A horizontal spring -block system of mass `2kg` executes `S.H.M` when the block is passing through its equilibrium positiojn an object of mass `1kg` is put on it the two move togather The new amplitude of vibration is `(A` being its initial amplitude)

Options

A. `sqrt((2)/(3))A`
B. `sqrt((3)/(2))A`
C. `sqrt(2)A`
D. `sqrt((A)/(sqrt(2))`

Answer

A

Explanation

Conserving momentum: `2V = 3V'`
`rArr V' = (2)/(3)V`
`E_(i) = (1)/(2) m_(1)V_(1)^(2) = (1)/(2) 2. V^(2) rArr (1)/(2) KA^(2) = V^(2)`
`E_(f) = (1)/(2) m_(2)V_(2)^(2) = (1)/(2) 3. (2)/(3).(2)/(3)V^(2) = (2)/(3) V^(2)`
`rArr (1)/(2)KA^(2) = (2)/(3)V^(2) = (2)/(3)E_(i)`
`(E_(i) = V^(2))` from above
`rArr (1)/(2)KA^(2) = (2)/(3)((1)/(2) KA^(2)) rArr K = sqrt((2)/(3))A`