Name _____________________________________________ Class ________ Date ________________
a. peak 1, x= _____ y= _____
b. peak 2, x= _____ y= _____ "yratio 2" = "ypeak 2" "ypeak 1" = _____
c. peak 3, x= _____ y= _____ "yratio 3" = "ypeak 3" "ypeak 2" = _____
d. peak 4, x= _____ y= _____ "yratio 4" = "ypeak 4" "ypeak 3" = _____
e. peak 5, x= _____ y= _____ "yratio 5" = "ypeak 5" "ypeak 4" = _____
5. Find the ratio for each successive peak change by dividing the previous peak time by the latter peak time. What did you find?
"xratio 2" = "xpeak 1" "xpeak 2" = _____ "xratio 3" = "xpeak 2" "xpeak 3" = _____ "xratio 4" = "xpeak 3" "xpeak 4" = _____ "xratio 5" = "xpeak 4" "xpeak 5" = _____
6. Increase Xmax in the viewing window, and determine at what elapsed time has the pendulum virtually stopped (peak is less than .05).
x= __________
1. To find the position for the damped pendulum for the given time you will need to use the Run Menu. Enter the Run Menu from the MAIN MENU by pressing
1 (RUN)or use the arrow keysto highlight RUN and press EXE. A blank screen should appear. If it does not, press AC/ON to clear the screen. Make sure the calculator is in radian mode by pressing SHIFT SETUP and arrow downto highlight Angle. Press F2(Rad) to select .
Press EXIT to exit the Set Up screen. Calculate the position for the damped pendulum at 2 seconds by calculating the expression:
0.852(1.03) sin (3.5 (2) - 1.8)
Enter and calculate this by pressing . 8 5 ^ 2 1 . 0 3 sin ( 3 . 5 2 – 1 . 8 ) EXE.
The position of the pendulum at x= 2 seconds is -0.657 feet or 0.657 feet to the left of center. Repeat for the other three times. Do this by pressing the right arrow keyto replay the expression and then press the right arrow keyto highlight the 2 and press 4 . Do this for both xor 2 values.
Press EXE to see the answer. Repeat for x= 6 and x= 8.
The position of the pendulum at x= 4 seconds is -0.19 feet or 0.19 feet to the left of center. The position of the pendulum at x= 6 seconds is 0.13 feet or .13 feet to the right of center.
The position of the pendulum at x= 8 seconds is 0.246 feet or 0.246 feet to the right of center.
2. To graph the model for the motion of the damped pendulum press MENU 5 (GRAPH) and enter the expression y= 0.85X 1.03 sin (3.5 X - 1.8) for Y1. Enter the expression by pressing . 8 5 ^ X,
,T 1 . 0 3 sin ( 3 . 5 X,,T – 1 . 8 ) EXE .To graph the model from x= -0.1 seconds to 10 seconds, and from y= -2 feet to 2 feet, press SHIFT F3 (VWIN) to access the View Window screen. Enter -0.1 for Xmin and 10 for Xmax by pressing (-) . 1 EXE 1 0 EXE 1 EXE. Enter -2 for Ymin and 2 for Ymax by pressing (-) 2 EXE 2 EXE 1 EXE.
Press EXITto exit the View Window screen and press F6 (DRAW) to view the graph.
3. To use the graphical solver feature to find the initial position of the pendulum, represented by the y-intercept of the curve, press F5(G-Solv) F4 (Y-ICPT) .
The initial position was 1 foot to the left of the center point.
4. To use the trace feature to find the first five peaks of the graph which represent the first five right-hand swing positions of the damped pendulum, press
F1 (Trace) to enter the trace mode. A tracer will appear on the curve. Press the
left arrow key to move the tracer to the first peak at x= 0.942 seconds. The peak occurs at .881 feet.
Press the right arrow keyto move the tracer to the second and third peaks. The peaks occur at x= 2.706 seconds and y= 0.652 feet and x= 4.549 seconds and
y= 0.492 feet respectively.
Press the right arrow key to move the tracer to the fourth and fifth peaks. The peaks occur at x= 6.313 seconds and y= 0.366 feet and x= 8.156 seconds and
y= 0.273 feet respectively.
To find the ratio for each successive peak change, return to the Run menu by pressing MENU 1 (RUN). Divide the "ypeak 2" = .652 by "ypeak 1" = .881 by pressing
.
6 5 2 . 8 8 1 EXE. The "yratio 2" = .74. Repeat for the other ratios.The resulting ratios are virtually the same, therefore the friction is damping the pendulum in a proportional manner. In other words, the pendulum is slowing down in a constant manner.
5. To find the ratio for each successive peak change, divide the previous peak time by the latter peak time. The first ratio (xratio 2) is found by dividing "xpeak 1" = 0.942 by "xpeak 2" = 2.706. Do this by pressing . 9 4 2 2 . 7 0 6
EXE . The "xratio 2" = .. Repeat for the other ratios.
The resulting ratios are increasing, which means the time interval between peaks is decreasing.
6. To find when the pendulum has virtually stopped, you must return to the graph. Do this by pressing MENU 5 (GRAPH). Increase the Xmax to 20 by pressing
SHIFT F3 (V-Window), pressing the down arrow keyand pressing 2 0 EXE.
Press EXE F6(DRAW) to view the graph.
Press F1(Trace) and a tracer will appear on the curve. Press the right arrow and find the peak values. The eleventh peak is y= 0.048.
The elapsed time in which the the pendulum has virtually stopped (a peak less than 0.05) is x= 18.883 seconds.