Đáp án:

a) $\dfrac{5}{9}s$

b) $\dfrac{{35}}{{36}}\left( s \right)$

Giải thích các bước giải:

a) Tại thời điểm ban đầu: ${x_0} = 4\cos \dfrac{\pi }{3} = 2cm$

Ta có: 

$\begin{array}{l} T = \dfrac{{2\pi }}{\omega } = \dfrac{{2\pi }}{{6\pi }} = \dfrac{1}{3}\left( s \right)\\ \alpha  = 2\pi  - \dfrac{\pi }{3} - \dfrac{\pi }{3} = \dfrac{{4\pi }}{3}\left( {rad} \right)\\ {t_1} = \dfrac{\alpha }{\omega } = \dfrac{{\dfrac{{4\pi }}{3}}}{{6\pi }} = \dfrac{2}{9}s\\  \Rightarrow {t_2} = {t_1} + T = \dfrac{2}{9} + \dfrac{1}{3} = \dfrac{5}{9}s \end{array}$

b) Ta có: 

$\begin{array}{l} \cos \beta  = \dfrac{x}{A} = \dfrac{{2\sqrt 3 }}{4} = \dfrac{{\sqrt 3 }}{2} \Rightarrow \beta  = \dfrac{\pi }{6}\left( {rad} \right)\\  \Rightarrow \gamma  = 2\pi  - \left( {\dfrac{\pi }{3} - \dfrac{\pi }{6}} \right) = \dfrac{{11\pi }}{6}\left( {rad} \right)\\ {t_3} = \dfrac{\gamma }{\omega } = \dfrac{{\dfrac{{11\pi }}{6}}}{{6\pi }} = \dfrac{{11}}{{36}}\left( s \right)\\ {t_4} = {t_3} + 2T = \dfrac{{11}}{{36}} + 2.\dfrac{1}{3} = \dfrac{{35}}{{36}}\left( s \right) \end{array}$