d(t) = 
where t is in seconds
WSMC High School State Competition
The LP Groove
Team Problem
April 21, 2001
Thomas Edison first recorded the human voice on a metal cylinder. As plastics were developed, vinyl became the choice for the recording medium. Then with the development of electronics, the medium changed to magnetic tape and now to laser reflections off compact disks. During the era of vinyl, the cylinder-shaped medium was replaced by a disk-shaped medium and they were called long-playing records or LP's for short. Machines that played the vinyl disks spun the disk at several speeds. In the 60's and 70's a popular speed was 33 1/3 revolutions per minute. A needle was vibrated in a spiraling groove that started near the outer edge and ended near the center completing the play back of the music on that side of the vinyl disk. The outside radius always was14.4 cm and the inside radius varied with the length of the music or talk recorded on it.
For this problem, consider a vinyl record that plays music for 18 minutes and 15 seconds and has an inside radius of 7.4 cm. Using this information find the following:
Your entire set of answers must fit on the front of the answer page and include any and all explanations and mathematical justifications.
The Scoring:
You will be scored on a four-point scale for each bulleted area:
Understands and Applies Mathematical Concepts and Procedures
• Algebraic Sense: Shows understanding of the functional relationships.
Measurement:
• Calculates rate and other indirect measurements.
• Understands how changes in dimension affect other measurements.
Uses Mathematical Reasoning
• Supports conclusions with applicable mathematics and clear explanations.
Communicates the Results
• Represents the mathematical processes and ideas in an effective format.
Problem Solving
• Shows understanding of the problem.
• Constructs valid solutions.
Total Points
Sample Solution
Average groove width:
33 1/3 rev/min • 18.25 min = 608 1/3 revolutions to play the side.
144 mm -74 mm = 70 mm is the change in the radius.
70 m/608 1/3 rev = 0.115 mm/rev which is the average width of the groove.
Length of the groove:
Since a circle approximates the path of the groove in one revolution and the circumference is directly proportional to the radius and the radius is a direct variation with the revolutions, we can think of the grooves as a trapezoid where one base is 2_144 mm. and the other is 2_74 mm. The height of the trapezoid is 608 1/3 grooves so the length of the groove is (2_144 + 2_74)/2 mm and this times 608 1/3 is 417,000 mm or approximately 417 meters.
Radius as a function of time:
This is a linear function with an intercept of 144 mm and a slope found by
0.115 mm/rev • 33 1/3 rev/min • 1 min/60 sec = 0.0639 mm/sec so
R(t) = 144 mm - (0.0639 mm/sec) t where t is time in seconds.
Speed as a function of time:
Speed = s(t) = change in distance / change in time = Æd/Æt.
Æd = 2_R(t) and Æt = 60sec/33 1/3 rev = 1.8 sec/rev
So s(t) = 2_(144mm - 0.0639 mm/sec • t)/1.8 sec = 503 mm/sec - 0.223 mm/sec/sec • t.
Distance traveled as a function of time:
Option 1:
d(t) =
where t is in seconds
Option 2:
Using a table and using the average speed times the time to get distances and then using a calculator for a quadratic of best fit:
time |
s(t) |
d(t) |
Calculation of d(t) |
0 |
503 |
0 |
|
100 |
481 |
49200 |
(503+481)/2 • 100 |
200 |
458 |
96100 |
(503+458)/2 • 200 |
300 |
436 |
140850 |
(503+436)/2 • 300 |
Quadratic regression yields d(t) = 503 mm/sec t + -0.111 mm/sec/sec t•t
Option 3:
503 mm/sec t + -0.111 mm/sec/sec t•t