Chapter 5, Problem 4PS

(a)

To determine

To Find: The sine components of $p_n\left(t\right)$ and then use the graphing utility to graph the components and then plot the graph for p .

Answer to Problem 4PS

The graph for the sine components of $p_n\left(t\right)$ is shown in Figure 1 and the graph for p is shown in Figure 2

Explanation of Solution

Given:

The given equation for the sound wave model is of the form,

  $p\left(t\right)=\frac{1}{4\pi }\left[p_1\left(t\right)+30p_2\left(t\right)+p_3\left(t\right)+p_5\left(t\right)+30p_6\left(t\right)\right]$

The value of $p_n\left(t\right)$ is $\frac{1}{n}\sin \left(524n\pi t\right)$ .

Calculation:

Consider the different values of $p_n\left(t\right)$ is obtained as,

For $n=1$

  $\begin{array}{c}{p}_1\left(t\right)=\frac{1}{n}\sin \left(524\pi t\right)\\ =\sin \left(524\pi t\right)\end{array}$

For $n=2$

  $\begin{array}{c}{p}_2\left(t\right)=\frac{1}{2}\sin \left(524\left(2\right)\pi t\right)\\ =\frac{1}{2}\sin \left(1048\pi t\right)\end{array}$

For $n=3$

  $\begin{array}{c}{p}_3\left(t\right)=\frac{1}{3}\sin \left(524\left(3\right)\pi t\right)\\ =\frac{1}{3}\sin \left(1572\pi t\right)\end{array}$

For $n=5$

  $\begin{array}{c}{p}_5\left(t\right)=\frac{1}{5}\sin \left(524\left(5\right)\pi t\right)\\ =\frac{1}{5}\sin \left(2620\pi t\right)\end{array}$

For $n=6$

  $\begin{array}{c}{p}_5\left(t\right)=\frac{1}{6}\sin \left(524\left(5\right)\pi t\right)\\ =\frac{1}{6}\sin \left(3144\pi t\right)\end{array}$

The graph for the above sine components is shown in Figure 1

  

Figure 1

Consider the given equation for sound wave model is of the form,

  $p\left(t\right)=\frac{1}{4\pi }\left[p_1\left(t\right)+30p_2\left(t\right)+p_3\left(t\right)+p_5\left(t\right)+30p_6\left(t\right)\right]$

Then,

  $p\left(t\right)=\frac{1}{4\pi }\left[\sin \left(524\pi t\right)+15\sin \left(1048\pi t\right)+\frac{1}{3}\sin \left(1572\pi t\right)+\frac{1}{5}\sin \left(2620\pi t\right)+5\sin \left(3144\pi t\right)\right]$

The graph for the above equation is shown in Figure 2

  

Figure 2

(b)

To determine

To Find: The period of each of the sine components of $p$ . Then find if p is periodic and if yes then determine the period.

Answer to Problem 4PS

The graph for $p\left(t\right)$ is periodic with the period of 0.0038.

Explanation of Solution

Consider the different values of $p_n\left(t\right)$ is obtained as,

For $n=1$

  $\begin{array}{c}{p}_1\left(t\right)=\frac{1}{n}\sin \left(524\pi t\right)\\ =\sin \left(524\pi t\right)\end{array}$

Then, the period is,

  $\begin{array}{c}\frac{2\pi }{524\pi }=\frac{1}{262}\\ =0.0038\end{array}$

For $n=2$

  $\begin{array}{c}{p}_2\left(t\right)=\frac{1}{2}\sin \left(524\left(2\right)\pi t\right)\\ =\frac{1}{2}\sin \left(1048\pi t\right)\end{array}$

Then, the period is,

  $\begin{array}{c}\frac{2\pi }{1048\pi }=\frac{1}{524}\\ =0.0019\end{array}$

For $n=3$

  $\begin{array}{c}{p}_3\left(t\right)=\frac{1}{3}\sin \left(524\left(3\right)\pi t\right)\\ =\frac{1}{3}\sin \left(1572\pi t\right)\end{array}$

Then, the period is,

  $\begin{array}{c}\frac{2\pi }{1572\pi }=\frac{1}{786}\\ =0.0013\end{array}$

For $n=5$

  $\begin{array}{c}{p}_5\left(t\right)=\frac{1}{5}\sin \left(524\left(5\right)\pi t\right)\\ =\frac{1}{5}\sin \left(2620\pi t\right)\end{array}$

Then, the period is,

  $\begin{array}{c}\frac{2\pi }{2620\pi }=\frac{1}{1310}\\ =0.00076\end{array}$

For $n=6$

  $\begin{array}{c}{p}_5\left(t\right)=\frac{1}{6}\sin \left(524\left(5\right)\pi t\right)\\ =\frac{1}{6}\sin \left(3144\pi t\right)\end{array}$

Then, the period is,

  $\begin{array}{c}\frac{2\pi }{3144\pi }=\frac{1}{1572}\\ =0.00064\end{array}$

Thus, the graph for $p\left(t\right)$ is periodic with the period of 0.0038.

(c)

To determine

To Find: The find the intercepts of the graph of p over one cycle.

Answer to Problem 4PS

The x intercept over one cycle of p are

  $\left(0,0\right),\left(0.00096,0\right),\left(0.00191\right),\left(0.00285,0\right)\text{and}\left(0.00382\right)$ .

Explanation of Solution

Consider the graph of p as shown in Figure 3

  

Figure 3

From the above graph the x intercept over one cycle of p are $\left(0,0\right),\left(0.00096,0\right),\left(0.00191\right),\left(0.00285,0\right)\text{and}\left(0.00382\right)$ .

(d)

To determine

To Find: The approximation for the absolute and the maximum values of p over one cycle.

Answer to Problem 4PS

The absolute maximum 1.195 and the absolute minimum is $-1.195$ .

Explanation of Solution

Consider the graph for p is shown in Figure 4

  

Figure 4

From the above graph it is clear that over one cycle of p the absolute maximum 1.195 and the absolute minimum is $-1.195$ .