Đáp án:

a) \(\varphi  =  - \dfrac{\pi }{2}\)

b) \(\varphi  = 0\)

c) \(\varphi  = \dfrac{\pi }{3}\)

d) \(\varphi  =  - \dfrac{\pi }{6}\)

Giải thích các bước giải:

Tại t = 0 thì: \(\left\{ \begin{array}{l}
x = 6\cos \varphi \\
v =  - 12\pi \sin \varphi 
\end{array} \right.\)

a) Ta có:

\(\left\{ \begin{array}{l}
x = 6\cos \varphi  = 0\\
v =  - 12\pi \sin \varphi  > 0
\end{array} \right. \Rightarrow \left\{ \begin{array}{l}
\cos \varphi  = 0\\
\sin \varphi  < 0
\end{array} \right. \Rightarrow \varphi  =  - \dfrac{\pi }{2}\)

b) Ta có:

\(\left\{ \begin{array}{l}
x = 6\cos \varphi  = 6\\
v =  - 12\pi \sin \varphi  = 0
\end{array} \right. \Rightarrow \left\{ \begin{array}{l}
\cos \varphi  = 1\\
\sin \varphi  = 0
\end{array} \right. \Rightarrow \varphi  = 0\)

c) Ta có:

\(\left\{ \begin{array}{l}
x = 6\cos \varphi  = 3\\
v =  - 12\pi \sin \varphi  < 0
\end{array} \right. \Rightarrow \left\{ \begin{array}{l}
\cos \varphi  = \dfrac{1}{2}\\
\sin \varphi  > 0
\end{array} \right. \Rightarrow \varphi  = \dfrac{\pi }{3}\)

d) Ta có:

\(\left\{ \begin{array}{l}
x = 6\cos \varphi  = 3\sqrt 3 \\
v =  - 12\pi \sin \varphi  > 0
\end{array} \right. \Rightarrow \left\{ \begin{array}{l}
\cos \varphi  = \dfrac{{\sqrt 3 }}{2}\\
\sin \varphi  < 0
\end{array} \right. \Rightarrow \varphi  =  - \dfrac{\pi }{6}\)