One end of a massless spring of spring constant k and natural length l 0 is fixed while the other end is…

The particle is moving in a horizontal circle, so it is accelerated towards the centre with magnitude $\frac{v^2}{r}$. The horizontal force on the particle is due to the spring and is given by $kx$, where $x$ is the elongation and $k$ is the spring constant.

$kx=\frac{m{v}^2}{r}=m{\omega }^{2}r=m{\omega }^2\left(l_0+x\right)$

$\Rightarrow \left(k-m{\omega }^2\right)x=m{\omega }^2{l}_0$

$\Rightarrow x=\frac{m{\omega }^2{l}_0}{k-m{\omega }^2}$