All the surfaces are frictionless and system is released from test when ideal spring of stiffness `k` is in its relaxed state. Choose the correct statement.

Options

A. The magnitude of maximum acceleration of particle `B` is `g/3`
B. The magnitude of maximum acceleration of particle `B` is `(2g)/3`
C. The maximum speed of particle `A` is `2gsqrt(m/(3k))`
D. The maximum speed of particle `A` is `gsqrt(m/(3k))`

Answer

B, C

Explanation

Consider the system at any time `t` when each block is moving with speed `v` and spring has an extension `x`.
`implies 2mg-T=2ma`……(1)
`impliesT-kx=ma`…(2)
From equation (1) & (2) we write

`(2g)/3-k/(3m)x=aimplies(d^(2)x)/(dt^(2))+(kx)/(3m)-(2g)/3=0`
`implies(d^(2)x)/(dt^(2))+k/(3m)[x-(2mg)/k]=0impliesx_(0)=(2mg)/k`
Amplitude of motion `=x_(0)-0=(2mg)/k`
`a_("max")=omega^(2)A=k/(3m)xx(2mg)/k=(2g)/3`
`v_("max")=omegaA=(2mg)/k sqrt(k/(3m))=2g sqrt(m/(3k))`
Second Method:
Using COME, we can write
`2mgx=(3mv^(2))/2+(kx^(2))/2`
For extreme position `v=0impliesx=0, (4mg)/k`
`Aimplies` Amplitude of motion `=(2mg)/k`
For maximum velocity `(d^(2)x)/(dt^(2))=0impliesx=(2mg)/kimplies` Equilibrium position
`implies 2mgxx(2mg)/k=(3mv_("max")^(2))+k(4m^(2)g^(2))/(2k)`
`implies(4m^(2)g^(2))/(2k)=(3mv_("max")^(2))/2impliesv_("max")^(2)=4g^(2)(m/(3k))`
`impliesv_("max")=2g sqrt(m/(3k))`
`a_("max")=|(2g)/3-k/(3m)x(4mg)/k|=(2g)/3`