Precalculus (6th Edition) Blitzer

Published by Pearson
ISBN 10: 0-13446-914-3
ISBN 13: 978-0-13446-914-0

Chapter 6 - Cumulative Review Exercises - Page 801: 15

Answer

a. $4\ m$ b. $\frac{5}{2\pi}\ Hz$ c. $\frac{2\pi}{5}\ sec$

Work Step by Step

a. Given the motion equation $d=4sin(5t)$, we can get the maximum displacement as the amplitude $A=4\ m$ b. The frequency $f$ can be found in the relation $\omega=2\pi f=5$. Thus $f=\frac{5}{2\pi}\ Hz$ c. The time required for one cycle is the period, which is given by $p=\frac{1}{f}=\frac{2\pi}{5}\ sec$
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