Physics for Scientists and Engineers: A Strategic Approach with Modern Physics (4th Edition)

Published by Pearson
ISBN 10: 0133942651
ISBN 13: 978-0-13394-265-1

Chapter 15 - Oscillations - Exercises and Problems: 37

Answer

$t = 0.096~s$

Work Step by Step

We can write an expression for the total energy in the system as: $E = \frac{1}{2}kA^2$ If the kinetic energy is twice the potential energy, then $K = \frac{2E}{3}$ and $U_s = \frac{E}{3}$. We then find and expression for $x$ when $U_s = \frac{E}{3}$; $U_s = \frac{E}{3}$ $\frac{1}{2}kx^2 = \frac{\frac{1}{2}kA^2}{3}$ $x^2 = \frac{A^2}{3}$ $x = \sqrt{\frac{1}{3}}~A$ We then find the time $t$ when $x = \sqrt{\frac{1}{3}}~A$: $x(t) = A~cos(10t)$ $\sqrt{\frac{1}{3}}~A = A~cos(10t)$ $cos(10t) = \sqrt{\frac{1}{3}}$ $t = \frac{arccos(\sqrt{\frac{1}{3}})}{10}$ $t = 0.096~s$
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