Chapter 5.6, Problem 7E

(a)

To determine

To find: The amplitude, period, and frequency of the motion.

Answer to Problem 7E

The amplitude, period, and frequency of the motion is $1.5$ , $\frac{4\pi }{3}$ , and $\frac{3}{4\pi }\text{Hz}$ .

Explanation of Solution

Given Information:

The function model displacement of an object is,

  $y=-0.25\cos \left(1.5t-\frac{\pi }{3}\right)$

Concept used:

The standard form of the simple harmonic motion for the maximum amplitude at $t=0$ is given by,

  $y=A\cos \omega t$

The angular velocity is given by:

  $\omega =\frac{2\pi }{T}$

The function of the simple harmonic motion in time period term is given by,

  $y=A\cos \left(\frac{2\pi t}{T}\right)$

The function of the simple harmonic motion is givenby,

  $y=A\cos \left(\frac{2\pi t}{T}+\varphi \right)$

The function model displacement of an object is,

  $y=-0.25\cos \left(1.5t-\frac{\pi }{3}\right)$

Compare both the equation to obtain,

  $\begin{array}{l}A=0.25\\ \frac{2\pi }{T}=1.5\\ T=\frac{4\pi }{3}\end{array}$

The frequency of the motion is calculated as,

  $\begin{array}{c}f=\frac{1}{T}\\ =\frac{3}{4\pi }\text{Hz}\end{array}$

Thus, the amplitude, period, and frequency of the motion is $1.5$ , $\frac{4\pi }{3}$ , and $\frac{3}{4\pi }\text{Hz}$ .

(b)

To determine

To find: The sketch of the motion.

Explanation of Solution

Given Information:

The function model displacement of an object is,

  $y=-0.25\cos \left(1.5t-\frac{\pi }{3}\right)$

The function model displacement of an object is,

  $y=-0.25\cos \left(1.5t-\frac{\pi }{3}\right)$

Consider the following graph of the displacement of the object over one complete cycle,