## Precalculus (6th Edition) Blitzer

Published by Pearson

# Chapter 7 - Section 7.4 - Systems of Nonlinear Equations in Two Variables - Exercise Set - Page 853: 84

#### Answer

a) The maximum displacement of the given simple harmonic motion is $6\text{ inches}\text{.}$. b) The frequency of the given simple harmonic motion is $0.75Hz\text{ or }\frac{3}{4}\text{ cycles per second}$. c) The time required for one cycle is $\frac{4}{3}\text{sec}$.

#### Work Step by Step

(a) The provided equation is $d=6\cos \left( \frac{3\pi }{2}t \right)$. Now, the maximum displacement of a simple harmonic function is given by its amplitude. Let the maximum displacement be ${{d}_{\max }}$. Here, $A=6$. Therefore, \begin{align} & {{d}_{\max }}=A \\ & {{d}_{\max }}=6 \\ \end{align} Thus, the maximum displacement in inches is $6$. (b) The given equation is $d=6\cos \left( \frac{3\pi }{2}t \right)$. The time period, T is given as \begin{align} & T=\frac{2\pi }{\frac{3\pi }{2}} \\ & T=\frac{4}{3} \\ \end{align} Thus, \begin{align} & f=\frac{1}{T} \\ & f=\frac{1}{\frac{4}{3}} \\ & f=\frac{3}{4} \\ & f=0.75 \\ \end{align} Therefore, the frequency of the simple harmonic motion is $0.75Hz\text{ or }\frac{3}{4}\text{ cycles per second}$. (c) The given equation is $d=6\cos \left( \frac{3\pi }{2}t \right)$. Comparing with $y=a\cos \left( bt \right)$, we get: \begin{align} & a=6 \\ & b=\frac{3\pi }{2} \end{align} The time required to complete one cycle of a simple harmonic motion is called its time period. Thus, the time period of the given simple harmonic motion is \begin{align} & T=\frac{2\pi }{\frac{3\pi }{2}} \\ & T=\frac{4}{3} \\ \end{align} Thus, the time required to complete one cycle is $\frac{4}{3}\text{sec}$.

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