vật lý 12

B

buichianh18896

$\begin{array}{l}
x = 20{\sin ^2}(4\pi t + \frac{\pi }{6}) = 10 - 10\cos (8\pi t + \frac{\pi }{3}) \\
\Rightarrow x = 10 - 10\cos (8\pi t + \frac{\pi }{3}) = 10 + 10\cos (8\pi t + \frac{\pi }{3} + \pi ) = 10 + 10\cos (8\pi t + \frac{{4\pi }}{3}) \\
t = 0,5 \\
\Rightarrow x = 5 \\
\Rightarrow v = x' = - 80\pi \sin (8\pi t + \frac{{4\pi }}{3}) = 40\pi \sqrt 3 \\
\Rightarrow a = x'' = - 640{\pi ^2}\cos (8\pi t + \frac{{4\pi }}{3}) = 230{\pi ^2} \\
\end{array}$
 
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