9.8 DISPLACEMENT EXCITATION

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In many cases, the designer seeks to isolate a device fromexternal vibrations

or motion of the support, as illustrated in Fig. 9-10. If we apply Newton’s

436 Chapter 9

Copyright © 2003 Marcel Dekker, Inc.

second law of motion to the system, the following differential equation is

obtained:

M

d2y2

dt2 ю RM

dy2

dt _

dy1

dt

_ _

ю KSр y2 _ y1Ю ј 0 (9-120)

The quantity y1 is the displacement of the base, and y2 is the displacement of

the main system. If we introduce the undamped natural frequency from Eq.

(9-2) and the damping ratio from Eq. (9-25), we obtain the following expression

for the equation of motion of the system:

d2y2

dt2 ю 2_!n

dy2

dt ю !2

ny2 ј 2_!n

dy1

dt ю !2

ny1 (9-121)

Let us consider the case for which the displacement of the base is

sinusoidal:

y1рtЮ ј y1m e j!t (9-122)

The right side of Eq. (9-121) may then be written in the following form:

2_!n

dy1

dt ю !2

ny1 ј р1 ю j2_rЮ!2

ny1m e j!t (9-123)

The quantity r ј f =fn is the frequency ratio. Because the right side of Eq. (9-

121) is sinusoidal for this case, the steady-state response (particular solution)

or the motion of the main system will also be sinusoidal:

y2рtЮ ј y2m e jр!t_ Ю (9-124)

Vibration Isolation for Noise Control 437

FIGURE 9-10 An SDOF spring–mass–damper system excited by the displacement

motion of the base.

Copyright © 2003 Marcel Dekker, Inc.

If we make the substitution from Eqs (9-123) and (9-124) into the

equation of motion, Eq. (9-121), the following result is obtained:

Ѕр1 _ r2Ю ю j2_r_ y2m e_j ј р1 ю j2_rЮ y1m (9-125)

The result may also be written in the following form:

Ѕр1 _ r2Ю2 ю р2_rЮ2_1=2 y2m e_j           2 ј Ѕ1 ю р2_rЮ2_1=2 y1m e_j          1 (9-126)

The phase angles are given as follows:

ј           2 _       1 (9-127)

tan       1 ј

2_r

1 _ r2 (9-128)

tan       2 ј 2_r (9-129)

The ratio of the system displacement to the displacement of the base

can be obtained from Eq. (9-126):

y2m

y1m ј Ѕ1 ю р2_rЮ2_1=2 e j

Ѕр1 _ r2Ю2 ю р2_rЮ2_1=2 (9-130)

The magnitude of this ratio is exactly the same as that found for the transmissibility,

as given by Eq. (9-103):

j y2mj ј jTrjj y1mj (9-131)

The same principles, as discussed previously for reduction of the force

transmitted to the base for a system excited by an external force, may be

applied for reduction of the displacement induced by motion of the base of a

system.

Example 9-7. An instrument package having a mass of 10 kg (22.0 lbm) is

supported by a spring–damper system having a damping ratio of _ ј 0:060.

The maximum displacement of the base to which the package is attached is

12mm (0.472 in), and the frequency of vibration of the base is 15 Hz.

Determine the spring constant for the support to limit the displacement

amplitude of the instrument package to 0.60mm (0.024 in).

The transmissibility for the system may be found from Eq. (9-131):

Tr ј

y2m

y1m ј

0:60

12:0 ј 0:050

Let us calculate the parameter from Eq. (9-109):

_ ј 1 ю р2Юр0:060Ю2р1 _ 0:0502Ю

р0:050Ю2 ј 1 ю 2:873 ј 3:873

438 Chapter 9

Copyright © 2003 Marcel Dekker, Inc.

The required frequency ratio may be found from Eq. (9-108):

рr2Ю2 _р2Юр3:873Юr2 _ р1_0:0502Ю

р0:050Ю2 ј 0

r2 ј 3:873юр3:8732 ю399Ю1=2 ј 24:22

r ј 4:921 ј f =fn

The undamped natural frequency for the system is as follows:

fn ј р15Ю=р4:921Ю ј 3:046Hz ј рKS=MЮ1=2=2_

The required spring constant for the support is found from the undamped

natural frequency:

KS ј р4_2Юр3:046Ю2р10Ю ј 3668N=m ј 3:668 kN=m р20:94lbf=inЮ

The required damping coefficient for the system may be calculated

from Eq. (9-25):

RM ј 2рKSMЮ1=2_ ј р2ЮЅр3668Юр10Ю_1=2р0:060Ю

RM ј 22:98N-s/m р0:1312 lbf -sec/in)

The maximum velocity of the system is found as follows:

vmax ј 2_fy2m ј р2_Юр15Юр0:60Ю ј 56:4mm=s р2:23in=secЮ

The maximum acceleration of the system is as follows:

amax ј 4_2f 2y2m ј р4_2Юр15Ю2р0:60Юр10_3Ю ј 5:33m=s2 р17:49 ft=sec2Ю

amax=g ј р5:33Ю=р9:806Ю ј 0:544g

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