Page 149 - Thomson, William Tyrrell-Theory of Vibration with Applications-Taylor _ Francis (2010)

136 Systems with Two or More Degrees of Freedom Chap. 5

available by differentiating Eq. (5.2-2) for the velocity:
0.732 -2.732
cos {(x)2t + ÎA2) (5.2-3)
^ 1 000 ' "2*^2 1.000
By letting / = 0 and specifying the initial conditions, the four constants can be
found.
Example 5.2-1
Determine the free vibration for the system of Fig. 5.1-1 for the initial conditions

■ïi(O) 2.0 i,( 0 )
and
^2(0) 4.0 ^2(0)
Substituting these initial conditions into Eqs. (5.2-2) and (5.2-3), we have
2.0 ;0 .7 3 2 \ . , ^ / - 2 .7 3 2 \ . , (5.2-2a)
4.0 l ooo)^'"'/'. 1.000/^'"'^2

{ o | = "|C|{*[^QQQ I COS.A, + o)2 C2 | i;o o o ) “ ^'^2 (5.2-3a)
To determine Cj sin we can multiply the second equation of Eq. (5.2-2a) by
2.732 and add the results to the first equation. To determine Cj sin multiply the
second equation of Eq. (5.2-2a) by —0.732 and add the results to the first equation. In
similar manner we can solve for cos ip^ and 0)2 ^ 2 cos ip2 to arrive at the following
four results:
12.928 - 3.464c^ sin

- 0.928 = -3 .4 6 4 c2 sin 1/^2
0 = 3.464iOjC| cos lAi
0 = - 3.464(0 2^2 cos 1A2

From the last two of the foregoing equations, it is seen that cos = cos 1A2 = 0» or
lAi = ip2 = 90°. Constants c, and C2 are then found from the first two of the foregoing
equations:
c, = 3.732
C2 = 0.268
and the equations for the free vibration of the system for the initial conditions stated
for the example become
/0 .732\ f - 2 J 3 2 \

^ > = 3.7321 j QQQ ) cos iO|i + 0.2681 j qqq >cos a>2 Î

7
3
2
-
I 2.132 \ (-0 .7 3 2 1 1
.
0
| 3 . 7 3 2 / + I 0 .2 6 8 /“ S " 2 '