9.4 FORCED VIBRATION

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If an external driving force is applied to the spring–mass–damper system, as

shown in Fig. 9-4, the governing equation of motion for the system may be

found from Newton’s second law of motion:

M

d2y

dt2 ю RM

dy

dt ю KSy ј FрtЮ (9-69)

Vibration Isolation for Noise Control 419

FIGURE 9-4 Forced vibratory system.

Copyright © 2003 Marcel Dekker, Inc.

If we introduce the undamped natural frequency from Eq. (9-2) and the

damping ratio from Eq. (9-25), we obtain the following form for the equation

of motion, Eq. (9-69):

d2y

dt2 ю 2_!n

dy

dt ю !2

ny ј

!2

nFрtЮ

KS

(9-70)

We find, in general, that the solution of Eq. (9-70) for the motion of

the system involves two components: (a) the transient part, which is the

solution of the homogeneous equation, FрtЮ ј 0, and (b) the steady-state

part, which is a particular solution of the complete equation. The transient

part is the same as that obtained for free motion, given by Eq. (9-21).

As we note from Eqs (9-30), (9-31), or (9-41), the transient portion of

the solution involves negative exponentials of the form expр__!ntЮ. These

terms generally decay to negligible values after a few cycles, unless the

damping is identically zero. Because all physical systems involve some

level of damping or energy dissipation, we see that the transient component

will approach zero for all mechanical vibrating systems. For example, if the

undamped natural frequency is fn ј 5 Hz and the damping ratio is _ ј 0:10,

we find the following numerical value for the argument of the exponential:

_!nt ј р0:10Юр2_Юр5ЮрtЮ ј 3:14t

The exponential term is less than 0.010 for an argument of _4:71 or larger

(absolute value). For a time of 1.50 seconds, we find the following value:

_!nt ј р3:14Юр1:50Ю ј 4:712

expр__!ntЮ ј e_4:712 ј 0:0090

Thus, for the conditions given in this example, the homogeneous solution or

transient component of the vibratory displacement has become negligible

only 1.50 seconds after the driving force has been applied, and only the

particular solution or steady-state component is of importance.

Let us examine the case in which the driving force is sinusoidal, or:

FрtЮ ј Fo cosр!tЮ (9-71)

We may also write the driving force in complex notation, keeping in mind

that only the real part has ‘‘real’’ physical significance:

FрtЮ ј Fo e j!t (9-72)

The quantity ! is the frequency for the applied force.

Because the right-hand side of Eq. (9-70) involves an exponential

function, in this case, the particular solution (steady-state solution) will

420 Chapter 9

Copyright © 2003 Marcel Dekker, Inc.

also involve exponential functions. Let us consider the steady-state solution

in the form:

yрtЮ ј C e jр!t_          Ю ј ymax e jр!t_         Ю (9-73)

The solution may also be written in the trigonometric form:

yрtЮ ј C cosр!t _       Ю (9-74)

The quantity      is the phase angle between the applied force and the displacement

of the system. We note that the forced-vibration system will

oscillate at the same frequency as that of the applied force, but the force

and displacement will not be exactly in-phase, in general.

If we make the substitution from Eq. (9-73) for yрtЮ into Eq. (9-70), the

governing equation of motion, the following result is obtained:

_C!2 e jр!t_     Ю ю j2C_!n! e jр!t_   Ю ю C!n e jр!t_          Ю ј

!2

nFo e j!t

KS

(9-75)

Let us define the frequency ratio as follows:

r _ !=!n ј f =fn (9-76)

If we divide both sides of Eq. (9-75) by ЅC!2

n e jр!t_           Ю_ and introduce the

frequency ratio, we obtain the following result:

_r2 ю j2_r ю 1 ј

Fo e j  

CKS ј

Fo e j  

ymaxKS

(9-77)

We may define the magnification factor (MF) as follows:

MF ј

ymax

рFo=KSЮ

(9-78)

The magnification factor is the ratio of the maximum amplitude of vibration

of the system to the static displacement of the system if the force Fo acts as a

static force. The quantity Fo is the maximum amplitude of the applied force.

We note that the peak-to-peak amplitude of motion yp ј 2ymax. The rms

amplitude of motion is yrms ј ymax=р21=2Ю.

We may write the left-hand side of Eq. (9-77) in the form involving a

magnitude ЅRe2 ю Im2_1=2 and a phase angle tan      ј Im=Re:

Ѕр1 _ r2Ю2 ю р2_rЮ2_1=2 e j        ј

e j       

MF

(9-79)

The phase angle between the displacement and the applied force is given by

the following:

tan        ј

2_r

1 _ r2 (9-80)

Vibration Isolation for Noise Control 421

Copyright © 2003 Marcel Dekker, Inc.

The magnification factor may be obtained from Eq. (9-79):

MF ј

1

Ѕр1 _ r2Ю2 ю р2_rЮ2_1=2 (9-81)

A plot of the magnification factor MF as a function of frequency ratio r ј f =fn for several different damping ratios _ is shown in Fig. 9-5.

There are several important observations that we can make from Fig.

9-5. First, if the forcing frequency f is less than the undamped natural

frequency fn, so that r < 1, the magnification ratio is always greater than

422 Chapter 9

FIGURE 9-5 Plot of the magnification factor MF ј KSymax=Fo vs. frequency ratio

r ј f =fn for an SDOF spring–mass–damper system with various values of the damping

ratio _ ј RM=RM;cr.

Copyright © 2003 Marcel Dekker, Inc.

unity for damping ratios _ < 1

2. If r < 1, for a damping ratio _ ј 1

2, we see

that the denominator of Eq. (9-81) is as follows:

Ѕр1 _ r2Ю2 ю р2_rЮ2_1=2 ј Ѕ1 _ r2 ю r4_1=2 < 1

and MF > 1 for this case.

Secondly, we observe that the magnification factor is always less than

unity for all frequencies if the damping ratio is larger than 2_1=2 ј 0:707.

For a damping ratio _ ј 1=21=2, the denominator of Eq. (9-81) is as follows:

Ѕр1 _ r2Ю2 ю р2_rЮ2_1=2 ј Ѕ1 ю r4_1=2 > 1

and MF < 1 for this case. If we wish to reduce the amplitude of motion of

the mass in the low-frequency region р0 < r < 1Ю, we must select that damping

ratio to be 0.707 or larger.

Finally, we note that the magnification factor is always less than unity

for any value of damping ratio if the frequency ratio r > 21=2 ј 1:414. For a

frequency ratio r ј 21=2, the denominator of Eq. (9-81) is as follows:

Ѕр1 _ r2Ю2 ю р2_rЮ2_1=2 ј Ѕ1 ю 8_2_1=2           1 for all _         0

and MF _ 1 for this case. This means that, if we are really serious about

reducing the motion significantly for a given frequency f , we must adjust the

undamped natural frequency of the system such that f > 21=2fn ј 1:414fn.

This feat may be accomplished for a given mass M by selecting the proper

value of the spring constant KS for the system, as illustrated by Eq. (9-15).

Example 9-3. A machine having a mass of 50 kg (110.2 lbm) is driven by a

sinusoidal force having a maximum amplitude of 40N (8.99 lbf ) and a frequency

of 10 Hz. The support system has a spring constant of 50 kN/m

(285 lbf /in) and a damping coefficient of 600 N-s/m (3.43 lbf -sec/in).

Determine the maximum amplitude of vibration for the system.

The undamped natural frequency for the system is determined from

Eq. (9-15):

fn ј рKS=MЮ1=2

2_ ј р50,000=50Ю1=2

р2_Ю ј 5:033 Hz

The frequency ratio is:

r ј f =fn ј р10Ю=р5:033Ю ј 1:987

The damping ratio may be found from Eq. (9-25):

_ ј

RM

2рKSMЮ1=2 ј р600Ю

р2ЮЅр50,000Юр50Ю_1=2 ј р600Ю

р3162:3Ю ј 0:1897

Vibration Isolation for Noise Control 423

Copyright © 2003 Marcel Dekker, Inc.

The magnification factor may be found from Eq. (9-81):

MF ј

1

fЅ1 _ 1:9872_2 ю Ѕр2Юр0:1897Юр1:987Ю_2g1=2 ј 0:3287 ј

KSymax

Fo

The maximum amplitude of motion for the system may be calculated:

ymax ј р0:3287Юр40Ю

р50,000Ю ј 0:263 _ 10_3 m ј 0:263mm р0:0104 inЮ

The peak-to-peak amplitude of motion is as follows:

yp ј 2ymax ј р2Юр0:263Ю ј 0:526mm р0:0207 inЮ

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