Thread: Pre-Cal Harmonic Functions

1. Pat and Terry are turning a long jump rope for Tracy, who is waiting to jump in. They turn the rope one time per second. The maximum height is 7 feet, while at its lowest, it jsut touches the ground. Assume the rope was on the ground at starting time.

a. Give a model (formula) that indicates the height of the jump rope at time t.

*I believe the formula is y=a sin omega t ----> y= a sin 2 pi t
I figured that omega must be equal to 2 pi because the period of the function was 1 second. How do I determine a (amplitude)? My professor said it is not 7 (what I would have used).

b. Sketch a graph of your model on the x and y axes below.

*My professor said that this sin function was going to be shifted up to 3.5 on the y axis. How did he come up with that? I also know that the period is 2 pi and the frequency is 1, correct?


c. Tracy is 5 feet 3 inches tall (5.25 feet). There is a time interval when she must jump in with the rope above her head. On your graph indicate the time interval when she must jump in. it is an interval on the t (horizontal) axis.

*Should the interval must be above 5'3" to 7 feet?


d. Determine to four decimal places the (split-second) length of the time interval Tracy has to jump in.

*I have no idea how to go about this... can anyone provide me with steps on figuring this out?

Any and all help is greatly appreciated! Thanks!

Hello,

You forgot that y>=0: the rope must be above ground level. Let the center of the circle be at height h. Then, y=h+a sin omega t. You have to determine h, a, omega. Your guess about omega=2pi is correct (since the rope turns one time per second.) Since -1<= sin omega t<=1 and 0<=y<=7, you get h and a.
For problem d, determine for which values of t you get y=5.25 (using calculators or tables for the sine function.)

Bye.