• question_answer The spring block system as shown in figure is in equilibrium. The string connecting blocks A and B is cut. The mass of all the three blocks is m and spring constant of both the spring is k. The amplitude of resulting oscillation of block A is (string massless) A) $\frac{mg}{k}$             B) $\frac{2mg}{k}$ C) $\frac{3mg}{k}$                       D) $\frac{4mg}{k}$

[B] Just after cutting the string extension in spring $=\frac{3mg}{k}$ The extension in the spring when block is in mean position $=\frac{mg}{k}$ $\therefore$ Amplitude of oscillation $A=\frac{3mg}{k}-\frac{mg}{k}=\frac{2mg}{k}.$