Câu hỏi 34

Ta có:

\(\left\{ \begin{array}{l}{x_1} = {A_1}cos\left( {\omega t + {\varphi _1}} \right)\\{x_2} = {A_2}cos\left( {\omega t + {\varphi _2}} \right)\\ - {x_2} = {A_2}cos\left( {\omega t + {\varphi _2} + \pi } \right)\end{array} \right.\)

+ Khoảng cách giữa hai chất điểm theo phương Ox: \(\Delta d = \left| {{x_1} - {x_2}} \right| = dcos\left( {\omega t + \varphi } \right)\)

Với \(\left\{ \begin{array}{l}\Delta \varphi  = {\varphi _1} - \left( {{\varphi _2} + \pi } \right)\\d = \sqrt {A_1^2 + A_2^2 + 2{A_1}{A_2}cos\Delta \varphi } \end{array} \right.\)

- Khi \({A_1} = 0 \Rightarrow d = {A_2} = 12cm\)

\(\begin{array}{l}{d^2} = A_1^2 + A_2^2 + 2{A_1}{A_2}cos\Delta \varphi \\ = {\left( {{A_1} + {A_2}cos\Delta \varphi } \right)^2} + A_2^2\left( {1 - co{s^2}\Delta \varphi } \right)\end{array}\)

\({d_{\min }}\) khi

\(\begin{array}{l}{A_1} + {A_2}cos\Delta \varphi  = 0 \Rightarrow {A_1} =  - {A_2}cos\Delta \varphi \\ \Leftrightarrow 9 =  - 12cos\Delta \varphi \\ \Rightarrow cos\Delta \varphi  = \dfrac{{ - 3}}{4}\end{array}\)

- Khi \(d = 10cm\), ta có \({d^2} = A_1^2 + A_2^2 + 2{A_1}{A_2}cos\Delta \varphi \)

\(\begin{array}{l} \Leftrightarrow {10^2} = A_1^2 + {12^2} + 2{A_1}.12.\left( { - \dfrac{3}{4}} \right)\\ \Rightarrow \left[ \begin{array}{l}{A_1} = 15,08 = {a_2}\\{A_1} = 2,92 = {a_1}\end{array} \right.\end{array}\)

Tỉ số cơ năng:

\(\begin{array}{l}\dfrac{{{{\rm{W}}_2}}}{{{{\rm{W}}_1}}} = \dfrac{{\dfrac{1}{2}m{\omega ^2}a_2^2 + \dfrac{1}{2}m{\omega ^2}A_2^2}}{{\dfrac{1}{2}m{\omega ^2}a_1^2 + \dfrac{1}{2}m{\omega ^2}A_2^2}}\\ = \dfrac{{a_2^2 + A_2^2}}{{a_1^2 + A_2^2}} = \dfrac{{15,{{08}^2} + {{12}^2}}}{{2,{{92}^2} + {{12}^2}}} = 2,435\end{array}\)

Chọn C