Page 208 - Thomson, William Tyrrell-Theory of Vibration with Applications-Taylor _ Francis (2010)

Sec. 6.11 Equal Roots 195

to be
Xi = <^l(-K,)^l(0 + <>iXi)Q2iO + 4>2ix,)q2it)
t
2
Thus, the time solution for any floor is composed of the normal modes used.
From the numerical information supplied on the normal modes, we now
determine the numerical values for the first equation, which can be rewritten as
^ ..:T2^(0
We have, for the first mode,
mu = = 5.2803m
— = = 0.299-v/ —
mjj \ m
/Cjl
iof = 0.02235 —
* m
^m (pi = 6.6912m
The equation for the first mode then becomes
q, + 0 . 2 9 9 - ^ + 0.02235^<7, = - 1.2672ii„(i)

Thus, given the values for k/m and ^j, the above equation can be solved for any
UqU).

6.11 EQUAL ROOTS

When equal roots are found in the characteristic equation, the corresponding
eigenvectors are not unique and a linear combination of such eigenvectors may
also satisfy the equation of motion. To illustrate this point, let (j)^ and (¡>2 be
eigenvectors belonging to a common eigenvalue Aq, and (f)^ be a third eigenvector
belonging to A3 that is different from Aq. We can then write
^4>i = Aq(/)i
A(f)2 ~ ^o<p2
A(f)2 = A3(^3
By multiplying the second equation by a constant b and adding it to the first, we
obtain another equation:
A((l>i + b(f)2) = Aq((/>i + b(j)2)
Thus, a new eigenvector (/>i2 = (</>i + b(j)2), which is a linear combination of the
first two, also satisfies the basic equation:
A(I>i2 = ^ o4*\2
and hence no unique mode exists for Aq.