+ Ta có: $\text{W}=\frac{1}{2}m{{\omega }^{2}}{{A}^{2}}\Leftrightarrow {{25.10}^{-3}}=\frac{1}{2}{{.50.10}^{-3}}{{.10}^{2}}.{{A}^{2}}\Rightarrow A=0,1m=10cm$
$\begin{array}{l} A_{}^2 = A_1^2 + A_2^2 + 2A_1^{}A_2^{}{\rm{cos}}\Delta \varphi \Leftrightarrow A_{}^2 = 2A_1^2\left( {1 + {\rm{cos}}\Delta \varphi } \right) \Leftrightarrow {10^2} = {2.10^2}.\left( {1 + {\rm{cos}}\Delta \varphi } \right)\\ \Leftrightarrow {\rm{cos}}\Delta \varphi = – \frac{1}{2} \Rightarrow \Delta \varphi = \frac{{2\pi }}{3} \end{array}$
