$F=\frac{m{{v}^{2}}}{r}=\frac{m{{(r\omega )}^{2}}}{r}=mr{{\omega }^{2}}$

then $r=l+$ elongation Given: elongation $=1 \;cm$ (in the first case)

or angular velocity $\omega$ the force required is

${{F}_{1}}=m(l+1){{\omega }^{2}}=kx=k\times 1=k$

$k=m(l+1){{\omega }^{2}}$ .......(i)

${{F}_{2}}=m(l+5){{(2\omega )}^{2}}=kx=k\times 5=5k$

$5k=4m(l+5){{\omega }^{2}}$.......(ii)

Now, dividing Eq. $(i)$ by Eq.$(ii)$, we get

$\frac{k}{5k}=\frac{m(l+1){{\omega }^{2}}}{4m(l+5){{\omega }^{2}}}$

$5(l+1)=4(l+5)$

$l=20-5=15\,cm$