## Trigonometry (11th Edition) Clone

Published by Pearson

# Chapter 4 - Graphs of the Circular Functions - Section 4.5 Harmonic Motion - 4.5 Exercises - Page 185: 7a

#### Answer

$s(t)=-4\cos\frac{2\pi}{3}t$

#### Work Step by Step

At $t=0$, the object is released and the distance of the object from the equilibrium position at this time is $4$ units below the equilibrium position. If we use $s(t)$ to model the motion, then $s(0)$ must equal $-4$. Therefore: $s(t)= a\cos wt$,$~~~~~~$with $a=-4$ The cosine function is chosen because at $t=0$: $a\cos wt=-5\cos w(0)=-5\cos 0=-5(1)=-5$ If the sine function would have been used, a phase shift would have been required. Also, since the time for one complete oscillation is 3 seconds, the period is 3 seconds. This value of the period can be used to solve for $w$ using the formula: Period$=\frac{2\pi}{w}$ $3=\frac{2\pi}{w}$ $w=\frac{2\pi}{3}$ Therefore, the equation of motion is $s(t)=-4\cos\frac{2\pi}{3}t$.

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