Physics for Scientists and Engineers: A Strategic Approach with Modern Physics (3rd Edition)

Published by Pearson
ISBN 10: 0321740904
ISBN 13: 978-0-32174-090-8

Chapter 14 - Oscillations - Conceptual Questions: 4

Answer

$E = \frac{1}{2}kA^2 = \frac{1}{2}mv_{max}^2$

Work Step by Step

The total mechanical energy $E$ in the system is constant. At all times during the motion, the sum of the potential energy and the kinetic energy is equal to the total mechanical energy. When all the energy in the system is stored as potential energy, $U = \frac{1}{2}kA^2$. When all the energy in the system is in the form of kinetic energy, $K = \frac{1}{2}mv_{max}^2$. $E = \frac{1}{2}kA^2 = \frac{1}{2}mv_{max}^2$
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