Một chất điểm dao động điều hoà với phương trình vận tốc \(v = 2\sqrt 2 {\rm{cos}}\left( {2t + \frac{{5\pi }}{6}} \right)\left( {{\rm{cm/s}}} \right)\) Tại thời điểm vật có vận tốc tức thời là 2 cm/s thì li độ của vật có thể là
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Suy ra, $v = V_{max}cos(\varphi_v)$ với $V_{max} = 2\sqrt{2}$ cm/s và $\varphi_v = 2t + \frac{5\pi}{6}$.
Khi $v = 2$ cm/s, ta có:
$2 = 2\sqrt{2} cos(\varphi_v) \Rightarrow cos(\varphi_v) = \frac{1}{\sqrt{2}} \Rightarrow \varphi_v = \pm \frac{\pi}{4} + k2\pi$.
Phương trình li độ có dạng: $x = Acos(\omega t + \varphi)$
Mối liên hệ giữa vận tốc và li độ: $v = -\omega Asin(\omega t + \varphi)$.
Ta có: $V_{max} = \omega A \Rightarrow A = \frac{V_{max}}{\omega} = \frac{2\sqrt{2}}{2} = \sqrt{2}$ cm.
Ngoài ra, $\varphi_v = \varphi_x + \frac{\pi}{2} \Rightarrow \varphi_x = \varphi_v - \frac{\pi}{2} = 2t + \frac{5\pi}{6} - \frac{\pi}{2} = 2t + \frac{\pi}{3}$.
Khi $v = 2$ cm/s, $cos(2t + \frac{5\pi}{6}) = \frac{1}{\sqrt{2}}$.
$v^2 + (\omega x)^2 = V_{max}^2 \Rightarrow (2)^2 + (2x)^2 = (2\sqrt{2})^2 \Rightarrow 4 + 4x^2 = 8 \Rightarrow 4x^2 = 4 \Rightarrow x^2 = 1 \Rightarrow x = \pm 1$.
Hoặc $x = Acos(\varphi_x) = \sqrt{2}cos(\varphi_v - \frac{\pi}{2}) = \sqrt{2}cos(\pm\frac{\pi}{4} + k2\pi - \frac{\pi}{2})$.
$x = A sin(\varphi_v) \Rightarrow sin(\varphi_v) = \pm \frac{\sqrt{2}}{2}$.
$x = \sqrt{2} (\pm \frac{\sqrt{2}}{2}) = \pm 1$ cm.
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