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

4

Explanation

N= mg
N+kx=3mg
mg +kx =3 mg
`x=(2mg)/k`
C.O.E
`1/2 k ((6 mg)/k)^(2)-1/2 k((2mg)/k)^(2)-(8mg).k mg`
`=1/2 mv^(2)`

`(36m^(2)g^(2))/k-(4m^(2)g^(2))/k-(16m^(2)g^(2))/k=1/2 mv^(2)`
`(16 m^(2)g^(2))/k=mv^(2) " " v=4g [m/k]^(-1//2)`