Trigonometry (11th Edition) Clone

Published by Pearson
ISBN 10: 978-0-13-421743-7
ISBN 13: 978-0-13421-743-7

Chapter 4 - Graphs of the Circular Functions - Section 4.5 Harmonic Motion - 4.5 Exercises - Page 186: 18d

Answer

$s(1.466)\approx2$ This means that after $t=1.466$ seconds, the weight is about 2 inches above the equilibrium position.

Work Step by Step

We calculate $s(1.466)$ by substituting $t=1.466$ into the equation and solving: $s(t)=-4\cos 10t$ $s(1.466)=-4\cos (10\times1.466)$ $s(1.466)=-4\cos (14.66)$ $s(1.466)=-4(-0.5)$ $s(1.466)=2$ We know that the motion starts from $-4$ inches. Since the value of $s$ is positive, this means that the weight is moving upwards and has passed the equilibrium position. Therefore, after $t=1.466$ seconds, the weight is about 2 inches above the equilibrium position.
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