Figure shown a block P of mass m resting on a smooth horizontal ground, attached to one and a spring of force constant k in netural length. If another block od same mass and moving with a velocity u towards right is placed on the block on which stick to it due to friction ,find the time it will take to reach its exterme position. Also find the amplitude of oscillation of the combined mass `2m`.

Explanation

When second mass stricks to the lower mass , due to such an inelastic collision, the velocity of combined block is reduced by half , that is `u//2`, to conserve momentum . Now at mean position the velocity of block can be written as
`(u)/(2) = A omega` (if A is amplitude of oscillation)
or `(u)/(2) = A sqrt((k)/(2m))`
[Here for combined block new angular frequency of SHM is `omega sqrt(k//2m)]`
or `A = (u)/(2) sqrt((2m)/(k)) = u sqrt((m)/(2k))`
As oscillation starts from mean position in reaching its extreme position , partical has to cover a phase of `pi//2` radians, thus time taken by the partical to reach its exterme position is `l = (pi//2)/(omega) = (pi)/(2) sqrt((k)/(2m))`