Chapter 4.6, Problem 68E

a.

To determine

To find: is y a function of t , explain.

Answer to Problem 68E

Yes. For each t there corresponds one and only one value of y.

Explanation of Solution

Given information: The motion of an oscillating weight suspended by a spring was

measured by a motion detector. The data were collected and the approximate maximum (positive and negative) displacements from equilibrium are shown in the below graph.

The displacement y is measured in centimeters and the time t is measured in seconds.

  

Calculation:

Yes. For each t there corresponds one and only one value of y.

b.

To determine

To approximate: the frequency of the oscillations.

Answer to Problem 68E

The frequency of the oscillations is 1.3 oscillations per second.

Explanation of Solution

Given information: The motion of an oscillating weight suspended by a spring was

measured by a motion detector. The data were collected and the approximate maximum (positive and negative) displacements from equilibrium are shown in the below graph.

The displacement y is measured in centimeters and the time t is measured in seconds.

  

Calculation:

One way to find the frequency is to note that the time between the first and second maximum points is t = 0.7622 − 0 = 0.7622. Thus, the frequency is approximate ${(0.7622)}^{-1}=1.3$ per second.

Therefore, the frequency of the oscillations is 1.3 oscillations per second.

c.

To determine

To fit : a model of the form $y=a{b}^t\cos ct$ to the data and use the result of the part (b) to approximate c and use the regression feature of the graphing utility to fit an exponential model to the positive maximum displacements of the weight.

Answer to Problem 68E

  $y=12{(0.221)}^t\cos (8.2t)$ .

c = 8.2.

Explanation of Solution

Given information:

Calculation:

From the given graph data:

  $y=12{(0.221)}^t\cos (8.2t)$ .

To obtain this model, first fit an exponential model $y=a{b}^t$ to the data points (0, 12), (0.7622, 3.76), and (1.5476, 1.16). These yields $y=12{(0.2210)}^t$ . Using $\frac{2\pi }{0.7622}\approx 8.2$ for the cosine term, obtain the model above.

d.

To determine

To rewrite: the model in the form $y=a{e}^{kt}\cos ct.$

Answer to Problem 68E

  $y=12e^{-1.51t}\cos (8.2t)$ .

Explanation of Solution

Given information:

Calculation:

  $\ln 0.221\approx -1.51\Rightarrow y=12e^{-1.51t}\cos (8.2t).$

e.

To determine

To graph: the model using the graphing utility and compare the result with the data in the graph above.

Answer to Problem 68E

Explanation of Solution

Given information:

Calculation:

The graph of the data using the graphing utility is shown below.