Physics: Principles with Applications (7th Edition)

Published by Pearson
ISBN 10: 0-32162-592-7
ISBN 13: 978-0-32162-592-2

Chapter 11 - Oscillations and Waves - Problems - Page 323: 20

Answer

(a) The amplitude is 0.055 meters. (b) The maximum speed is 0.59 m/s

Work Step by Step

(a) The total energy during the motion is constant. We can find the amplitude of the motion as|: $\frac{1}{2}kA^2 = \frac{1}{2}kx^2+\frac{1}{2}mv^2$ $A^2 = x^2+\frac{mv^2}{k}$ $A = \sqrt{x^2+\frac{mv^2}{k}}$ $A = \sqrt{(0.020~m)^2+\frac{(2.7~kg)(0.55~m/s)^2}{310~N/m}}$ $A = 0.055~m$ The amplitude is 0.055 meters. (b) We can find the maximum speed as: $v_{max} = \sqrt{\frac{k}{m}}~A$ $v_{max} = \sqrt{\frac{310~N/m}{2.7~kg}}~(0.055~m)$ $v_{max} = 0.59~m/s$ The maximum speed is 0.59 m/s.
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