vibs lecture 2

8/7/2019 Vibs Lecture 2

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Always the Same Equation of Motion

02

2

! Kxdt

xd

M

Mass m on a spring:

Pendulum:

02

2

! k xdt

xd m

0)/(2

2

! UU

Lmg dt

d m

04

52

2

! k xdt

xd mTwo-mass/Pulley system:

Equation of Motion always of the form:

M = ³mass-like´ quantity

K = ³spring-like´ quantityx = ³position-like´ variable

02

2

! x

M

K

dt

xd General solution can only depend on

the combined quantity K/M

Rotating bar: 02

21

2

2

! UU

kLdt

d I c



)(sin)( ] [ ! t X t x n

2

2

¹¹ º

¸©©ª

¨!

n

oo

v x X

[ ¹¹

º

¸©©

ª

¨!

o

no

v

x [] 1tan

X

-X

ov

]

Period of oscillation:

2

n

T

X [ ! (seconds)

Natural Frequency:n[ (radians/s) (Hz or cycles per second)

2

nn

f [

T !

(Amplitude) (phase angle)

So«..always the Same solution:



Strategy for Problem Solving:

2

n[

Very straightfor ward ± the subtleties are in getting the initial conditions

and the equilibrium position, but these come from other considerations

Apply initial conditions to determine amplitude and phase

Draw F.B.D.

0)/(2

2

! UU

Lm g dt

d mGet Equation of Motion

0)/(2

2

! UU

L g

dt

d Identify Natural Frequency of Vibration

)(sin)( ] [U 5! t t nWrite down general solution for motion



Damped Vibrations

�Force depends on

Velocity

� Pull quickly larger resistance

� Pull slowly small resistance

� Force opposes direction of (linear) motion

� Mechanical Energy is lost; dissipated in moving viscous fluid

All systems have some damping ± friction, air resistance, etc.

Here, we consider damping by ³linear viscous dampers´, i.e. ³dashpots´

Dashpot (viscous damper ):

Piston filledwith a viscous fluid

Resisting force cv F !

Velocity v

c = viscous damping coefficient



A true dashpot consists of a plunger inside a fluid-filled cylinder.

³c´ then depends on the fluid viscosity and plunger dimension

� If A}1 cm2:

� cair =2.6x10-8 N-s/m

� cw

ater =10 x c

air

� coil=103 x cwater

� cmaple syrup=10 x coil

�cchoc syrup=

10x cmaple syrup

� cpeanut butter =10 x cchoc syr

(Other types of damping exist and obey different equations, e.g. friction, but

we stick with a linear damping ± usual assumption for many engineering

systems)



Damped Vibrations: Equation of Motion

x measured from the

equilibrium position

Equation of motion:2

2

dt xd m

dt dxck x !

Again introduced a dimensionless time t t m

k t n[ !!~

n

k

m[ !0

2

2

! xm

k

d

t

d x

m

c

d

t

xd

0~~2

2

!¹¹

º

¸©©

ª

¨ x

t d

d x

m

c

t d

xd

n[

Another

dimensionless

parameter: nm

c

[

1/

/

1

/!!

sk g

sk g

sk g

m N s



n

k

m[ !0

2

2

! x

m

k

dt

dx

m

c

dt

xd

02

2

2

2

! xdt

dx

dt

xd

nn [ :[ nm

c

zeta [ : 2"" !!

³Viscous Damping Factor´

Characterizes strength of damping

Solution must be a function of time with two parameters: ),;( : [nt x



Damped Vibrations: Solution for Motion

( ) t x t CeP!Still try solution:

2 22 0t t t

n ne e eP P PP ^ [ P [ !

General solution is a sum of the two possibilities:

n

k

m

[ !

nm

c

[

:

2!02 2

2

2

! xd t

d x

d t

xd

nn

[: [

Plug in:

2 22 0n nP ^ [ P [ !

Characteristic

Equation

2 2 2

22 4 4 12

n n n

n n^ [ ^ [ [P ^ [ [ ^ s ! ! s

Solution of Characteristic Equation:

t t nnnn

eet x

)1(

2

)1(

1

22

)(

!

^ [^ [^ [^ [

M ATL AB TIME ±

let¶s just see what

comes out««

DEMO TIME ±

let¶s just seewhat r eally

happens««



t t nnnn eC eC t x)1(

2

)1(

1

22

)(

!^ [ ^ [ ^ [ ^ [

][)( 1

2

1

1

22 t t t nnn eeet x

!^ [^ [^ [

^ can vary from 0 to g can be positive, negative or zero21^

1^ "³Overdamped´:

12 ^ real

012

1 ! ^ [^ [P nn

0122 ! ^ [ ^ [ P nn

Damping is ³strong´

Motion decays exponentially ± no oscillations at all !

t t eet x 21

21)(PP

!



012 !^

n[ P !1 n[ P !2

t t t nnn eeet x[[[

!! )()( 2121

Only one constant C1+C2 ± not a general solution

³Critical Damping´: Damping just at a ³Goldilocks´ value1!^

Special C ase ± actual sol uti on is t net t x [! )()( 21

Why critical? Exponent is more negative than in the overdamped case !

12 ^ [ ^ [ [ nnn

³ Cr itical ́ :

F ast est r eturn of moti on t o z ero positi on



Damped

natural

frequency

12 ^ imaginary

x t e e en n nt i t i t ( ) !

^ [ ^ [ ^ [

1

1

2

12 2

Damping is ³weak´ ± most typical case³Underdamped´: 10 ^

][)( 1

2

1

1

22 t it it nnn eeet x^ [^ [^ [

!

21d n[ ^ [ ! ][)(

2

1

t it it d d n eC eC et x[ [ ^ [

!

Looks like undamped case but

with different frequency

)sin()( ] [^ [ ! t X et xd

t n

Oscillatory

motion

Decaying

exponential

0)( pgp t xt Motion always decays to zero; dashpot is

continually sucking energy out of the system



Underdamped Case 1^

( ) sinnt

d x t Xe t

^ [[ ]

!

2d

d

T X

[ !



Underdamped Case:

Observ e motion, ext r act system parameters

1

2

11

2 2

sin

sin

n

n

t

d

t

d

Xe t x

x Xe t

^ [

^ [

[ ]

[ ]

!

Experiment: Apply initial disturbance and measure response

Look at successive peak values at times t1 and t2=t1+X d

2 1 1sin sin 2 sin

d d d t t t [ ] [ T ] [ ] ! !

1

2 22

2 2ln

1 1n d n

n

x

x

T T ̂ H ^ [ X ^ [

[ ^ ^ ! ! ! !

³Logarithmic Decrement´:

1

1

1

2

n

n d

n d

t

t

x ee

x e

^ [^ [ X

^ [ X

! !

22

)2( T H

H ^

!

21

2

^ X

T [

!

d

n



Total weight W = 3400 lb

Weight supported by rear end W/2 = 1700 lb

Test #1: Apply 100 lb force; observe 3´ displacement

Test #2: Release force, measure rebound:

Rises then falls to 0.5´ below equilibrium, then oscillates back toward zero

1 spring k, 1 shock absorber c on each side

What are values of k and c?

-3.5

-3

-2.5

-2

-1.5

-1

-0.5

0

0.5

1

1.5

0 2 4 6 8

Task: Determine Degradation of Truck Suspension System



k

c c

k

Idealize system: mass on two springs,

two viscous dampers

Static deflection gives k: 01002 ! lbk st H

100200 /

2(0.25 )

lbk lb ft

ft ! !

2

2/75.2

/2.32/1700

/400)2( srad

s ft lb

ft lb

m

k n !!![

2 dashpots in parallel:

Rebound gives H ^ c: 79.1)"5.0

"3ln( !

!H

nm

c

[^

2

)2(!

274.0)2( 22

!

!T H

H ^

ft slb s

s ft

lbm

c n /8.39)274.0)(/75.2(/2.32

1700 2

2 !!! ^ [



Strategy for Problem Solving:

2

n[

Apply initial conditions to determine constants

Draw F.B.D.

2

2 0

d x dx

m ck xdt dt !Get Equation of Motion

2

2

0d x c dx k

xdt m dt m

!Identify parameters [n and ^ :

Write down general solution, for appropriate value of ^:

Put equation in standard form:

2

20

d x c dx k x

dt m dt m !

n^ [ 2

)(sin)( ] [^ [

!

t Xet x n

t n³Underdamped´: 10 ^

³Critical Damping´: 1!^ t n

et t x[

! )()( 21

³Overdamped´: 1^ t t

eet x 21

21)(PP

!

vibs lecture 2

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