Principle and reasoning. Work from the governing physical relation and apply it to every stated part.
For a mass–spring oscillator, ω=sqrt(\(\frac{k}{m}\)), a_max=ω²A and v_max=ωA.
\[\begin{aligned}\omega &= \sqrt{\frac{1200}{3}}\\&= 20.0\ \mathrm{rad\,s^{-1}}\end{aligned}\]
\[\begin{aligned}f &= \frac{\omega}{2\pi}\\&= 3.18\ \mathrm{Hz}\end{aligned}\]
\[\begin{aligned}a_{\max} &= \omega^2A\\&= (20)^2(0.020)\\&= 8.00\ \mathrm{m\,s^{-2}}\end{aligned}\]
\[\begin{aligned}v_{\max} &= \omega A\\&= (20)(0.020)\\&= 0.400\ \mathrm{m\,s^{-1}}\end{aligned}\]
Final answer\[f=3.18\ \mathrm{Hz},\quad a_{\max}=8.00\ \mathrm{m\,s^{-2}},\quad v_{\max}=0.400\ \mathrm{m\,s^{-1}}\]