Blocks A and B shown in the figure are having equal masses m. The system is released from rest with the spring unstretched. The string between A and ground is cut, when there is maximum extension in the spring. The acceleration of centre of mass of the two blocks at this instant is

Options

A. `a_(2)-a_(1)=a_(1)-a_(2)`
B. At an extension of `(x_(0))/(4)`of the spring, the magnitude of acceleration of the block connected to the spring is `(3g)/(10)`
C. `x_(0)=(4Mg)/(k)`
D. When spring achieves an extension of `(x_(0))/(2)` for the first time, the speed of the block connected to the spring is `3g(M)/(5K)`

Answer

A

Explanation

`Mg-T=Ma_(2)` ...(1)
`2Mg-T=3Ma_(3)` ...(2)
`2T-Kx=2Ma_(1)` ...(3)
From (1) and
`M(a_(2)+a_(3))="Mg-T+Mg"(-T)/(2)=2Mg-(3T)/(2)|`
implies `ubrace(2Ma_(1)=2Mg-(3T)/(2))_(xx(4)/(3)) implies(8Ma_(1))/(3)=(3Ma)/(3)=(3Ma)/(3)-2T` ...(4)
`(3)+(4)implies(8Mg)/(3)-kx=(2M+(8M)/(3))a_(1)`
`implies (14M)/(3)a_(1)=-K[x-(8Mg)/(3K)] implies a_(1)=-(3K)/(14M)[x-(8Mg)/(3K)]`
`omega=sqrt((3K)/(14M))`,`A=(8Mg)/(3K)(x_(0))/(2)=2A`[Maximum elongation=2A]
`V_(max)=Aomega=(8Mg)/(3K)sqrt((3K)/(14M))` `a_(1)atx=(x_(0))/(4)=-(3K)/(14M)xx(4Mg)(3K)=(2g)/(7)` Correct answer is (A)