College Physics (4th Edition)

Published by McGraw-Hill Education
ISBN 10: 0073512141
ISBN 13: 978-0-07351-214-3

Chapter 10 - Problems - Page 401: 66


The amplitude is $3.0~cm$

Work Step by Step

We can find the amplitude: $E = \frac{1}{2}mv_m^2$ $E = \frac{1}{2}m(A~\omega)^2$ $A^2 = \frac{2E}{m\omega^2}$ $A = \sqrt{\frac{2E}{m~\omega^2}}$ $A = \sqrt{\frac{2E}{m~(\sqrt{\frac{g}{L}})^2}}$ $A = \sqrt{\frac{2EL}{mg}}$ $A = \sqrt{\frac{(2)(0.015~J)(0.75~m)}{(2.5~kg)~(9.80~m/s^2)}}$ $A = 0.030~m$ $A = 3.0~cm$ The amplitude is $3.0~cm$
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