In the figure shown the spring constant is K the mass of the upper disc is m and that of the lower disc is 3m the upper block is depressed down from its equilibrium position by a distance `delta=5mg//k` and released at t=0 find the velocity of m when normal reaction on 3m is mg

Options

A. zero
B. `g[m//k]^(1//2)`
C. `2g[m//k]^(1//2)`
D. `4g[m//k]^(1//2)`

Answer

D

Explanation

`N=mg`
`N+kx=3mg`
`mg+kx=3mg`
`x=(2mg)/k`
`C.O.E 1/2k((6mg)/k)^(2)-1/2k((2mg)/k)^(2)-(8mg)/k mg=1/2mv^(2)`
`(36m^(2)g^(2))/k-(4m^(2)g^(2))/k-(16m^(2)g^(2))/k=1/2mv^(2)`
`(16m^(2)g^(2))/k=mv^(2)v=4g[m/k]^(1//2)`