Chapter 4.8, Problem 31E

Solution

a.

The value of a .

  $a=25\text{ ft}$

Given:

Given, the Ferris wheel shown below makes one complete turn every 20 sec. A rider’s height h , above the ground can be modeled by the equation $h=a\sin \omega t+k$ , where h and k are given in feet and t is given in seconds.

  

Calculation:

The height of the rider changes from 25 feet below to 25 above i.e., -25 to 25. Hence the amplitude of the motion is 25.

Hence,

  $a=25\text{ ft}$

Conclusion:

  $a=25\text{ ft}$

b.

The value of k .

  $k=33\text{ ft}$

Given:

Given, the Ferris wheel shown below makes one complete turn every 20 sec. A rider’s height h , above the ground can be modeled by the equation $h=a\sin \omega t+k$ , where h and k are given in feet and t is given in seconds.

  

Calculation:

k is the vertical shift of the motion above the ground.

Here the shift of center of wheel above the ground is 8+25=33 ft.

Hence,

  $k=33\text{ ft}$

Conclusion:

  $k=33\text{ ft}$

c.

The value of $\omega$

  $\omega =\frac{\pi }{10}\text{ radians/sec}$

Given:

Given, the Ferris wheel shown below makes one complete turn every 20 sec. A rider’s height h , above the ground can be modeled by the equation $h=a\sin \omega t+k$ , where h and k are given in feet and t is given in seconds.

  

Calculation:

Given one complete turn takes 20 seconds i.e., period is 20 sec.

The frequency is given by:

  $\begin{array}{c}f=\frac{1}{p}\\ \frac{\omega }{2\pi }=\frac{1}{20}\\ \omega =\frac{2\pi }{20}\\ \omega =\frac{\pi }{10}\end{array}$

Hence,

  $\omega =\frac{\pi }{10}\text{ radians/sec}$ .

Conclusion:

  $\omega =\frac{\pi }{10}\text{ radians/sec}$