July 7, 2020

Silent Engineering(Lecture 3)

Forced Vibration Analysis of Plate and Formation of Sound Field

Due to Vibrating Source- Force Vibration and

Point sound source and wave equation -

Tokyo Institute of TechnologyDept. of Mechanical EngineeringSchool of Engineering

Prof. Nobuyuki Iwatsuki

１. Examples of Forced Vibration Based on Approximated Modal Analysis withRayleigh-Ritz Method

1.1 Vibration responses of peripherally clamped thin rectangular plate

Thickness: hDensity: ρYoung’s modulus: EPoison’s ratio: νPoint driving force: P(t)=P0cosωt

xa

b

O

y

Driving point

Response point

(xd,yd)

(xr,yr)

( )( ) ( )

)1(2tan

2

cos),(),,(

221

2222

Ni

tyxWQtyxw

i

iii

iiii

iii

～　　　　　　

　

=−

=

+−

−=

−

∑

ωωωωζφ

ωωζωω

φω

Vibration displacement:

(1)

),()( ddii yxWtPQ =Modal load for point excitation:

(2)Because the plate is excited by a force, P(t), at a certain point(xd,yd) , the distributed external force, p(x,y,t) can be represented as

　

)()()(),,( dd yyxxtPtyxp −−= δδwhere δ denotes Dirac’s delta function

),()(),( ddS iii yxWtPdsyxpWQ ∫ ==

Therefore Explained in the 1st lecture

Vibration velocity:( )

( ) ( )∑

+−

−−=

iiii

iii tyxWQtyxw2222 2

sin),(),,(ωωζωω

φωω　 (3)

( )( ) ( )

∑+−

−−=

iiii

iii tyxWQtyxw2222

2

2

cos),(),,(ωωζωω

φωω　

Vibration acceleration:

(4)

where

−−−⋅

−−−=

⋅=

∑

∑

by

by

by

by

ax

ax

ax

ax

C

yWxWCyxW

jyjyjy

jyjy

ixixix

ixix

jiij

jyji

ixij

,,,

,,

,,,

,,

,

,,

,

sinsinhcoscosh

sinsinhcoscosh

)()(),(

λλα

λλ

λλα

λλ

　　　　　　　　

　　　　

(5)i λi

1 4.7302 7.8533 10.996: :

Coefficients, Cij, are set so that modal mass becomes unity.

Acceleration distribution, w(x,y,t), can be calculated.

The following frequency spectra of transfer functions are also calculated.

Compliance :)()()(

ωωω

jFjXjD = (6)

)()()(

ωωωjFjVjY = (7)Mobility :

)()()(

ωωωjFjAjG = (8)Accelerance :

Modal damping ratio, ζi, can be given as the experimentally measuredor estimated values.

FFT Analyzer

Modal Analysis SoftwareLX-cada

Accelerometer

ExciterForcesensor

Setup for Experimental Modal Analysis

Mode ζ

(0,0) 0.0222(1,0) 0.0149(2,0) 0.0242(0,1) 0.0541(1,1) 0.0248(3,0) 0.0142

Mode ζ

(2,1) 0.0120(4,0) 0.0130(3,1) 0.0095(0,2) 0.0100(1,2) 0.0211(4,1) 0.0109

Mode ζ

(5,0) 0.0254(2,2) 0.0148(3,2) 0.0115(5,1) 0.0120(6,0) 0.0491(4,2) 0.0106

The experimentally measured modal damping ratio

Frequency spectra of acceleranceof peripherally clamped rectangular thin plate

Driving point: (xd,yd)=(113.4,149.25)Response point: (xr,yr)=(1100.0,450.0)

(2,0)(1,0)

(0,0)

(0,1)(1,1)(2,1)

(4,0)(3,1)

(4,1)

(2,2) (3,2)

(5,1)

(4,2)

Driving point: (xd,yd)=(113.4,149.25)Response point: (xr,yr)=(663.4,374.75) Node

(0,0)(2,0)

(4,0)(2,2) (4,2)

(0,2)

1.2 Examples of calculation

Accelerance takes its peak values at natural frequency

In case where response point is located on nodal line, corresponding mode is not induced.

Frequency spectra of acceleranceof peripherally clamped rectangular thin plate

Driving point: (xd,yd)=(663.4,374.75)Response point: (xr,yr)=(1100.0,450.0)

Driving point: (xd,yd)=(113.4.4,374.75)Response point: (xr,yr)=(1100.0,450.0)

NodeNode

In case where driving point is located on nodal line, corresponding mode is not induced.

2. Sound Generation Mechanism2.1 Fundamental equation of sound wave

Sound pressure applying to infinitesimal element of medium

Sound pressure : p(x,y,z)Density of medium : ρDisplacement of medium ; d(ξ,η,ζ)

By propagating sound to the element, density, ρ, pressure at a point (x,y,z) and volume,V, were changed as ρ0, P0 , V0 → ρ, P, V

The sound pressure at the point can be defined as

)(),,,( 0 tPPtzyxp −= (6)The volume is calculated as

zyxzyx

zyxzyx

V

∆⋅∆⋅∆

∂∂

+∂∂

+∂∂

+≅

∆⋅∆⋅∆

∂∂

+

∂∂

+

∂∂

+=

ζηξ

ζηξ

1

111

　 (7)

The ratio of volume change is given as

　

dd •∇==∂∂

+∂∂

+∂∂

=−

=∆ div0

0

zyxVVV ζηξ (8)

Since mass of the elements should be kept, the following equation holds.VV ρρ =00 (9)

Therefore )1(0 ∆+= ρρ (10))1(0 s+= ρρ (11)

:∆ dilatation

s: condensation

∆−=s (12)

( )　

ζηξ ,,u =

By differentiating Eq.(8) with respect to time, we obtain

　

ud divdiv =∂∂

=∂∆∂

ttwhere particle velocity is

(13)

(14)From Eqs.(10) and (13), we obtain

　

ud divdiv0 ==∂∂ ρρ

t(15)

Equation of continuity

Equation of motion in x-direction can be expressed as

( )

　

　　　　　　　　 zyxxp

zyxxppzyx

xpp

tzyx

∆⋅∆⋅∆∂∂

−=

∆⋅∆

∆

∂∂

+−∆⋅∆

∆

∂∂

−=∂∂

∆⋅∆⋅∆21

21ξρ

Namely

　

xp

t ∂∂

−=∂∂ξρ (16)

Also in y- and z- direction, we obtain

　

p-pt

∇=−=∂∂ graduρ (17)

In the infinitesimal element, the following equation holds

　

tts

t ∂∂

≅∂∂

+=∂∂ uuu

00 )1( ρρρ (18)

2.2 Wave equation

　

∆−== KKspBy using bulk modulus, K,

(20)

By differentiating Eq.(19) with respect to time, we obtain

　

　　　　　　

　　　　　　

udivKzyx

K

zyxtK

tK

tp

−=

∂∂

+∂∂

+∂∂

−=

∂∂

+∂∂

+∂∂

∂∂

−=∂∆∂

−=∂∂

ζηξ

ζηξ

(19)

By differentiating Eq.(20) with respect to time, we obtain

　

　　　　　　

　　　　　

　　

　

pKzp

yp

xpKpK

tK

tp

2

2

2

2

2

2

2

2

2

grad1-divdiv

∇=

∂∂

+∂∂

+∂∂

=

−=

∂∂

−=∂∂

ρ

ρρu

(21)

　

　　　　　　

　　　　　

pctp 222

2

∇=∂∂

Wave equation

By setting sound speed as

　

　　　　　　

　　　　　

ρKc = (22)

(23)

The wave equation of particle velocity is also given as

　

　　　　　　

　　　　　

uu divgrad2

2

•=∂∂

ρK

t(24)

Wave equation with respect to particle velocity

2.3 Velocity potential

By using velocity potential, φ, we obtain

tp

ztytxt ∂∂

=∂∂

=∂∂

∂∂

=∂∂

∂∂

=∂∂ φρφζφηφξ

　　　 ,,, (25)

The wave equation can be rewritten as

　

　　　　　　

　　　　　

∂∂

+∂∂

+∂∂

=∇=∂∂

2

2

2

2

2

2222

2

2

zyxcc

tφφφφφ (26)

Wave equation with velocity potential

2.4 Solution of wave equationPlane sound wave:

　

　　　　　　

　　　　　

2

22

2

2

xc

t ∂∂

=∂∂ φφ

(27)

Wave equation:

Solution:

　

　　　　　　

　　　　　

　　　 )()(

)cos()cos(),(kxtjkxtj BeAe

kxtBkxtAtx−− +=

++−=ωω

ωωφ

(28)

where

　

　　　　　　

　　　　　

λπω 2

==c

k wavenumber (29)

Spherical sound wave:Sound wave radiating from spherical sound source

Wave equation:

　

　　　　　　

　　　　　

∂∂

+∂∂

+∂∂

=∂∂

2

2

2

2

2

22

2

2

zyxc

tφφφφ

Spherical coordinate system

By substituting coordinate transform as

　

　　　　　　

　　　　　

θϕθϕθ

cossinsincossin

rzryrx

===

(30)

into the wave equation, we obtain

　

　　　　　　

　　　　　

∂∂

+

∂∂

∂∂

+∂∂

+∂∂

=∂∂

2

2

2222

22

2

2

sin1sin

sin12

ϕφ

θθφθ

θθφφφ

rrrrrc

t (31)

Since velocity potential is a function of the radius, we obtain

　

　　　　　　

　　　　　

2

22

2

2 )()(rrc

tr

∂∂

=∂

∂ φφ(32)

Solution:

　

　　　　　　

　　 )(

)cos(krtjAe

krtAr−=

−=ω

ωφConsider only divergent wave

　

　　　　　　

　　 )(

)cos(

krtjerA

krtrA

−=

−=

ω

ωφTherefore

(33)

Sound pressure and particle velocity can be given as

　

　　　　　　

)(

)(

1 krtjr

krtj

erAjk

rr

erAj

tp

−

−

+=

∂∂

−=

=∂∂

=

ω

ω

φξ

ωρφρ

(34)

(35)

Sound wave from pulsating sphere:Point sound source

Let’s consider the harmonic vibration in radial direction on sphericalsound source with radius r0.

tjr e ωξξ 0

=

Since particle velocity on spherical surface is equal to the harmonic vibration velocity,the following equation holds.

　

　　　　　　

tjkrtjr ee

rAjk

rωω ξξ 0

)(

00

01 =

+= − (36)

Therefore

　

　　　　　　

00

20 1

0

jkrerA

jkr

+= ξ (37)

If

　

　　　　　　

10 <<kr

　

　　　　　　

　　　　 02

00

0

44

ξππ

rA

AA

=

=

where

(38)

　

　　　　　　

)(0

)(0

)(0

41

4

4

krtjr

krtj

krtj

er

Ajkrr

er

Ajt

p

er

A

−

−

−

+=

∂∂

−=

=∂∂

=

=

ω

ω

ω

πφξ

πωρφρ

πφ

Therefore, velocity potential, sound pressure and particle velocity can be represented as

(39)

(40)

(41)

Important equations for point sound source

Point sound source on a plane:

Low of mirrors

Low of mirrors

Since a rigid plane does not vibratethe particle velocity on the reflectionpoint on the plane should be zero and there is no change of velocitypotential based on the low of mirrors.

Point sound source of mirror image

Actual point sound source

Reflection point

Rigid plane

Observationpoint

r1

r2

Therefore the velocity potential at an observation point P can be given as

　

　　　　　　

)(

2

0)(

1

0 21

44krtjkrtj e

rAe

rA −− += ωω

ππφ (42)

Therefore let’s consider two point sound sources with same sound intensity and phase just at both side of the rigid plane.

Observationpoint

Two point sound sources at both side of rigid plane

The following conditions are takeninto account.

　

　　　　　　

θ

θ

sin2

sin2

2

1

drr

drr

+≅

−≅r>>d

(43)

By substituting Eqs.(43) into Eq.(42), we obtain

　

　　　　　　

+=

−− θθω

πφ

sin2

1

sin2

1

)(0 114

kdjkdjkrtj er

er

eA (44)

　

　　　　　　

rrrr ≅≅ 21 ,By assuming that ,

　

)sin2

cos(2

)(0 θπ

φ ω kder

A krtj −= (45)Directivity

Let assume that d→0

　

)(0

2krtje

rA −= ω

πφ (46)

Sound radiation from vibrating plate:Let’s assume that many point sound source are located on the vibrating plate and radiate sound.

Infinite rigid plane

Vibratingplate

　

　　 )(0

)(0

2

2krtj

krtj

er

dS

er

Ad

−

−

=

=

ω

ω

πξπ

φ

Velocity potential at an observationpoint due to a point sound sourceof infinitesimal element onthe vibrating plate can begiven as

(47)Velocity potential at an observationpoint can then be given as

　

　　

１∫ −=S

)(0

2dSe

rkrtj ωξ

πφ

(48)

　

　　

　　　　 ∫

∫

−

−

=

=∂∂

=

S

)(0

S

)(0

2

2

dSer

dSer

jt

p

krtj

krtj

ω

ω

ξπρ

ξωπρφρ

Sound pressure at the observation point can then be given as

Sound pressure can be calculated with vibration acceleration distribution

on vibrating plate

(49)

2.5 An example of calculationLet’s consider a simple case where a peripherally simply supported thin rectangular plate vibrates with a certain vibration mode shape at its natural frequency. All vibration modes have the same maximum vibration amplitudeof 10mm.

Calculated in last week.

NnMmyb

yna

xmCtyxw mn ,...,2,1,,...,2,1,cossinsin),,( === 　　ωππ

(50)

a=1326.8mm b=749.5mm h=4.5mm E=205GPaν=0.3ρ=7860kg/m3

　　

　　　　　　　 dxdyd

tyxw

dSd

tyxwtzyxp

b aair

Sair

PPP

∫ ∫

∫

=

=

0 0

)',,(2

2)',,(),,,(

πρ

πρ

( ) ( )

　　

air

PPP

vdttyyyxxd

/'

222

−=

+−+−=

Sound pressure at the observation point:

where

xy

z

O

dxdy

),,( PPP zyxObservation point P:

d

ab

(1,1)

(3,1) (5,1)

(1,3)(3,3)

(7,1)(5,3)

(2,1)

(1,2)

(2,2)

(4,1)(3,2)

(4,2)

(2,3)

(5,2)

(6,1)

(6,1)

(6,2)

(7,2)

(6,3)

SPL of the mode having oddnumber nodal lines is small.

Pa20,log20 10 µ== refref

RMS pppSPL 　　

Sound pressure level:(50)

Observation point: )0.1000,75.474,4.763(),,( =OOO zyxA little bit far from the center of plate

Effect of mode shape:

＋

＋

－－

＋

＋

－－

＋＋

－－

＋

A vibration mode having odd number nodal lines

A vibration mode having even number nodal lines

Sound pressures formed by vibration area with opposite phases tend to offset each other.

ds

ds

P P

d

d ds

ds

d

dw‥

-w‥

w‥

w‥

Interference of two soundwaves with opposite phase minimize SPL.

Interference of two soundwaves with the same phase make SPL twice.

(7,1) mode at 323.51Hz, The observation point :(0.0, 0.0, zO)

DISTANCE zO m

Distance

SPL is affected by the distance between sound source and observation point.

Effect of distance:

z

xy

O

θ

ϕ

Rd

Observation point P:

)cos,sinsin,cossin(),,(

　

　　

θϕθϕθ RRRzyx PPP

=

Vibrating rectangular

Acceleration w(x,y)‥

Sound pressure p

Effect of direction:

Azimuth

Zenith angle

θ=0° 10°

20°

30° 40°50°

60°

70°

80°

(7,1) mode at 323.51Hz, Radius of the observation point is 1m

SPL is affected by the dirction angle from sound source to observation point.

Please note that these results are obtained froma simple vibration such as only a certain mode shapeand that we will have to consider the response as thesummation of all mode of vibration and the responsecharacteristics which should be determined by excitingpoints.

3. Concluding remarksExamples of forced vibration analysis and soundgeneration mechanism are explained. (1)Acceleration distribution on vibrating plate can

be calculated based on modal analysis of the plate.

(2)Frequency spectrum of accelerance takes peakvalues at natural angular frequency.

(3)Vibration response is strongly affected by positions of excitation and response.

(4)Sound radiation can be calculated by solvingsound wave equation of velocity potential.

(5)Sound pressure radiating from a vibrating platecan be calculated with accelerationdistribution of vibrating plate.

(6)Sound pressure radiating from vibratingplate is strongly affected by mode shapeand distance and direction between sound source and observation point.

3. Concluding remarks