Câu 2 : \(x=10\sqrt{2}cos\left(5\pi t-\dfrac{\pi}{2}\right)\left(cm\right)\)a) Biên độ: \(A=10\sqrt{2}\left(cm\right)\)Pha ban đầu: \(\phi=-\dfrac{\pi}{2}\)Tần số góc: : \(\omega=5\pi\left(rad/s\right)\)Chu kì: \(T=\dfrac{2\pi}{\omega}=\dfrac{2\pi}{5\pi}=0,4\left(s\right)\)Tần số : \(f=\dfrac{1}{T}=\dfrac{1}{0,4}=2,5\left(Hz\right)\)b) \(t_o=\dfrac{1}{5}\left(s\right)\Rightarrow\Phi_o=5\pi.\dfrac{1}{5}-\dfrac{\pi}{2}=\dfrac{\pi}{2}\left(rad\right)\)c) \(t=2\left(s\right)\Rightarrow x=10\sqrt{2}cos\left(5\pi.0,2-\dfrac{\pi}{2}\right)=0\left(cm\right)\)d) \(v_{max}=A\omega=10\sqrt{2}.5\pi=50\pi\sqrt{2}\sim222\left(cm/s\right)\)\(a_{max}=A\omega^2=10.\sqrt[]{2}.\left(5\pi\right)^2=2500\sqrt{2}\sim3536\left(cm/s^2\right)\)e) \(v=-10\sqrt{2}.5\pi sin\left(5\pi t-\dfrac{\pi}{2}\right)=-50\pi\sqrt{2}sin\left(5\pi t-\dfrac{\pi}{2}\right)\left(cm/s\right)\)\(a=-10\sqrt{2}.\left(5\pi\right)^2cos\left(5\pi t-\dfrac{\pi}{2}\right)=-2500\sqrt{2}cos\left(5\pi t-\dfrac{\pi}{2}\right)\left(cm/s^2\right)\)Câu trả lời: Câu 2 : \(x=10\sqrt{2}cos\left(5\pi t-\dfrac{\pi}{2}\right)\left(cm\right)\)a) Biên độ: \(A=10\sqrt{2}\left(cm\right)\)Pha ban đầu: \(\phi=-\dfrac{\pi}{2}\)Tần số góc: : \(\omega=5\pi\left(rad/s\right)\)Chu kì: \(T=\dfrac{2\pi}{\omega}=\dfrac{2\pi}{5\pi}=0,4\left(s\right)\)Tần số : \(f=\dfrac{1}{T}=\dfrac{1}{0,4}=2,5\left(Hz\right)\)b) \(t_o=\dfrac{1}{5}\left(s\right)\Rightarrow\Phi_o=5\pi.\dfrac{1}{5}-\dfrac{\pi}{2}=\dfrac{\pi}{2}\left(rad\right)\)c) \(t=2\left(s\right)\Rightarrow x=10\sqrt{2}cos\left(5\pi.0,2-\dfrac{\pi}{2}\right)=0\left(cm\right)\)d) \(v_{max}=A\omega=10\sqrt{2}.5\pi=50\pi\sqrt{2}\sim222\left(cm/s\right)\)\(a_{max}=A\omega^2=10.\sqrt[]{2}.\left(5\pi\right)^2=2500\sqrt{2}\sim3536\left(cm/s^2\right)\)e) \(v=-10\sqrt{2}.5\pi sin\left(5\pi t-\dfrac{\pi}{2}\right)=-50\pi\sqrt{2}sin\left(5\pi t-\dfrac{\pi}{2}\right)\left(cm/s\right)\)\(a=-10\sqrt{2}.\left(5\pi\right)^2cos\left(5\pi t-\dfrac{\pi}{2}\right)=-2500\sqrt{2}cos\left(5\pi t-\dfrac{\pi}{2}\right)\left(cm/s^2\right)\)