Đáp án:
a) \(\left\{ \begin{array}{l}
x = 25{\pi ^2}\\
\omega = 5\pi \Rightarrow T = 0,4s \Rightarrow f = 2,5Hz\\
\varphi = - \dfrac{\pi }{2}
\end{array} \right.\)
b) \(\left\{ \begin{array}{l}
A = 50\pi \\
\omega = 5\pi \Rightarrow T = 0,4s \Rightarrow f = 2,5Hz\\
\varphi = \dfrac{{5\pi }}{6}
\end{array} \right.\)
Giải thích các bước giải:
a) Ta có:
\(\begin{array}{l}
x = - 25{\pi ^2}\cos \left( {5\pi t + \dfrac{\pi }{2}} \right) = 25{\pi ^2}\cos \left( {\pi - 5\pi t - \dfrac{\pi }{2}} \right)\\
\Rightarrow x = 25{\pi ^2}\cos \left( { - 5\pi t + \dfrac{\pi }{2}} \right) = 25{\pi ^2}\cos \left( {5\pi t - \dfrac{\pi }{2}} \right)
\end{array}\)
Vậy: \(\left\{ \begin{array}{l}
x = 25{\pi ^2}\\
\omega = 5\pi \Rightarrow T = 0,4s \Rightarrow f = 2,5Hz\\
\varphi = - \dfrac{\pi }{2}
\end{array} \right.\)
b) Ta có:
\(\begin{array}{l}
x = - 50{\pi ^2}\sin \left( {5\pi t + \dfrac{\pi }{3}} \right) = 50\pi \sin \left( { - 5\pi t - \dfrac{\pi }{3}} \right)\\
\Rightarrow x = 50\pi \cos \left( {\dfrac{\pi }{2} + 5\pi t + \dfrac{\pi }{3}} \right) = 50\pi \cos \left( {5\pi t + \dfrac{{5\pi }}{6}} \right)
\end{array}\)
Vậy \(\left\{ \begin{array}{l}
A = 50\pi \\
\omega = 5\pi \Rightarrow T = 0,4s \Rightarrow f = 2,5Hz\\
\varphi = \dfrac{{5\pi }}{6}
\end{array} \right.\)
