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

Options

A. `sqrt((2)/(3))`A
B. `sqrt((3)/(2))A`
C. `sqrt2A`
D. `(A)/(sqrt2)`

Answer

A

Explanation

Conserving momentum `2V=3V'`
`V'=(2)/(3)V`
`E_i=(1)/(2)m_1v_1^2=(1)/(2)xx2V^2=V^2=(1)/(2)KA^2`
`E_f=(1)/(2)xxm_2V_2^2=(1)/(2)xx3xx(2)/(3)xx(2)/(3)V^2=(2)/(3)V^2`
`(1)/(2)KA^2=(2)/(3)V^2=(2)/(3)E_i` (because `E_i=V^2` from above)
`(1)/(2)KA^2=(2)/(3)((1)/(2)KA^2)impliesA'=sqrt((2)/(3))A`