An ideal spring with spring - constant `K` is bung from the colling and `a` block of mass `M` is attached to its lower end the mass is released with the spring initally unstetched . Then the maximum exlemsion in the spring is

Options

A. `(4 Mg)/(k)`
B. `(2Mg)/(k)`
C. `(Mg)/(k)`
D. `(Mg)/(2k)`

Answer

B

Explanation

Loss in `P.E` = Gain in `K.E + P.E` stored in spring
`MgX_(max) =0+(1)/(2)kX_(max)^(2) , X_(max)=(2Mg)/(k)`.