Chapter 4.8, Problem 76E

a.

To determine

To calculate:The amplitude of the given simple harmonic function.

Answer to Problem 76E

The amplitude of the given simple harmonic function is $A=4$ .

Explanation of Solution

Given information:The graph of the simple harmonic function is given below.

  

Formula used:

The amplitude of a harmonic function is

  $A=\frac{y_{\max }-y_{\min }}{2}$

Calculation:

From the above graph of the given function it can be observed that

The maximum value of yis $y_{\max }=+4$ .

The minimum value of yis $y_{\min }=-4$ .

Therefore the amplitude of the given function is

  $\begin{array}{c}A=\frac{y_{\max }-y_{\min }}{2}\\ =\frac{4-\left(-4\right)}{2}=\frac{8}{2}\\ A=4\end{array}$

Hence the amplitude of the given simple harmonic function is $A=4$ .

b.

To determine

To calculate:The time period of the given simple harmonic function.

Answer to Problem 76E

The time period of the given simple harmonic function is $T=3$ .

Explanation of Solution

Given information:The graph of the simple harmonic function is given below.

  

Calculation:

The time periodof a function is the value of x after which the cycle repeats itself.

From the graph above it can be observed that since the value $y=4$ at $x=0$ repeats after $x=3$ ,the time period of the function is $T=3$ .

Hence the time period of the given simple harmonic function is $T=3$ .

c.

To determine

To identify: The form of the given simple harmonic function.

Answer to Problem 76E

The simple harmonic function is of the form $d=a\cos \left(wt\right)$ .

Explanation of Solution

Given information: The graph of the simple harmonic function is given below.

  

Concept used:

In the function of the form

  $\begin{array}{l}d=a\sin \left(wt\right)\\ \text{at}t=0\to \sin \left(wt\right)=0\\ d=a\sin \left(wt\right)=0\\ \text{therefore},\text{at}t=0\to d=0\end{array}$

In the function of the form

  $\begin{array}{l}d=a\cos \left(wt\right)\\ \text{at}t=0\to \cos \left(wt\right)=1\\ d=a\cos \left(wt\right)=a\\ \text{therefore},\text{at}t=0\to d\ne 0\end{array}$

Calculation:

The simple harmonic function can be of the form $d=a\sin wt\text{or}d=\cos wt$ .

From the graph above it can be observed that

  $att=0,d=4\ne 0$ .

Also if the function is of the form

  $\begin{array}{l}d=a\cos \left(wt\right)\\ \text{at}t=0\to \cos \left(wt\right)=1\\ d=a\cos \left(wt\right)=a\\ \text{therefore},\text{at}t=0\to d\ne 0\end{array}$

Therefore it can be said that since $att=0,d=4\ne 0$ , the given function is of the form $d=a\cos \left(wt\right)$ .

Hence the simple harmonic function is of the form $d=a\cos \left(wt\right)$ .