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

Published by Pearson

# Chapter 13 - Exercises and Problems - Page 240: 67

#### Answer

a) .065 meters b) .51 seconds

#### Work Step by Step

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

After you claim an answer you’ll have 24 hours to send in a draft. An editor will review the submission and either publish your submission or provide feedback.