Three point masses `m`, `2m` and `m`, connected with ideal spring (of spring constant `k`) and ideal string as shown in the figure, are placed on a smooth horizontal surface
At `t=0`, three constant forces `F`, `2F` and `3F` start acting on the point masses `m`, `2m` and `m` respectively, as shown in figure. Find the maximum extension in the spring

Options

A. `(2F)/(m)`
B. `(5F)/(2m)`
C. `(3F)/(m)`
D. `(8F)/(3m)`

Explanation

Let any time `t_(1)` extension in spring is `x`, so
`rArr F-kx=ma_(1)rArr a_(1)=(F-kx)/(m)`….(`1`)
`rArr T-kx-2F=2ma_(2)`……(`2`)
`rArr3F-T=ma_(2)`.....(`3`)
With the help of equations (`2`) and (`3`), we have
`F=kx=3ma_(2)rArr(F-kx)/(3m)=a_(2)`.......(`4`)
With the help of equations (`1`) and (`4`), we have
`veca_(12)=veca_(1)-veca_(2)=a_(1)(-hati)-a_(2)(hati)=(a_(1)+a_(2))(hati)`
`V_(12)(dV_(12))/(dx)=a_(12)=(4(F-kx))/(3m)rArrint_(0)^(0)V_(12)dV_(12)=(4)/(3m)int_(0)^(x)(F-kx)dx`
`rArr0=(4)/(3m)[Fx-(kx^(2))/(2)]rArrX_(max)=(2F)/(k)`
Second method : System is equivalent to
Because
Above system behaves as single unit `