Cơ năng: \(W=\dfrac{1}{2}mv_{max}^2\Rightarrow v_{max}=0,2(m/s)\)
Do vận tốc vuông pha với gia tốc nên: \((\dfrac{v}{v_{max}})^2+(\dfrac{a}{a_{max}})^2=1\)
\(\Rightarrow (\dfrac{0,1}{0,2})^2+(\dfrac{-\sqrt 3}{a_{max}})^2=1\)
\(\Rightarrow a_{max}=2(m/s^2)\)
\(v_{max}=\omega.A\)
\(a_{max}=\omega^2A\)
Suy ra: \(\omega=\dfrac{a_{max}}{v_{max}}=\dfrac{2}{0,2}=10(rad/s)\)
\(\Rightarrow A = \dfrac{v_{max}}{\omega}=2(cm)\)
Ban đầu, có: \(x_0=-\dfrac{a}{\omega^2}=\sqrt 3 (cm)\)
Pha ban đầu: \(\cos\varphi=\dfrac{x_0}{A}=\dfrac{\sqrt 3}{2}\)
Do \(v>0 \Rightarrow \varphi <0\) \(\Rightarrow \varphi =-\dfrac{\pi}{6}\)
Vậy \(x=2\cos(10t-\dfrac{\pi}{6})(cm)\)
