Dựa vào đồ thị ta có `W_(d_max)=W=16(mJ)`

Thời điểm ban đầu `W_d=4(mJ)=>W_t=12(mJ)`

Xét `W_t/W=(1/2m omega^2x^2)/(1/2m omega^2 A^2)=12/16=3/4`

`=> x^2/A^2=3/4=> x=+-(Asqrt3)/2`

\(t=\dfrac{5}{6}(s)=\dfrac{30}{360}.T+\dfrac{T}{2}\Rightarrow T=\dfrac{10}{7}(s)\\\Rightarrow \omega=\dfrac{7\pi}{5}(rad/s)\\W=16=\dfrac{1}{2}m\omega^2A^2\Rightarrow A=\dfrac{5\sqrt{10}}{7}(cm)\\t=12(s)\Rightarrow \alpha=8.360^o+144^o\\\Rightarrow x=A.\sin(120^o-144^o)\approx 0,92(cm)\\\Rightarrow W_d=W-W_t=16-\dfrac{1}{2}.0,32.\left(\dfrac{7\pi}{5}\right)^2.0,92^2\approx 13,3(mJ)\\\longrightarrow A.\)