Page 131 - Thomson, William Tyrrell-Theory of Vibration with Applications-Taylor _ Francis (2010)

118 Transient Vibration Chap. 4

These quantities are then used in the following recurrence formula:
h
g(y, + 27^ + 273+ n ) (4.8-5)
h
y,+i = 7 + 6 ^^2 + P4) (4.8-6)
where it is recognized that the four values of Y divided by 6 represent an average
slope dx/dt and the four values of F divided by 6 result in an average of dy/dt as
defined by Eqs. (4.8-4).
Example 4.8-1
Solve Example 4.5-1 by the R unge-K utta method.
Solution: The differential equation of motion is
= i/( 0 - 500x
Let y = i ; then
y == F { x , t ) = 1 / ( 0 - 500x
With h = 0.02, the following table is calculated:
t JC y = i /
L = 0 0 0 25
0 .0 1 0 0.25 25
0 .0 1 0.0025 0.25 23.75
h ^ 0 .0 2 0.0050 0.475 22.50
The calculation for JC2 and y 2 follows:

X2 = 0 + ^ ( 0 -H 0.50 0.50 -h 0.475) = 0.00491667

^2 = 0 + ^ ^ ( 2 5 -h 50 47.50 22.50) = 0.4833333
To continue to point 3, we repeat the foregoing table:
t2 = 0 .0 2 0.00491667 0.4833333 22.541665
0.03 0.0097500 0.70874997 20.12500
0.03 0.01200417 0.6845833 18.997915
^3 = 0.04 0.01860834 0.8632913 15.695830

We then calculate JC3 and
JC3 = 0.00491667
+ ^ ^ (0 .4 8 3 3 3 3 + 1.4174999 + 1.3691666 + 0.8632913)
= 0.00491667 + 0.01377764 = 0.01869431
y 3 = 0.483333 -f 0.38827775 = 0.87161075
To complete the calculation, the example was programmed on a digital com­
puter, and the results showed excellent accuracy. Table 4.8-1 gives the numerical