body of mass m oscillate about mean position. Suppose, a body reaches at lower extreme point and the maximum extension of spring is x, The decrease in potential energy of body = mgx Increase in elastic potential energy of spring `=(1)/(2) kx^(2)` Now mechanical energy is conserved `therefore mgx = (1)/(2) kx^(2)` `therefore x= (2mg)/(k)"""........."(1)` When the forces below and above the ball suspended at the end of spring are equal, the position of ball BECOMES meanposition. Suppose the support of hand is removed, the extension of spring x. increases, the resultant force will be zero `therefore F= +kx.` `therefore mg= kx.` `therefore x.= (mg)/(x)"""........."(2)` Taking ratio of equation (1) and (2), `(x)/(x.)= 2` `therefore x= 2X.` but x= 4 CM (is given) `therefore 4= 2x.` `therefore x.= 2cm`, will be the amplitude of oscillation.