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0593_C13_fm Page 453 Monday, May 6, 2002 3:21 PM

Introduction to Vibrations 453

or

[ −ω 2 + ( 2 km )] [ { −ω 2 + ( 2 km )] (km ) 2 } −(km ) −ω 2 + ( 2 km )] = 0 (13.7.17)
[
2
2
−
Solving for ω we obtain:
2
ω = ( ) ω = ) ω = ( )
2
2
2
2
2
1 2 − )(km , 2 2(km , 3 2 + )(km (13.7.18)
Observe that instead of obtaining one solution we have three solutions. That is, there
are three positive frequencies (three natural frequencies) that make the determinant of Eq.
(13.7.16) equal to zero and thus allow a nonzero solution of Eqs. (13.7.13), (13.7.14), and
(13.7.15) to occur. This in turn means that we can expect to find three sets of amplitudes
solving Eqs. (13.7.13), (13.7.14), and (13.7.15), one set for each frequency ω (i = 1, 2, 3).
i
To find these amplitude solutions, we can select one of the ω (say, the smallest of the
i
ω ), substitute it into Eqs. (13.7.13) to (13.7.15), and thus solve for the amplitudes A (i =
i
i
1, 2, 3). We can then repeat the process for the other two values of ω . Notice, however,
i
that although there are three equations for the three amplitudes, the equations are not
independent in view of Eq. (13.7.16). That is, there are at most two independent equations,
thus we need another equation to obtain a unique solution for the amplitudes. Such an
equation can be obtained by arbitrarily specifying the magnitude of one or more of the
amplitudes. For example, we could “normalize” the amplitudes such that
A + A + A = 1 (13.7.19)
2
2
2
1 2 3
Then, Eqs. (13.7.13), (13.7.14), (13.7.15), and (13.7.19) are equivalent to an independent set
of three equations, enabling us to determine unique values of the amplitudes.
To this end, let us select the smallest of the ω (ω ) and substitute it (that is, {(2 – ) 2
1
i
(k/m)} 1/2 ) into Eqs. (13.7.13), (13.7.14), and (13.7.15). This produces the equations:
2 A − A = 0 (13.7.20)
1 2

−A + 2 A − A = 0 (13.7.21)
1 2 3

−A + 2 A = 0 (13.7.22)
2 3
Observe that these equations are dependent. (If we multiply the first and third equations
by 2 and add them we obtain a multiple of the second equation.) Thus, by selecting two
(say, the first two) of these equations and by combining them with Eq. (13.7.19), we obtain
upon solving the expressions:

A = 12 , A = 2 2 , A = 12 (13.7.23)
1 2 3