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2.2 Response to Harmonic Excitations

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Consider a simple oscillator with an excitation

force f ðtÞ; as shown in Figure 2.1.

The equation of motion is given by

mx€ þ bx_ þ kx ¼ f ðtÞ ð2:1Þ

Suppose that f ðtÞ is sinusoidal (i.e., harmonic).

Pick the time reference such that

f ðtÞ ¼ f0 cos vt ð2:2Þ

where

v ¼ excitation frequency

f0 ¼ forcing excitation amplitude

For a system subjected to a forcing excitation, we have

Total Response

T ¼ Homogeneous Response xh

ðNatural ResponseÞ

þ Particular Response xp

ðEnforced ResponseÞ

¼ Free Response

ðDepends only on initial conditionsÞ

does not contain enforced response

and depends entirely on the

natural=homogeneous response

þ Forced Response

ðDepends only on f0 Þ but contains a natural=homogeneous component

Using these concepts, we analyze the forced problem, which may be written as

x€ þ

x_ þ

x ¼

cos vt ¼ uðtÞ ð2:3Þ

x€ þ 2zvnx_ þv2

n x ¼ a cos vt ¼ uðtÞ ð2:4Þ

where uðtÞ is the modified excitation. Also,

vn ¼ undamped natural frequency

z ¼ damping ratio

The total response is given by

x ¼ xh þ xp ð2:5Þ

Spring

Viscous Damper

Mass

FIGURE 2.1 A forced simple oscillator.

2-2 Vibration and Shock Handbook

with

xh ¼ C1 el1 þ C2e l2 t ð2:6Þ

The particular solution xp; by definition, is one solution that satisfies Equation 2.4. It should be

intuitively clear that this will be of the form

xp ¼ a1 cos vt þ a2 sin vt {except for the case: z ¼ 0 and v ¼ vn} ð2:7Þ

where the constants a1 and a2 are determined by substituting Equation 2.7 into the system Equation 2.4

and equating the like coefficient. This is known as the method of undetermined coefficients.

We will consider several important cases.

2.2.1 Response Characteristics

Case 1: Undamped oscillator with excitation frequency 􀀐 natural frequency

We have

x€ þv2

nx ¼ a cos vt with v – vn ð2:8Þ

Homogeneous solution:

xh ¼ A1 cos vnt þ A2 sin vnt ð2:9Þ

Particular solution:

xp ¼

ðvn

2 2 v2Þ

cos vt ð2:10Þ

It can be easily verified that xp given by Equation 2.10 satisfies the forced system Equation 2.4, with

z ¼ 0: Hence, it is a particular solution.

Complete solution:

|xffl¼fflfflAfflffl1fflcfflofflsfflfflvffl{nzt fflþfflfflAffl2fflfflsifflnfflfflvfflffln}t

Satisfies the homogeneous equation

þ a

ðv2

n 2 v2Þ

cos vt

|fflfflfflfflfflffl{zfflfflfflfflfflffl}

Satisfies the equation with input

ð2:11Þ

Now A1 and A2 are determined using the initial conditions:

xð0Þ ¼ x0 and x_ð0Þ ¼ v0 ð2:12Þ

Specifically, we obtain

x0 ¼ A1 þ

n 2 v2 ð2:13aÞ

v0 ¼ A2vn ð2:13bÞ

Frequency-Domain Analysis 2-3

Hence, the complete response is

x ¼ x0 2 a

ðv2

n 2 v2Þ

􀀒 􀀓

cos vnt þ v0

sin vnt

|fflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflffl{zfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflffl}

Homogeneous solution

þ a

n 2 v2 cos vt

|fflfflfflfflfflffl{zfflfflfflfflfflffl}

Particular solution

ð2:14aÞ

¼ x0 cos vnt þ v0

sin vnt

|fflfflfflfflfflfflfflfflffl{zfflfflfflfflfflfflfflfflffl}

Free response

ðdepends only on initial conditionsÞ comes from xh

pSinusodal at vn

þ a

ðv2

n 2v2Þ ½|cfflofflsfflfflvffltfflffl2{zcfflofflsfflfflvfflnfflfflt}􀀉

2 sin ðvn þvÞ 2 t sin ðvn 2vÞ 2 t

|fflfflfflfflfflfflfflfflfflfflfflfflffl{zfflfflfflfflfflfflfflfflfflfflfflfflffl}

Forced response ðdepends on inputÞ comes from both xh and xp

pWill exhibit a beat phenomenon for vn 2v; i:e:;

ðvn þvÞ=2 wave modulated by ðvn 2vÞ=2 wave

ð2:14bÞ

This is a stable response in the sense of bounded-input bounded-output (BIBO) stability, as it is

bounded and does not increase steadily.

Note: If there is no forcing excitation, the homogeneous solution H and the free response X will be

identical. With a forcing input, the natural response (the homogeneous solution) will be influenced by it

in general, as is clear from Equation 2.14b.

Case 2: Undamped oscillator with v 5 vn (resonant condition)

In this case, the xp that was used before is no longer valid. This is the degenerate case, because otherwise

the particular solution cannot be distinguished from the homogeneous solution and the former will be

completely absorbed into the latter. Instead, in view of the “double-integration” nature of the forced

system equation when v ¼ vn; we use the particular solution ðPÞ:

xp ¼

sin vt ð2:15Þ

This choice of particular solution is strictly justified by the fact that it satisfies the forced system

equation.

Complete solution:

x ¼ A1 cos vt þ A2 sin vt þ

sin vt ð2:16Þ

Initial conditions:

xð0Þ ¼ x0 and x_ð0Þ ¼ v0:

We obtain

x0 ¼ A1 ð2:17aÞ

v0 ¼ vA2 ð2:17bÞ

The total response is

x ¼ x0 cos vt þ v0

v sin vt |fflfflfflfflfflfflfflfflfflffl{zfflfflfflfflfflfflfflfflfflffl}

Free response ðdepends on initial conditionsÞ pSinusodal at v

þ at

2v sin vt

|fflfflffl{zfflfflffl}

Forced response ðdepends on inputÞ pAmplitude increases linearly

ð2:18Þ

Since the forced response increases steadily, this is an unstable response in the BIBO sense.

Furthermore, the homogeneous solution H and the free response X are identical, and the particular

solution P is identical to the forced response F in this case.

2-4 Vibration and Shock Handbook

Note that the same system (undamped oscillator) gives a bounded response for some excitations, while

producing an unstable (steady linear increase) response when the excitation frequency is equal to its

natural frequency. Hence, the system is not quite unstable, but is not quite stable either. In fact, the

undamped oscillator is said to be marginally stable. When the excitation frequency is equal to the natural

frequency it is reasonable for the system to respond in a complementary and steadily increasing manner

because this corresponds to the most “receptive” excitation. Specifically, in this case, the excitation

complements the natural response of the system. In other words, the system is “in resonance” with the

excitation, and the condition is called a resonance. We will address this aspect for the more general case of

a damped oscillator, in the sequel.

Figure 2.2 shows typical forced responses of an undamped oscillator when there is a large difference

between the excitation and the natural frequencies (Case 1); a small difference between the excitation and

the natural frequencies when a beat phenomenon is clearly manifested (also Case 1); and for the resonant

case (Case 2).

Case 3: Damped oscillator

The equation of forced motion is

x€ þ 2zvnx_ þv2

nx ¼ a cos vt ð2:19Þ

Particular Solution (Method 1):

Since derivatives of both odd order and even order are present in this equation, the

particular solution should have terms corresponding to odd and even derivatives of the

(a)

(b)

(c)

Time t

FIGURE 2.2 Forced response of a harmonically excited undamped simple oscillator: (a) for a large frequency

difference; (b) for a small frequency difference (beat phenomenon); (c) response at resonance.

Frequency-Domain Analysis 2-5

forcing function (i.e., sin vt and cos vt). Hence, the appropriate particular solution will be of the

form:

xp ¼ a1 cos vt þ a2 sin vt ð2:20Þ

Substitute Equation 2.20 into Equation 2.19. We obtain

2v2a1 cos vt 2 v2a2 sin vt þ 2zvn½2va1 sin vt þ va2 cos vt􀀉 þ v2

n½a1 cos vt þ a2 sin vt􀀉 ¼ a cos vt

Equate like coefficients:

2v2a1 þ 2zvnva2 þ v2

na1 ¼ a

2v2a2 2 2zvnva1 þ v2

na2 ¼ 0

Hence, we have

ðv2

n 2 v2Þa1 þ 2zvnva2 ¼ a ð2:21aÞ

22zvnva1 þ ðv2

n 2 v2Þa2 ¼ 0 ð2:21bÞ

This can be written in the matrix – vector form:

ðv2

n 2 v2Þ 2zvnv

22zvnv ðv2

n 2 v2Þ

" #

ð2:21cÞ

The solution is

" #

ðv2

n 2 v2Þ 22zvnv

2zvnv ðv2

n 2 v2Þ

" #

ð2:22Þ

with the determinant

D ¼ ðv2

n 2 v2Þ2 þ ð2zvnvÞ2 ð2:23Þ

On simplification, we obtain

a1 ¼ ðv2

n 2 v2Þ

a ð2:24aÞ

a2 ¼

2zvnv

a ð2:24bÞ

This is the method of undetermined coefficients.

Particular Solution (Method 2): Complex Function Method

Consider

x€ þ 2zvnx_ þv2

nx ¼ a e jvt ð2:25Þ

where the excitation is complex. (Note: e jvt ¼ cos vt þ j sin vt.)

The resulting complex particular solution is

xp ¼ XðjvÞe jvt ð2:26Þ

Note that we should take the real part of this solution as the true particular solution.

First substitute Equation 2.26 into Equation 2.25:

􀀑

2 v2 þ 2zvnjv þ v2

􀀜

e jvt ¼ ae jvt

2-6 Vibration and Shock Handbook

Hence, since ejvt – 0 in general,

X ¼

½2v2 þ 2zvnjv þ v2

n􀀉 ð2:27Þ

The characteristic polynomial of the system is

DðlÞ ¼ l2 þ 2zvnl þ v2

n ð2:28aÞ

or, with the Laplace variable s;

DðsÞ ¼ s2 þ 2zvns þ v2

n ð2:28bÞ

If we set s ¼ jv; we have

DðjvÞ ¼ 2v2 þ 2zvnjv þ v2

n ð2:28cÞ

Note that Equation 2.28c is the denominator of Equation 2.27. Hence, Equation 2.27 can be written

X ¼

DðjvÞ ð2:29Þ

It follows from Equation 2.26 that the complex particular solution is

xp ¼

DðjvÞ

e jvt ð2:30Þ

Next let

DðjvÞ ¼ lDle jf ð2:31Þ

Then, by substituting Equation 2.31 in Equation 2.30, we obtain

xp ¼

lDl e jðvt2fÞ ð2:32Þ

where it is clear from Equation 2.31 that

lDl ¼ magnitude of D ðjvÞ

f ¼ phase angle of DðjvÞ

The actual real particular solution is the real part of Equation 2.32 and is given by

xp ¼

lDl cosðvt 2 fÞ ð2:33Þ

It can easily be verified that this result is identical to what was obtained previously by Method 1, as

given by Equation 2.20 together with Equations 2.23 and 2.24.

In passing, we will note here that the frequency-domain transfer function (i.e., response and excitation

in the frequency domain) of the system Equation 2.19 is:

GðjvÞ ¼

D ¼

s2 þ 2zvns þ v2

􀀒 􀀓

s¼jv ð2:34Þ

This frequency transfer function (also known as the frequency-response function) is obtained

from the Laplace transfer function GðsÞ by setting s ¼ jv: We will discuss this aspect in more

detail later.

The particular section ðPÞ is equal to the steady-state solution, because the homogeneous solution dies

out due to damping.

The particular solution (Equation 2.33) has the following characteristics:

1. The frequency is the same as the excitation frequency v:

2. The amplitude is amplified by the magnitude 1

􀀋

lDl ¼ lGðjvÞl:

Frequency-Domain Analysis 2-7

3. The response is “lagged” by the phase angle f of D (or “led” by the phase angle of GðjvÞ; denoted

by ,GðjvÞÞ:

4. Since the homogeneous solution of a stable system decays to zero, the particular solution is also

the steady-state solution.

Resonance

The amplification

􀀄

lGðjvÞl ¼ 1

􀀋

lDl

􀀅

is maximum (i.e., resonance) when lDl is a minimum or lDl2 is a

minimum. As noted earlier, this condition of peak amplification of a system when excited by a sinusoidal

input is called resonance and the associated frequency of excitation is called resonant frequency. We can

determine the resonance of the system (Equation 2.19) as follows:

Equation 2.28c is D ¼ v2

n 2 v2 þ 2zvnv j

Hence,

lDl2 ¼ ðv2

n 2 v2Þ2 þ ð2zvnvÞ2 ¼ D ð2:35Þ

The resonance corresponds to a minimum value of D; or

dv ¼ 2ðv2

n 2 v2Þð22vÞ þ 2ð2zvnÞ2v ¼ 0 For a minimum: ð2:36Þ

Hence, with straightforward algebra, the required condition for resonance is

2v2

n þ v2 þ 2z2v2

n ¼ 0

v2 ¼ ð1 2 2z2Þv2

v ¼

ffiffiffiffiffiffiffiffiffiffi

1 2 2z2

This is the resonant frequency, and is denoted as

vr ¼

ffiffiffiffiffiffiffiffiffiffi

1 2 2z2

vn ð2:37Þ

Note that vr # vd # vn, where vd is the damped natural frequency given by vd ¼

ffiffiffiffiffiffiffiffi

1 2 z2

vn: These

three frequencies (resonant frequency, damped natural frequency, and undamped natural frequency) are

almost equal for small z (i.e., for light damping).

The magnitude and the phase angle plots of GðjvÞ are shown in Figure 2.3. These curves correspond to

the amplification and the phase change of the particular response (the steady-state response) with respect

to the excitation input. This pair, the magnitude and phase angle plots of a transfer function with respect

to frequency, is termed a Bode plot. Usually, logarithmic scales are used for both magnitude (e.g.,

decibels) and frequency (e.g., decades). In summary, the steady-state response of a linear system to a

sinusoidal excitation is completely determined by the frequency transfer function of the system. The total

response is determined by adding H to P and substituting initial conditions, as usual.

For an undamped oscillator ðz ¼ 0Þ; we notice from Equation 2.34 that the magnitude of GðjvÞ

becomes infinity when the excitation frequency is equal to the natural frequency ðvnÞ of the oscillator.

This frequency ðvnÞ is clearly the resonant frequency (as well as natural frequency) of the oscillator. This

fact has been further supported by the nature of the corresponding time response (see Equation 2.18 and

Figure 2.2(c)), which grows (linearly) with time.

2.2.2 Measurement of Damping Ratio (Q-Factor Method)

The frequency transfer function of a simple oscillator (Equation 2.19) may be used to determine the

damping ratio. This frequency-domain method is also termed the half-power point method, for reasons

that should become clear from the following development.

2-8 Vibration and Shock Handbook

First we assume that z , 1=

ffiffi

2 p : Strictly speaking, we should assume that z , 1=2

ffiffi

2 p :

Without loss of generality, consider the normalized (or, non-dimensionalized) transfer function

GðjvÞ ¼

s2 þ 2zvns þ v2

" #

s¼jv ¼

n 2 v2 þ 2zvnvj ð2:38Þ

As noted before, the transfer function GðsÞ; where s is the Laplace variable, can be converted into the

corresponding frequency transfer function simply by setting s ¼ jv: Its value at the undamped natural

frequency is

GðjvÞlv¼vn ¼

2zj ð2:39aÞ

Hence, the magnitude of GðjvÞ (amplification) at v ¼ vn is

lGðjvÞlv¼vn ¼

2z ð2:39bÞ

For small z we have vr ø vn: Hence, 1=2z is approximately the peak magnitude at resonance (the

resonant peak). The actual peak is slightly larger.

It is clear from Equation 2.39a that the phase angle of GðjvÞ at v ¼ vn is 2p=2:

When power is half of the peak power value (e.g., because the displacement squared is proportional to

potential energy), then the velocity squared is proportional to kinetic energy, and power is the rate of

Frequency w

wr wn

Phase Lead

(−f)

Amplification

−180

−90

G D

FIGURE 2.3 Magnitude and phase angle curves of a simple oscillator (A Bode plot).

Frequency-Domain Analysis 2-9

change of energy. Therefore, when the amplification is 1=

ffiffi

2 p of the peak value we have half-power points,

given by:

1ffiffi

2 p

2z ¼

n 2 v2 þ 2jzvnv

􀀈 􀀈 􀀈 􀀈 􀀈

1 2

􀀏 􀀐2

þ 2jz

􀀏 􀀐

􀀈 􀀈 􀀈 􀀈 􀀈 􀀈 􀀈 􀀈 􀀈

ð2:40Þ

Square Equation 2.40:

2 £ 4z 2 ¼

1 2

􀀏 􀀐2 􀀒 􀀓2

þ 4z 2 v

􀀏 􀀐2

Hence,

􀀏 􀀐4

2 2

􀀏 􀀐2

þ 1 þ 4z 2 v

􀀏 􀀐2

¼ 8z 2

􀀏 􀀐4

2 2ð1 2 2z 2Þ

􀀏 􀀐2

þ ð1 2 8z 2 |fflfflffl{zfflfflffl}Þ

¼ 0 ð2:41Þ

Now assume that z2 , 1=8 or z , 1=2

ffiffi

2 p : Otherwise, we will not obtain two positive roots for

ðv=vnÞ2. Solve for ðv=vnÞ2, which will give two roots v21

and v22

for v2: Next, assume v22

. v21

Compare ðv2 2 v21

Þðv2 2 v22

Þ ¼ 0 with Equation 2.41.

Sum of roots:

v22

þ v21

n ¼ 2ð1 2 2z 2Þ ð2:42Þ

Product of roots:

2v21

n ¼ ð1 2 8z 2Þ ð2:43Þ

Hence,

v2 2 v1

􀀏 􀀐2

v22

þv21

2 2v 2v1

􀁻 !

¼ 2ð1 2 2z 2Þ 2 2

ffiffiffiffiffiffiffiffiffiffi

1 2 8z 2

¼ 2 2 4z2 2 2 1 2

2 £ 8z2 þ Oðz 4Þ

􀀒 􀀓

ø 2 2 4z2 2 2 þ 8z2 {because Oðz 4Þ ! 0 for small z }

ø 4z 2

v2 2 v1

ø 2z

Hence, the damping ratio

z ø ðv2 2 v1Þ

2vn ¼

2vn

ø v2 2 v1

v2 þv1 ð2:44Þ

It follows that, once the magnitude of the frequency-response function GðjvÞ is experimentally

determined, the damping ratio can be estimated from Equation 2.44, as illustrated in Figure 2.4.

2-10 Vibration and Shock Handbook

The Q-factor, which measures the sharpness of resonant peak, is defined by

Q-factor ¼

Dv ¼

2z ð2:45Þ

The term originated from the field of electrical tuning circuits where a sharp resonant peak is a

desirable thing (quality factor). Some useful results on the frequency-response of a simple oscillator are

summarized in Box 2.1.

Example 2.1

A dynamic model of a fluid coupling system is shown in Figure 2.5. The fluid coupler is represented by a

rotatory viscous damper with damping constant b: It is connected to a rotatory load of moment of

inertia J; restrained by a torsional spring of stiffness k; as shown. We now obtain the frequency transfer

function of the system that relates the restraining torque t of the spring to the angular displacement

excitation aðtÞ that is applied at the free end of the fluid coupler. If aðtÞ ¼ a0 sin vt; what is the

magnitude (i.e., amplitude) of t at steady state?

Solution

Newton’s Second Law gives

Ju€ ¼ bða_ 2u_Þ 2 ku

Hence,

Ju€þ bu_ þ ku_ ¼ ba_ ðiÞ

Motion transfer function is

a ¼

Js2 þ bs þ k ðiiÞ

Note that the frequency transfer function is obtained simply by setting s ¼ jv:

The restraining torque of the spring is t ¼ ku: Hence,

a ¼

kbs

Js2 þ bs þ k ¼ GðsÞ ðiiiÞ

Frequency w

Amplification

w1wnw2

Δw

FIGURE 2.4 The Q-factor method of damping measurement.

Frequency-Domain Analysis 2-11

Then, the corresponding frequency-response function (frequency transfer function) is

GðjvÞ ¼

kbjv

ðk 2 Jv2Þ þ bjv ðivÞ

For a harmonic excitation of

a ¼ a0 e jvt ðvÞ

Box 2.1

HARMONIC RESPONSE OF A SIMPLE

OSCILLATOR

Undamped Oscillator:

x€ þv2

nx ¼ a cos vt; xð0Þ ¼ x0; x_ð0Þ ¼ v0

x ¼ x0 cos vnt þ v0

sin vnt

|fflfflfflfflfflfflfflfflffl{zfflfflfflfflfflfflfflfflffl}

þ a

n 2 v2 ½cos vt 2 cos vnt􀀉 |fflfflfflfflfflfflfflfflfflfflffl{zfflfflfflfflfflfflfflfflfflfflffl}

for v – vn

x ¼ Same X þ

sin vt for v ¼ vn

ðresonanceÞ

Damped Oscillator:

x€ þ 2zvnx_ þv2

nx ¼ a cos vt

x ¼ H þ

lv2

n 2 v2 þ 2jzvnvl cosðvt 2 fÞ

|fflfflfflfflfflfflfflfflfflfflfflfflfflffl{zfflfflfflfflfflfflfflfflfflfflfflfflfflffl}

where P

tan f ¼

2zvnv

n 2 v2 ; f ¼ phase lag:

Particular solution P is also the steady-state response.

Homogeneous solution H ¼ A1 el1 t þ A2 el2 t

where l1 and l2 are roots of l2 þ 2zvnl þ v2

n ¼ 0

ðcharacteristic equationÞ

A1 and A2 are determined from initial conditions: xð0Þ ¼ x0; x_ð0Þ ¼ v0

Resonant Frequency: vr ¼

ffiffiffiffiffiffiffiffiffiffi

1 2 2z2

The magnitude of P will peak at resonance.

Damping Ratio:

z ¼

2vn ¼

v2 2 v1

v2 þ v1

for low damping

where, Dv ¼ half-power bandwidth ¼ v2 2 v1

Note: Q-factor ¼

Dv ¼

for low damping.

2-12 Vibration and Shock Handbook

we have

t ¼ lGðjvÞla0 eðjvtþfÞ ðviÞ

at steady state.

Here, the phase lead of t with respect to a is

f ¼ /GðjvÞ ¼ /jv 2 /ðk 2 Jv2 þ bjvÞ

2 tan21 bv

ðk 2 Jv2Þ ðviiÞ

The magnitude of the restraining torque, at

steady state, is

t0 ¼ a0lGðjvÞl ¼ a0

kbv

lk 2 Jv2 þ bjvl

a0kbv ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi

ðk 2 Jv2Þ2 þ b2v2

Hence,

t0 ¼

Ja0v2

ffiffiffiffiffiffiffiffiffiffinffivffiffiffi£ffiffiffi2ffiffizffivffiffinffiffiffiffiffiffiffi

ðv2

n 2 v2Þ2 þ ð2zvnvÞ2

p ðviiiÞ

t0 ¼

a0k2zvnv ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi

ðv2

n 2 v2Þ2 þ ð2zvnvÞ2

p ðixÞ

with k=J ¼ v2

n; b=J ¼ 2zvn; and vn equal to the undamped natural frequency of the load.

Now, define the normalized frequency

r ¼

vn ðxÞ

Then, from (ix) we have

t0 ¼

2ka0zr ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi

ð1 2 r2Þ2 þ ð2zrÞ2

p ðxiÞ

For r ¼ 1:

t0 ¼

2ka0z

2z ¼ ka0 ðxiiÞ

This means, at resonance, the applied twist is directly transmitted to the load spring.

For small r:

t0 ¼

2ka0zr

1 ðxiiiÞ

which is small, and becomes zero at r ¼ 0: Hence, at low frequencies, the transmitted torque is small.

For larger r:

t0 ¼

2ka0zr

r2 ¼

2ka0z

r ðxivÞ

which is small and goes to zero. Hence, at high frequencies as well, the transmitted torque is small.

The variation of t0 with the frequency ratio r is sketched in Figure 2.6.

Restraining

Spring

Fluid

Coupling

Load

Inertia

a(t) q

FIGURE 2.5 A fluid coupling system.

Frequency-Domain Analysis 2-13

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