A block of mass M is attached with a spring of spring constant k. The whole arrangement is placed on the frictionless surface of a vehicle as shown in figure. If the vehicle starts moving towards right with an acceleration a (there is no friction anywhere), then

According to Work Energy Theorem, work done by all the forces is equal to change in kinetic energy, so

$\begin{array}{l} {\mathrm{W}}_{\text{total }}=\mathrm{\Delta K} \\ \Rightarrow {\mathrm{W}}_{\text{ext }}+{\mathrm{W}}_{\mathrm{nc}}=\mathrm{\Delta U}+\mathrm{\Delta K}\end{array}$

Since, ${\mathrm{W}}_{\mathrm{nc}}=0$, because friction is absent

So, ${\mathrm{W}}_{\text{ext }}=\mathrm{\Delta U}+\mathrm{\Delta K}$

$\begin{array}{l}\Rightarrow \left(\mathrm{Ma}\right){\mathrm{x}}_{max}=\frac{1}{2}{\mathrm{kx}}_{max}^2\\ \Rightarrow {\mathrm{x}}_{max}=\frac{2\mathrm{Ma}}{\mathrm{k}}\end{array}$

The block will execute SHM about its equilibrium position and its initial position will be the one of the extreme positions i.e., block returns to its initial position at the end of the cycle and compression in the spring is zero.
So, (4) is also correct.