Một vật dao động điều hòa trên quỹ đạo dài $8cm$ với chu kì $T=2s$. Chọn gốc thời gian là lúc vật đi qua vị trí cân bằng theo chiều dương. Phương trình dao động của vật là :

$x = 5cos(2πt – π/2) (cm)$

$x = 5cos(2πt + π/2) (cm)$

$x = 5cos(πt – π/2) (cm)$

$x = 5cos(πt + π/2) (cm)$

${\text{x}} = 16c{\text{os}}\left( {2\pi t - \frac{\pi }{2}} \right)(cm)$

${\text{x}} = 16c{\text{os}}\left( {2\pi t + \frac{\pi }{2}} \right)(cm)$

${\text{x}} = 8c{\text{os}}\left( {2\pi t + \frac{\pi }{2}} \right)(cm)$

${\text{x}} = 8c{\text{os}}\left( {2\pi t - \frac{\pi }{2}} \right)(cm)$

\(a = - 2\pi cos\left( {2\pi t + \dfrac{\pi }{6}} \right){\text{ }}cm/{s^2}\)

\(a = 40sin\left( {2\pi t + \dfrac{\pi }{6}} \right){\text{ }}cm/{s^2}\)

\(a = - 40cos\left( {2\pi t + \dfrac{\pi }{6}} \right){\text{ }}cm/{s^2}\)

\(a = 2\pi sin\left( {2\pi t + \dfrac{\pi }{6}} \right){\text{ }}cm/{s^2}\)

\(\frac{\pi }{2} < \varphi  < \pi \)

$\frac{\pi }{2} < \varphi  < 0$

$ - \pi  < \varphi  <  - \frac{\pi }{2}$ 

$0 < \varphi  < \frac{\pi }{2}$

\(x{\text{ }} = {\text{ }}5\sqrt 2 sin(2\pi t{\text{ }} + \frac{\pi }{4}){\text{ }}cm\)

\(x = 5\cos (2\pi t - \frac{\pi }{6})cm\)

\(x{\text{ }} = {\text{ }}5sin(2\pi t{\text{ }} + \frac{\pi }{4}){\text{ }}cm\)

\(x{\text{ }} = {\text{ }}5\sqrt 2 sin(2\pi t{\text{ }} - \frac{\pi }{6}){\text{ }}cm\)

\(x = 2c{\rm{os}}\left( {\pi t - \frac{\pi }{2}} \right)cm\)

\(x = 4c{\rm{os}}\left( {2\pi t - \frac{\pi }{2}} \right)cm\)

\(x = 2c{\rm{os}}\left( {\pi t + \frac{\pi }{2}} \right)cm\)

\(x = 4c{\rm{os}}\left( {\pi t - \frac{\pi }{2}} \right)cm\)

\(x = 4c{\rm{os}}\left( {\frac{\pi }{3}t - \frac{\pi }{3}} \right)cm\)   

 \(x = 4c{\rm{os}}\left( {\frac{\pi }{3}t + \frac{\pi }{6}} \right)cm\)

\(x = 4c{\rm{os}}\left( {\pi t - \frac{\pi }{6}} \right)cm\)

\(x = 4c{\rm{os}}\left( {\frac{{2\pi }}{3}t - \frac{{5\pi }}{6}} \right)cm\)