Given ${\rm{m}}{\omega ^2}\left( {l + {{\rm{x}}_1}} \right) = {\rm{k}}{{\rm{x}}_1}$

$m{(2\omega )^2}\left( {l + {x_2}} \right) = k{x_2}$

where $x_{1}=1 \mathrm{cm} \quad \mathrm{x}_{2}=5 \mathrm{cm}$

$\therefore \frac{{4\left( {l + {{\rm{x}}_2}} \right)}}{{l + {{\rm{x}}_1}}} = \frac{{{{\rm{x}}_2}}}{{{{\rm{x}}_1}}}$

$4{x_1}\left( {l + {x_2}} \right) = {x_2}\left( {l + {x_1}} \right)$

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

$l = 15{\rm{cm}}$