Two masses m 1 and m 2 are suspended together by a massless spring of constant K . When the masses are in equilibrium, m 1 is removed without disturbing the system. The amplitude of oscillations is

Answer

`sqrt(k/m2),m1g/k`

Explanation

`sqrt((k)/(m_(2)))`,`(m_(1)g)/(k)`
Let `Deltax_(1)` and `Deltax_(2)`be extension in spring with `m_(1)` and `m_(2)` attached and `m_(2)` attached
`Deltax_(1)=((m_(1)+m_(2))g)/(k)`
`Deltax_(2)=(m_(2)g)/(k)`
`omegasqrt((k)/(m_(2)))`
`A=Deltax_(1)-Deltax_(2)`
`A=(m_(1)g)/(k)` (distance between extreme position of mean position)