One end of an ideal spring is fixed to a wall at origin `O` and axis of spring is parallel to x-axis. A block of mass `m = 1kg` is attached to free end of the spring and it is performing SHM. Equation of position of the block in co-ordinate system shown in figure is `x = 10 + 3 sin (10t)`. Here, t is in second and `x` in `cm`. Another block of mass `M = 3kg`, moving towards the origin with velocity `30 cm//s` collides with the block performing SHM at `t = 0` and gets stuck to it. Calculate

(a) new amplitude of oscillations,
(b) neweqution for position of the combined body,
( c) loss of energy during collision. Neglect friction.

Explanation

`x = 10 + 3 sin (10 t)`
Since the block is oscillating in horizontal direction, its equilibrium position corresponds to natural length of the sprin. Hence, natural length of spring is `l_(0) = 10 cm`
Displacement of block from equilibrium position is
`s = x - l_(0)`
`s = 3 sin (10 t) cm or s = 0.03 sin (10 t) m` (i)
But displacement of a block performing SHM is given by `x = a sin ( omega t)`.
Comparing it with Eq. (i) `A = 0.03m and omega = 10 rad//s` But angular frequency of block , performing SHM and tiedwith an ideal spring is `omega = sqrt((k)/(m))`
Spring constant `K = omega^(2) m = (10)^(2) xx 1 = 100 N//m`
Velocity of the block , (performing SHM) is given by `v = (dx)/(dt) = 30 cos (10 t) cm`
Its velocity at time `t = 0 is v = 30 cm//s or 0.30 m//s`
Velocity of the block (just before colliding) is `0.6m//s` (left ward) : therefore it is to taken `(- 0.6 m//s)`
Since masses of two colliding blocks are equal and collision is elastic , therefore , velocity of oscillating block (just after collision) becomes equal to `(- 6.0 m//s)`