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. `2 Mg//K`
C. `Mg//K`
D. `Mg//2K`

Answer

B

Explanation

Let `x` be the maximum extension of the spring from energy conservation

Lass in gravitational potential energy
= Gain in potential energy of spring
`Mgx = (1)/(2) Kx^(2)`
`rArr x = (2Mg)/(K)`