a,
$T=1(s)$
$\to \omega=\dfrac{2\pi}{T}=2\pi(rad/s)$
Khi $x=3(cm)$ thì $v=80(cm/s)$
Ta có: $x^2+\dfrac{v^2}{\omega^2}=A^2$
$\to 3^2+\dfrac{80^2}{10.2^2}=A^2$
$\to A=13(cm)$
$\cos\varphi=\dfrac{2,5}{13}=\dfrac{5}{26}$
$\to \varphi=-\arccos\dfrac{5}{26}$
Vậy $x=13\cos\left(2\pi t-\arccos\dfrac{5}{26}\right)$
b,
Quãng đường 1 chu kì:
$s=4A=52(cm)$
Chiều dài quỹ đạo:
$2A=26(cm)$
c,
Góc quét: $\Delta\varphi=\dfrac{3\pi}{2}$
$\to \Delta t=\dfrac{\Delta\varphi}{\omega}=\dfrac{3}{4}s$