Giải chi tiết câu 54, 55 giúp mình với ạ

\(54)\\A=4(cm)\\a)\\W=W_{d_{max}}=80(mJ)=0,08(J)\\b)\\W_d=\dfrac{1}{2}mv^2\\\Rightarrow W_{d_{max}}=\dfrac{1}{2}mv_{max}^2=\dfrac{1}{2}.0,4.v^2_{max}=0,08\\\Rightarrow v_{max}=\dfrac{\sqrt{10}}{5}(cm/s)\\c)\\v_{max}=A\omega=4\omega=\dfrac{\sqrt{10}}{5}\\\Rightarrow \omega=\dfrac{\sqrt{10}}{20}(rad/s)\\W_t=\dfrac{1}{2}m\omega^2x^2=\dfrac{1}{2}.0,4.\left(\dfrac{\sqrt{20}}{10}\right)^2.0,02^2=1,6.10^{-5}(J)\\55)\\a)\\\dfrac{T}{2}=0,6\Rightarrow T=1,2(s)\\\omega=\dfrac{2\pi}{T}=\dfrac{2\pi}{1,2}=\dfrac{5\pi}{3}(rad/s)\\b)\)

Dựa vào đồ thị `v_(max)=0,35(m/s)`

\(c)\\W=W_{d_{max}}=\dfrac{1}{2}mv_{max}^2=\dfrac{1}{2}.0,2.0,35^2=0,01225(J)\\d)\\v_{max}=A\omega=0,35\Rightarrow A=\dfrac{0,35}{\omega}=\dfrac{35}{\dfrac{5\pi}{3}}\approx0,066(m)=6,6(cm)\)