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.

Answer

A, B, C

Explanation

(a) Initially, `omega^(2) = (k)/(m)`

At `t = 0`, block of mass m is at mean position `(x = 10 cm)` and moving towards positive x-direction with velocity `A omega` or `30cm//s`
From conservation of liner momentum,
`(M + m)v = M(30)`
Substituting the values, we have
v = velocity of combined mass just after collision `= 15 cm//s` or `0.15 m//s`
From conservation of mechanical energy,
`(1)/(2)(M + m)v^(2) = (1)/(2) k (A' )^(2)`
Here, A' = new amplitude of oscillation of combined mass
`:. A = (sqrt((M + m)/(k)))v = ((4)/(100))^(1//2)(0.15) = 0.03m`
or `A = 3cm`

(b) New angular frequency `omega = sqrt((k)/(M +m)) = sqrt((100)/(4)) = 5 rad//s`
`:. x = 10 - 3 sin 5t` ( c) Loss of mechanical energy `E_(f) - E_(i)`
` = (1)/(2)(1)(0.3)^(2) + (1)/(2)(3)(0.3)^(2) - (1)/(2)(4)(0.15)^(2)`
`0.135J`.