5. Vibration under General Forcing Conditions

1. IntroductionIf the excitation is periodic but not harmonic, it can be replaced by a sum of harmonic functions

using the harmonic analysis procedure as discussed. By the principle of superposition, the response of the system can then be determined by superposing the responses due to individual harmonic forcing functions. On the other hand, if the system is subjected to a suddenly applied nonperiodic force, the response will be transient, since steady-state vibrations are not usually produced. The transient response of a system can be found using what is known as the convolution integral.

2. Response Under a General Periodic ForceWhen the external force F(t) is periodic with period = 2/, it can be expanded in a Fourier

series:

F ( t )=a02

+∑j=1

∞a jcos jwt+∑

j=1

∞b j sin jwt

(1)where

a j=2τ∫0

τF (t ) cos jωt dt

, j = 0, 1, 2 . . . (2)and

b j=2τ∫0

τF (t ) sin jωt dt

, j = 1, 2 . . . (3)The equation of motion of the system can be expressed as

m x+c x+kx=F ( t )=a02

+∑j=1

∞a j cos jwt+∑

j=1

∞b j sin jwt

(4)The right-hand side of this equation is a constant plus a sum of harmonic functions. Using the principle of superposition, the steady-state solution of Eq. (4) is the sum of the steady-state solution of the following equations:

m x+c x+kx=F ( t )=a02 (5)

m x+c x+kx=F (t )=a jcos jωt (6)m x+c x+kx=F ( t )=b j sin jωt (7)

Noting that the solution of Eq. (5) is given by

x p ( t )=a02k (8)

and we can express the solutions of Eqs. (6) and (7), respectively, as

x p (t )=(a j /k )

√ (1− j2r2)2+ (2 ς jr )2cos ( jωt−φ j )

(9)

x p (t )=(b j /k )

√ (1− j2r2)2+ (2 ς jr )2sin ( jωt−φ j )

(10)

1

where

φ j=tan−1( 2 ς jr

1− j2r2 ) (11)and

r= ωωn (12)

Thus the complete steady-state solution of Eq. (4) is given by

x p ( t )=a02k

+∑j=1

∞ (a j/k )

√(1− j2r2 )2+(2 ς jr )2cos ( jωt−φ j )+∑

j=1

∞ (b j/ k )

√(1− j2r 2)2+ (2ς jr )2sin ( jωt−φ j )

(13)

It can be seen from the solution, Eq. (13), that the amplitude and phase shift corresponding to the jth term depend on j. If j = n, for any j, the amplitude of the corresponding harmonic will be comparatively large. This will be particularly true for small values of j and. Further, as j becomes larger, the amplitude becomes smaller and the corresponding terms tend to zero. Thus the first few terms are usually sufficient to obtain the response with reasonable accuracy.

The solution given by Eq. (13) denotes the steady-state response of the system. The transient part of the solution arising from the initial conditions can also be included to find the complete solution. To find the complete solution, we need to evaluate the arbitrary constants by setting the value of the complete solution and its derivative to the specified values of initial displacement x(0) and the initial velocity x(0). This results in a complicated expression for the transient part of the total solution.

Example No. 1 – Periodic Vibration of Hydraulic ValveIn the study of vibrations of valves used in hydraulic control systems, the valve and its elastic stem are modeled as a damped spring-mass systems as shown in Fig. 1(a). In addition to the spring force and damping force, there is a fluid pressure force on the valve that changes with the amount of opening or closing of the valve. Find the steady-state response of the valve when the pressure in the chamber varies as indicated in Fig. 1(b). Assume k = 2500 N/m, c = 10 N-s/m., and m = 0.25 kg.

2

Given: Hydraulic control valve with m = 0.25, k = 2500 N/m. and c = 10 N-s/m and pressure on the valve as given in Fig. 1(b).Required: Steady-state response of the valve, xp(t).Solution:Find the Fourier series expansion of the force acting on the valve. Add the responses due to individual harmonic force components.

The valve can be considered as a mass connected to a spring and a damper on one side and subjected to a forcing function F(t) on the other side. The forcing function;

F ( t )=Ap ( t ) (a)where A is the cross sectional area of the chamber, given by

A=π (50 )2

4=625 π mm2=0 .000625 m2

(b)and p(t) is the pressure acting on the valve at any instant t. Since p(t) is periodic with period = 2 seconds and A is a constant. F(t) is also a periodic function of period = 2 seconds. The frequency of the forcing function is = (2/) = rad/sec.F(t) can be expressed in a Fourier series as:

3

F (t )=a02

+a1cosωt+a2cos 2ωt+⋯+b1sinωt+b2sin 2ωt+⋯(c)

where ai and bi are given by Eqs. (2) and (3). Since the function F(t) is given by

F ( t )={50000 At for 0≤t≤ τ2

50000 A (2−t ) forτ2≤t≤τ

(d)the Fourier coefficients ai and bi can be computed with the help of Eqs (2) and (3):

a0=22 [∫01 50000 Atdt+∫1

250000 A (2−t ) dt ]=50000 A

(e)

a1=22 [∫01 50000 At cos π tdt+∫1

250000 A (2−t ) cos π tdt ]=−2×10

5 Aπ2 (f)

b1=22 [∫01 50000 At sin π tdt+∫1

250000 A (2−t )sin π tdt ]=0

(g)

a2=22 [∫01 50000At cos2 π tdt+∫1

250000A (2−t ) cos2π tdt ]=0

(h)

b2=22 [∫01 50000 At sin 2π tdt+∫1

250000 A (2−t ) sin 2π tdt ]=0

(i)

a3=22 [∫01 50000 At cos3 π tdt+∫1

250000 A (2−t ) cos3 π tdt ]=−2×10

5 A9 π2 (j)

b3=22 [∫01 50000 At sin 3 π tdt+∫1

250000 A (2−t )sin 3 π tdt ]=0

(k)Likewise, a4 = a6 = . . . = b4 = b5 = b6 = . . . = 0. By considering only the first three harmonics, the forcing function:

F ( t )≃25000 A−2×105 A

π2cosωt−2×10

5A9 π2

cos 3ωt(l)

The steady-state response of the valve to the forcing of Eq. (m) :

x p (t )=25000 Ak

−2×105A / (kπ2)

√(1−r2 )2+(2 ςr )2cos (ωt−φ1)−

2×105 A /(9kπ2 )

√(1−9 r2 )2+(6 ςr )2cos (3ωt−φ3 )

(m)Natural frequency of the valve:

ωn=√ km

=√25000.25=100 rad /sec

(n)Forcing frequency :

ω=2πτ

=2π2

=π rad /sec(o)

Thus, frequency ratio:

4

r= ωωn

= π100

=0 .031416(p)

and damping ratio:

ς= ccc

= c2mωn

=102 (0 .25 ) (100 )

=0 .2(q)

The phase angles 1 and 3 can be computed as follows:

φ1=tan−1( 2ςr1−r2 )=tan−1[ 2 (0 .2 ) (0.031416 )

1−(0 .031416 )2 ]=0 .0125664 rad(r)

and

φ3=tan−1 ( 6 ςr

1−9 r2 )=tan−1[ 6 (0 .2 ) (0 .031416 )1−9 (0.031416 )2 ]=0 .0380483 rad

(s)In view of Eqs. (b) and (n) to (s), the solution can be written as

x p ( t )=0 .019635−0.015930cos (πt−0 .0125664 )−0 .0017828cos (3πt−0 .0380483 ) (t)

3. Response Under a Periodic Force of Irregular FormIn some case, the force acting on a system may be quite irregular and may be determined only

experimentally. Examples of such forces include wing-and earthquake-induced forces. In such cases, the forces will be available in graphical form and no analytic expression can be found to describe F(t). Sometimes, the value of F(t) may be available only at a number of discrete points t1, t2, . . ., tN. In all these cases, it is possible to find the Fourier coefficients by using a numerical integration procedure, If F1, F2, . . . , FN denote the values of F(t) at t1, t2, . . ., tN, respectively, where N denotes an even number of equidistant points in one time period ( = Nt), as shown in Fig. 2, the application of trapezoidal rule gives

a0=2N ∑

j=1

∞Fi

(14)

a j=2N∑j=1

∞F jcos

2 jπt j

τ , j = 1,2, . . . (15)

b j=2N∑j=1

∞F j sin

2 jπt j

τ , j = 1,2, . . . (16)

5

Once the Fourier coefficients a0, ai, and bi are known, the steady-state response of the system can be found using Eq. (13) with

r= 2πτωn

4. Response Under Nonperiodic ForceThe periodic forces of any general wave form can be represented by Fourier series as a

superposition of harmonic components of various frequencies. The response of a linear system is then found by superposing the harmonic response to each of the exciting forces. When the exciting force F(t) is nonperiodic, such as that due to blast from an explosion, a different method of calculating the response is required. Various methods can be used to find the response of the system to an arbitrary excitation. Some of these methods are as follows:a. By representing the excitation by a Fourier integral;b. By using the method of convolution integral;c. By using the method of Laplace transformation;d. By first approximating F(t) by a suitable interpolation model and then using a numerical

procedure; ande. By numerically integrating the equations of motion.

5. Convolution IntegralA nonperiodic exciting force usually has a magnitude that varies with time; it acts for a specified

period of time and then stops. The simplest form of such a force is the impulsive force. An impulsive force is one that has a large magnitude F and acts for a very short period of time t. From dynamics we know that impulse can be measured by finding the change in momentum of the system caused by it. If x1 and x2 denote the velocities of the mass m before and after the application of the impulse, we have

Im pulse=FΔt=m x2−m x1 (17)

By designating the magnitude of the impulse Ft by F, we can write, in general,

6

F−=∫t

t+ΔtFdt

(18)A unit impulse (f) is defined as

f−= lim

Δt→0∫t

t+ΔtFdt=Fdt=1

(19)It can be seen that in order for Fdt to have a finite value, F tends to infinity (since dt tends to zero). Although the unit impulse function has no physical meaning, it is a convenient tool in our present analysis.

5.1 Response to an impulse First consider the response of a single degree of freedom system to an impulse excitation; this case is important in studying the response under more general excitations. Consider a viscously damped spring-mass system subjected to a unit impulse at t = 0, as shown in Fig. 3(a) and (b).

For an underdamped system, the solution of the equation of motionm x+c x+kx=0 (20)

is given as follows:

x (t )=e−ςωn t {x0cosωd t+

x0+ςωn x0ωd

sinωd t}(21)

where

ς= c2mωn (22)

ωd=ωn√1−ς2=√ km

−( c2m )

2

(23)

ωn=√ km (24)

7

If the mass is at rest before the unit impulse is applied (x = x = 0 for t < 0 or at t = 0), we obtain, from the impulse-momentum relation,

Im pulse=f−=1=m x ( t=0 )~m x (t=0− )=m x0

(25)Thus the initial conditions are given by

x (t=0 )=x0=0

x (t=0 )= x0=1m (26)

In view of Eq. (26), Eq. (21) reduces to

x (t )=g (t )= e−ςωn t

mωd

sinωd t(27)

Equation (27) gives the response of a single degree of freedom system to a unit impulse, which is also known as the impulse response function, denoted by g(t). The function g(t), Eq. (27), is shown in Fig. 3(c).

If the magnitude of the impulse is F instead of unity, the initial velocity x0 is F/m and the response of the system becomes

x (t )=F−e−ςωn t

mωd

sinωd t=F−g (t )

(28)If the impulse F is applied at an arbitrary time t = , as shown in Fig. 4(a), it will change the velocity at t = by an amount F/m. Assuming that x = 0 until the impulse is applied, the displacement x at any subsequent time t, caused by a change in the velocity at time t, is given by Eq. (28) with t replaced by the time elapsed after the application of the impulse, that is, t – .

8

Thus we obtainx (t )=F

−g ( t−τ )

(29)This is shown in Fig. 4(b).

5.2 Response to general forcing conditionNow consider the response of the system under an arbitrary external force F(t), shown in Fig. 5

9

This force may be assumed to be made up of a series of impulses of varying magnitude. Assuming that at time , the force F() acts on the system for a short period of time t, the impulse acting at t = is given by F(). At any time t, the elapsed time since the impulse is t – , so the response of the system at t due to this impulse alone is given by Eq. (29) with F = F():

Δx ( t )=F ( τ ) Δτg ( t−τ ) (30)The total response at time t can be found by summing all the responses due to the elementary impulses acting at all times :

x (t )=∑ F (τ ) g ( t−τ ) Δτ (31)Letting → 0 and replacing the summation by integration, we obtain

x (t )=∫0tF ( τ ) g ( t−τ )dτ

(32)By substituting Eq. (27) into Eq. (32), we obtain

x (t )= 1mωd

∫0tF ( τ ) e−ςω n ( t−τ ) sinωd (t−τ ) dτ

(33)which represents the response of an underdamped single degree of freedom system to the arbitrary excitation F(t). Note that Eq. (33) does not consider the effect of initial conditions of the system. The integral in Eq. (32) or Eq. (33) is called the convolution or Duhamel integral. In many cases the function F(t) has a form that permits an explicit integration of Eq. (33). In case such integration is not possible, it can be evaluated numerically without much difficulty.

5.3 Response to base excitation

10

If a spring-mass damper system is subjected to an arbitrary base excitation described by its displacement, velocity, or acceleration, the equation of motion can be expressed in terms of the relative displacement of the mass z = x – y as follows:

m z+c z+kz=−m y (34)This equation is similar to the equation

m x+c x+kx=F (35)with variable z replacing x and the term –m yreplacing the forcing function F. Hence all of the results derived for the force-excited system are applicable to the base-excited system also for z when the form F is replaced by –m y. For an underdamped system subjected to base excitation, the relative displacement can be found from Eq. (33).

Example No. 2 – Compacting Machine Under Linear ForceDetermine the response of the compacting machine shown in Fig. 6(a) when a linearly varying force (shown in Fig. 6(b) is applied due to the motion of the cam

11

Given:Compacting machine, modeled as single degree of freedom system subjected to ramp function.Required: Response of the systemSolution:Evaluate the Duhamel integral with F(t) = F t∙The linearly varying force shown in Fig. 6(b) is known as the ramp function. The forcing function can be represented as F() = F∙, where dF denotes the rate of increase of force F per unit time. By substituting this into Eq. (33), then

x (t )= δFmωd

∫0

tτe

−ςωn ( t−τ )sinωd ( t−τ ) dτ

x (t )= δFmωd

∫0

t(t−τ ) e−ςωn ( t−τ )

sinωd (t−τ ) (−dτ )− δF⋅ymωd

∫0

te−ςωn ( t−τ )

sinωd (t−τ ) (−dτ )

x (t )= δFk {t−2 ςωn

+e−ςωn t [ 2 ςωn

cosωd t−(ωd2−ς2ωn

2

ωn2ωd

)sinωd t ]}(answer)

For an undamped system,

x (t )= δFωnk

(ωn t−sinωn t )

Figure 6(c) shows the response.

6. Response SpectrumThe graph showing the variation of the maximum response (maximum displacement, velocity,

acceleration, or any other quantity) with the natural frequency (or natural period) of a single degree of freedom system to a specified forcing function is known as the response spectrum. Since the maximum response is plotted against the natural frequency (or natural period), the response spectrum gives the maximum response of all possible single degree of freedom systems. The response spectrum is widely used in earthquake engineering design.

Once the response spectrum corresponding to a specified forcing function is available, we need to know just the natural frequency of the system to find its maximum response.

Example No. 3 – Response Spectrum of Sinusoidal PulseFind the undamped response spectrum for the sinusoidal pulse force shown in Fig. 7(a) using the initial conditions x(0) = x(0) = 0.

12

Given:Single degree of freedom undamped system subjected to one-half period of a sinusoidal force.Required:Response spectrumSolution:Find the response and express its maximum value in terms of its natural time period

The equation of motion of an undamped system:

m x+kx=F (t )={F0sinωt , 0≤t≤t00 , t>t0 (a)

where

ω= πt0 (b)

The solution of Eq. (a) can be obtained by superposing the homogeneous solution xh(t) and the particular solution xp(t) as

x (t )=xh (t )+x p (t ) (c)that is,

x (t )=A cosωn t+A sinωn t+( F 0

k−mω2 )sinωt(d)

where A and B are constants and n is the natural frequency of the system:

ωn=2 πτ n

=√ km (e)

13

Using the initial conditions x(0) = x(0) = 0 in Eq. (d), find the constants A and B as

A=0 , B=−

F0ω

ωn (k−mω2) (f)Thus the solution becomes

x (t )=F0 /k

1−(ω /ωn )2 (sinωt− ωωn

sinωn t ), 0≤t≤t 0 (g)

which can be rewritten as

x (t )δ st

= 1

1−( τn

2 t0 )2 (sin πt

t0−

τn

2t0sin2 πtτn

), 0≤t≤t 0 (h) (answer)

where

δ st=F0k (i)

The solution given by Eq. (h) is valid only during the period of force application 0 ≤ t ≤ t0. Since there is no force applied for t > t0, the solution can be expressed as a free vibration solution:

x (t )=A 'cosωn t+B ' sinωn t , t > t0 (j)where the constants A’ and B’ can be found by using the values of x(t = t0) and x(t = t0) given by Eq. (8), as initial conditions for the duration t > t0. This gives

x ( t=t0 )=α(− τn

2 t 0sin2πt0τn

)=A 'cosωn t 0+B' sinωn t 0 (k)

x ( t=t0 )=α( πt 0

− πt 0cos2 πt0τn

)=−ωn A' sinωn t 0+ωn B

'cosωn t 0 (l)

where

α=δst

1−( τn

2 t 0 )2

(m)Equations (k) and (l) can be solved to find A’ and B’ as

A'= απωn t0

sinωn t0 ,

B'=− απωn t0

(1+cosωn t0 )(n)

Equations (n) can be substituted into Eq. (j) to obtain

x (t )δ st

=(τ n/ t0)

2 [1−(τ n/2 t0 )2 ] [sin2 π ( t0τn

− tτn

)−sin 2π ( t0τn)]

t ≥ t0 (o) (answer)

Equations (h) and (o) give the response of the system in nondimensional form, that is x/st is expressed in terms of t/n. Thus any specified value of t0/n, the maximum value of x/st can be found.

14

The maximum value of x/st, when plotted against t0/n, gives the response spectrum shown in Figure 7(b). It can be observed that the maximum value of (x/st) max = 175 occurs at value of (t0/n) = 0.75 (answer).

==In the example, the input force is simple and hence a closed form solution has obtained for the

response spectrum. However, if the input force is arbitrary, we can find the response spectrum only numerically. In such a case, Eq. (33) can be used to express the peak response of an undamped single degree of freedom system due to an arbitrary input force F(t) as

x (t )|max=1

mωn∫0

tF (τ ) sinωn (t−τ ) dτ|max

(36)6.1 Response spectrum for base excitation

In the design of machinery and structures subjected to a ground shock, such as that caused by an earthquake, the response spectrum corresponding to the base excitation is useful. If the base of a damped single degree of freedom system is subject to an acceleration y(t), the equation of motion, in terms of the relative displacement z = x – y, is given by Eq., (34) and the response z(t) by Eq. 36. In the case of a ground shock, the velocity response spectrum is generally used. The displacement and acceleration spectra are then expressed in terms of the velocity spectrum. For a harmonic oscillator (an undamped system under free vibration), we notice that

x|max=−ωn2 x|max and x|max=ωn

2 x|max (37)Thus the acceleration and displacement spectra Sa and Sd can be obtained in terms of the velocity spectrum (Sv):

Sd=Sv

ωn

,Sa=ωn Sv (38)

To consider damping in the system, if we assume that the maximum relative displacement occurs after the shock pulse has passed, the subsequent motion must be harmonic. In such a case, we can use Eq. (38). The fictitious velocity associated with this apparent harmonic motion is called the pseudo velocity and its response spectrum, Sv, is called the pseudo spectrum. The velocity spectra of damped systems are used extensively in earthquake analysis.

To find the relative velocity spectrum, we differentiate Eq. (36) and obtain

z (t )=− 1ωd

∫0

ty (τ ) e−ςωn ( t−τ ) [−ςωd sin ωd (t−τ )+ωd cosωd (t−τ ) ]dτ

(39)Equation (39) can be rewritten as

z (t )= e−ςωn t

√1−ς 2√P2+Q2 sin (ωd t−φ )

(40)where

P=∫0

ty ( t ) eςωn τ

cosωd τdτ (41)

15

Q=∫0

ty ( t ) eςωnτ sinωd τdτ (42)

and

φ=tan−1[−(P√1−ς2+Qς )Pς−Q√1−ς2 ]

(43)The velocity response spectrum, Sv, can be obtained from Eq. (40)

Sv=|z ( t )|max=| e−ςωn t

√1−ς 2√P2+Q2|

max (44)Thus the pseudo response spectra are given by

Sd=|z|max=Sv

ωn

; Sv=|z|max ; Sa=|z|max=ωnSv (45)

Example No. 4 – Water Tank Subjected to Base AccelerationThe water tank, shown in Fig. 8(a), is subjected to a linearly varying ground acceleration as shown in Fig. 8(b) due to an earthquake. The mass of the tank is m, the stiffness of the column is k, and damping is negligible. Find the response spectrum for the relative displacement, z = x – y, of the water tank.

16

Given:Water tank subjected to the base acceleration shown in Fig. 8(b).Required:Response spectrum of relative displacement of the tank.Solution:Model the water tank as an undamped single degree of freedom system. Find the maximum relative displacement of the tank and express it as a function of n.

The base acceleration

y (t )= ymax (1− tt0 ) for 0 ≤ t ≤ 2t0 (a)

y (t )=0 for t > 2t0 (b)

Response during 0 ≤ t ≤ 2t0: By substituting Eq. (a) into Eq. (36), the response for an undamped system:

z (t )=− 1ωn

ymax [∫0t (1− τt0 ) (sinωn t cosωn τ−cosωn t sinωn τ )dτ ]

(c)

z (t )=−ymaxωn2 [1− t

t0−cosωn t+

1ωn t0

sinωn t ](d)

Maximum response zmax :

z (t )=−ymaxt0ωn

2 [−1+ωn t0sinωn t+cosωn t ]=0(e)

This equation gives the time tm at which zmax occurs:

tm=2ωn

tan−1 (ωn t0)(f)

By substituting Eq. (f) into Eq. (d), the maximum response of the tank :

zmax=−ymaxωn2 [1− tm

t0−cosωn tm+ 1

ωn t 0sinωn tm]

(g) (answer)

Response during t > 2t0: Since there is no excitation during this time,

z (t )=z0 cosωn t+( z0ωn

)sinωn t(h)

Provided that we take the initial displacement and initial velocity asz0=z ( t=2t0) and z0= z ( t=2t0) (i)using Eq. (g). The maximum of z(t) given by Eq. (8) can be identified as

17

zmax=[z02+( z0ωn

)2]1/2

(j) (answer)where z0 and z0 are computed as indicated in Eq. (i).

7. Laplace Transformation The Laplace transform method can be used to find the response of a system under any type of excitation, including the harmonic and periodic types. This method can be used for efficient solution of linear differential equations, particularly those with constant coefficients. It permits the conversion of differential equations into algebraic ones, which are easier to manipulate. The major advantages of the method are that it can treat discontinuous functions without any particular difficulty and that it automatically takes into account the initial conditions.The Laplace transform of a function x(t), denoted symbolically asx(s) = Lx(t), is defined as

x (t )=Lx (t )=∫0

∞e−st x (t ) dt

(46)where s is, in general, a complex quantity and is called the subsidiary variable. The function e -st is called the kernel of the transformation. Since the integration is with respect to t, the transformation gives a function of s. In order to solve a vibration problem using the Laplace transform method, the following steps are necessary:a. Write the equation of motion of the system.b. Transform each term of the equation, using known initial conditions.c. Solve for the transformed response of the system.d. Obtain the desired solution (response) by using inverse Laplace transformation.

In order to solve the forced vibration equation

m x+c x+kx=F (t ) (47)by the Laplace transform method, it is necessary to find the transforms of the derivatives

x (t )=dxdt

(t )and

x (t )=d2 xdt 2

( t )

These can be found as follows:

Ldxdt

(t )=∫0∞e−st dx

dt(t ) dt

(48)This can be integrated by parts to obtain

Ldxdt

( t )=e−st x ( t )|0∞+s∫0

∞e−st x ( t ) dt=s x ( s )−x (0 )

(49)where x(0) = x0 is the initial displacement of the mass m. Similarly, the Laplace transform of the second derivative of x(t) can be obtained:

Ld2 xdt 2

( t )=∫0∞e−st d

2 xdt 2

( t ) dt=s2 x ( s)−sx (0 )− x (0 )(50)

18

where x(0) = x0 is the initial velocity of the mass m. Since the Laplace transform of the force F(t) is given by

F ( s )=LF (t )=∫0∞e−st F (t ) dt

(51)we can transform both sides of Eq. (47) and obtain, using Eqs. (46) and (48) to (51)

mL x (t )+cL x (t )+kLx (t )=LF (t )or

(ms2+cs+k ) x ( s )=F ( s )+m x (0 )+ (ms+c ) x (0 ) (52)where the right-hand side of Eq. (52) can be regarded as a generalized transformed excitation.

For the present, we take x(0) and x(0) as zero, which is equivalent to ignoring the homogeneous solution of the differential equation (47). Then the ratio of the transformed excitation to the transformed responseZ(s) can be expressed as

Z ( s)= F (s )x ( s)

=ms2+cs+k(53)

The function Z(s) is known as the generalized impedance of the system. The reciprocal of the function Z(s) is called the admittance or transfer function of the system and is denoted as Y(s):

Y (s )= 1Z ( s)

=x (s )F (s )

= 1

ms2+cs+k= 1

m ( s2+2 ςωns+ωn2) (54)

It can be seen that by letting s = i in Y(s) and multiplying by k, we obtain the complex frequency response H(i) defined in Eq. (54). Equation (54) can also be expressed as

x (s )=Y (s ) F (s ) (55)which indicates that the transfer function can be regarded as an algebraic operator that operates on the transformed force to yield the transformed response.

To find the desired response x(t) from x(s), we have to take the inverse Laplace transform of x(s), which can be defined symbolically as

x (t )=L−1 x (s )=L−1 Y ( s) F ( s) (56)In general, the operator L-1 involves a line integral in the complex domain. Fortunately, we need to evaluate these integrals separately for each problem; such integrations have been carried out for various common forms of the function F(t) and tabulated. In order to find the solution using Eq. (56), we usually look for ways of decomposing x(s) into a combination of simple functions whose inverse transformations are available in Laplace transform tables. We can decompose x(s) conveniently by the method of partial fractions.

In the above discussion, we ignored the homogeneous solution by assuming x(0) and x(0) as zero. We now consider the general solution by taking the initial conditions as x(0) = x0 and x(0) =x0. From Eq. (52), the transformed response x(s) can be obtained:

x (s )= F (s )m (s2+2 ςωn s+ωn

2 )+

s+2 ςωn

s2+2 ςωn s+ωn2x0+

1s2+2 ςωn s+ωn

2x0

(57)We can obtain the inverse transform of x(s) by considering each term on the right side of Eq. (57) separately. We also make use of the following relation:

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L−1 f 1 (s ) f 2 ( s)=∫0tf 1 (τ ) f 2 ( t−τ ) dτ

(58)By considering the first term on the right side of Eq. (57) as f 1(s)f 2(s), where

f 1 ( s)=F (s ) andf 2 ( s)= 1

m (s2+2ςωn s+ωn2)

and by noting that f1(t) = L-1f 1(s) = F(t), we obtain

L−1 f 1 ( s) f 2 (s )= 1mωd

∫0

tF (t ) e−ςωn ( t−τ )

sinωd ( t−τ ) dτ(59)

Consider the second term on the right side of Eq. (57), we find the inverse transform of the coefficient of x0 from the table of Laplace transform.

L−1( s+2 ςωn

s2+2 ςωns+ωn2 )= 1

√1−ς2e−ςωn t sin (ωdt+φ1)

(60)where

φ1=cos−1 (ς ) (61)

Finally, the inverse transform of the coefficient of x 0 in the third term on the right side of Eq. (57) can be obtained from table of Laplace transform:

L−1( 1s2+2 ςωns+ωn

2 )= 1ωd

e−ςωn t sinωd t

(62)Using Eqs. (47), (59), (60), and (62), the general solution of Eq. (47) can be expressed as

x (t )=x0

(1−ς2 )1/2e−ςωnt sin (ωd t+φ1)+

x0ωd

e−ςω

n sinωd t+1

mωd∫0

tF (τ ) e

−ςωn

(t−τ )sinωd ( t−τ ) dt

(63)

END

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