Đáp án:
em tham khảo nhé
\(\omega = \dfrac{{2\pi }}{T} = 2\pi rad/s.\)
Ta có \(A = \sqrt {x_0^2 + {{(\dfrac{{{v_{_0}}}}{\omega })}^2}} = \sqrt {{{( - 5\sqrt 2 )}^2} + {{(\dfrac{{10\pi \sqrt 2 }}{{2\pi }})}^2}} = 10cm.\)
\(t = 0 \to \left\{ \begin{array}{l}10\cos \varphi = - 5\sqrt 2 \\{v_0} < 0 \Rightarrow \sin \varphi > 0\end{array} \right. \Rightarrow \varphi = \dfrac{{3\pi }}{4}rad\)
\( \Rightarrow x = 10\cos (2\pi t + \dfrac{{3\pi }}{4})(cm).\)