Đáp án:
1.
$\begin{array}{l}
A = 2cm;\omega = 2\pi \left( {rad/s} \right);T = 1s;f = 1Hz\\
\varphi = \frac{\pi }{4};{x_{max}} = 2cm;{a_{max}} = 80\left( {cm/{s^2}} \right)
\end{array}$
2.
$\begin{array}{l}
{Z_L} = 820\Omega ;{Z_C} = 10\Omega ;Z = 810,06\Omega \\
\varphi = 1,56\\
u = 4582,4\cos \left( {100\pi t + 1,56} \right)V
\end{array}$
Giải thích các bước giải:
1.
Theo đề bài ta có:
$\begin{array}{l}
A = 2cm\\
\omega = 2\pi \left( {rad/s} \right)\\
T = \frac{{2\pi }}{\omega } = 1s\\
f = \frac{1}{T} = 1Hz\\
\varphi = \frac{\pi }{4}\\
{x_{max}} = A = 2cm\\
{a_{max}} = {\omega ^2}A = {\left( {2\pi } \right)^2}.2 = 80\left( {cm/{s^2}} \right)
\end{array}$
2. Trở kháng
$\begin{array}{l}
{Z_L} = L\omega = \frac{{8,2}}{\pi }.100\pi = 820\Omega \\
{Z_C} = \frac{1}{{C\omega }} = \frac{1}{{\frac{1}{{1000\pi }}.100\pi }} = 10\Omega \\
Z = \sqrt {{R^2} + {{\left( {{Z_L} - {Z_C}} \right)}^2}} = \sqrt {{{10}^2} + {{\left( {820 - 10} \right)}^2}} = 810,06\Omega
\end{array}$
Độ lệch pha
$\tan \varphi = \frac{{{Z_L} - {Z_C}}}{R} = \frac{{820 - 10}}{{10}} \Rightarrow \varphi = 1,56$
Phương trình u
$\begin{array}{l}
\varphi = {\varphi _u} - {\varphi _i} \Rightarrow 1,56 = {\varphi _u} - 0 \Rightarrow {\varphi _u} = 1,56\\
{U_0} = {I_0}.Z = 4\sqrt 2 .810,06 = 4582,4V\\
u = 4582,4\cos \left( {100\pi t + 1,56} \right)V
\end{array}$
