Chapter 12.6, Problem 26PPS

a.

To determine

To find: Period of the function as a fraction of a second.

Answer to Problem 26PPS

Period of a function as a fraction of a second is $\frac{1}{150}$ second

Explanation of Solution

Given information:

Point R on the piston rod rotates $150$ times per second.

Given

Point R on the piston rod rotates $150$ times per second.

We period is the time it takes to complete one rotation.

It means

Period $=\frac{1}{150}$

Therefore,

Period of a function as a fraction of a second is $\frac{1}{150}$ second

b.

To determine

To find: Graph of the function that represents the given situation.

Explanation of Solution

Given information:

The shortest distance $d$ is $0.5$ inch

The longest distance $d$ is $3.5$ inch Point R on the piston rod rotates $150$ times per second.

Given

Point R on the piston rod rotates $150$ times per second.

We period is the time it takes to complete one rotation.

It means

Period $=\frac{1}{150}$

  $\begin{array}{l}\frac{2\pi }{b}=\frac{1}{150}\\ \Rightarrow b=2\pi \times 150\\ \Rightarrow b=300\pi \end{array}$

Given

The shortest distance $d$ is $0.5$ inch

The longest distance $d$ is $3.5$ inch We know

Amplitude is the difference between longest distance and shortest distance divided by $2$

Therefore,

Amplitude $=\frac{3.5-0.5}{2}$

  $\Rightarrow$ Amplitude $=\frac{3}{2}$

Graph of above equation is