Đáp án:
a)
+ Tần số góc \(\omega = \sqrt {\dfrac{k}{m}} = \sqrt {\dfrac{{10}}{{0,1}}} = 10(rad/s)\)
+ Ta có \({{\rm{W}}_t} = \dfrac{1}{2}k{x^2}\)
\({{\rm{W}}_d} = \dfrac{1}{2}k({A^2} - {x^2})\)
\({{\rm{W}}_d} = {{\rm{W}}_t} \Rightarrow \dfrac{1}{2}k({A^2} - {x^2}) = \dfrac{1}{2}k{x^2} \Rightarrow x = \pm \dfrac{A}{{\sqrt 2 }}\)
Vậy \(x = 3\sqrt 2 = \dfrac{A}{{\sqrt 2 }} \Rightarrow A = 6cm\)
+\(t = 0:\left\{ \begin{array}{l}{x_0} = A\cos \varphi \\v = - A\omega \sin \varphi < 0\end{array} \right. \Leftrightarrow \left\{ \begin{array}{l}\cos \varphi = \dfrac{{{x_0}}}{A} = \dfrac{{3\sqrt 2 }}{6} = \dfrac{1}{{\sqrt 2 }}\\\sin \varphi > 0\end{array} \right. \Rightarrow \varphi = \dfrac{\pi }{4}rad\)
Vậy phương trình dao động điều hòa \(x = 6\cos (10t + \dfrac{\pi }{4})(cm)\)
b) Ta có \({{\rm{W}}_t} = \dfrac{1}{2}k{x^2} = \dfrac{1}{2}.10.0,{03^2} = 4,{5.10^{ - 3}}J\)
\({{\rm{W}}_d} = \dfrac{1}{2}k({A^2} - {x^2}) = \dfrac{1}{2}.10.(0,{06^2} - 0,{03^2}) = 0,0135J\)