Chapter 4.8, Problem 57E

a.

To determine

Graph the function.

Answer to Problem 57E

  

Explanation of Solution

Given information:

A ball that is bobbing up and down on the end of a spring has a maximum displacement of $3$ inches. Its motion (in ideal conditions) is modelled by $y=\frac{1}{4}\cos 16t,t>0$ , where $y$ is measured in feet and $t$ is the time in seconds

Graph the function.

Calculation:

Consider a ball that is bobbing up and down on the end of a spring has a maximum displacement of $3$ inches.

  $y=\frac{1}{4}\cos 16t,t>0$

The above motion can be representing graphically.

  

Hence the results shown in plot.

b.

To determine

What is the period of the oscillations?

Answer to Problem 57E

  $\boxed{\frac{\pi }{8}}$

Explanation of Solution

Given information:

A ball that is bobbing up and down on the end of a spring has a maximum displacement of $3$ inches. Its motion (in ideal conditions) is modelled by $y=\frac{1}{4}\cos 16t,t>0$ , where $y$ is measured in feet and $t$ is the time in seconds

What is the period of the oscillations?

Calculation:

In general a simple harmonic motion is modelled by equation,

  $d=a\cos \omega t$ we know $\omega =16$

So, the period of oscillation is,

  $\begin{array}{l}T=\frac{2\pi }{\omega }\\ =\frac{2\pi }{16}\\ =\frac{\pi }{8}\end{array}$

Hence, the period of the oscillations is $\boxed{\frac{\pi }{8}}$ .

c.

To determine

Determine the first time the weight passes the point of equilibrium ( $y=0$ ).

Answer to Problem 57E

  $\boxed{t=\frac{\pi }{32}}$

Explanation of Solution

Given information:

A ball that is bobbing up and down on the end of a spring has a maximum displacement of $3$ inches. Its motion (in ideal conditions) is modelled by $y=\frac{1}{4}\cos 16t,t>0$ , where $y$ is measured in feet and $t$ is the time in seconds.

Determine the first time the weight passes the point of equilibrium ( $y=0$ ).

Calculation:

The first time the weight passes the point of equilibrium ( $y=0$ ).

  $\begin{array}{l}y=\frac{1}{4}\cos 16t\\ \cos 16t=0\\ \cos 16t=\cos \frac{\pi }{2},\cos \frac{3\pi }{2},\cos \frac{5\pi }{2}\end{array}$

Since, the weight passes the point of equilibrium,

  $\begin{array}{l}\cos 16t=\cos \frac{\pi }{2}\\ 16t=\cos \frac{\pi }{2}\\ 16t={\cos }^{-1}\left(\cos \frac{\pi }{2}\right)\\ t=\frac{\pi }{32}\end{array}$

Hence, the first time the weight passes the point of equilibrium $\boxed{t=\frac{\pi }{32}}$ .