Answer
(a) $a_{max}=0.0495g$
(b) $v_{max}=0.128m/s$
(c) minimum
Work Step by Step
(a) We can find the maximum acceleration as follows:
$a_{max}=A(\frac{2\pi}{T})^2$
We plug in the known values to obtain:
$a_{max}=(0.0335)(\frac{2\pi}{1.65})^2$
$a_{max}=0.486m/s^2$
Now we can express the acceleration as a fraction of the acceleration of gravity as
$a_{max}=(0.486m/s^2)(\frac{g}{9.81m/s^2})$
$a_{max}=0.0495g$
(b) We can find the maximum speed as
$v_{max}=A(\frac{2\pi}{T})$
We plug in the known values to obtain:
$v_{max}=(0.0335)(\frac{2\pi}{1.65})$
$v_{max}=0.128m/s$
(c) We know that the speed is minimum when the acceleration is maximum as the
maximum acceleration occurs when the branch is farthest from the equilibrium.
