In the two block spring system, force constant of spring is k = 6N/m. Spring is stretched by 12 cm and then left. Match the following.

Answer

`(A rarrq , B rarr s, C rarr p)`

Explanation

`omega=sqrt((k)/(mu))`
Here, `mu`=reduced mass
`=(m_(1)m_(2))/(m_(1)+m_(2))=(2)/(3)kg`
`therefore omega=sqrt((6)/(2//3))=3 "rads"^(-1)`
Amplitude 12 cm distance in inverse ratio of mass,
`therefore A_(1)=8 cm` and `A_(2)=4 cm`
Now, maximum kinetic energy
`K_(1)=(1)/(2)m_(1)omegaA_(1)^(2)`
`=(1)/(2)xx1xx3xx(8xx10^(-2))^(2)=9.6 mJ`
`K_(2)=(1)/(2)xx2xx3xx(4xx10^(-2))^(2)=4.8 mJ`