Answer
$v_{max}=0.021\frac{m}{s}$
$a_{max}=4239\frac{m}{s^2}$
Work Step by Step
We can find the angular frequency as follows:
$\omega=2\pi f$
We plug in the known values to obtain:
$\omega=2(3.1416)(32768)$
$\omega=205887\frac{rad}{s}$
Now, we can determine the maximum velocity and maximum acceleration
$v_{max}=A\omega$
$\implies v_{max}=(100\times 10^{-9})(205887)$
$v_{max}=0.021\frac{m}{s}$
$a_{max}=A\omega^2$
$\implies a_{max}=(100\times 10^{-9})(205887)^2$
$a_{max}=4239\frac{m}{s^2}$
