Page 68 - Thomson, William Tyrrell-Theory of Vibration with Applications-Taylor _ Francis (2010)

Sec. 3.1 Forced Harmonic Vibration 55

Complex frequency response. From the vector force polygon of Fig.
3.1-2, it is easily seen that the terms of Eq. (3.1-1) are projections of the vectors on
the vertical axis. If the force had been F q cos cot instead of sin cot, the vector
force polygon would be unchanged and the terms of the equation then would have
been the projections of the vectors on the horizontal axis. Taking note of this, we
could let the harmonic force be represented by
F()(cos cot + i sin cot) = (3.1-12)
This would be equivalent to multiplying the quantities along the vertical axis by
i = - 1 and using complex vectors. The displacement can then be written as
X - = Xe‘^' (3.1-13)
where is a complex displacement vector:
X = Xe-^^ (3.1-14)
Substituting into the differential equation and canceling from each side of the
equation give the results
( -co^m + icco + k) X =
and
Fo/k
^ _______ _ __________ IF___________ /^3 t-t5\
{ k - ( o ^ m ) + i { c o j ) \ - + i { 2 C o ) / i o „ )
It is now convenient to introduce the complex frequency response Fi{co)
defined as the output divided by the input:
\/k
H{co) = ^ (3.1-16)
1 ( w / i o j F i2 ic x )/ (D ^
(Often the factor 1 /k is considered together with the force, leaving the frequency
response a nondimensional quantity.) Thus, H{co) depends only on the frequency
ratio and the damping factor.
The real and imaginary parts of H{co) can be identified by multiplying and
dividing Eq. (3.1-16) by the complex conjugate of the denominator. The result is
1 - (w/w„)

H{ay) = —I-
1 - ((o/oj„f\ + [2^u)/(0„f [l - (a)/co„f] + [2ia)/(o„f
(3.1-17)
This equation shows that at resonance, the real part is zero and the response is
given by the imaginary part, which is
1
H(co) = - i (3.1-18)
It is easily seen that the phase angle is
tan (f) =

1 - {o,/co„Ÿ