The system shown in the figure can move on a smooth surface. They are initially compressed by `6cm` and then released, then choose the correct options.

(a) The system performs, SHM with time period `(pi)/(10)s`
(b) The block of mass `3kg` perform SHM with amplitude `4 cm`
( c) The block of mass `6kg` will have maximum momentum of `2.40kg - m//s`
(d) The time periods of two blocks are in the ratio of `1:sqrt(2)`

Explanation

a. `T = 2 pisqrt((mu)/(k))`
Here, `mu = (m_(1) m_(2))/(mm_(1) + m_(2)) = (3 xx 3)/(9) = 2 kg`
`T = 2 pi sqrt((2)/(800)) = 2 pi xx (1)/(20)`
`T = (pi)/(10) sec`
b. Here compression of spring `x_(0) = 6 cm`
Let displacement of block (1) and (2) w.r.t. center of mass be `x_(1) and x_(2)`
But `x_(1) + x_(2) = x_(0) and m_(1) = m_(2)x_(2)`
or `3 x_(1) = 6 x_(2)` or `x_(1) + 2 x_(2)`
` x_(1) + x_(2) = x_(0) or x_(1) + (x_(1))/(2) = x_(0)`
or `(3)/(2) x_(1) = x_(0) implies x_(1) = (2)/(3) x_(0) = (2)/(3) xx 6 = 4 cm`
c. Appling conservation principal of momentum
`3 xx v_(1) - 6 v_(2) = 0`
`3 v_(1)=- 6 v_(2)`
`v_(1) = 2 v_(2)`
Appling mechaincal energy conservation , we get
`(1)/(2) xx 3 xx v_(1)^(2) + (1)/(2) xx 6 xx v_(2)^(2) = (1)/(2) kx_(0)^(2)`
or `(3)/(2) v_(1)^(2) + 3v_(2)^(2) = (1)/(2) xx 800 xx (0.06)^(2)`
or `(3)/(2) (2 v_(2))^(2) + 3v_(2)^(2) = 400 xx 36 xx 10^(-4)`
or `v_(2) = (1.2)/(3) = 0.4 m//s`
`:. P_(2 max) = 6 xx 0.4 = 2.4 kg m//s`