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: 43

Answer

(a) $m = 55~kg$ (b) $v = 0.72~m/s$

Work Step by Step

(a) From the graph, we can see that the period is 3.0 seconds. We can find the mass. $T = 2\pi~\sqrt{\frac{m}{k}}$ $m = \frac{T^2~k}{(2\pi)^2}$ $m = \frac{(3.0~s)^2(240~N/m)}{(2\pi)^2}$ $m = 55~kg$ (b) From the graph we can see that the length oscillates between 0.6 meters and 1.4 meters. Thus the natural length of the spring is 1.0 meter and the amplitude is 0.4 meters. We can use conservation of energy to find the speed when the spring's length is 1.2 meters, that is, when the spring is stretched by 0.2 meters. $\frac{1}{2}mv^2+\frac{1}{2}kx^2 = \frac{1}{2}kA^2$ $v^2 = \frac{k~(A^2-x^2)}{m}$ $v = \sqrt{\frac{k~(A^2-x^2)}{m}}$ $v = \sqrt{\frac{(240~N/m)~[(0.4~m)^2-(0.2~m)^2]}{55~kg}}$ $v = 0.72~m/s$
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