Answer
(a) $\omega=19\frac{rad}{s}$
(b) $T=0.33s$
(c) $a_{max}=92\frac{m}{s^2}$
Work Step by Step
(a) The angular frequency is given as
$\omega=\frac{v_{max}}{A}$
We plug in the known values to obtain:
$\omega=\frac{4.8}{0.25}$
$\omega=19\frac{rad}{s}$
(b) The time period can be calculated as
$T=\frac{2\pi}{T}$
$\implies T=\frac{2(3.1416)}{19.2}$
$T=0.33s$
(c) The maximum acceleration is given as
$a_{max}=v_{max}\omega$
$\implies a_{max}=(4.8)(19)$
$a_{max}=92\frac{m}{s^2}$
