A spring is loaded with two blocks `m_(1)` and `m_(2)` where `m_(1)` is rigidly fixed with the spring and `m_(2)` is just kept on the block `m_(1)`. The maximum energy of oscillation is possible for the system having the block `m_(2)` in constant with `m_(1)` is

Options

A. `(m_(1)^(2)g^(2))/(k)`
B. `(m_(1)g^(2))/(2k)`
C. `(m_(2)^(2)g^(2))/(2k)`
D. `((m_(1)+m_(2))^(2)g^(2))/(2k)`

Answer

D

Explanation

let x be the required amplitude of oscillation for this amplitude of oscillation the normal contact force between the blocks can be given by
`R-m_(2)g=m_(2)a`
where, acceleration
`a=-omega^(2)x`
`thereforeR-m_(2)g=-m_(2)omega^(2)x`
For maximum amplitude of
oscillation without loosing contact
of block `m_(2)` with block `m_(1)`, R=0
we get, `x=g//omega^(2)`
Where `omega` is the angular frequency of oscillation of `(m_(1)+m_(2))` and spring. therefore
`omega=sqrt((k)/((m_(1)+m_(2))))` or `omega^(2)=(k)/((m_(1)+m_(2)))`
`impliesx=((m_(1)+m_(2))g)/(k)`
Hence, the maximum energy of oscillation
`u_(max)=(1)/(2)kx^(2)=((m_(1)+m_(2))^(2)g^(2))/(2k)`