$k=73.5 \,Nm ^{-1} \quad \text { Force }=5 \times 9.8$

In figure $(1)$

$5 \times 9.8=(2 k) x_1$

$\therefore x_1=\frac{5 \times 9.8}{2 \times 73.5}=\frac{1}{3}$

In figure $(2)$

$5 \times 9.8=\frac{k \times k}{k+k} \times x_2$

or $5 \times 9.8=\frac{k}{2} \times x_2$

$x_2=\frac{98}{73.5}=\frac{4}{3}$

In figure $(3)$

$5 \times 9.8=k x_3$

$x_3=\frac{5 \times 9.8}{73.5}=\frac{2}{3}$