+ Ta có: \(\left\{ {\begin{array}{*{20}{l}} {{x_1} = {A_1}.\cos \left( {\omega t + {\varphi _1}} \right)}\\ {{x_2} = {A_2}.\cos \left( {\omega t + {\varphi _2}} \right)} \end{array}} \right. \Rightarrow \left\{ {\begin{array}{*{20}{l}} {{x_1} = {A_1}.\cos \left( {\omega t + {\varphi _1}} \right)}\\ { - {x_2} = {A_2}.\cos \left( {\omega t + {\varphi _2} + \pi } \right)} \end{array}} \right.\)

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

Với: \(\left\{ {\begin{array}{*{20}{l}} {\Delta \varphi = {\varphi _1} - \left( {{\varphi _2} + \pi } \right)}\\ {d = \sqrt {A_1^2 + A_2^2 + 2{A_1}{A_2}.\cos \Delta \varphi } } \end{array}} \right.\)

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

\( \Leftrightarrow \sqrt {{0^2} + A_2^2 + 2.0.{A_2}.\cos \Delta \varphi } = 12cm \Rightarrow {A_2} = 12cm\)

+ Lại có: \({d^2} = A_1^2 + A_2^2 + 2{A_1}{A_2}.\cos \Delta \varphi \)

\( \Rightarrow {d^2} = {\left( {{A_1} + {A_2}.\cos \Delta \varphi } \right)^2} + A_2^2\left( {1 - {{\cos }^2}\Delta \varphi } \right)\)

\( \Rightarrow {d_{\min }} \Leftrightarrow {A_1} + {A_2}.\cos \Delta \varphi = 0 \Rightarrow \cos \Delta \varphi = - \frac{{{A_1}}}{{{A_2}}}\)

Mà \({d_{\min }} \Leftrightarrow {A_1} = 9cm \Rightarrow \cos \Delta \varphi = - \frac{9}{{12}} = - \frac{3}{4}\)

+ Khi \(d = 10cm\) ta có:

\({d^2} = A_1^2 + A_2^2 + 2{A_1}{A_2}.\cos \Delta \varphi \Leftrightarrow {10^2} = A_1^2 + {12^2} + 2{A_1}.12.\left( { - \frac{3}{4}} \right)\)

\( \Leftrightarrow \left[ {\begin{array}{*{20}{l}} {{A_1} = 2,92cm = {a_1}}\\ {{A_1} = 15,08cm = {a_2}} \end{array}} \right.\)

Tỉ số cơ năng: 

\({\mkern 1mu} {\mkern 1mu} {\mkern 1mu} {\mkern 1mu} {\mkern 1mu} {\mkern 1mu} {\mkern 1mu} {\mkern 1mu} \frac{{{W_1}}}{{{W_2}}} = \frac{{\frac{1}{2}m{\omega ^2}a_1^2 + \frac{1}{2}m{\omega ^2}A_2^2}}{{\frac{1}{2}m{\omega ^2}a_2^2 + \frac{1}{2}m{\omega ^2}A_2^2}} = \frac{{a_1^2 + A_2^2}}{{a_2^2 + A_2^2}} = \frac{{15,{{08}^2} + {{12}^2}}}{{2,{{92}^2} + {{12}^2}}} = 0,4\)