Chapter 4.4, Problem 89E

Solution

a.

To identify: What is the period P of this function Note A piano tuner strikes a tuning fork for the note middle C and creates a sound wave that can be modeled by $y=1.5sin524\pi t$ where t is the time in seconds.

The period P of this function $\frac{1}{262}$

Given information:

The equation is $y=1.5sin524\pi t$

Explanation:

Consider the given information.

Compare the given equation with the standard equation $y=a\sin \left(bx+h\right)$

We get $a=1.5,b=524\pi$

Now the periods is

  $\begin{array}{c}Periods=\frac{2\pi }{b}\\ =\frac{2\pi }{524\pi }\\ =\frac{1}{262}\end{array}$

So the period P of this function $\frac{1}{262}$

b.

To identify: The what is the frequency of this note.

The frequency is $262Hertz$

Given information:

The equation is $y=1.5sin524\pi t$

Explanation:

Consider the given information.

From the part (a) we got the period as $P=\frac{1}{262}$

Now the frequency is

  $\begin{array}{c}Frequency=\frac{1}{P}\\ =\frac{1}{\frac{1}{262}}\\ =262Hertz\end{array}$

The frequency is $262Hertz$

c.

To Graph: Graph the function $y=1.5sin524\pi t$

The graph is

  

Given information:

The equation is $y=1.5sin524\pi t$

Explanation:

Consider the given information.

The graph of the function $y=1.5sin524\pi t$ is given below