Question

Given that the equation of motion of a mass is $x=0.20\sin \left(3.0t\right)$ m. Find the velocity and acceleration of the mass when the object is 5 cm from its equilibrium position. Repeat for $x=0$.

Answer 1

$\omega =3$ rad/s
A=0.2 m
At X=5 cm
$v=\pm \omega \sqrt{A^2-X^2}$
$=\pm 3.0\sqrt{{\left(0.2\right)}^2-{\left(0.05\right)}^2}$
$=\pm 0.58$ m/s
$a=-{\omega }^{2}X=-{\left(3\right)}^2\left(0.05\right)$
$=-0.45$ m/s${}^2$
At X=0
$v=\pm \omega A$
$=\pm \left(3.0\right)\left(0.2\right)$
$=\pm 0.6$ m/s
$a=-{\omega }^{2}x=-{\left(3\right)}^2\left(0\right)$
$=0$