Block `A` of the mass `M` in the figure is released from rest when the extension in the spring is `x_(0)`. The maximum downwards displacement of the block is (assume `x_(0) lt Mg//k`).

Options

A. `2 ((Mg)/(k) - x_(0))`
B. `(Mg)/(2k) + x_(0)`
C. `(2 Mg)/(k) - x_(0)`
D. `(2 Mg)/(k) + x_(0)`

Answer

A

Explanation

Loss in gravitational potential energy of block is equal to gain in elastic potential energy of spring
`:. Mgx = (1)/(2) k (x + x_(0))^(2) - (1)/(2) kx_(0)^(2)`
On solving `x = 2 ((Mg)/(k) - x_(0))`