Correct Answer - D
We know that `eta=(FL)/(Ax)` , where x is the displacement.
As, `A=L^(2)` So, `eta=(FL)/(L^(2)x)=(F)/(Lx)`, so shearing force
`F=eta L x` …….(i)
It means F `prop` x and this F is directed towards mean popsition, hence after the force is withdrawn, the block will execute liner S.H.M. Here spring factor `=eta L` and inertia factor = mass = M. ltBrgt As time period.
`T=2pi sqrt(("inertia factor")/("spring factor"))=2pi sqrt((M)/(eta L))`