Chapter 4.8, Problem 67E

a.

To determine

To graph:The given function.

Explanation of Solution

Given information:A ball is attached to a spring and its motion is given by the function $y=\frac{1}{4}\cos 16t,t>0$ .

Graph:

The given function is

  $y=\frac{1}{4}\cos 16t,t>0$ .......(1)

The graph of the function $y=\frac{1}{4}\cos 16t,t>0$ is given below.

  

b.

To determine

To calculate:The time period of the given function.

Answer to Problem 67E

The time period of the given function is $T=\frac{\pi }{8}$ .

Explanation of Solution

Given information:A ball is attached to a spring and its motion is given by the function $y=\frac{1}{4}\cos 16t,t>0$ .

Formula used:

The generalized equation of cosine function is

  $y=A\cos \left(Bx+C\right)+D$

Here

  $\begin{array}{l}\text{Amplitude}=Amp=\text{|}A\text{|}\\ \text{Period}\text{of}\text{function}=T=\frac{2\pi }{B}\\ \text{Horizontal/Phase}\text{shift}=Bx+C=0\to x=-\frac{C}{B}\\ \text{Vertical}\text{shift}=D\end{array}$

Calculation:

The generalized equation of cosine function is

  $y=A\cos \left(Bx+C\right)+D$......(1)

The given function is

  $y=\frac{1}{4}\cos 16t,t>0$......(2)

By comparing equation (1) & (2)

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

Hence the time period of the given functionis $T=\frac{\pi }{8}$ .

c.

To determine

To calculate:The first time $\left(t\right)$ the ball passes the point of equilibrium ( $y=0$ ).

Answer to Problem 67E

The first time the ball passes the point of equilibrium ( $y=0$ ) is at $t=\frac{\pi }{32}$ .

Explanation of Solution

Given information:A ball is attached to a spring and its motion is given by the function $y=\frac{1}{4}\cos 16t,t>0$ .

Formula used:

For a cosine function

  $\begin{array}{l}y=a\cos \left(wt\right),t>0\\ \\ y=0\text{when}\cos \left(wt\right)=0\\ \cos \left(wt\right)=0\\ wt=\frac{n\pi }{2},n\text{is}\text{a}\text{natural}\text{number}\end{array}$

Calculation:

The given function is

  $y=\frac{1}{4}\cos 16t,t>0$......(2)

The first time the ball passes the equilibrium $\left(y=0\right)$ .

  $\begin{array}{l}\text{Put}y=0\text{in}\text{equation}\left(1\right)\\ 0=\frac{1}{4}\cos \left(16t\right)\\ \cos \left(16t\right)=0\\ 16t=\frac{n\pi }{2},n\text{is}\text{a}\text{natural}\text{number}\\ \\ \text{since}\text{it}\text{is}\text{the}\text{first}\text{time}\text{,put}n=1\\ 16t=\frac{\pi }{2}\\ t=\frac{\pi }{32}\end{array}$

Hence the first timethe ball passes the point of equilibrium ( $y=0$ ) is at $t=\frac{\pi }{32}$ .