Answer
a) $f=0.384Hz$
b) $v_{max}=0.125m/s$
c) $A=5.20cm$
Work Step by Step
(a) We know that
$f=\frac{\omega}{2\pi}$
We plug in the known values to obtain:
$f=\frac{2.41rad/s}{2\pi}$
$f=0.384Hz$
(b) The maximum speed of the block is given as
$v_{max}=\frac{\alpha_{max}}{\omega}$
$v_{max}=\frac{0.302m/s^2}{2.41rad/s}$
$v_{max}=0.125m/s$
(c) The amplitude can be determined as
$A=\frac{\alpha_{max}}{\omega}^2$
We plug in the known values to obtain:
$A=\frac{0.302m/s^2}{(2.41rad/s)^2}$
$A=5.20cm$
