Câu hỏi 14

Theo bài ra ta có \(\left\{ \begin{array}{l}l = 4\frac{\lambda }{2} = 40cm =  > \lambda  = 20cm\\\Delta t = {t_2} - {t_1} = \frac{1}{{6f}} = \frac{T}{6}\end{array} \right.\)

\(\left\{ \begin{array}{l}6 = {A_b}\cos \alpha \\5 = {A_b}\cos \left( {120 - \alpha } \right)\end{array} \right. =  > {A_b} = 11cm\)

\(\left\{ \begin{array}{l}\frac{{{u_M}}}{{{u_B}}} =  - \frac{{{A_M}}}{{{A_b}}} =  > {A_M} =  - {A_b}\frac{{{u_M}}}{{{u_B}}} =  - 12\frac{3}{{ - 5}} = 6,6mm\\\delta  = \frac{v}{{{v_{max}}}} = \frac{{\frac{\lambda }{T}}}{{{A_b}\frac{{2\pi }}{T}}} = \frac{\lambda }{{2\pi {A_b}}} = \frac{{200}}{{2\pi .7,2}} \approx 4,8\end{array} \right.\)

Chọn D