Two balls with masses `m_(1) =1kg` and `m_(2) = 2kg` are slipped on a thin smooth horizontal rod as shown. The balls are interconnected by a light spring of stiffess `K = 24N//m`. The left hand ball `(m_(1))` is imparted the initial velocity `12cm//s` towards second ball. the amplitude of oscillations made by the arrangement is............

Options

A. `3cm`
B. `2cm`
C. `1cm`
D. `4cm`

Answer

B

Explanation

Total mechanical energy of system =kinetic energy of mass `m_(1) = (1)/(2) m_(1)v_(1)^(2) …………(1)`
But total mechanical energy = kinetic energy due to motion of centre of mass + energy of oscillation.
`rArr` energy of oscillation
`=(1)/(2) m_(1)v_(1)^(2) - (1)/(2) (m_(1)+m_(2)) [(m_(1)v_(1))/(m_(1)m_(2))]^(2) = (1)/(2) mu v_(1)^(2) = (1)/(2) mu (Aw)^(2)`
`rArr v_(1) = Aw rArr A = (v_(1))/(w) , v_(1) =12 cm//s w =sqrt((k)/(mu))`: solve for `A`.