## Physics for Scientists and Engineers: A Strategic Approach with Modern Physics (3rd Edition)

Published by Pearson

# Chapter 14 - Oscillations - Exercises and Problems - Page 403: 35

#### 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 as: $T = 2\pi~\sqrt{\frac{m}{k}}$ $m = \frac{T^2~k}{(2\pi)^2}$ $m = \frac{(3.0~s)^2(240~N/m)}{(2\pi)^2}$ $m = 55~kg$ (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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