Question

Đã trả lời bởi chuyên gia

Answer 1

Chọn đáp án A

Ta có: $\{\begin{cases} x_{12} = x_{1} + x_{2} = 3 \sqrt{3} ∠ \frac{π}{2} (1) \\ x_{23} = x_{2} + x_{3} = 3 ∠ 0 (2) \end{cases} \to_{}^{(1), (2)} x_{1} - x_{3} = 6 ∠ \frac{2 π}{3} (3)$

Vì $x_{1}$ và $x_{2}$ ngược pha nhau nên: $\frac{x_{1}}{x_{3}} = - \frac{A_{1}}{A_{3}} \Rightarrow x_{1} = - \frac{A_{1}}{A_{3}} x_{3} (4)$

Thay (4) vào (3) ta có: $- \frac{A_{1}}{A_{3}} x_{3} - x_{3} = 6 ∠ \frac{2 π}{3} = 6 \cos (ω t + \frac{2 π}{3})$

$\Leftrightarrow \frac{A_{1}}{A_{3}} x_{3} + x_{3} - 6 \cos (ω t + \frac{2 π}{3}) = 6 \cos (ω t - \frac{π}{3}) (5)$

Lại có: $x_{3} = A_{3} \cos (ω t + φ_{3}) \to_{}^{(5)} \frac{A_{1}}{A_{3}} . A_{3} \cos (ω t + φ_{3}) + A_{3} \cos (ω t + φ_{3}) = 6 \cos (ω t - \frac{π}{3})$

$\Rightarrow φ_{3} = - \frac{π}{3} \Rightarrow x_{3} = A_{3} \cos (ω t - \frac{π}{3})$

ta có: $x_{3} = A_{3} \cos (ω t + φ_{3}) \to_{}^{(5)} \frac{A_{1}}{A_{3}} . A_{3} \cos (ω t + φ_{3}) + A_{3} \cos (ω t + φ_{3}) = 6 \cos (ω t - \frac{π}{3})$

$\Rightarrow A_{2}^{2} = A_{3}^{2} + 3^{2} - 2.3. A_{3} . \frac{1}{2} = A_{3}^{2} - 2 A_{3} . \frac{3}{2} + \frac{9}{4} + \frac{27}{4} = {(A_{3} - \frac{3}{2})}^{2} + \frac{27}{4}$

Nhận thấy $A_{2 m i n} \Leftrightarrow {(A_{3} - \frac{3}{2})}^{2} = \min = 0 \Rightarrow A_{2}^{2} = \frac{27}{4} \Rightarrow A_{2} = \frac{3 \sqrt{3}}{2} (c m)$

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Answer 2