Two block of equal mass m are connected by an unstreatched spring and the system is kept at rest on a frictionless horizontal suface. A constant force F is applied on the first block pulling it away from the other as shown in figure.
If the extension of the speing is xo at time t, then the displacement of the first block at this instanti is

Options

A. `(Ft^(2))/(2m)-x_(0)`
B. `1/2((Ft^(2))/(2m)+x_(0))`
C. `1/2((2F^(2))/m-x_(0))`
D. `1/2((Ft^(2))/(2m)-x_(0))`

Answer

D

Explanation

Suppose the displacement of the first block is `x_(1)` and that of the second is `x_(2)` then
`x=(mx_(1)+mx_(2))/(2m), (Ft^(2))/(4m)=(x_(1)+x_(2))/2`
or `x_(1)+x_(2)=(Ft^(2))/(2m)`……….i
Further the extension of the spring is `x_(1)-x_(2)`. Therefore
`x_(1)-x_(2)=x_(0)`.....ii
From eqn i and ii
`x_(1)=1/2((Ft^(2))/(2m)+x_(0))impliesx_(2)=1/2((Ft^(2))/(2m)-x_(0))`