Cho D1, D2 và D3 là ba dao động điều hòa cùng phương, cùng tần số

* Hướng dẫn giải

Chọn đáp án A

Ta có: $\left\{\begin{array}{l}{x}_{12}=x_1+x_2=3\sqrt{3}\angle \frac{\pi }{2}\left(1\right)\\ x_{23}=x_2+x_3=3\angle 0\left(2\right)\end{array}\right.\underset{}{\overset{\left(1\right),\left(2\right)}{\to }}{x}_1-x_3=6\angle \frac{2\pi }{3}\left(3\right)$

Vì $x_1$ và $x_2$ ngược pha nhau nên: $\frac{x_1}{x_3}=-\frac{A_1}{A_3}\Rightarrow x_1=-\frac{A_1}{A_3}{x}_3\left(4\right)$

Thay (4) vào (3) ta có: $-\frac{A_1}{A_3}{x}_3-x_3=6\angle \frac{2\pi }{3}=6\cos \left(\omega t+\frac{2\pi }{3}\right)$

$\iff \frac{A_1}{A_3}{x}_3+x_3-6\cos \left(\omega t+\frac{2\pi }{3}\right)=6\cos \left(\omega t-\frac{\pi }{3}\right)\left(5\right)$

Lại có: $x_3=A_3\cos \left(\omega t+{\phi }_3\right)\underset{}{\overset{\left(5\right)}{\to }}\frac{A_1}{A_3}.A_3\cos \left(\omega t+{\phi }_3\right)+A_3\cos \left(\omega t+{\phi }_3\right)=6\cos \left(\omega t-\frac{\pi }{3}\right)$

$\Rightarrow {\phi }_3=-\frac{\pi }{3}\Rightarrow x_3=A_3\cos \left(\omega t-\frac{\pi }{3}\right)$

ta có: $x_3=A_3\cos \left(\omega t+{\phi }_3\right)\underset{}{\overset{\left(5\right)}{\to }}\frac{A_1}{A_3}.A_3\cos \left(\omega t+{\phi }_3\right)+A_3\cos \left(\omega t+{\phi }_3\right)=6\cos \left(\omega t-\frac{\pi }{3}\right)$

$\Rightarrow A_2^2=A_3^2+3^2-\mathrm{2.3.}{A}_3.\frac{1}{2}=A_3^2-2A_3.\frac{3}{2}+\frac{9}{4}+\frac{27}{4}={\left(A_3-\frac{3}{2}\right)}^2+\frac{27}{4}$

Nhận thấy $A_{2 min}\iff {\left(A_3-\frac{3}{2}\right)}^2=\min =0\Rightarrow A_2^2=\frac{27}{4}\Rightarrow A_2=\frac{3\sqrt{3}}{2}\left(cm\right)$