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 - Page 401: 5


28.3 cm

Work Step by Step

Potential energy is equal to $\frac{1}{2}kx^2$. If $A$ is the amplitude, the total energy is $\frac{1}{2}kA^2$. If $\sqrt{2}A$ is the new amplitude, the new energy is $\frac{1}{2}(2)kA^2=kA^2$, double the original energy. Therefore, the new amplitude must be $(20cm)(\sqrt{2})=28.3 cm$
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