Answer
(a) $1.877s$
(b) $0.23m$
(c) $0.77m/s$
(d) $2.58m/s^2$
Work Step by Step
(a) We can find the time period as follows:
$T=\frac{2\pi r}{v}$
We plug in the known values to obtain:
$T=\frac{2\pi (0.23m)}{0.77m/s}$
$T=1.877s$
(b) We know that amplitude is the maximum distance from mean position. Thus, $A=0.23m$
(c) We can find the maximum speed as follows:
$v_{max}=A(\frac{v}{r})$
as $A=r$
$v_{max}=\frac{rv}{r}=v$
$v_{max}=0.77m/s$
(d) We can find the magnitude of the maximum acceleration as follows:
$a_{max}=\frac{v^2}{r}$
We plug in the known values to obtain:
$a_{max}=\frac{(0.77m/s)^2}{0.23m}$
$a_{max}=2.58m/s^2$