University Physics with Modern Physics (14th Edition)

Published by Pearson
ISBN 10: 0321973615
ISBN 13: 978-0-32197-361-0

Chapter 14 - Periodic Motion - Problems - Exercises - Page 461: 14.36

Answer

$x = 0.707~A$

Work Step by Step

We can write an expression for the total energy in the system. $E = \frac{1}{2}kA^2$ When the elastic potential energy equals the kinetic energy, then each form of energy will be equal to $\frac{E}{2}$. We can find $x$ when the elastic potential energy is equal to $\frac{E}{2}$. $\frac{1}{2}kx^2 = \frac{E}{2}$ $\frac{1}{2}kx^2 = \frac{\frac{1}{2}kA^2}{2}$ $x^2 = \frac{A^2}{2}$ $x = \sqrt{\frac{1}{2}}~A$ $x = 0.707~A$
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