G. G. Parfitt 73 
A second improvement in a mount is to reduce the height of the basic reso- 
nance peak in its transmissibility curve by damping, or otherwise. One obvious 
reason for this is that, particularly inthe presence of random vibrational forces, 
it may not be entirely possible to avoid excitation at the resonance frequency. 
Other secondary reasons are the avoidance of temporary resonance during the 
run-up or stopping of a rotary machine, the minimization of oscillation following 
shock or other stray disturbance, and, as will be shown below, a possible damping 
of foundation resonances. 
A third beneficial step is to increase the rate at which isolation increases 
above the resonance, so as, for instance, to reduce the levels attained by the 
foundation resonance peaks of Fig. 4.5. 
Such ends can be achieved by damping or by adding mass to the system in 
various ways, and the resulting effects will be considered in what follows. 
4.3. DAMPING OF SPRING ISOLATORS 
If the isolator spring is damped, its performance for sinusoidal forces can 
be represented by treating its stiffness as a complex quantity S*, viz., 
S* = S(1+ 7) 
where 6 is the damping coefficient. Both S$ and 6 may be functions of frequency 
in general. One important case, to which many fairly lightly damped solid ma- 
terials conform approximately, is that where S and 6 are constants. Another 
is the case of pure viscous damping, in which S is constant and 6 is proportional 
to frequency. Writing 
5 
GeHE& (6) 
®o 
then defines the constant 5, the damping ratio, which is equal to the ratio of 
the actual damping present to that required to give critical damping of the sys- 
tem at its resonant frequency. Ghul 
The transmissibility 7 for a damped mount on a rigid foundation is given 
in general by 
T? nN 1+ 5? 
(1 — w?So/w2s)? + 8? (7) 
where § is the stiffness at frequency o and S, that at the resonance frequency 
@. This equation is plotted for 5= 0 in Fig. 4.1b. 
Introduction of heavy damping obviously will reduce the resonance peak, 
but in general will detract from the high-frequency isolation. Thus, with S = So 
equal to a constant and 5 constant, Eq. (7) reduces for w > wo, to 
2 
e 2% (0 8 
Pash (E3) (8) 
or for S and Sconstant [Eq. (6)] to 
T= 25(=2) (9) 
