Essential University Physics: Volume 1 (3rd Edition)

Published by Pearson
ISBN 10: 0321993721
ISBN 13: 978-0-32199-372-4

Chapter 13 - Exercises and Problems: 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: $T = 2 \pi \sqrt{\frac{m}{k}} = 2 \pi \sqrt{\frac{.49}{74}} =\fbox{.51 seconds}$
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