Đáp án:
a) \( \pm \dfrac{\pi }{2}\)
b) \(\left[ \begin{array}{l}
\omega t + \varphi = 0\\
\omega t + \varphi = \pi
\end{array} \right.\)
c) \( \pm \dfrac{{2\pi }}{3}\)
Giải thích các bước giải:
a) Ở VTCB:
\(\begin{array}{l}
x = 0 \Rightarrow A\cos \left( {\omega t + \varphi } \right) = 0\\
\Rightarrow \cos \left( {\omega t + \varphi } \right) = 0 \Rightarrow \omega t + \varphi = \dfrac{\pi }{2}
\end{array}\)
b) Ở biên:
\(\begin{array}{l}
x = \pm A \Rightarrow A\cos \left( {\omega t + \varphi } \right) = \pm A\\
\Rightarrow \cos \left( {\omega t + \varphi } \right) = \pm 1 \Rightarrow \left[ \begin{array}{l}
\omega t + \varphi = 0\\
\omega t + \varphi = \pi
\end{array} \right.
\end{array}\)
c) Ta có:
\(\begin{array}{l}
x = - \dfrac{A}{2} \Rightarrow A\cos \left( {\omega t + \varphi } \right) = - \dfrac{A}{2}\\
\Rightarrow \cos \left( {\omega t + \varphi } \right) = - \dfrac{1}{2} \Rightarrow \omega t + \varphi = \pm \dfrac{{2\pi }}{3}
\end{array}\)
