Essential University Physics: Volume 1 (3rd Edition)

Published by Pearson
ISBN 10: 0321993721
ISBN 13: 978-0-32199-372-4

Chapter 13 - Exercises and Problems: 77

Answer

$.54 Hz; .22 m; -6^{\circ}$

Work Step by Step

We find the velocity of the two blocks after the collision: $v = \frac{.8\times1.7}{2}=.68 \ m/s$ We first find the amplitude increase. We know that the new block will add energy to the system. Thus, we find: $\frac{1}{2}kx_0^2 +\frac{1}{2}mv^2= \frac{1}{2}kx_f^2$ $\frac{1}{2}(23)(.1)^2 +\frac{1}{2}(2)(.68)^2= \frac{1}{2}(23)x_f^2$ Thus, the amplitude is: $x_f=22.4 \ cm = .22 \ m$ We now find the new frequency: $f=\frac{2\pi}{\sqrt{\frac{k}{m+M}}}$ $f=\frac{\sqrt{\frac{23}{2}}}{2\pi}=.54\ Hz$ We now find the phase constant: $\phi = tan^{-1}(\frac{v}{2\pi fA_0})-90\sqrt{\frac{M}{m+M}}$ $\phi = tan^{-1}(\frac{.68}{2\pi (.54)(.1)})-90\sqrt{\frac{1.2}{2}}=-6^{\circ}$
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