1. THIS IS THE DATAPLOT PROGRAM FILE     SPRINGS.DP
2. PURPOSE--XXX
3.
4. -----START POINT-----------------------------------
5.
6. PURPOSE--DETERMINE THE NATURAL VIBRATING FREQUENCY
7.          OF A VERTICAL SPRING SYSTEM.
8. ANALYSIS TECHNIQUE--DETERMINE THE EIGENVALUES
9.                     OF A NON-SYMMETRIC MATRIX
10. APPLICATION--SPRING VIBRATION
11. SOURCE (PROBLEM)--FOGIEL, THE LINEAR ALGEBRA PROBLEM SOLVER,
12.                   RESEARCH AND EDUCATION ASSOCIATION, 1980
13.                   PAGE 693
14. GIVEN--THE SYSTEM IS VERTICAL,
15.        ATTACHED AT TOP AND BOTTOM,
16.        IN THE ORDER (FROM TOP TO BOTTOM)
17.        SPRING 1, MASS 1, SPRING 2, MASS 2, SPRING 3
18.           SPRING 1 CONSTANT K1 = 1
19.           SPRING 2 CONSTANT K2 = 4
20.           SPRING 3 CONSTANT K3 = 4
21.           MASS 1            M1 = 1
22.           MASS 2            M1 = 4
23. TO FIND--THE NATURAL FREQUENCIES OF THE SYSTEM
24. NOTE--TO SET UP THE 2 EQUATIONS OF MASS
25.       (1 EQUATION FOR EACH OF THE 2 MASSES)
26.       WE APPLY NEWTON'S SECOND LAW OF MATION
27.          (CHANGE OF MOMENTUM = SUM OF FORCES ACTING ON PARTICLE)
28.       AND HOOKE'S LAW
29.          (MAGNITUDE OF RESTORING FORCE OF A SPRING
30.          IS PROPROTIONAL TO DISPLACEMENT AND
31.          TO THE SPRING CONSTANT).
32. NOTE--THE SPECIFICS ARE AS FOLLOWS--
33.       LET Y1 = DISPLACEMENT OF MASS 1 FROM ITS EQUILIBRIUM POSITION
34.       LET Y2 = DISPLACEMENT OF MASS 2 FROM ITS EQUILIBRIUM POSITION
35.       THE 2 DIFFERENTIAL EQUATIONS ARE THEREFORE
36.             M1*Y1'' = -K1*Y1 - K2*(Y1-Y2)
37.             M2*Y2'' = K2*(Y1-Y2) - K3*Y2
38.       FOR HARMONIC MOTION--
39.             Y1'' = -F**2 * Y1
40.             Y2'' = -F**2 * Y2
41.       THUS HAVE
42.             M1*(-F**2 * Y1) = -K1*Y1 - K2*(Y1-Y2)
43.             M2*(-F**2 * Y2) = K2*(Y1-Y2) - K3*Y2
44.       SUBSTITUTING FOR K1, K2, K3, M1, AND M2 YIELDS
45.             -F**2 * Y1 = -5*Y1 + 4*Y2
46.             -F**2 * Y2 =    Y1 - 2*Y2
47.       WRITING F**2 AS LAMBDA AND TRANSPOSING, WE HAVE
48.             5*Y1 - 4*Y2 = LAMBDA*Y1
49.              -Y1 + 2*Y2 = LAMBDA*Y2
50.       THE NATURAL FREQUENCIES ARE THUS SEEN
51.       TO BE THE EIGENVALUES OF THE COEFFICIENT MATRIX.
52. NOTE--FOR TESTING PURPOSES, THE SOLUTION IS
53.       1 AND SQRT(6) = 2.449
54.
55. -----START POINT-----
56.
57ECHO
58DIMENSION 100 VARIABLES
59.
60.      STEP 1--
61.      DEFINE THE MATRIX
62.
63READ MATRIX A
645 -4
65-1 2
66END OF DATA
67PRINT A
68.
69.      STEP 2--
70.      DETERMINE THE EIGENVALUES
71.
72LET E = MATRIX EIGENVALUES A
73PRINT E
74