Chapter 13.6, Problem 27PPS

a.

To determine

To find: Period of the function in seconds.

Answer to Problem 27PPS

Period of given situation is $24$ seconds

Explanation of Solution

Given information:

A tarnado siren makes $2.5$ rotations per minute.

Radius of beam of sound $=1$

Given

A tarnado siren makes $2.5$ rotations per minute.

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

It means

Period $=\frac{60}{2.5}$

  $\Rightarrow$ Period $=24$ seconds.

Therefore,

Period of given situation is $24$ seconds

b.

To determine

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

Explanation of Solution

Given information:

A tarnado siren makes $2.5$ rotations per minute.

Radius of beam of sound $=1$

Ms.Miller’s house is $1$ mile from the siren.

Given

A tarnado siren makes $2.5$ rotations per minute.

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

It means

Period $=\frac{60}{2.5}$

  $\Rightarrow$ Period $=24$ seconds.

  $\begin{array}{l}\frac{2\pi }{b}=24\\ \Rightarrow b=\frac{2\pi }{24}\\ \Rightarrow b=\frac{\pi }{12}\end{array}$

Radius of beam of sound $=1$

Ms.Miller’s house is $1$ mile from the siren.

Therefore,

Amplitude $=1+1$

  $\Rightarrow$ Amplitude $=a=2$

Given that

Here at $t=0$ distance of sound $d=0$

Therefore,

It is a sine function

On substituting values

We get

  $\begin{array}{l}d=a\sin bt\\ \Rightarrow d=2\sin \left(\frac{\pi }{12}t\right)\\ \Rightarrow d=2\sin \frac{\pi }{12}t\end{array}$

Therefore,

Equation that represents given situation is

  $d=2\sin \frac{\pi }{12}t$

Graph of above equation is