## Essential University Physics: Volume 1 (3rd Edition)

a) We find: $A = \frac{mg}{k} = \frac{(.49)(9.81)}{74}=\fbox{.065 meters}$ b) We use the equation for period to find: $T = 2 \pi \sqrt{\frac{m}{k}} = 2 \pi \sqrt{\frac{.49}{74}} =\fbox{.51 seconds}$