An Example Vibration of a String
A string 50 cm long and weighing 0.5 g is under a tension of 33 kg. Initially the mid-point of the string is displaced 0.5 cm from its equilibrium position and released. We want to calculate the displacement as a function of time at 5 cm intervals along the length of the string, using equation 12-34. From equation 1235 the At must be 8.8 x 10~5 seconds.
The spreadsheet shown in Figure 12-14 illustrates the solution of the vibrating string problem. Column B contains time in increments of At from zero to 2.8 x 1 (T3 seconds (only part of the spreadsheet is shown). The first row of displacement values (row 12, values shown in bold on the spreadsheet) are the initial conditions. The values in the second row (row 13, values in italics) are calculated according to equation 12-36; the formula in cell D13 is
Values in subsequent rows (rows 14-27 in Figure 12-14; rows 14-44 on the CD-ROM) are calculated according to equation 12-34; the formula in cell D14 is
=C13+E13-D12
|
A |
B |
C |
D |
E |
F |
G |
H |
I |
J |
K |
Ll |
M | |
|
1 |
The Wave Equation: Vibration of <i String | ||||||||||||
|
2 |
length, cm |
I |
50 |
(LI | |||||||||
|
3 |
tension, g |
33000 |
CT) | ||||||||||
|
4 |
weight,g |
0.5 |
CM) | ||||||||||
|
5 |
weight per unit length, g/cm |
0.01 |
(W) | ||||||||||
|
6 |
gravitational constant, cm/sec2 |
980 |
(g) | ||||||||||
|
7 |
Dx |
5 |
(Dx) | ||||||||||
|
8 |
Dt |
................... |
8.79E-0S |
(Dt) | |||||||||
|
1Ü |
(lisia |
nee x, cm | |||||||||||
|
11 |
0 |
5 |
10 |
15 |
20 |
26 |
30 |
35 |
40 |
45 |
50 | ||
|
12 |
8 |
6 |
0.1 |
0.2 |
0.3 |
«.4 |
0,5 |
0.4 |
0.3 |
0.2 |
0.1 |
0 I | |
|
13 |
8.8E-05 |
0 |
0.1 |
0.2 |
0,3 |
0.4 |
0.4 |
04 |
0.3 |
0.2 |
0.1 |
0 | |
|
14 |
1 8E-04 |
0 |
0.1 |
0.2 |
0.3 |
0.3 |
0.3 |
0.3 |
0.3 |
0.2 |
0.1 |
0 | |
|
15 |
2.6E-04 |
0 |
0.1 |
0.2 |
0.2 |
0.2 |
0.2 |
0.2 |
0.2 |
0.2 |
0.1 |
0 I | |
|
16 |
3.5E-04 |
0 |
0,1 |
0.1 |
0.1 |
0.1 |
01 |
0.1 |
0.1 |
0.1 |
01 |
0 | |
|
17 |
4.4E-04 |
0 |
0.0 |
0.0 |
0.0 |
00 |
0.0 |
0,0 |
0.0 |
0.0 |
0.0 |
0 | |
|
18 |
1 S.3E-04 |
0 |
-0.1 |
-0,1 |
-0.1 |
-0.1 |
-0.1 |
-0,1 |
-01 |
-0,1 |
-0.1 |
0 | |
|
19 |
6-2E-04 |
0 |
-0.1 |
-0.2 |
-0,2 |
-0.2 |
-0,2 |
-0.2 |
-0.2 |
-0,2 |
-0.1 |
0 | |
|
20 |
I 7.0E-04 |
0 |
-0.1 |
-0.2 |
-0,3 |
-0.3 |
-0.3 |
-0.3 |
-0.3 |
-0.2 |
-0,1 |
0 | |
|
21 |
7.9E-04 |
0 |
-0.1 |
-0.2 |
-0.3 |
-0.4 |
-0.4 |
-0.4 |
-0.3 |
-0.2 |
-0.1 |
0 | |
|
22 |
8.8E-Q4 |
0 |
-0.1 |
-0.2 |
-0.3 |
-0.4 |
-0.5 |
-0.4 |
-0.3 |
-0.2 |
-0.1 |
4 | |
|
23 |
9.7£~04 |
0 |
-0.1 |
-0.2 |
-0.3 |
-0.4 |
-0.4 |
-0.4 |
-0.3 |
-0,2 |
-0.1 |
0 | |
|
24 |
1.1E-03 |
0 |
-0.1 |
-0.2 |
-0.3 |
-0,3 |
-0,3 |
-0.3 |
-0.3 |
-0.2 |
-0.1 |
0 | |
|
25 |
1.1 £403 |
» |
-0.1 |
-0.2 |
-0.2 |
-0.2 |
-0.2 |
-0.2 |
-0.2 |
-0.2 |
-0.1 |
0 | |
|
26 |
1-2E-03 |
0 |
-0.1 |
-0.1 |
-0.1 |
-0.1 |
-0.1 |
-0.1 |
-0.1 |
-0.1 |
-0,1 |
0 | |
|
27 |
1.3E-03 |
0 |
0.0 |
0.0 |
0,0 |
0.0 |
0,0 |
0.0 |
0.0 |
0.0 |
0.0 |
0 | |
Figure 12-14. A spreadsheet layout for solving a hyperbolic PDE. (folder 'Chapter 12 (PDE) Examples, workbook 'Hyperbolic PDE', sheet 'Sheetl')
If you examine the values in the table, you will see that 20 time increments constitute a complete cycle of vibration. This vibration time, 0.001758 seconds, corresponds to a frequency of 569 s_1, and agrees exactly with the value calculated by the formula f (12-37)
The above procedure can be expanded to model vibrations in two space dimensions.
d2F dt2
dx2 dy2
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