Response of a Stable Second Order System

7/29/2019 Response of a Stable Second Order System

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Response of a STABLE Second Order SystemSeon Han (3/29/2011)

In this document, we will consider a simple m-c-k system shown below.

Genetic m-c-k system

The equation of motion is given bymx + cx + kx = F(t) .

The part of this document is REVIEW of Chapter 3 of Palm and the rest is based on Chapter 8. Here,I will address the following topics. The system we consider

1. Free System (Chapter 3)

- Undamped system

- Underdamped, overdamped, critically damped

2. Forced System (Sections 8.1-8.3, 13.1-13.2)

- Impulsive Force

- Arbitrary Force

- Harmonic Force

- Base Excited Motion

- Rotating Unbalance

1 Free Response Types

Consider a simple mass-damper-spring system subject to initial conditions only. This is the case whenF(t) = 0. The equation of motion is given by

mx + cx + kx = 0 with given x (0) and x (0) .

When m, c, and k are POSITIVE (or of the same sign), the response is stable. The response will eventuallydie out to zero. There are three ways that this can happen. The response can decay exponentially, decayexponentially while oscillating, or the borderline between the first two.

There are two characteristic roots for this system (because it is a second order). When the characteristic

roots are both real and negative, you say that the system is OVERDAPMED. If the characteristic rootsare complex (with negative real part), the system is UNDERDAMPED and decays while oscillating. If thecharacteristic roots are real, negative and repeated, the system is CRITICALLY DAMPED and the responselooks similar to that of the overdamped system. Here are some examples:

1. x + 3x + 2x = t2 has characteristic roots s = 1 and 2. The system is overdamped.2. x + 2x + 1x = sin t has characteristic roots s = 1 and 1. The system is critically damped.3. x + 2x + 2x = sin t has characteristic roots s = 1 i. The system is overdamped.

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1.1 Undamped System (c = 0)

Consider a system with c = 0, mx + kx = 0. The characteristic roots are

s = i

k/m,

and the system is neutrally stable. The solution is given by

x (t) = C1 sin

k/mt + C1 cos

k/mt.

The mass oscillates at a period of

T =2k/m

[s] ,

has a frequency of

f =1

T=

k/m

2[Hz ] ,

and has an angular frequency of = 2f =

k/m [rad/s] .

The coefficient of t in the sinusoidal function is the angular frequency. This frequency is unique to a system

(depends on m and k only) and is called the NATURAL FREQUENCY of the system (therefore, the subscriptn). From now on, we drop angular, and the frequency is measured in [rad/s] , never never [Hz ] .

You realize quickly that the systems with the same mass to stiffness ratio subject to the same initialconditions have the SAME response. That is, the following systems have the same solution x (t) :

x + 2x = 0

2x + 4x = 0

3x + 6x = 0.

Therefore, we do not need to know the individual values ofm and k, but the natural frequency n is sufficientenough to find x (t) . Therefore, we often use the NORMALIZED EOM.

1.2 Normalization of EOM

Consider an underdamped system, mx + cx + kx = 0. The characteristic roots are

s = cm

(c2 4mk) /m2,

where c2 4mk is a negative number if the system is underdamped.We typically divide the equation by m so that we have

x +c

mx +

k

mx = 0.

We recognize k/m as 2n. We write c/m as 2n such that

x + 2nx + 2nx = 0.

This is the NORMALIZED form of the EOM. We call the DAMPING RATIO. You can figure out that is c/2

mk

c

m= 2n = 2

k

m

=1

2

c

m

m

k

=c

2

mk.

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The characteristic roots are (in terms of and n)

s = n

2 1n.The system is overdamped if c2 > 4mk or > 1. The system is critically damped if c2 = 4mk or = 1.The system is underdamped if c2 < 4mk or < 1. Now do you see why we use ? Whether the system is

underdamped or overdamped can be determined from a single non-dimensional parameter instead ofm, c,and k.We introduce ONE more nondimensional parameter called damped natural frequency, d, which is useful

for underdamped system only. It is defined as

d =

1 2n for < 1.

The underdamped system response is written as

x (t) = ent (C1 cos dt + C2 sin dt) .

The damped natural frequency is the ACTUAL frequency that x (t) is oscillating at.

2 Forced Response to an Impulsive forceThe response of a system to a unit impulse is VERY important in dynamics. Consider a system model(equation of motion) given by

mx + kx = fo(t),

which is a simple mass-spring system subject to impulsive force fo(t) . In this case, fo(t) has a unit of[N] . The Dirac delta function has a unit of [1/s] , and therefore, fo has a unit of [N s] not [N] . WHATHAS A UNIT OF [N s]? IMPULSE has a unit of [N s] . fo is the impulse, not the force, applied to themass.

Just to remind you, recall

f(t) =d

dt(mv)

f(t) dt = (mv) ,where

f(t) dt is called IMPULSE and mv is called momentum. The principle of linear momentum

says the impulse is equal to change in momentum. The strength of an impact force is measured bythe amount of impulse applied, not the magnitude of force itself. The magnitude of the force applied, afterall, is fo . It is the impulse that determines the how much momentum the system gains after the forceis gone.

Now, lets find the response, x (t) , due to this impulsive force (use zero initial conditions). The easiestway is to use the Laplace transform. We find

X(s) =fo

ms2 + k

x (t) = L1 [X(s)] =fom sin

kmt.

The response to the unit impulse (fo = 1) is SO important that we give another name, g (t) , instead of usingx (t) . That is g (t) is the solution to

mg + kg = (t) with zero ICs

G (s) =1

ms2 + k

g (t) =1

msin

k

mt (1)

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We call g (t) the impulse response function. IN SOME textbooks, the impulse response function mayrefer to the impulse response when m = 1 ( g + 2ng + 2ng = (t)) so that

g (t) = sin

k

mt.

Now, it is YOUR TURN to find the impulse response functions for the damped system (for all three

underdamped, overdamped, and critically damped).

3 Forced Response to an Arbitrary Forcing

The response to an unit impulse (when fo = 1) is VERY important because it allows us to find the responseto an arbitrary force f(t) . Consider the same system subject to f(t) .

mx + kx = f(t).

After taking the Laplace transform, we have

X(s) =1

ms2 + kF(s) .

The transfer function is given byX(s)

F(s)=

1

ms2 + k,

and, surprise and surprise, it is the SAME as the impulse response function in the s domain, G (s)!!!!!!!From X(s) = G (s) F(s) , we use the CONVOLUTION theorem to find the the response in the time

domain,

x (t) =

t0

g (t ) f() d ort0

f(t ) g () d,

where we already know g (t) = 1m

sin

km

t. This integral may not be easy to evaluate by hand but is certainly

doable numerically (remember numerical integration using trapezoidal or Simpsons rule from Colleen Bergsclass?). This integral says you can find the response to ANY force, f(t) , once you find the impulse response

function g (t) . You really should be impressed.

Example 1 Consider a mass-spring system subject to F(t) , F (t) shown in the figure. Find the displacementof the mass, x (t) . Use zero initial conditions.

k= 100 N/m

m = 1 kg

x t( )

f t( )

f t( ) [N]

t[s]5

10

Solution: We already found that the impulse response function for a mass-spring system is (Equation 1)

g (t) =1

msin

k

mt = sin 10t.

The force is given by

f(t) = 2t for 0 < t < 5

= 10 for t > 5.

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The response is given by (If your force is discontinuous, it is easier to use the first equation.)

x (t) =

t0

g (t ) f() d.

I need to convert the force as a function of ,

f() = 2 for 0 < 5= 10 for > 5.

f() is discontinuous and we dont have a single function that we can plug intot0 g (t ) f() d. The

dummy variable,, is varied from 0 to t.For t 5, f() can be expressed in terms of a continuous function,

f() = 2 for 0 < t.For t > 5, we can write f() in two parts

f() = 2 for 0 < 5= 10 for 5 < < t.

Then, for t 5,x (t) =

t0

g (t ) 2d

=

t0

sin10(t ) 2 d

= 150

sin(10t) +t

5.

For t > 5,

x (t) =

50

g (t ) 2 d+t5

g (t ) 10d

= 150

sin(10t) + 150

sin(10t 50) + 1.

4 Forced Response to a Harmonic Force (Chapter 8)

Now, consider the case when F(t) is harmonic. The equation of motion is given by

mx + cx + kx = Fcos ft.

Lets do an example with numbers.

Example 2 Find the response of a m-c-k system given below

x + 2x + 5x = 7cos 3t

The system is subject to initial conditions given by x (0) = 1 and x (0) = 1.Time Domain Solution: We find the particular solution, homogeneous solution, and apply the initial

conditions to the total solution. The particular solution has the form of

xp (t) = A cos3t + B sin3t.

We substitute this particular solution back into xp + 2xp + 5xp = 7cos 3t to obtain

9 (A cos3t + B sin3t) + 2(3) (A sin3t + B cos3t)+5 (A cos3t + B sin3t) = 7cos 3t.

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Equating the coefficient of cos3t, we have

9A + 6B + 5A = 7

Equating the coefficient of sin3t, we have

9B

6A + 5B = 0.

Solving the two simultaneous equations, we find

A = 7/13B = 21/26.

Now, we have the particular solution, xp (t) = 713 cos3t + 2126 sin3t.Next, the homogeneous solution is obtained fromx+2x+5x = 0. The characteristic roots are s = 12i.

Then, the homogenous solution is given by

xh (t) = et (C1 cos2t + C2 sin2t) .

Now, we assemble the total solution,

x (t) = xh (t) + xp (t)

= et (C1 cos2t + C2 sin2t) 713

cos3t +21

26sin3t.

Finally, the initial conditions (x (0) = 1 and x (0) = 1) are applied to find C1 = 20/13 and C2 = 3/52.Laplace-Inverse Laplace Transform Method: You couldve gotten the same results by taking a

Laplace transform and taking the inverse Laplace transform. The Laplace transform of the equation ofmotion gives

s2X(s) sx (0) x (0) + 2 (sX(s) x (0)) + 5X(s) = 7

s2 + 9

s2 + 2s + 5X(s) s 3 =7

s2

+ 9X(s) =

1

(s2 + 2s + 5)

7

s2 + 9+

s + 3

(s2 + 2s + 5).

After rewriting it in a more friendly form, you will then take the inverse Laplace transform to find

x (t) = et

20

13cos2t +

3

52sin2t

7

13cos3t +

21

26sin3t.

4.1 Steady-State Solution to the Harmonically Forced m-c-k System

Note that the homogenous solution dies out given enough time, and only the particular solution will remain.The particular solution in this case is the steady-state solution. If the system is stable (so that thehomogeneous solution is guaranteed to disappear), the particular solution is the steady-state

solution. In many cases, you may care ONLY about the amplitude of the steady-state solution. For theexample above, the steady-state solution can be rewritten in the amplitude-phase form such that

713

cos3t +21

26sin3t = Ccos(7t + ) .

I am using cosine because the FORCING is cosine. NOTE that YOU CANNOT find the particularsolution easily using the Laplace transform method (You cant simply let the initial conditionszeros). You must solve the problem in the time domain to find the particular solution only (at least untilI show you the frequency domain method).

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Lets continue to find the amplitude and the phase lag. Rewrite the right hand side by expandingcos(7t + ) = cos7t cos sin7t sin . Then, for the two sides to be equal, we must have

713

= Ccos

21

26

=

Csin . (2)

By squaring the first and second equations and summing them, we have7

13

2+

21

26

2= C2.

We always take the positive C such that

C =

7

13

2+

21

26

2= 0.97

Now, we find the phase angle. Going back to Equation 2, with C being positive, we find that cos is negative

and sin is also negative. Then, the phase angle must be in the third quadrant. Lets divide the secondequation by the first to get

tan =21/267/13 .

Your calculator will give you = 0.9828 (56.3). This is not the correct answer. You must find the answerthat has the same reference angle (angle measured from the x axis) and in the third quadrant. The actualanswer is = + 0.983 = 4.12 rad or Then,

713

cos3t +21

26sin3t = 0.97cos(7t + 4.124) .

The particular solution, 0.97cos(7t + 4.124) , is ahead of forcing, 7cos3t, by 4.124 radians or 4.124/7 = 0.308seconds. Another way of thinking is that the particular solution lags behind the forcing by 24.124 radiansor 0.308 seconds.

4.2 Frequency Domain Method for Amplitude-Phase Form of the Particularsolution (8.1-8.3 of Palm)

You and I agree that it is very labor intensive to find the particular solution using the time domain method.Here is another method that I recommend you using. Lets go back to

mx + cx + kx = Fcos ft,

Note that cos ft is the real part of eift. We can solve a similar problem where the system is the

forced by eift instead of cos ft and take the real part of the solution to be the solution to thecurrent problem. That is,

mx + cx + kx = F eift.

We specifically look for the solution that oscillates at the forcing frequency (the particular solution). Thesolution that we look for will have the form of

xp (t) = Xpeift.

We substitute back in to find2fm + icf + kXpeift = F eiftXp =

Fk 2fm + icf

. (3)

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In terms of non-dimensional parameters, we can write

Xp =F

k

11 2f/2n + i2f/n

(4)Note that Xp is complex. Write this in the amplitude phase form,

Xp =F

k 2fm2

+ (cf)2ei

=

Fk 2fm

2+ (cf)

2

ei,

Response amplitude |Xp|

where

= tan1cf

k 2fm = tan1 cf

k 2fm= tan1 cfk + 2fm

.

Note that you have to choose on the first and the second quadrant (0 because of the positiveimaginary part and + or real part). Therefore, is between 0 and -. When you use the calculator, thecalculator will only give you the answer between /2 and /2. YOU must find the answer in the correctquadrant. Then,

xp (t) = Re

F

k 2fm + icfeieift

xp (t) =F

k 2fm2 + (cf)2cos(ft + ) .

Lets redo the example from the previous section. The eom is given by x + 2x + 5x = 7cos 3t. Then,

|Xp| = Fk 2fm

2+ (cf)

2

=7

(5 321)2 + (2 3)2= 0.971

= tan1cf

2fm k= tan1

2 3321 5 = 0.98279

= 0.982 is in the first quadrant. This is incorrect because the possible value for must be in the third orfourth. Since the argument of tan1 is positive, the answer is in the third quadrant and = + 0.982 = 4.12rad or =

(

0.982) =

2.16. Then,

xp (t) = 0.971cos(3t + 4.12)

or

= 0.971cos(3t 2.16)IMPORTANT: Take a look at the ratio, |Xp| /F,

|Xp|F

=1

k 2fm2

+ (cf)2

.

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Compare this with the transfer function of this system,

T(s) =1

ms2 + cs + k

Replace s with if, and find the magnitude of T(if)

|T(if)| = 1k 2fm

2+ (cf)

2

,

which is IDENTICAL to |Xp| /F. Therefore, the amplitude of the particular solution (which happens to bethe steady-state response also because the system is stable) when the forcing frequency is f can be writtenas

|Xp| = F T(if) .We call, |Xp| /F, the magnitude ratio and denote it as M. This book is quite confusing because it definesM to be

M = |Xp| /F or k |Xp| /F.The BODE plots are the plots of M and . The magnitude ratio M is drawn in the logarithmic scale in

Figure 8.2.4. Note that the logmagnitude ratio, m, in decibels is

m = 20 log M,

where M has no unit and m has a unit of dB.

Example 3 Book Example: 8.2.1. The model is given by

6x + 12x + 174x = 15f(t)

f(t) = 5 sin 7t

The transfer function is given by

T(s) =15

6s2 + 12s + 174

T(if) =15

62f

+ 12 (if) + 174

=15

120 + 84i .

The magnitude ratio isM = |T(i7)| = 0.1024,

or if you are using the second definition of M,

M = |kT(i7)| = 174 0.1024.

The phase is = 1 (120 + 84i) = 0 tan1(84/ 120) = 0 2.53 rad. What does it mean to haveM = 0.1024 and a phase of 2.53? This means that the response is (particular solution)

xp (t) = 0.1024 5sin(7t 2.53)= 0.1024 5sin

7

t 2.53

7

.

The response amplitude is 0.1024 times the forcing amplitude and the peak of the response lags behind thepeak of the input force by 2.53/7 seconds.

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4.3 Frequency Response Plots (to HARMONIC INPUT only)

The frequency response plots refer to the plots of the AMPLITUDE of the PARTICULAR solution and thePHASE of the particular solution as functions of the input frequency. Of course, it would not make anysense if the input is not harmonic.

For our current case where the system is forced harmonically, you may want to know how the responseamplitude changes with varying input frequency. You would want to plot

|Xp

|vs f, where

|Xp

|is given in

Equation 3. The amplitude plot is always accompanied by the phase plot, vs f.If you know the values of m, c, and k, use Equation 3 to plot |Xp| as a function of b. However, if you

are a designer, your job is to find m, c, and/or k! It is better if you have the response plots. It is betterto plot the nondimensional version of the amplitude (note that phase already does not have units). Whatdo you think |Xp| should be normalized with respect to? Once you write |Xp| in terms of the normalizedfrequency or frequency ratio, f/n, it becomes obvious how |Xp| should be normalized. Rewrite Equation4 in terms of r, and you will obtain

|Xp| = Fk

11 (f/n)2 + i2(f/n) . (5)

It is natural to normalize|Xp

|with respect to F/k. The term, F/k, has a physical meaning. It is the static

deflection due to a constant force of F [N] . Therefore, |Xp| / (F/k) is the ratio of the dynamic amplitude tothe static amplitude. |Xp| / (F/k) = 3 means that the amplitude of the dynamic response (particular solutiononly) is three times of the static deflection if the force was not harmonically varying. There are VARIOUSnames for |Xp| / (F/k) . Your book calls it magnitude ratio, M. My old textbook called it magnificationfactor.

Starting from Equation 5, we have

M =k |Xp|

F=

1(1 r2 + i2r)

=1

(1 r2)2 + (2r)2, (6)

where we denoter = f/n

and call it frequency ratio. The frequency ratio, r, is the NORMALIZED forcing frequency. The non-dimensional magnitude ratio is plotted Figure 8.2.4. Note that the y axis is in decibels and x in log scale.Here is my version in Figure 1. I plotted |Xp| / (F/k) for 0 < < 2. You should note that this is only HALFof the whole picture (more important half). The other half of the frequency response plot is the PHASEplot. You will see that the phase varies from 0 to - (Figure 8.2.4 in your text).

From the frequency response plot, you should notice the followings:

1. The dynamic response amplitude is equal to the static deflection iff = 0 for any value of . This isno brainer. If the static force is applied (this is what it means to have f = 0), the displacement ofthe mass is, of course, the static deflection.

2. As f increases, the dynamic amplitude approaches zero for any value of .

From Equation 6, we find that M has the maximum value of 1/2

1 2 at the forcing frequencyr =

1 22. SEE TABLE 8.2.2.

Mpeak =1

2

1 2

rpeak =

1 22 valid for 0.707 and the maximum steady-state response occurs when the forcing frequency iszero (or constant force).

4.4 Resonance (true resonance when c = 0 and f = n)

Consider an undamped system that is forced at the natural frequency. That is

x + 2nx = Fcos nt.

The solution is given by

x (t) = C1 sin nt + C2 cos nt + 12

F2n

t sin nt,

where C1 and C2 are determined from the initial conditions. The last term is the particular solution, andit grows LINEARLY with time. It takes infinite time for the response to grow infinity. When the forcingfrequency matches the natural frequency, we say we have resonant response or resonance. Your book expandsthe definition of resonance to the damped system.

4.5 Beating (Near Resonance) (Section 8.3)

Consider an undamped system forced by a cosine forcing with forcing frequency f. Let us use zero initialconditions (for your convenience), and then the response is given by

x (t) =F

k (1 r2)cos ft

F

k (1 r2)cos nt

=F

k (1 r2) (cos ft cos nt) .

Note that your book uses sine forcing (page 442). This solution is valid for any f that is NOT equal ton. Note that sum or difference of sinusoidal functions with different frequency can be written as

cos a + cos b = 2 cosa + b

2cos

a b2

cos a cos b = 2sin a + b2

sina b

2

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We can rewrite the solution as

x (t) = 2 Fk (1 r2) sin

f n2

sinf + n

2,

where sinfn

2 is SLOWLY varying. The term, 2F

k(1r2) sinfn

2 , controls the envelop, and sinf+n

2

the fast oscillations inside the envelop (See figure 8.3.2 on pg444). Due to the symmetry, the beat frequencyis twice the envelop frequency orbeat = |f n| .

4.6 OTHER Frequency Response Plots: (1) Force transmitted to the wall, (2)acceleration of the mass

The plot of what I call the magnification ratio or Xp/ (F/k) is a powerful design tool. You can select sothat the steady-state amplitude (Xp) is under some threshold. Why? Maybe your m-c-k system is housedin an enclosure and you are worried that the mass may hit the enclosure.

Unfortunately, the response amplitude may not be the only design criteria you need to satisfy. Forexample, you may be worried that the wall that the m-c-k system is attached to may break or you may beworried that the sensitive equipment (block, m) that is mounted on the vibration isolator (c and k) may

break due to excessive acceleration.Lets consider these two cases.

Force Transmitted to the Wall The force transmitted to the wall is

Ftr = kx + cx

= Re

(k + cif) Xpeift

.

The amplitude of the force transmitted (of course the particular solution only) is (use Equation 3)

|Ftr| = |(k + cif) Xp|

= (k + cif) Fk 2fm + icf .If you are given c and k and F, then use above expression to find the force transmitted. Otherwise, it ismuch better to use the normalized expression.

What do you think is the reasonable quantity to normalize the force transmitted? It should have a unit offorce and input amplitude should be included. The amplitude of the input force, of course. The normalizedforce transmitted (If you want to name this, I would call it force transmissibility) is

|Ftr|F

=

(k + cif)k 2fm + icf

= (1 + i2r)

(1 r2

+ i2r) =

1 + (2r)2

(1 r2)2 + (2r)2.

In class, I showed a different way to arrive at this. (I substituted x = Xp cos(wft + ) into Ftr = kx + cx,and found the amplitude. You can see that it is much easier to use eift instead of sine and cosine.)

I have the plot of the normalized amplitude of the force transmitted (Figure 2).

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Figure 2: The normalized amplitude of force transmitted, |Ftr| /F, for 0 < < 2.

Acceleration of the Mass You will come across a statement such as The maximum force that a compo-nent can withstand is 3g, which is not incorrect but imprecise. What they mean is that the maximum forcethat a component can withstand is 3gm where m is the mass of the component and g is the gravitationalacceleration, or the maximum acceleration that the component can withstand is 3g. The acceleration is

a = xp (t) = Re

d2

dt2

Xp (if) eift

= Re

Xp (if) 2feift=

Xp (if) 2f cos(ft + )The amplitude of the acceleration (complex amplitude that has the phase information, too) is

A = Xp (if) 2f

=F 2f

k 2fm + icf .

The phase angle is = A

What is the reasonable quantity that we can use to normalize the amplitude? It should have a unit ofacceleration and should have the input amplitude. I would use F/m, where F is the force amplitude and mis the mass. Then,

|A|F/m

=

r2

1 r2 + i2r

= r2(1 r2)2 + (2r)2

.

I dont have this plot.

5 Base Excited (harmonically excited) Motion

Consider a m-c-k system that is mounted on a moving base. The motion of the base is known and is givenby y (t) = Y cos bt.

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The equation of motion is given by

mx + cx + kx = cy + ky,

where x (t) is the displacement of the mass measured from the inertial frame1.Then, the particular solution is given by

xp (t) = Re T(ib) Y eibt

or |T(ib) Y| cos(bt + ) ,where

T(s) =Xp (s)

Y (s)=

cs + k

ms2 + cs + k.

The amplitude of the response (steady-state response) is

|Xp| = c (ib) + km2b + c (ib) + k Y

. (7)As a designer, you may be interested in frequency response plots such as

1. the response amplitude, |Xp|,2. the relative response amplitude

|Zp

|, or

3. the amplitude of the force transmitted to the ground.

If you know the values ofm, c, and k, use Equation 7 to plot Xp as a function of b. However, if you area designer, your job is to find m, c, and/or k! It is better if you have the dimensionless frequency responseplots. The response amplitudes are normalized by the input amplitudes, |Xp| /Y or |Zp| /Y , and the forceamplitude is normalized by kY.

The normalized response amplitude, |Xp| /Y, has a good name. It is called the DISPLACEMENTTRANSMISSIBILITY. It is obvious from the name that it is the ratio of the dynamic amplitude to theamplitude of the base motion.

5.1 Displacement Transmissibility, |Xp| /Y

From 7, the displacement transmissibility is given by|Xp|

Y=

c (ib) + km2b + c (ib) + k .

1 It is, however, possible to write the equation of motion in terms of RELATIVE displacement, z (t) = x (t) y (t) , observedfrom the moving base. The equation of motion in terms of the relative displacement is

mz + cz + kz = my.

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Figure 3: Base Excited Motion, Displacement Transmissibility

I will keep the complex form until the last moment. Because the LHS is dimensionless, the RHS also doesnot have units. Divide the top and bottom by m, and divide the top and bottom by 2n again.

|Xp|Y

=

c (ib) + km2b + c (ib) + k mm =

2n (ib) + 2n2b + 2n (ib) + 2n

=

2n (ib) + 2n2b + 2n (ib) + 2n 2n2n =

i2r + 1r2 + i2r + 1

=

1 + (2r)2

(1 r2)2 + (2r)2.

Note that the phase angle is = (i2r + 1)

1 r2 + i2r .

You must be able to find the phase angle using a calculator. In matlab, use the command angle.m.Here is the plot of displacement transmissibility in Figure 3.Note that as increases, the displacement transmissibility flattens out to 1. WHY WHY WHY? Does

this make sense to you? As increases (which means c increases), the damper acts more and more like arigid connection. Therefore, the response of the mass will become same as the base motion.

How about when k ? You know intuitively that |Xp| /Y should be 1 again. Where does it say thatin Figure 3? As k increases, n increases, and r decreases. That is, as k , r 0. At r = 0, thedisplacement transmissibility is 1 for ANY VALUES OF .

5.2 Force Transmitted

The force transmitted to the ground is

Ftr = k (x y) + c (x y) .

From EOM, we can also writeFtr = mx.

Either expression will work. Starting from

x (t) = Xpeibt

y = Y eibt,

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we findFtr = (k (Xp Y) + ibc (Xp Y)) eibt or m2bXpeibt,

where Xp is complex and Y is a real number. The amplitude of the force transmitted is

|Ftr| = |k (Xp Y) + ibc (Xp Y)| orm2bXp

,

where we already know

Xp =c (ib) + k

m2b + c (ib) + kY

=i2r + 1

r2 + i2r + 1 Y.

You can use either |k (Xp Y) + c (Xp Y)| orm2bXp to find the force transmitted. I will use the first

expression (just to torture myself and prove to you that you get the same Ftr), the normalized force is

FtrkY

=

XpY

1

+ ibc

k

XpY

1= 1 + ib 2n2n XpY 1

=

(1 + i2r)

i2r + 1

r2 + i2r + 1 1

=

(1 + i2r)

r2

1 r2 + i2r

= r2

1 + (2r)2

(1 r2) + (2r)2.

Why dont you use |Ftr| =m2bXp

and confirm that the results agree?Anyway, I look at this expression carefully, and I realized as r the force transmitted also goes to

INFINITY. Remember when the mass is subject to harmonic force, the force transmitted goes to zero asr . However, if the mck system is subject to harmonic base motion, the force transmitted to the wallINCREASES (quadratically) with increasing r.

5.3 Relative Displacement

I am going to leave it up to you to figure out the amplitude of the relative displacement (z (t) = x (t)y (t)).

6 Rotating Unbalance (Washer with unbalanced load)

The equation of motion is given by

x + 2n

x + 2n

x =me

M2r

sin r

t.

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Figure 4: Amplitude ratio, |X| / (me/M) , for a system with rotating unbalance.

The particular solution is given by

xp (t) = Re

T(ir)me

M2re

irt

= Re

1

2r + i2nr + 2nme

M2re

irt

=m2re/M

(2n 2r)2 + (2nr)2sin(rt + )

=me

M

r2(1 r2)2 + (2r)2

response amplitude, |Xp|sin(rt + ) , (8)

where

r =rn

and = T(ir) = arctan 2r1 r2 ( 0).

The amplitude of vibration (particular solution only) normalized by me/M is given by

|Xp|me/M

=r2

(1 r2)2 + (2r)2,

which is shown in Figure ??. The maximum amplitude occurs when r = 1/

1 22 (note that this valueis slightly greater than 1) and the maximum value is given by

|Xp|max =me

M

1

2

1 2 .

6.0.1 Force transmitted to the ground due to the rotation of unbalance

The force is transmitted through the spring and damper. The force transmitted to the ground is given by

Ftr = kx + cx.

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Then, the amplitude of the force transmitted to the ground is given by (using Equation 8)

|Ftr| = |kXp + cirXp|=

k2 + (cr)

2 |Xp|

=me

M k

r21 + (2r)2(1 r2)2 + (2r)2 .

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Response of a Stable Second Order System

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