Physics - NLM Question with Solution | TestHub

In equilibrium the spring tension is $T=k{x}_0$ and $2T=mg$ (movable pulley).

Pulling $m$ down by a further $x$ stretches the spring by $2x$: $T^{\prime }=k(x_0+2x)$.

For the mass: $mg-2T^{\prime }=ma\Rightarrow mg-2k{x}_0-4kx=ma$. Since $2k{x}_0=mg$, $a=-\frac{4kx}{m}$, i.e. magnitude $\frac{4kx}{m}$ (directed back up).