A block of mass `m=1kg` is attached to a free end of a spring whose one end is fixed with as shown in the figure. The block is performing simple harmonic motion. The position of the block from `O` is given by `x=2+(1)/(sqrt(2))sin2t`, where `x` is in meter and `t` is in second. A particle of same mass is released from the circular path at a height `h=80cm`. then particle collidic and stricks to the block. The collision takes place when velocity of the block is zero and spring is elongated. Find
(`i`) Time period of new `SHM`.
(`ii`) Maximum velocity of the system

Explanation

`x=2+(1)/(sqrt(2))sin2t`
`rArr 2=sqrt((k)/(m)) rArr k=4N//m`
`omegaL=(1)/(omegaC)`
Just before collision, velocity of the shell,
`2omegaL=(1)/(2omegaC')`
Now , `P_(i)=P_(r)`
`0+mx4=(m+m)xv_(0)rArrv_(0)=2m//s`
Using conservation of energy,
`:. C'=(1)/(4omega^(2)L)=(LC)/(4L)=(C )/(4)kx^(2)+ :. (DeltaC)/(C )=(3)/(4)(2m)(2)^(2)`
`=(1)/(f_(1))=((1.5-1))/(R )+((4)/(3)-1)(-(1)/(R )-(1)/(R ))+((1.5-1)/(R ))=(1)/(R )-(2)/(3R)=(1)/(3R)(2m)v'^(2)`
`=(1)/(f_(2))=((1.5-1))/(R )+((4)/(3)-1)(-(1)/(R ))+((1.5-1)/(R ))=(1)/(R )-(1)/(3R)=(2)/(3R)`