Câu hỏi 14

Đáp án B

\(\begin{array}{l}{\omega _0} = \sqrt {{\omega _1}{\omega _2}}  = \sqrt {100\pi .120\pi }  = 20\pi \sqrt {30} \,rad/s\\{U_C} = \frac{U}{{\sqrt {{{\left( {\frac{\omega }{{{\omega _0}}}} \right)}^4} - 2{n^{ - 1}}{{\left( {\frac{\omega }{{{\omega _0}}}} \right)}^2} + 1} }} =  > {\left( {\frac{\omega }{{{\omega _0}}}} \right)^4} - 2{n^{ - 1}}{\left( {\frac{\omega }{{{\omega _0}}}} \right)^2} + 1 - {\left( {\frac{U}{{{U_C}}}} \right)^2} = 0\end{array}\)

Theo định lý Vi – ét

\(\left\{ \matrix{
{\left( {{{{\omega _3}} \over {{\omega _0}}}} \right)^2} + {\left( {{{{\omega _3} + 40\pi } \over {{\omega _0}}}} \right)^2} = 2{n^{ - 1}} \hfill \cr
{\left( {{{{\omega _3}} \over {{\omega _0}}}} \right)^2}.{\left( {{{{\omega _3} + 40\pi } \over {{\omega _0}}}} \right)^2} = {{24} \over {49}} \hfill \cr} \right.\buildrel {{\omega _0} = 20\pi \sqrt {30} } \over
\longrightarrow \left\{ \matrix{
{\omega _0} = 232 \hfill \cr
n = 1,3 \hfill \cr} \right. = > {\omega _C} = {{{\omega _0}} \over {\sqrt n }} \approx 96\pi \)