\(W_{d_{max}}=4W_{d_{t_1}}\\\Rightarrow \dfrac{1}{2}kA^2=4.\dfrac{1}{2}kx_1^2\\\Rightarrow x^2=\dfrac{A^2}{4}\\\Rightarrow x=\pm\dfrac{A}{2}\\\cos\alpha=\dfrac{x}{A}=\pm\dfrac{\dfrac{A}{2}}{A}==\pm\dfrac{1}{2}\)   Mà tại `t_1` vật vẫn chưa đổi chiều nên \(\Rightarrow \left[\begin{array}{l}\alpha=-\dfrac{\pi}{6}\\\alpha=\dfrac{2\pi}{3}\end{array}\right.\\TH1: \alpha=-\dfrac{\pi}{6}\\\Rightarrow t_1=\dfrac{\alpha}{2\pi}.T=\dfrac{-\dfrac{\pi}{6}}{2\pi}.T=\dfrac{\pi}{6}\\\Rightarrow T=-2\pi<0 (\text{loai})\\TH2:\alpha=\dfrac{2\pi}{3}\\\Rightarrow t_1=\dfrac{\alpha}{2\pi}.T=\dfrac{\dfrac{2\pi}{3}}{2\pi}.T=\dfrac{\pi}{6}\\\Rightarrow T=\dfrac{\pi}{2}>0(\text{nhan})\\\Rightarrow \omega =\dfrac{2\pi}{T}=4(rad/s)\\t_2=\dfrac{5\pi}{12}\\\Rightarrow \Delta t=t_2-t_1=\dfrac{5\pi}{12}-\dfrac{\pi}{6}=\dfrac{\pi}{4}=\dfrac{T}{2}\\\Rightarrow S=2A=12\\\Rightarrow A=6\\\)

Thời điểm ban đầu, vận ở vị trí cân bằng nên tốc độ đạt cực đại

\(\Rightarrow |v|_{max}=A\omega =6.4=24(cm/s)\\\longrightarrow A.\)