solution

First let's compute $\omega$. The period $T ~=~ s$, so

$$\omega ~=~ {2\pi\over T} ~=~ 1 ~ s^{-1} (= 1 /second)$$ (1.11)

Now since $\omega^2 ~=~ k/m$,

$$k ~=~ m\omega^2 ~=~ 1kg ~ 1^2/s^2 ~=~ 1 N/m$$ (1.12)