Answer
(a) 45 cm
(b) 3.31 rad/s
(c) 1.49 m/s
Work Step by Step
(a) Amplitude A of the motion equals the distance from the equilibrium position to the point of maximum height. So,
$A=45 cm$
(b) Let's apply equation 10.6 $\omega=\frac{2\pi}{T}$ to find the angular velocity of the motion.
$\omega=\frac{2\pi}{T}$ ; Let's plug known values into this equation.
$\omega=\frac{2\pi\space rad}{1.9\space s}\approx3.31\space rad/s$
(c) Let's apply equation 10.8 $V_{max}=a\omega$ to find the maximum speed of the motion.
$V_{max}=A\omega$ ; Let's plug known values into this equation.
$V_{max}=0.45\space m\times 3.31\space rad/s\approx1.49 m/s$
