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Module: 5Lecture: 1

Units

Force N (kg m / s 2)Displacement x (m)Acceleration x (m / s 2)Mass m (kg)Damping coefficient C (N-s / m)Velocity x (m / s)Linear Stiffness k (N / m → kg -m / s 2 / m → kg /s 2)

Angular Velocity 1/sMoment of Force N-mAcceleration due to gravity, g = 9.81 m / s 2

F

k displacement when force F pulls the spring

k f

1y)flexibilit(

mk

n natural frequency of a single degree freedom system

Characteristic of Harmonic Motion:

)(sinsin

)(sin

cos

sin

2

2

2

t At A x

t A

t A x

t A x

m

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Response diagram of displacement (x), velocity ( ) and acceleration ( )

Forced Vibration of a Single Degree of Freedom System

- Excitation force becomes periodic (reciprocating engine).- Harmonic or periodic forces or rotating or reciprocating parts is exciting in nature.- Exciting force can also be shock / transient force due to landing of a/c, firing of missile,

ejecting of external stores, launch of space vehicle etc.- Sustained non-periodic random aerodynamic forces experienced by an a/c due to

turbulence of the atmosphere.

Response of First - Order Systems to Harmonic Excitation Frequency ResponseThe differential equation for a first – order system in the form of a damper – spring system isshown below

cx t k x t F t (1)where c and k are damping coefficient and stiffness. The homogeneous solution of Eq. (1), can

be obtained by letting 0F t . In this section we focus our attention on the particular solution,

forceexcitation)( t F

m

t

x x

x

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which represents the response to external forces. First, we consider the simplest case, namely, theresponse to harmonic excitation. To this end, it is convenient to let the force F t have the form

cosF t kf t kA t (2)

Where is the excitation frequency, sometimes referred to as the driving frequency, Note that f t and A have units of displacement. The reason for writing the excitation in the form (2) is

so as to permit expressing the response in terms of a non-dimensional ratio, as we shall seeshortly. No dimensional ratios often enhance the usefulness of a solution by extending itsapplicability to a large variety of cases. Inserting Eq. (2) into Eq. (1) and dividing through by c ,we obtain

cos x t ax t Aa t (3)Where

1k ac

(4)

in which is the time constant.The solution of the homogeneous differential equation, obtained by letting 0 A in Eq. (3) ,decays exponentially with time, for which reason it is called the transient solution. On the otherhand, the particular solution does not vanish as time and is known as the steady – state solutionto the harmonic excitation in question. By virtue of the fact that the system is linear, the principleof superposition holds, so that the homogeneous solution and the particular solution can beobtained separately and then combined linearly to obtain the complete solution.Because the excitation force is harmonic, it can be verified easily that the steady – state responseis also harmonic and has same frequency . Moreover, because Eq. (3) involves the function

x t and its first derivative x t , the response must contain not only cos t but also sin t .Hence, let us assume that the steady – state solution of Eq. (3) has the form

1 2sin cos x t C t C t (5)Where 1C and 2C are constants yet to be determined. Inserting solution (5) into Eq. (3), weobtain

1 2 1 2cos sin sin cos cosC t C t a C t C t Aa t (6)Equation (6) can be satisfied only if the coefficients of sin t on the one hand and thecoefficients of cos t on the other hand are the same on both sides of the equation. This, in turn,requires the satisfaction of the equations

1 2

1 2

0aC C C aC Aa

(7)

which represent two algebraic equations in the unknowns 1C and 2C . Their solution is2

1 22 2 2 2

Aa AaC C

a a

(8)

Introducing Eqs. (8) into Eq. (5), we obtain the steady – state solution

2 2 sin cos Aa

x t t a t a

(9)

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Solution (9) can be expressed in a more convenient form. To this end, let us introduce thenotation

1 1

2 2 2 22 2sin cosa

a a

(10)

Then, Eq. (9) can be written as cos x t X t (11)

where

1 221

A X

a

(12)

Is the amplitude and

1

tan a

(13)is the phase angle. Both X and are functions of the excitation frequency .The response to harmonic excitation can be obtained more conveniently by using complex vectorrepresentation of the excitation and the response. From Sec. 1.6, we recall that

cos sini t e t i t (14)Where 1i so that Eq. (2) can be rewritten as

cos Re i t F t kf t kA t kAe (15a)Where Re denotes the real part of the function. Similarly in the case of sinusoidal excitation wecan write

sin Im i t F t kf t kA t kAe (15b)

Where Im denotes the imaginary part of the function. Hence, we can rewrite Eq. (3) in the form i t x t ax t aAe (16)

Then, if the excitation is given by Eq. (15a), we retain the real part of the response and if theexcitation is given by Eq. (15b), we retain the imaginary part of the response.Concentrating once again on the steady – state response, we write the solution of Eq. (16) in theform

i t x t X i e (17)Inserting Eq. (17) into Eq. (16)

we obtain i t i t Z i X i e aAe (18)Where

Z i a i (19)Is the impedance function for this first – order system. Dividing Eq. (18) through by i t e andsolving for X i , we obtain

1

aA aA A X i

Z i a i i

(20)

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Where 1 a c k is the time constant. It will prove convenient to introduce the nondimensional ratio

21 1

1 1 X i iG i

A i

(21)

Where G i is known as the frequency response. Inserting Eq. (21) into Eq. (17) , we canwrite the harmonic response in the general form

i t x t AG i e (22)But, the frequency response G i , as any complex function, can be expressed as

iG i G i e (23)where G i is the magnitude and is the phase angle of G i . Introducing Eq. (23) into

Eq. (22) , we obtain i t x t A G i e (24)

So that if the excitation is in the form of Eq. (15a) , the response is the real part of Eq.(24) , or

cos x t A G i t (25a)and if the excitation is in the form of Eq. (15b), the response is the imaginery part of Eq. (24), or

sin x t A G i t (25b)From Eqs. (25) it follows that, if the excitation is harmonic with the frequency , the response isalso harmonic and has the same frequency. Hence, in studying the nature of the response,

plotting the response as a function of time will not be very rewarding. Considerably more insight

into the system behaviour can be gained by examining how the system responds as the drivingfrequency varies. In particular plots of the magnitude G i and of the phase angle versus the frequency are very revealing. From complex algebra, if we consider Eq. (21) , thenwe can write

1 22 21

2 2

1Re Im

1G i G i G i

(26)

And we note from Eq. (12) that G i X A . The plot G i versus is shown inFig. 1. We observe from Fig. 1 that for small driving frequencies the magnitude G i is closeto 1 and for high frequencies the magnitude approaches 0. Hence, the system permits low –frequency harmonics to go through undistorted, but it attenuates greatly high – frequencyharmonics. For this reason a first – order system is known as a low-pass filter. To obtain the

phase angle, we recall first that cos sinie i . Then using Eqs. (21) and (23), we can write

1 1Imtan tanRe

G i

G i

(27)

which checks with Eq. (13). The plot versus is shown in Fig. 2. The plots G i versus and versus are known as frequency – response plots.

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Figure 1

Figure 2

The magnitude G i of the frequency response can be interpreted geometrically by observingfrom Eq. (24) that the magnitude of the force in the spring is

sF t k x t kA G i (28)Moreover, from Eqs. (15), the magnitude of the harmonic excitation is

F t kA (29)Hence, combining Eqs. (28) n and (29) , we can write

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sF t

G iF t

(30)

Or, the magnitude of the frequency response is equal to ratio of the magnitude of the springforce sF t to the magnitude of the excitation force F t .