Answer
(a) $1.01\frac{m}{s}$
(b) $808.5\frac{m}{s^2},82.4g$
Work Step by Step
(a) We know that
$v_{max}=A(2\pi f)$
We plug in the known values to obtain:
$v_{max}=(0.00125)(2\pi \times 128)=1.01\frac{m}{s}$
(b) The maximum acceleration can be calculated as
$a_{max}=A(2\pi f)^2$
We plug in the known values to obtain:
$a_{max}=(1.25)(2\pi \times 128)^2=808.5\frac{m}{s^2}$
Now we can write this acceleration as a multiple of g
$a_{max}=808.5(\frac{g}{9.81})=82.4g$
