A small block of mass m is fixed at upper end of a massive vertical spring of spring constant `k=(2mg)/(L)` and natural length `10L` The lower end of spring is free and is at a height L from fixed horizontal floor as shown. The spring is initially unstressed and the spring block system is released from rest in the shown position.
Q. Till the blocks reaches its lowest position for the first time, the time duration for which the spring remains compressed is

Options

A. `pisqrt((L)/(2g))+sqrt((L)/(4g))sin^-1(1)/(3)`
B. `(pi)/(4)sqrt((L)/(g))+sqrt((L)/(4g))sin^-1(1)/(3)`
C. `pisqrt((L)/(2g))+sqrt((L)/(4g))sin^-1(2)/(3)`
D. `(pi)/(2)sqrt((L)/(2g))+sqrt((L)/(4g))sin^-1(2)/(3)`

Answer

B

Explanation

`V_(max)=(3)/(2)sqrt(gL)` and `omega=sqrt((k)/(m))=2sqrt((g)/(L))`
`A=(V_(max))/(omega)=(3)/(4)L`
Hence time taken t, from start of compression till the block reaches mean position is given by
`x=Asinomegat_0` where `x=(L)/(4)`
`t_0=sqrt((L)/(4g))sin^-1((1)/(3))`
Time taken by the block to reach from mean position to bottom most position is `(2pi)/(4omega)=(pi)/(4)sqrt((L)/(g))`
Hence the required time `=(pi)/(4)sqrt((L)/(g))+sqrt((L)/(4g))sin^-1(1)/(3)`