Answer
a) $k=58.8N/m$
b) $\omega=11.43rad/s$
Work Step by Step
a) The energy is conserved, so $$E_0=E_f$$
In this exercise, we only need to look at elastic energy and gravitational energy. Since the block drops $0.15m$, its height changes from $h_0=0.15m$ to $h_f=0m$. In reverse, the spring's position changes from $x_0=0m$ to $x_f=0.15m$
$$mgh_0=\frac{1}{2}kx_f^2$$ $$k=\frac{2mgh_0}{x_f^2}$$
The block's mass $m=0.45kg$, so $$k=58.8N/m$$
b) The angular frequency is $$\omega=\sqrt{\frac{k}{m}}=11.43rad/s$$
