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Dr. Millerjothi, BITS Pilani, Dubai Campus Forced system analysis Need for Vibration isolation and critical speed and resonance Construction and functioning of measuring instruments Chapter 3

MV - Chap 3

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Page 1: MV - Chap 3

Dr. Millerjothi, BITS Pilani, Dubai Campus

�Forced system analysis

�Need for Vibration isolation and critical

speed and resonance

�Construction and functioning of

measuring instruments

Chapter 3

Page 2: MV - Chap 3

Dr. Millerjothi, BITS Pilani, Dubai Campus

• When a system is subjected to harmonic excitation, it is forced

to vibrate at the same frequency as that of the excitation.

• Common sources:

�Unbalance in rotating machines

�Forces produced by reciprocating machines

�The motion of the machine itself

• Resonance is to be avoided in most cases with dampers and

absorbers

Page 3: MV - Chap 3

Dr. Millerjothi, BITS Pilani, Dubai Campus

• Harmonic excitation may be in the form of a force or

displacement of some point in the system.

• Single DOF with viscous damping:

- Excited by

-

- Complementary (homogeneous)

and particular solutions

- The particular solution is a steady-state oscillation of

the same frequency as that of the excitation.

- The particular solution can be assumed to be

FORCED HARMONIC VIBRATION

tFkxxcxm ωsin0=++ &&&

( )φω −= tXx sin

Page 4: MV - Chap 3

Dr. Millerjothi, BITS Pilani, Dubai Campus

• X is the amplitude of oscillation and ϕ is the phase of the

displacement with respect to the exciting force.

• In harmonic motion, the phases of

the velocity and acceleration are ahead

of the displacement by 90o and 180o,

respectively, as in the figure given.

( ) ( )

k

m

k

c

k

c

k

m

k

F

mk

c

cmk

FX

222

2

0

2

1

222

0

1

tan ,

1

tan ,

ω

ω

φωω

ωω

φωω

−=

+

=

−=

+−= −

Page 5: MV - Chap 3

Dr. Millerjothi, BITS Pilani, Dubai Campus

In terms of the following quantities:

n

c

c

c

nc

n

k

c

c

c

k

c

factordampingc

c

dampingcriticalmc

nsoscillatioundampedoffrequencynaturalm

k

ωω

ζωω

ζ

ω

ω

2

2

==

==

==

==

Page 6: MV - Chap 3

Dr. Millerjothi, BITS Pilani, Dubai Campus

The non-dimensional expressions for the amplitude and

phase become

2

2220

1

2

tan

,

21

1

=

+

=

n

n

nn

F

Xk

ωω

ωω

ζφ

ωω

ζωω

Page 7: MV - Chap 3

Dr. Millerjothi, BITS Pilani, Dubai Campus

Xk/Fo and ϕ are functions

only of the frequency ratio

ω/ωn and the damping

ratio ζ. This can be plotted

as shown.

Page 8: MV - Chap 3

Dr. Millerjothi, BITS Pilani, Dubai Campus

For ω/ωn <<1, the impressed force is nearly equal to the

spring force, as in (a)

For ω/ωn =1, the phase angle is 90o and the force diagram is as in (b)

For ω/ωn >>1, ϕ approaches 180o and the impressed force is expended almost entirely in overcoming the large inertia force

Page 9: MV - Chap 3

Dr. Millerjothi, BITS Pilani, Dubai Campus

The differential equation and the complete solution

( )

( )1

2

1

222

0

02

1sin

21

)sin(

sin2

φωζ

ωω

ζωω

φω

ωωζω

ζω +−+

+

−=

=++

−teX

t

k

Ftx

tm

Fxxx

n

t

nn

nn

n

&&&

Page 10: MV - Chap 3

Dr. Millerjothi, BITS Pilani, Dubai Campus

Let harmonic force be represented by

The displacement can then be written as

Then

and

Complex frequency response

ti

oo eFtitF ωωω =+ )sin(cos

( ) ( )complexX

eXeXeXex titiiti

=

=== −− ,ωωφφω

( )

+

=

=++−

nn

i

k

F

X

FXkicm

ωζω

ωω

ωω

21

2

0

0

2

Page 11: MV - Chap 3

Dr. Millerjothi, BITS Pilani, Dubai Campus

The complex frequency response H(ω)

Often the factor 1/k is considered together with the force,

leaving the frequency response a non-dimensional quantity.

Thus H(ω) depends only on the frequency ratio and the

damping ratio.

( )

+

==

nn

i

k

F

XH

ωζω

ωω

ω2

1

1

2

0

Page 12: MV - Chap 3

Dr. Millerjothi, BITS Pilani, Dubai Campus

The real and imaginary parts of H(ω)

( ) ,

21

2

21

1

22

222

2

2

+

+

=

nn

n

nn

niH

ωζω

ωω

ωζω

ωζω

ωω

ωω

ω

Page 13: MV - Chap 3

Dr. Millerjothi, BITS Pilani, Dubai Campus

At resonance, the real part is zero and the response is

And the phase angle

( )ζ

ωωω2

1, iHn −==when

2

1

2

tan

=

n

n

ωω

ωζω

φ

Page 14: MV - Chap 3

Dr. Millerjothi, BITS Pilani, Dubai Campus

Unbalance in rotating machine:

� A spring-mass system constrained

to move in the vertical direction

� The unbalance is represented by

an eccentric mass m with eccentricity

e that is rotating with angular velocity ω

The displacement of the non-rotating mass (M – m) from the static

equilibrium position

ROTATING UNBALANCE

tex ωsin+

Page 15: MV - Chap 3

Dr. Millerjothi, BITS Pilani, Dubai Campus

The equation of motion

Since it is identical to tFkxxcxM

tmekxxcxM

xckxtexdt

dmxmM

ωωω

ω

sin

sin)(

0)sin()(

0

2

2

2

=++

=++

=++++−

&&&

&&&

&&&

222

2

02220

2222

0

]2[])(1[

)1

(

]2[])(1[

1

tan,)()(

nn

n

nn

F

MX

F

Xk

Mk

c

cMk

FX

ωω

ζωω

ω

ωω

ζωω

ωω

φωω

+−

=⇒

+−

=

−=

+−=

Page 16: MV - Chap 3

Dr. Millerjothi, BITS Pilani, Dubai Campus

Hence, the steady-state solution is

2222

2

2222

2

)(1

)(25

tan,

]2[])(1[

)(

tan,)()(

n

n

nn

n

me

MX

Mk

c

cMk

meX

ωωωω

φ

ωω

ζωω

ωω

ωω

φωω

ω

−=

+−

=

−=

+−=

Page 17: MV - Chap 3

Dr. Millerjothi, BITS Pilani, Dubai Campus

The complete solution is ( ) ( )

[ ])sin(

)(

1sin

222

2

1

2

1

φωωω

ω

φωζζω

−+−

+

+−= −

t

cMk

me

teXtx n

tn

me

MX

Page 18: MV - Chap 3

Dr. Millerjothi, BITS Pilani, Dubai Campus

Page 19: MV - Chap 3

Dr. Millerjothi, BITS Pilani, Dubai Campus

Page 20: MV - Chap 3

Dr. Millerjothi, BITS Pilani, Dubai Campus

Static unbalance:

When the unbalanced masses all lie in a single plane, as in the c

ase of a thin rotor disk, the resultant unbalance is a single radial

force.

The static unbalance can be detected by a static test

ROTOR UNBALANCE

System with static balance System with dynamic balance

Page 21: MV - Chap 3

Dr. Millerjothi, BITS Pilani, Dubai Campus

Dynamic unbalance:

When the unbalance appears in more than one plane, the

resultant is a force and a rocking moment, which is referred to as

dynamic unbalance

A static test may detect the resultant force, but the rocking

moment cannot be detected without spinning the rotor.

If the two unbalanced masses are equal and 180 apart, the rotor

will be statically balanced about the axis of the shaft. However,

when the rotor is spinning, each unbalanced disk would set up

a rotating centrifugal force, tending to rock the shaft on its

bearings.

Page 22: MV - Chap 3

Dr. Millerjothi, BITS Pilani, Dubai Campus

Machines to detect and correct the rotor unbalance

The balancing machine consists of supporting bearings

that are spring-mounted so as to detect the unbalanced

forces by their motion.

Balancing machines

Page 23: MV - Chap 3

Dr. Millerjothi, BITS Pilani, Dubai Campus

Whirling:

The rotating of the plane made by the bent shaft and the line of c

enters of the bearings.

Results from such various causes as mass unbalance, hysteresis

damping in the shaft, gyroscopic forces, fluid friction in bearings

and so on.

Can take place in the same or opposite direction as that of the

rotation of the shaft and the whirling speed may or may not be

equal to the rotating speed

WHIRLING OF ROTATING SHAFTS

Page 24: MV - Chap 3

Dr. Millerjothi, BITS Pilani, Dubai Campus

Consider a single disk of mass m symmetrically located on a sha

ft supported by two bearings. (see Figure)

The center of mass G of the disk is at a distance e, eccentricity

from the geometric center S of the disk.

Page 25: MV - Chap 3

Dr. Millerjothi, BITS Pilani, Dubai Campus

The centerline of the bearings intersects the plane of the disk O

and the shaft center is deflected by r = OS

Assume the shaft (i.e., the line e = SG) be rotating at a constant

speed ω, and in the general case, the line r = OS to be whirling

at speed dθ/dt, that is not equal to ω.

For the equation of motion, we can develop the acceleration of the

mass center as follows:

SGGG aaa

erOG

/+=

+=

Page 26: MV - Chap 3

Dr. Millerjothi, BITS Pilani, Dubai Campus

Resolving aG in the radial and tangential directions

[ ][ ] θθωωθθ

θωωθ

θωθθ

θωωθ

θωθω

&&&&&

&&&&

&&&&

&&&

crterrm

rckrterrm

jtewrr

iterra

jteiteirr

G

G

−=−−−

−−=−−−

−−++

−−−=

−+−+=

)sin()2(

)cos()(

)]sin()2[

)]cos()[(

)sin()cos(

2

22

2

22

Page 27: MV - Chap 3

Dr. Millerjothi, BITS Pilani, Dubai Campus

Which can be rearranged as

The general case of whirl comes under the classification of self-

excited motion, where the exciting forces inducing the motion

are controlled by the motion itself.

In the steady-state synchronous whirl, where and

the problem reduces to that of 1 DOF and 2

DOF.

)sin(2

)cos(

2

22

θωωθθ

θωωθ

−=

−+

−=

−++

terrm

cr

term

kr

m

cr

&&&&

&&&&

ωθ =&

0=== rr &&&&&θ

Page 28: MV - Chap 3

Dr. Millerjothi, BITS Pilani, Dubai Campus

Assume that and with integration , where φis the phase angle between e and r, which is now a constant.

With

For phase angle

Synchronous whirl:

ωθ =& φωθ −= t

0=== rr &&&&&θ

2

2

22

1

2

tan

sin

cos)(

=

=

=−

n

n

erm

c

erm

k

ωω

ωω

ζφ

φωω

φωω

Page 29: MV - Chap 3

Dr. Millerjothi, BITS Pilani, Dubai Campus

Noting from the vector triangle,

222

2

222

2

21

)()(

+

=

+−=

nn

n

e

cmk

mer

ωω

ζωω

ωω

ωω

ω

22

2

2

cos

+

=

ωω

ωωφ

m

c

m

k

k

Page 30: MV - Chap 3

Dr. Millerjothi, BITS Pilani, Dubai Campus

The eccentricity line e = SG leads the displacement line r = OS

by the phase angle φ.

When the rotation speed coincides with the critical speed

or the natural frequency of the shaft in lateral

vibration, a condition of resonance is encountered.

mkn =ω

Page 31: MV - Chap 3

Dr. Millerjothi, BITS Pilani, Dubai Campus

Figure shows the disk-shaft system under three different

conditions

At very high speed, the center of mass G tends to approach th

e fixed point O, and the shaft center S rotates about it in a

circle of radius e

Page 32: MV - Chap 3

Dr. Millerjothi, BITS Pilani, Dubai Campus

The equation of motion

SUPPORT MOTION

)sin(sin

,

0)()(

2 tYytYmymkzzczm

thenyxzLet

yxcyxkxm

ωωω ==−=++

−=

=−+−+

&&&&&

&&&&

Page 33: MV - Chap 3

Dr. Millerjothi, BITS Pilani, Dubai Campus

The solution can be written as

For absolute motion

2222

2

tan,)()(

)sin(

ωω

φωω

ω

φω

mk

c

cmk

YmZ

tZz

−=

+−=

−=

tiiti

tiiti

ti

eXeXex

eZeZeZ

Yey

yzxx

ωϕϕω

ωφφω

ω

−−

−−

==

==

=

+=

)(

)(

,

Page 34: MV - Chap 3

Dr. Millerjothi, BITS Pilani, Dubai Campus

ti

ti

tii

i

titii

titiitiitii

Yeicmk

ick

eYicmk

Ym

eYZex

icmk

YmZe

YemeZeicmk

YemeZekeZeiceZem

ω

ω

ωφ

φ

ωωφ

ωωφωφωφ

ωωω

ωωω

ωωω

ωωω

ωωωω

)(

)(

)(

)(

)()()(

2

2

2

2

2

22

222

+−+

=

++−

=

+=+−

=

=+−

=++−

−−−

since

Page 35: MV - Chap 3

Dr. Millerjothi, BITS Pilani, Dubai Campus

( )( ) ( )

( ) ( )22

3

22

2

2

222

22

tan

,

21

21

cmkk

mc

cmk

ck

Y

X

nn

n

ωωω

ϕ

ωζω

ωω

ωζω

ωω

ω

+−=

+

+

=+−

+=

Page 36: MV - Chap 3

Dr. Millerjothi, BITS Pilani, Dubai Campus

| X/Y | =1 at the frequency ω/ωn = √2

( ) ( )22

3

22

2

2

tan

21

21

cmkk

mc

Y

X

nn

n

ωωω

ϕ

ωζω

ωω

ωζω

+−=

+

+

=

Page 37: MV - Chap 3

Dr. Millerjothi, BITS Pilani, Dubai Campus

Isolation System

To protect a delicate object from excessive vibration transmitted to

it from its supporting structure, or

To prevent vibratory forces generated by machines from being

transmitted to its surroundings

The motion transmitted from the supporting structure to the mass

m is less than 1 when the ratio ω/ωn is greater than √2.

This can be done by using a soft spring.

Reducing the transmitted force:

VIBRATION ISOLATION

Page 38: MV - Chap 3

Dr. Millerjothi, BITS Pilani, Dubai Campus

The force to be isolated is transmitted through

the spring and damper, as shown in Fig.

12 <→>Y

X

nωω

when2

k

2

kc

F

Xcω

kX

φF

Xm 2ω( ) ( )

22

2

0

0

2

22

21

,sin

21

+

==

+=+=

nn

n

T

kF

XtFF

kXXckXF

ωςω

ωω

ω

ωςω

ω

when

Page 39: MV - Chap 3

Dr. Millerjothi, BITS Pilani, Dubai Campus

The transmissibility, defined as the ratio of the transmitted

force to that of the disturbing force.

( ) 12

1

1

1,

21

21

,

2

2

2

22

2

2

−∆

=→∆

==

=

+

+

==

2

gf

TRg

m

k

TR

F

FTR

n

n

n

n

D

T

πω

ωω

ωςω

ωω

ωςω

dampingNeglecting

bilityTransmissi

Page 40: MV - Chap 3

Dr. Millerjothi, BITS Pilani, Dubai Campus

Comparison of the preceding equation with the steady state

amplitude and phase

To reduce the amplitude X of the isolated mass m without

changing TR, m is often mounted on large mass M, as shown in

Figure. The stiffness k then be

increased to the ratio k/(m+M) constant.

The amplitude X is reduced because k

appears in the denominator

Y

X

F

FTR T ==

0

Page 41: MV - Chap 3

Dr. Millerjothi, BITS Pilani, Dubai Campus

Page 42: MV - Chap 3

Dr. Millerjothi, BITS Pilani, Dubai Campus

Page 43: MV - Chap 3

Dr. Millerjothi, BITS Pilani, Dubai Campus

Energy in a vibrating system is either dissipated into heat

or radiated away.

Energy dissipation:

• Usually determined under conditions of cyclic oscillations

• Force-Displacement relationship

• The force-displacement curve will enclose an area as

hysteresis loop

• Proportional to the energy lost per cycle

• The energy lost per cycle due to a damping force Fd

ENERGY DISSIPATED BY DAMPING

Page 44: MV - Chap 3

Dr. Millerjothi, BITS Pilani, Dubai Campus

Damping is present in all system

internal molecular friction – Structural damping

sliding friction – Coulomb damping

fluid resistance – Viscous damping

Since mathematical description of damping is quite complicate, simplified damping modal such as viscous has been developed to evaluate the system response.

Hysteresis loop : area enclosed by force-displacement curve.

proportional to the energy lost per cycle.

where is the energy lost per cycle due to a camping force

ENERGY DISSIPATED BY DAMPING

∫= dxFW dddW

dF

Page 45: MV - Chap 3

Dr. Millerjothi, BITS Pilani, Dubai Campus

In general, depends on many factors, such as temperature,

frequency, or amplitude.

Case of spring-mass system with viscous damping:

The steady-state displacement and velocity

The energy dissipated per cycle

dW

xcFd&=

( )φω −= tXx sin

( )φωω −= tXx cos&

22

0

222

2

)(cos XcdttXc

dtxcdxxcWd

ωπφωωωπ

∫∫=−=

== &&

Page 46: MV - Chap 3

Dr. Millerjothi, BITS Pilani, Dubai Campus

Of particular interest is the energy dissipated in forced vibration at

resonance.

With and ,

Then

and the damping force becomes

By rearranging the foregoing equation to

mkn =ω kmc ζ2= 22 kXWd ζπ=

Page 47: MV - Chap 3

Dr. Millerjothi, BITS Pilani, Dubai Campus

The energy dissipated per cycle

is the area enclosed by the

ellipse.

If we add to the force kx,

the hysteresis loop is rotated as

shown in Fig ⇒Voight model

Specific Damping Capacity:

The energy loss per cycle divided by the peak potential energy

Loss Coefficient: The ratio of damping energy loss per radian

divided by the peak potential or strain energy

dF

π2/dW

Page 48: MV - Chap 3

Dr. Millerjothi, BITS Pilani, Dubai Campus

Equivalent damping Ceq: equate the energy dissipated by the

viscous damping to that of the nonviscous damping force

with assumed harmonic motion

Equivalent Viscous Damping

n

deqeqd

c

FX

X

WCXCW

ω

πωωπ

0

2

2

=

=→=

resonant&dampingviscous

Page 49: MV - Chap 3

Dr. Millerjothi, BITS Pilani, Dubai Campus

From experiments of most structural metals, damping is

independent of the frequency and proportional to the square

of the amplitude

Structural Damping

tFkxxxm

CXC

XW

eqeq

d

ωπωα

πωα

ωπ

α

sin0

2

2

=+

+

=→=

=

&&&

Page 50: MV - Chap 3

Dr. Millerjothi, BITS Pilani, Dubai Campus

Complex stiffness

( )( )

( ) kimk

FX

ikkeFxikxm

eFxikxm

xieXixeXx

ti

ti

titi

γω

γπα

γγ

πα

ωω

ω

ω

ωω

+−=

+

==++

=

++

===

20

0

0

:1

:,

,

stiffnesscomplex

factordampingstructural

motionharmonicassumestiffnesscomplex

&&

&&

&

Page 51: MV - Chap 3

Dr. Millerjothi, BITS Pilani, Dubai Campus

Compare to viscous damping at resonance

k

FX

γ0, =resonanceAt

ςφς

22

0 =∴⇒=k

FX

( ) ( )

( ) ( ) 222222

2

2

0

1,

1

1

1

1

,

γ

γ

γ

φ

ωω

+−

−=

+−

−=

+=+−

==

=

ry

r

rx

iyxirF

XrH

rn

where

FRF

Let

0

γ2

1

H(r)

-y

x

Page 52: MV - Chap 3

Dr. Millerjothi, BITS Pilani, Dubai Campus

The following equation is that of a circle with radius,

and centre,

( )γ

γ

γγ

irH

yxr

yx

−=

−===

=

++

1,0,1,

2

1

2

122

2

resonanceAt

Page 53: MV - Chap 3

Dr. Millerjothi, BITS Pilani, Dubai Campus

In forced vibration, there is a quantity Q related to damping that is

a measure of the sharpness of resonance.

Assume viscous damping

SHARPNESS OF RESONANCE

k

FX

F

Xk

tFkxxcxm

res

n

n

n

nn

ςωω

ωω

ωω

ςφ

ωω

ςωω

ω

2,1

1

2

tan,

21

1

sin

0

222

20

0

==

=

+

=

=++

when

&&&

Page 54: MV - Chap 3

Dr. Millerjothi, BITS Pilani, Dubai Campus

( )pointspowerhalfSidebands −⇒

= resres XXX

2

1707.0:

0X

X

1ω 2ω

ζ2

1

ζ2ζ22

1

nωω

Page 55: MV - Chap 3

Dr. Millerjothi, BITS Pilani, Dubai Campus

( ) 22

2

2

2

22

2

222

00

1221

418

21

1

22

1

2

ςςςωω

ωω

ςωω

ς

ωω

ςωω

ς

−±−=

+

−=⇒

+

===

n

nn

nn

res

F

kX

F

Xk

Page 56: MV - Chap 3

Dr. Millerjothi, BITS Pilani, Dubai Campus

n

nnn

n

n

n

ωωω

ωωω

ωωω

ωωω

ς

ςωω

ςωω

ςωω

ς

12

1212

2

2

1

2

2

2

2

2

1

2

2

4

21

21

21

1

−≅

+−

−=

−=

+=

−=

±=

<<Let

γ

ςωωω

1dampingstructuralFor

2

1

1212

=

=−

=−

=

Q

ff

fQ nn

Page 57: MV - Chap 3

Dr. Millerjothi, BITS Pilani, Dubai Campus

Seismic Unit of Fig

Displacement, velocity, or acceleration is indicated by the relative

motion of the suspended mass with respect to the case.

Vibration Measuring Instruments

m

k

c

x

y

tYmkzzczm

tYyyxz

yxkyxcxm

ωω

ω

sin

sin,

0)()(

2=++

=−=

=−+−+

&&&

&&&&

Let

motion suppot as Same

Page 58: MV - Chap 3

Dr. Millerjothi, BITS Pilani, Dubai Campus

In support motion x/y is of interest,

z/y is of interest for measuring instruments.

( ) ( ) ( )

=−

=

+

=+−

=

22

2

2

2

2

222

2

1

2

tan

21

n

n

nn

n

mk

cand

Y

cmk

YmZ

ωω

ωω

ζ

ωω

φ

ωω

ζωω

ωω

ωω

ω

Page 59: MV - Chap 3

Dr. Millerjothi, BITS Pilani, Dubai Campus

Response of a vibration-measuring instrument

Page 60: MV - Chap 3

Dr. Millerjothi, BITS Pilani, Dubai Campus

Seismometer is an instrument with low natural frequency

When the natural frequency is low in comparison to the vibration

frequency to be measured Z approaches Y regardless of the

damping ζ.

The mass m then remains stationary while the supporting case

moves with the vibrating body.

Such instruments are called seismometers

One of the disadvantages of the seismometer is its large size.

Seismometer

size large Reguire 1 , 2.5 r seismomete →≈

⟩=

y

z

nωω

Page 61: MV - Chap 3

Dr. Millerjothi, BITS Pilani, Dubai Campus

Accelerometer: instrument with high natural frequency

When the natural frequency of the instrument is high compared

to that of the vibration to be measured, the instrument

indicates acceleration.

The factor approaches unity for

ω/ωn → 0, so that

Accelerometer

Page 62: MV - Chap 3

Dr. Millerjothi, BITS Pilani, Dubai Campus

The useful range of the accelerometer can be seen from figure

below for various values of ζ

Page 63: MV - Chap 3

Dr. Millerjothi, BITS Pilani, Dubai Campus

The useful frequency range of the undamped accelerometer is

somewhat limited.

With ζ = 0.7, the useful frequency range is 0 ≤ ω/ωn ≤ 0.20

with a maximum error less than 0.01 percent. Thus, an instrument

with a natural frequency of 100 Hz has a useful frequency

range from 0 to 20Hz with negligible error.

Page 64: MV - Chap 3

Dr. Millerjothi, BITS Pilani, Dubai Campus

To reproduce a complex wave without changing its shape, the

phase of all harmonic components must remain unchanged with

respect to the fundamental.

This requires that the phase angle be zero or that all the

harmonic components must be shifted equally.

The first case of zero phase shift corresponds to ζ = 0 for

ω/ωn < 1.

The second case of an equal time- wise shift of all harmonics is

nearly satisfied for ζ = 0.7 for ω/ωn = 1.

When ζ = 0.7 for ω/ωn < 1 can be expressed as

Phase distortion