Đáp án:
$\begin{align}
& a)x=2,5cm;v=5\sqrt{3}\pi (cm/s);a=-100m/{{s}^{2}};F=-10N \\
& b)x=-2,5cm;v=-5\sqrt{3}\pi (cm/s);a=100m/{{s}^{2}};F=10N \\
\end{align}$
Giải thích các bước giải:
$m=0,1kg;x=5\sin (2\pi t+\frac{\pi }{6})$
Li độ:
$\begin{align}
& x=5.\sin (2\pi t+\dfrac{\pi }{6}) \\
& =5.cos(2\pi t+\dfrac{\pi }{6}-\dfrac{\pi }{2}) \\
& =5.cos(2\pi t-\dfrac{\pi }{3}) \\
\end{align}$
Vận tốc:
$\begin{align}
& v=-2\pi .5.\sin (2\pi t-\dfrac{\pi }{3}) \\
& =-10\pi .\sin (2\pi t-\dfrac{\pi }{3}) \\
\end{align}$
gia tốc:
$\begin{align}
& a=-{{\left( 2\pi \right)}^{2}}.5.cos(2\pi t-\dfrac{\pi }{3}) \\
& =-200.cos(2\pi t-\dfrac{\pi }{3}) \\
\end{align}$
Lực phục hồi:
$\begin{align}
& F=m.a=0,1.(-200).cos(2\pi t-\frac{\pi }{3}) \\
& =-20.cos(2\pi t-\frac{\pi }{3}) \\
\end{align}$
a) t=5s
$\begin{align}
& x=5.cos(2\pi .5-\frac{\pi }{3})=2,5cm \\
& v=-10\pi .\sin (2\pi .5-\frac{\pi }{3})=5\sqrt{3}\pi (cm/s) \\
& a=-200.cos(2\pi .5-\frac{\pi }{3})=-100m/{{s}^{2}} \\
& F=-20.cos(2\pi .5-\frac{\pi }{3})=-10N \\
\end{align}$
b) pha dao động 120 độ
$\begin{align}
& x=5.cos(\dfrac{2\pi }{3})=-2,5cm \\
& v=-10\pi .\sin (\dfrac{2\pi }{3})=-5\sqrt{3}\pi (cm/s) \\
& a=-200.cos(\dfrac{2\pi }{3})=100m/{{s}^{2}} \\
& F=-20.cos(\dfrac{2\pi }{3})=10N \\
\end{align}$
