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