A. $6\sqrt{3}$   m/s². 

* Độ lệch pha: $\Delta \phi =\frac{2\pi MN}{\lambda }=\frac{8\pi }{3}=2\pi +\frac{2\pi }{3}\left\{\begin{array}{l}{A}_M=6\left(mm\right)\\ A_N=6\left|\cos \Delta \phi \right|=3\left(mm\right)\end{array}\right.$

* Chọn điểm bụng M làm gốc thì $\left\{\begin{array}{l}{v}_M=20\pi .6\cos 20\pi t\left(mm/s\right)\\ v_N=-20\pi .3\cos 20\pi t\left(mm/s\right)\\ a_N=v_N^{\text{'}}={\left(20\pi \right)}^2.3\sin 20\pi t\left(mm/s^2\right)\end{array}\right.$

* Khi  $v_M=\pm 60\pi \sqrt{2}\left(mm/s\right)\Rightarrow \cos 20\pi t=\pm \frac{1}{\sqrt{2}}\Rightarrow \sin 20\pi t=\frac{\pm 1}{\sqrt{2}}$

$\Rightarrow a_N=\pm 6000\sqrt{2}\left(mm/s^2\right)\Rightarrow$Chọn B