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: 

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