Hai vật dao động điều hòa cùng tần số góc ω

Ta có: \({A_1} + {A_2} \ge 2\sqrt {{A_1}{A_2}}  \Rightarrow {A_1}{A_2} \le {\left( {\frac{{10}}{2}} \right)^2} = 25\) 

+ Lại có:  

+ Theo Bất đẳng thức Bu-nhi-a, ta có:

\(\left\{ \begin{array}{l}
{A_1} = \sqrt {x_1^2 + \frac{{v_1^2}}{{{\omega ^2}}}} \\
{A_2} = \sqrt {x_2^2 + \frac{{v_2^2}}{{{\omega ^2}}}} 
\end{array} \right. \Rightarrow \sqrt {x_1^2 + \frac{{v_1^2}}{{{\omega ^2}}}} \sqrt {x_2^2 + \frac{{v_2^2}}{{{\omega ^2}}}}  \le 25\) 

 \(\begin{array}{l}
\left( {{a^2} + {b^2}} \right)\left( {{c^2} + {d^2}} \right) \ge {\left( {ac + bd} \right)^2}\\
 \Rightarrow \left( {x_1^2 + \frac{{v_1^2}}{{{\omega ^2}}}} \right)\left( {x_2^2 + \frac{{v_2^2}}{{{\omega ^2}}}} \right) \ge {\left( {{x_1}.\frac{{{v_2}}}{\omega } + \frac{{{v_1}}}{\omega }{x_2}} \right)^2}\\
 \Rightarrow {\left( {{x_1}.\frac{{{v_2}}}{\omega } + \frac{{{v_1}}}{\omega }{x_2}} \right)^2} \le {25^2}\\
 \Leftrightarrow {x_1}.\frac{{{v_2}}}{\omega } + \frac{{{v_1}}}{\omega }{x_2} \le 25\frac{{10}}{\omega } \le 25\\
 \Rightarrow \omega  \ge 0,4 \Rightarrow {\omega _{\min }} = 0,4rad/s
\end{array}\)