A body of mass `m` is attached by an inelastic string to a suspended spring of spring constant `k`. Both the string and the spring have negligible mass and the string is inextensible and of length `L`. Initially, the mass `m` is at rest.
Q. If the mass `m` is now raised up to point `A` (the top end of the string see fig. and allowed to fall from rest, the maximum extension of the spring in the subsequent motion will be

Options

A. `L`
B. `(mg)/(k)`
C. `(mg)/(k)sqrt(1+(2kL)/(mg))`
D. `(mg)/(k)[1+sqrt(1+(2kL)/(mg))]`

Answer

D

Explanation

Let `x` be the maximum extension of the spring. The mass falls from rest through the vertical is `E_k=mg(L+x)`. This must be stored in the spring as elastic energy at tis maximum extension Hence `E_k=(1)/(2)kx^2=mg(L+x)`
This is quadratic, `x^2-(2mg)/(k)x(2mgL)/(k)=0`
Hence, `x=((2mg)/(k)+-sqrt(((2mg)/(k))^2+(8mgL)/(k)))/(2)`
Maximum value of `x=(mg)/(k)+(mg)/(k)sqrt(1+(2kL)/(mg))`
`=(mg)/(k)(1+sqrt(1+(2kL)/(mg)))`