\[\]

Áp dụng hệ thức vuông pha:
\[\begin{align}
& {{\left( \frac{v}{{{v}_{\max }}} \right)}^{2}}+{{\left( \frac{a}{{{a}_{\max }}} \right)}^{2}}=1\xrightarrow{A=10cm}{{\left( \frac{50\sqrt{3}}{10\omega } \right)}^{2}}+{{\left( \frac{500}{10{{\omega }^{2}}} \right)}^{2}}=1 \\
& \to \omega =10\left( rad/s \right)\to {{v}_{\max }}=A\omega =0,1.10=1\left( m/s \right) \\
\end{align}\]