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 185: 8


(a) $s(t)=-6\cos\frac{\pi}{2}t$ (b) $2.30$ units (c) $\frac{1}{4}$ oscillations per second

Work Step by Step

(a) At $t=0$, the object is released and the distance of the object from the equilibrium position at this time is $6$ units below the equilibrium position. If we use $s(t)$ to model the motion, then $s(0)$ must equal $-6$. Therefore: $s(t)= a\cos wt$,$~~~~~~$with $a=-6$ The cosine function is chosen because at $t=0$: $a\cos wt=-6\cos w(0)=-6\cos 0=-6(1)=-6$ If the sine function would have been used, a phase shift would have been required. Also, since the time for one complete oscillation is 4 seconds, the period is 4 seconds. This value of the period can be used to solve for $w$ using the formula: Period$=\frac{2\pi}{w}$ $4=\frac{2\pi}{w}$ $w=\frac{2\pi}{4}$ $w=\frac{\pi}{2}$ Therefore, the equation of motion is $s(t)=-6\cos\frac{\pi}{2}t$. (b) To determine the position at $t=1.25$, we need to substitute $t=1.25$ into the equation and solve: $s(t)=-6\cos\frac{\pi}{2}t$ $s(t)=-6\cos(\frac{\pi}{2}\times1.25)$ $s(t)=-6\cos(1.963)$ $s(t)=-6(-0.383)$ $s(t)=2.30$ units (c) Frequency is the reciprocal of the period. Therefore, Frequency$=\frac{1}{4}$ oscillations per second
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