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 - Exercises and Problems: 15

Answer

(a) $A = 0.10~m$ (b) $v = 0.35~m/s$

Work Step by Step

(a) $\frac{1}{2}kA^2 = \frac{1}{2}mv_{max}^2$ $A = \sqrt{\frac{m}{k}}~v_{max}$ $A = \sqrt{\frac{1.0~kg}{16~N/m}}~(0.40~m/s)$ $A = 0.10~m$ (b) We can use conservation of energy to find the speed. $\frac{1}{2}mv^2+\frac{1}{2}k(\frac{A}{2})^2=\frac{1}{2}mv_{max}^2$ $v^2=v_{max}^2-(\frac{k}{m})(\frac{A}{2})^2$ $v = \sqrt{v_{max}^2-(\frac{k}{m})(\frac{A}{2})^2}$ $v = \sqrt{(0.40~m/s)^2-(\frac{16~N/m}{1.0~kg})(\frac{0.10~m}{2})^2}$ $v = 0.35~m/s$
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