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extF

)sin(

00)/(

2

2

on

n

n

tCxxx

xxxmkx

xmkx

1

A. Free vibration = 0

Chapter 8 Vibration

Undamped free vibration mk

x

x

t

C0x T

n= (k/m)1/2: natural frequencyC : amplitude o : initial phase anglexo : initial displacement, xo= C sin(o )T : period, such that Tn =2, or T = 2 /n

f : frequency = 1/T

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2

Damped free vibration :

xmxckxF

0 20)/()/(

0

2

xxxxmkxmcx

kxxcxm

nn

nm

c

2factor damping viscousDefine

Define viscous damping constant c (N s/m)

...because... ,e e issolution general The

e and e :solutions twoare thereHence,)1( :are s' possible Two

02or 0)e()e(2)e(

21

2211

2

22

22

tt

ttn

nn

tn

tn

t

AAxAxAx

AAA

Use a trial solution x = A et :

k

0x

m

x)(dashpot

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3

solution. a is Hence

0)(d

)(d 2

d

)(d0 2 and 0 2

:solutions are and If

:note Mathematic

21

21221

221

22

2221

211

21

xx

xxt

xx

t

xxxxxxxx

xx

nn

nnnn

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4

Case 1 : Overdamped when > 1, such that + and - < 0

tt eAeAx

2

1

x decays to zero without oscillation

t

x

1,

overdamped1Critical

damped

1 underdamping

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5

Case 2: Critical damping when = 1 + = - = -n

General solution : x = (A1 + A2 t )exp(-nt)x approaches zero quickly without oscillation.

R.H.S.0

e)2 2( e2e e2e)(

e)ee(2d

)ee(d

)e(d

)e(d2

d

)e(d L.H.S./

02 02 From

solution. a also is e that prove toisnext The

solution. a is ee

22

22

2

22

2

2

2

2

22

111

tnnnn

tn

tn

tn

tn

tn

ttnn

ttn

tn

t

n

tnn

nn

t

tt

n

nnnn

nnn

nn

n

nn

n

n

tttt

ttt

t

tt

t

t

tA

xxxxxx

tAxAAx

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6

Case 3: Underdamped < 1

! real )( cos C)( (C/2)2cos

])sin( i-)( cos )sin( i)( cos[(C/2))((C/2) )ee((C/2)

(C/2)e and (C/2)e conjugate.complex are and that Assume number. real a is

)( 1frequency natural damped Define

)(

1)1(

d

d

dddd

)()(

i-i

i-2

i1

21

21

2d

12

11

21

22

dd

dd

dd

22

teet

etttteeeeeex

AAAAx

eeAeAx

eeAeA

eAeAxi

t

t

t

ttiti

ttiti

ttitin

ttiti

ttnnn

n

n

n

n

n

n

nnn

d

1x 2x

1t 2t

x

tPeriod d = 2/d

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7

22

21

1

2

1

2/decrement clogarithmi Define

n

ndn

xxn

Experimental guideline: • Measure and n

• Calculate • Calculate viscous damping constant c according to = c/(2mn)

dn

dn

n

eCe

Ce

x

xt

t

)(2

1

1

1

cycles successive twoof ntsdisplaceme theof ratio The

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8

xmtFxckxF o sin

B. Forced vibration of particles :

The equation of motion :

tFkxxcxm o sin

tmFxxx o

nn sin2 2

kcx

0x)(dashpot

tFF O sin

)(spring

m

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9

sin2)cos

/ (2) From

(1) From

(2) 0sincos2sin(1) /cos sin2cos

0cos]sincos2sin[sin]/cos sin2cos[

sin)/()sin( )cos(2)sin(sin)/(2

)sin(

)cos( then),sin( Let

22

22

22

22

22

22

2

2

nn

o

nn

onn

nn

onn

onn

onn

p

p

p

mFC

CCCmFCCC

tCCCtmFCCC

tmFtCtCtCtmFxxx

tCx

tCxtCx

222 )(

22tan

n

n

n

n

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n2

2)( n

222 /2/1 nn

222 )(

22tan

n

n

n

n

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2222

2222

222

222

22

(4])(-[1

)4])(-[1)/(

)2

2])(-)[1

)4])(-[1)/(

sin2)cos

/ (2) From

nn

nnno

nn

nn

nno

nn

o

mF

mF

mFC

222

222

/2/1

1

/factor ion Magnificat

/2/1

/

nno

nn

o

kF

CM

kFC

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12

M

n /

1

1

01.0

2.01

Maximum M occurs at: 0 })2

(])(1{[d

d 222 nn

nres 21 2The resonance frequency is

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13

n /1)(resonance

02.0

1

0

2/

Consider the following regions:

(1) is small, tan > 0, 0+, xp in phase with the driving force(2) is large, tan < 0, 0-, = , xp lags the driving force by 90o

(3) n-, tan +, /2(-)

n+, tan -, /2(+)

tan = 2n

n

)(

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14

If the driving force is not applied to the mass, but is applied to the base of the system:

tsin2 22nn bxxx

If b2 is replaced by Fo/m:

)( Bxxmkxxc

k

c

x0x

tbxB sin

m

tsin2 2nn

m

Fxxx o

This can be used as a device to detect earthquake.

Bxmkxxcxm

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15

Determine : (a) steady-state displacement (b) max. force transmitted to the base.

k c

m

pA

m =45 kg, k = 35 kN/m, c = 1250 N.s/m, p = 4000 sin (30 t) Pa, A= 50 x 10-3 m2.

Example

)(498.0)9.27452/(1250)2/(/9.2745/1035/ 3

dunderdampemcsradmk

n

n

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16

The amplitude of the steady-state vibration is:

m

nn

koFX

00528.02/12

79.2/30498.0222

9.27/301

31035/

310504000

2/12

/222

/1

/

rad

n

n716.12

9.27/301

9.27/30498.021tan2

/1

/21tan

#)716.130sin(00528.0)sin( mttXxP

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17

The force transmitted to the base is :

)(cos)(sin

tXctkXxckxF pptr

For max Ftr :

kct

tddFtr

1tan0

o

3

31 9.46

10351030351250tan

t

#

o

o3

max

2719.46sin00528.0301250

9.46cos00528.01035

N

Ftr