Đáp án:

1. \( - 3cm\)

2. \(12\pi \sqrt 3 \left( {cm/s} \right)\)

3. \(\begin{array}{l}
{v_{\max }} = 24\pi \left( {cm/s} \right)\\
{a_{\max }} = 96{\pi ^2}\left( {cm/{s^2}} \right)
\end{array}\)

Giải thích các bước giải:

1. Ta có: 

\(\begin{array}{l}
T = \dfrac{{60}}{{120}} = 0,5s\\
\omega  = \dfrac{{2\pi }}{T} = \dfrac{{2\pi }}{{0,5}} = 4\pi \left( {rad/s} \right)\\
A = \dfrac{{12}}{2} = 6\left( {cm} \right)\\
\cos \varphi  = \dfrac{x}{A} = \dfrac{{ - \dfrac{A}{2}}}{A} =  - \dfrac{1}{2} \Rightarrow \varphi  =  - \dfrac{{2\pi }}{3}\\
 \Rightarrow x = 6\cos \left( {4\pi t - \dfrac{{2\pi }}{3}} \right) = 6\cos \left( {4\pi .60 - \dfrac{{2\pi }}{3}} \right) =  - 3cm
\end{array}\)

2. Ta có: 

\(\begin{array}{l}
v =  - 24\pi \sin \left( {4\pi t - \dfrac{{2\pi }}{3}} \right) =  - 24\pi \sin \left( {4\pi .2 - \dfrac{{2\pi }}{3}} \right) = 12\pi \sqrt 3 \left( {cm/s} \right)\\
a =  - 96{\pi ^2}\cos \left( {4\pi t - \dfrac{{2\pi }}{3}} \right) =  - 96{\pi ^2}\cos \left( {4\pi .2 - \dfrac{{2\pi }}{3}} \right) = 48{\pi ^2}\left( {cm/{s^2}} \right)
\end{array}\)

3. Ta có: 

\(\begin{array}{l}
{v_{\max }} = 24\pi \left( {cm/s} \right)\\
{a_{\max }} = 96{\pi ^2}\left( {cm/{s^2}} \right)
\end{array}\)