damping lab 1

8/17/2019 damping lab 1

1/60

1)Nomenclature

Table 1 Nomenclature

Symbol Meanings Unitω Angular velocity ran

meq Equivalent mass kg

ζ Damping ratio rad/s

q Damping coefficient kg/s

f Frequency Hz

I 0 Inertia of beam about pivot O kgm

!

m "ass of load kg

ms "ass of spring kg

f N #atural frequency Hz

T $eriod s

V %elocity m/s


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2)Introduction:

In mec&anical vibration natural frequency and stiffness of materials in generatingvibration ' ( %ibration can be divide into t&ree group )

a( natural vibration

• natural motion generated from ob*ect containing some internal

energy suc& as teacup+ t&e energy come generated due to flickering

of t&e finger b( forced vibration

• %ibration due to forced vibration on a system E,ample+ unbalance

-&eel -ill cause ve&icle to vibrate equal to rotational speed of t&e

-&eelc( unstable vibration

• %ibration due to vibration motion -it&out e,ternal e,citation as t&e

ob*ect produce it o-n vibration force maintaining t&e motion

E,ample of unstable vibration is aerodynamic flutter

.&erefore+ understanding mec&anical vibration is vital to study t&e properties+ impact and

ability to govern t&e vibration in t&e design and operation of t&e mec&anical plant

.&e ob*ective of t&is e,periment -as to learn &o- spring stiffness+ natural frequency

and damping effect on system -it& single degree of freedom. this lab report are subdivided into four parts which are parts A, B, C and D. Part A is to measure

the spring stiness. Besides, for part B, the aim is to determine the

natural frequency of oscillation with and without lumped mass

correction. Furthermore in part C, the experiment is conducted on

iscous damping and its consequence to the natural frequency. !astly

part D, determine error analysis.


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3)Objectives:.&e aim of t&is one degree of freedom system e,periment are

a .o appro,imate t&e natural frequency of system equivalent to lumped parameter

system by using t&e formulaf N =

1

2 π √ k

m+ms3

= 1

2π √ k

meq

'Formula for lumped0parameter system frequency+ f lump= 1

2π √ k m ('1niversity !23!(

b Attest t&eory of damping ratio ' ζ (+ derived from t&e equation

( f f N )2

+ζ 2=1+ -&ic& states t&at small damping ratio -ill result in damped

natural oscillation frequency value is nearly t&e same as t&e undammed natural

oscillation frequency

c .o acquire t&e stiffness of a &elical spring

d .o analyze t&e uncertainties measurement t&at caused by error analysis

4)ApparatusApparatus -&ic& are used to conduct t&is e,periment are .ecquipment universal

vibration+ digital stop -atc&es+ accurate balance+ and ruler


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5)Part A: Measurin t!e sti"ness o# a sprin

5.1. Introduction

$art A is to resolve t&e spring stiffness of t&e spring -&ic& -as used during t&e e,periment-&ic& is used in t&e calculation of ot&er calculation .&e t&eory of Hooke4s la-+ mass added to

t&e spring s&ould not e,ceed t&e proportional limit

5.. !b"ectives

• .o determine t&e stiffness of t&e spring e,ert by different mass

• Evaluate value t&roug& calculation and t&e spring stiffness obtained from t&e grap&

5.#. Theory $ac%ground

5$3$1$ %oo&e's (a

Hooke4s 5a- states t&at t&e deforming force applied is directly proportional to t&e size

of deformation or elongation as long as do not e,ceed elastic limit

'&ttps//---3umnedu/s&ips/modules/p&ys/&ooke/&ooke&tm( Hooke4s 5a- can be

define mat&ematically as

F =kδ &quation 1

6&ere F 7 applied load or force '#(δ 7 deflection or deformation 'm(

k 7 spring stiffness '#/m(

8rap& of force against deflection %9 t&e stiffness plotted is essential in identifying t&e gradient

of t&e grap&

:;! 'east Squares method


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utilised t&e coordinates on a scatter plot s&o-n in &quation . Therefore( it -as used to

determine t&e gradient+ m and t&e spring stiffness+1

m

https)**+++.utdallas.edu*,herve*-bdi'eastSquares/0pretty.pdf

m=∑ xy−(∑ x ) (∑ y )

n

∑ x2−(∑ x )

2

n

&quation

5.. 2rocedure

"# $he hanger and spring is attached to the tecquipment uniersal i%ration

apparatus and at end of the spring.

$he initial length of the spring is measured using ruler and recorded.

'# A "() weight is slotted to onto the hanger and the elongation of the

length of the spring is measured and recorded.

*# +tep ' is repeated for loads of !2#+ ;2#+ and =2#.

# -raph of force ersus deection is plotted to conclude t&e spring stiffness.

5.1. 3esult and calculations


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Table 2art - &4perimental 3esult

i .otal suspended

mass+ mi 'kg(

.otal

force+ Fi 7

mig '#(

9cale

>eading+ ?i

Deflection+

?i @ ?o

Increment in

deflection 'mm(

'mm

(

'm( 'mm

(

'm( 'mm( 'm(

2 2 2 B 22B 2 2 2 2

3 323C 32 22 B 222

B

B 222B

! !2;C !2 B= 22B= 3 223

B 222B

; ;2:B ;2 C! 22C! != 22!

=

B 222B

= =2 =2 CB 22CB ;2 22;

2

222

Assuming gravitational force+ g 7CB3 m/s!

( "( " &( & '( ' *( *

(

(.("

(.("

(.(&

(.(&

(.('

(.('

(.(*

(oad *+$ ,e-ection

.orce/ N

,e-ection/ m

Error >eference source not found

Table # alculation of graph equation using the method of least squares(

i Deflection+ ?i @ ?o .otal force+ Fi 7 mig '#( F i2 '?i @ ?o

¿ ¿ F i

2 2 2 2 2


7/60

3 222B 32 322 22B

! 223 !2 =22 2;!

; 22!= ;2 C22 2!

= 22;2 =2 322 3!

total∑ ¿

∑ ¿22B 322 ;222 !;!

According to t&e general formula for linear equation+

y=mx+c

8radient+ m

xn

∑ ¿¿¿

n .∑ xn2−¿

m=n .∑ xn y n−∑ xn∑ yn

¿

m=5 (2.32 )−(100)(0.078)

5 (3000 )−(100)2

¿7.6×10−4

y0intercept+ c

x

xn

∑ ¿¿¿

n .∑ x n2−¿(¿¿ n)2−¿∑ xn y n.∑ xn

¿∑ y n∑ ¿

c=¿


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c=0.078 (3000 )−2.32(100)

5 (3000 )−10000

a=0.0004

.&erefore equation of slope

y=0.00076 x+0.0004

6&ere t&e spring stiffness+ k

k = 1

0.00076

¿1315.79 N / m

¿1.315kN /m

1.1. 6iscussion

eference source not found+ proof t&at deflection

of t&e spring is proportional to t&e applied force


9/60

Furt&ermore+ random error cause by inappropriate met&od obtaining t&e measurement -ill lead

to error in t&e reading E,ample of suc& an error are taking t&e reading -&en t&e spring moving

slig&tly -&ic& -ill c&ange t&e lengt& of t&e spring ommon error suc& as paralla, error is also

common -&en no effort is taken into taking t&e measurement .&ese problem can be abstain by

practicing greater care in taking measurement -it&out contact -it& t&e spring and assign a

person to take t&e reading so t&at same reference point is used in taking t&e measurement /t is

also %etter to attach a 0xed measurement tool %eside the spring so that

reading readings can %e o%tain without haing to hold the ruler

.o obtain &ig&er accuracy in reading several reading must be taken to minimise t&e c&ances of

error to t&e minimal

part b


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5.5. Introduction

In section < of t&e e,periment+ t&e natural frequency of t&e spring oscillation -as obtain

by applying varies load to t&e spring .&e e,perimental data -as obtained and recorded

.&eoretical value -it& and -it&out lumped mass ad*ustment -ere identified and compare

5.0. !b"ectives• .o study and identify t&e natural frequency of oscillation of a spring -it& and

-it&out lumped mass ad*ustment

• ompare t&e grap& obtained t&eoretical and e,perimental result+ associated -it&

period of oscillations and varies load on t&e &anger

• .o verify t&e correctness of t&e e,perimental result ontained

5.7. Theoretical $ac%ground

5$0$1$ Natural .reuenc #atural frequency is t&e frequency of a system oscillate in t&e none,istence of any

damping -&ic& is also kno-n as simple &armonic motion+ -&ic& is a sinusodial motion

&ttp//---mat&psuedu/tseng/class/"at&!:3/#otes0"ec&%pdf

Figure 1: Free Body Diagra of spring mass system.

Figure 3 s&o-s a spring mass system oscillate in t&e direction of gravitational force in y

a,is of moment + -&ic& is categorise as one degree of freedom

$oint A in t&e Figure 3 s&o- actual lengt& of t&e spring -&ere no force is e,ert on it

6&en t&e mass applied + it -ill lengt&en to static equilibrium position at point


11/60

Different load applied on t&e spring -ill results different position of point


12/60

5$0$2$ (umped Mass +stem

5umped mass systems are systems t&at -it& restricted number of freedom

Allo-ed for t&e mass of spring an energy met&od to determine t&e natural frequency of

t&e system >ayleig&4s Energy is adopted in t&is case -&ic& assumes

• motion is simple &armonic and vibration mode

• "a,imum kinetic energy is equal to ma,imum strain energy

• .&e mode s&ape4 is assumed

'1niversity !23!(

>eferring to figure bello-

8igure 1 System +ith 1 d.o.f.

#ote t&at+

c7 t&e free lengt& of t&e spring

m7 mass of ob*ectv7 velocity at c

%7 velocity at c G dc

ms7mass of spring

6e can assume +


13/60

inetic energy of t&e body+ m+ !

1=

1

2m V

2

inetic energy of t&e dc element of spring+ ms+

"c

#

ms¿ $¿

!2=

1

2¿

For t&e identical triangle 'mode s&ape(+$

c=

V

# +$2=( cV # )

2

5astly + t&e kinetic energy for t&e -&ole spring+ ms+

!=1

2

ms

#

V 2

# 2∫0

#

c2

"c

&quation

¿1

2

ms

3 V

2


14/60

f s= 1

2π √ k

m+ms3


15/60

5.9. 2rocedure

8igure 6amped rotational system +ith single d.o.f.

i# 1ass of the spring and hanger are measured and recorded.

ii# $he spring is attached to the tecquipment uniersal i%ration

apparatus with hanger hanged at the end of the spring.iii# 2ertical oscillations with "(mm of amplitude is exert and '(

num%er of cycles is timed using digital stopwatch are measured

and recorded.

i# +tep iii is repeated for (, "() , &() ,'() and *().

# $hen graph of experimental and theoretical period are plotted

against arious masses.

1.1. 3esult and calculations

"ass of t&e spring+ms=0.034kg

"ass of t&e &anger+m%=0.01kg

Induced amplitude+ & '10mm=0.01m

Table 2art $ e4perimental data

.otal mass of

-eig&ts on

&anger+mi

#umber

of cycles+

N

"ean time of

N

oscillations+

$eriod of

oscillation+√ mi +m%

√mi+m%+ms3

+

ms

mi


16/60

t́

'sec(

(1 sec

T = t́ N

'sec(

(kg )1

2

(kg )1

2

( N ) (kg)

2 2 ;2 2 2 23222 23=3 2

32 323C= ;2 ;=;; 2!33= 323= 32!23

22;;=

!2 !2;B ;2 ==22 2!=B2 3=;3; 3=;:;

223

;2 ;2:B3 ;2 C=!2 2;3=3 3:3 3:=B

22333

=2 =2: ;2 32B!22 2;2 !2!3 !2!=

222B;

note that gravity 7CB3 m /s2


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Table 5 Theoretical relationship based on formulaT =( 2π √ k )√ )

9quare root of totalmass -it&out spring+

√ mi+m% 'kg(

9quare root of total mass

-it& spring+

√mi+m%+ms3

'kg($eriod '-it&out spring

mass(+.3 'sec(

$eriod '-it& spring

mass(+.! 'sec(

23222 23=3 223;!3 22!:;2

323= 32!23 23:== 23C

3=;3; 3=;:; 2!=C!; 2!=B3

3:3 3:=B 2;2;=2= 2;2;C:B

!2!3 !2!= 2;:23BC 2;:2C!

#ote"3 7

mi+m%

"! 7mi+m%+

ms

3

(.3 " ".& ".* ".4 ".3 & &.&

(

(.(

(."

(."

(.&

(.&

(.'

(.'

(.*

5xperimental data

assuming spring is

massless

$heoretical data assuming

spring is massless

!inear 6$heoretical data

assuming spring is

massless#

Mass &)

Period

8igure # omparison bet+een theoretical data and e4perimental data :assuming spring is massless;


18/60

(.3 " ".& ".* ".4 ".3 & &.&

(

(.(

(."

(."

(.&

(.&

(.'

(.'

(.*

$heoretical dataconsidering spring with

mass

!inear 6$heoretical data

considering spring with

mass#

5xperimental data with

spring mass consideration

Mass &)

Period

8igure 5 omparison bet+een theoretical data and e4perimental data :assuming spring is have mass;

1.1. 6iscussion

Figure ; omparison bet-een t&eoretical data and e,perimental data 'assuming spring is

massless( in figure C and figure 32 proof t&e difference bet-een t&e value obtain bet-een

e,perimental data and t&e t&eoretical .&e common bet-een t&ese t-o grap& s&o-ed t&at is t&e

spring is elastic limits p&ase


19/60

From t&e grap&s plotted + t&e e,perimental data c&anges in small degree and does not line up in

a straig&t line .&e cause is due to paralla, error t&is can be e,plain due to fast oscillation of t&e

spring and time keeper of t&e stop-atc& .&us+ indirectly cause t&e period of oscillation in

e,periment is &as &ig&er value as compare to t&eoretical value -&ic& are clearly s&o-n on t&e

grap&s plotted to avoid suc& an error laser oscillator counter s&ould be used

furt&ermore condition of t&e string s&ould also take into consideration t&is because induced

e,tension of t&e spring at t&e initial mig&t varies from t&e recommended e,tension In addition

mass attac& to t&e &anger must be firm to abstain certain affect on t&e oscillation Ot&er t&an

t&at+ t&e spring mig&t not be in a good condition since it is not a ne- spring

In nuts&ell+ obtaining mean value in t&e e,periment are vital to minimise error in t&e

e,periment


20/60

part c

5.ao

!22=(

.&e mat&ematical model of viscous damping can be e,press belo-+&quation 0

F =−q ´ x

&quation 7

T =−q* *́

.o study t&e effects of viscous damping+ -e apply #e-ton4s 9econd la- to t&e follo-ing

system+


21/60

8igure 0 6amped 1 d.o.f. system

8igure 7 8ree body diagram

Applying #e-ton II +

∑ F x=m ´ x

−kx−q ´ x=m ´ x

&quation 9

´ x+ q

m ´ x +

k

m x=0

.&e resultant equation is a linear+ second order+ &omogeneous+ constant coefficient+ ordinary

differential equation 9uc& equations can be tackled by assuming a solution in t&e form of a

comple, e,ponential '1niversity !23!( Hence+ letJs assume t&at

&quation <

x (t )= + est

9ince bot& K and s can be comple, 9ubstituting x (t )= + est

and´ x+

q

m ´ x+

k

m x=0

and simplify yields t&e c&aracteristic equation+

&quation 1/

s2+

q

m s+

k

m=0

And roots of t&is equation -ill be+

&quation 11


22/60

s1

s2

}=−q2m [1(√1−4 mk q2 ]


23/60

Ho-ever+ in t&is e,periment in $art for t&e e,perimental setup as belo-+

8igure 9 Schematic diagram of damped rotational system +ith single d.o.f.

.&e #e-ton 9econd 5a- can -ritten as+

&quation 1

*́+q a

2

I 0*́+

k c2

I 0*=0

6&ere

*́ 7angular acceleration

*́ 7 angular velocity

*

7 angle c&ange

q 7 damping coefficient

k

7 spring stiffnessa 7 distance from pivot to damper

c 7 distance from pivot to &elical spring

I 0 7 moment of inertial of t&e arm along t&e pivot O+


24/60

ecalling t&at t&e logarit&mic decrement

&quation 15

33

!

!

!ln

3

!

2

−=

=

ζ

π δ

n y

y

n

-&ic&+ if solved for

ζ

and t&en replaced -it& t&e e,pression for

!ζ

above fives t&e

damping co0efficient

&quation 10

3(!/!ln'

!

!!!

+

=

no

o

y y

n

kI

a

cq

π


25/60

6&ere

y2 7 t&e amplitude of t&e vibration at t&e beginning

yn 7 t&e amplitude of t&e vibration at n cycle of oscillation

n 7 oscillation number

If t&e das&pot is drained of its oil+ it -ill gives out q72 and so t&en Equation 3= becomes

&quation 17

k f

c I

I

k c f

N

o

o

N

!

!

!

=⇒=

π π

Substitution of &quation 17 into &quation 10 gives(

&quation 19

3

(!/!ln'

!!

!

+

=

no

N

y y

n

a

c

f

k

q

π

π

As long t&ere is natural frequency of oscillation andn

o

y

y

e,ist+ Equation 3B can be

evaluated to determined t&e damping coefficient


26/60

5.11. 2rocedure

8igure < Schematic diagram of damped rotational system +ith single d.o.f.

3 .&e apparatus -as set up as s&o-n above

! A empty das&pot -as fi,ed at a distance a from t&e pivot O and t&e &elical spring used in

$art A and $art < -as fi,ed at a distance of c from t&e pivot O

;


27/60

1.1. 3esult and alculations

8igure 1/) Typical time history of viscously damped oscillation

"easurements of Apparatus '>efer to Figure ! Damped rotational system -it& single dof(

a=150mm

c=600mm

Table 0 &4perimental data for 2art

ategory Das&pot -it&out oil

'undamped(

Das&pot -it& oil 'damped(

.otal number of full cycle

oscillations+ n

: =!

.ime taken for n4

oscillation 's(

B=3 B=

$eriod 'n

s ( +. 's(23:2 23;

Frequency 'Hz( :C 3=22 y

0 'mm(;3 !

2 y n 'mm( ! !

Initial deflection angle+

*0 'rad(

;! !

alculation)

LLL#ote .&e value spring stiffness k + is taken from part A calculation


28/60

Moment of inertia of arm about !( I 0 (

.&e data from t&e e,periment -it&out oil in das&pot 'undamped( is used Equation

I 0=( c2π f N )

2

k + is used

6&ere+

f N =6.659 ,-

k =1315.79 N /m

c=0.6 m

I 0=( 0.62π ×6.659 )

2

(1315.79 )

I 0=0.2706 Nm s2

6amping oefficient( q :using &quation 10;

q=2×0.60

0.152

√1315.79×0.2706

[ 2 π (15)

ln( 3.12.6 ) ]2

+1

q=1.878kg / s


29/60

&4perimental damped freeoscillation(

f q=6.659 ,-

Theoretical damped freeoscillation :using &quation 1;(

f "ampe"= 1

2π √ 1315.79×0.62

0.2706 (1− 1.8782×0.15

4

4×1315.79×0.2706×0.62 )

f q=f "ampe"=6.659 ,-

omparison bet+een e4perimental and theoreticalf q is equivalent to e,perimental

f q

':CHz(

alculating e4perimental

f q

f N ∧ζ :using &quation 17 and &quation 19;(

f q

f N =

6.659

6.659


30/60

f q

f N =1

+

ζ =

√1−

( f q

f N

)

2

ζ =√ 1−(1 )2

ζ =0 + -&ic& is ζ M 3

alculating theoretical result of

f q

f N ∧ζ

:using &quation 17 and&quation 19;(

f q

f N =

6.140

6.659

f q

f N =0.90899

+ -&ic& is very close to 3

ζ =√1−( f q

f N )2

ζ =√ 1−(0.922 )2

ζ =0.387 + -&ic& is ζ M 3


31/60

1.. 6iscussion


32/60

2$ Part ,: rror Analsis

Introduction

.&e purpose of t&e error analysis is to determine t&e accuracy of t&e value -e obtain during t&e

e,periment %alue of t&e parameter can be ac&ieved by substitute t&e measured value obtain

from t&e lab+ -&ic& is e,posed to certain error . This will lead to measured uantities and

function of algebraic might be the cause of the error in new parameter. .&ere are t-o

types of uncertainties -&ic& are syst=ematic errors and random uncertainties

Theory


33/60

=√∑i=1 N

( x i− x )2

N −1

In general case+ t&e magnitude of a parameter is determined by t&e substitution of measured

quantities+ eac& sub*ect to error+ in an algebraic e,pression 9o t&e error in t&e ne- parameter can

be some function of t&e algebraic e,pression+ t&e measured quantities and t&eir respective

uncertainties

.&e absolute error can be defined in quantity , as d, -&ile t&e fractional error can be defined in

quantity as

δx x

= "x x

5.1. 2art 6)

5$12$1$ rror Analsis

.&e evaluation of data gained from e,periment al-ays carries a certain level of

uncertainties .&e -ill to reduce t&e uncertainties must be balanced -it& t&e

necessity to ac&ieve ma,imum efficiency of information collected and

e,perimental yield -it& t&e resources available'1niversity !23!(

.o determine t&e uncertainty of a function t&at is made up of several parameters

t&e formula belo- is used

For function /=f ( w 0 x 0 y 1 ) + its uncertainty can be defined as+

&quation 1<

+∂∂

+∂∂

+∂∂

±= dy y

dx x

dww

d β β β

β

In t&is e,periment+ t&e uncertainty of t&e damping coefficient q+ is to be

determined From


34/60

3(!/!ln'

!!

!

+

=

no

N

y y

n

a

c

f

k

q

π

π

.&erefore+ t&e uncertainty of q for damped spring can be defined as+

&quation /

∂∂

+∂∂

+∂∂

+∂∂

+∂∂

+∂∂

±= dk k

qdf

f

qdy

y

qdy

y

qdc

c

qda

a

qdq N

N

n

n

o

o


35/60

5.1#. 2art ) =iscous damping and its effects on natural frequency

5.1. 2art 6) &rror -nalysis

∂∂

+∂∂

+∂∂

+∂∂

+∂∂

+∂∂

±= dk k

qdf

f

qdy

y

qdy

y

qdc

c

qda

a

qdq N

N

n

n

o

o

.o determine t&e

uncertainty of t&e damping coefficient q + Equation !! is used

.o solve t&is long equation at ease it -as broken into smaller parts for solving and later

combined toget&er to determine t&e final result

#ote t&e calculation is only conducted for t&e damped system as t&e undamped system is

assumed t&at no damping took place

5$14$1$ 6ncertaint o# 7a8

2 y0

2 y n¿2+1

¿¿ ¿2πn

¿√ ¿

−2

k

π f n (#

2

a3 )

¿2 q

2 a 2 a=¿


36/60

3.1

2.6¿2+1

¿¿¿

2π (15 )¿

√ ¿

−2

1315.79

π 6.659 ( 0.62

0.153 )

¿2 q

2 a 2 a=¿

722!:2=3

5$14$2$ 6ncertaint o# 7c8

2232

3((A!/3;ln'

3:!'

3:2A:CA

A2DC3;3:

!

3((!!ln'

!

'

!

!

!

!

!

×+

××××

=∂∂

+

=∂∂

π

π

π

π

dcc

q

dc

y y

n

a

c

f

k

dcc

q

no

N


37/60

2%

2 c "c=6.259 x103

5$14$3$ 6ncertaint o# 7o8

( )

( ) ( )

2232

3(!/!ln'

!(!/!ln'

=

2232

3(!/!ln'

!

(!/!ln'!(!('

!

3'

22323(!/!ln'

!

3((!!ln'

!'

!

;!

;

!!

!

;!

3!!

!

!

3!!

!

!

×

+

××

=∂∂

×

+

×××

×−−×

=

∂∂

×

+

××

=

∂∂

+

=∂∂

−

−−

−

noono

N o

o

no

o

no

N

o

o

no N

o

o

o

no

N o

o

y y

n y y y

a

c

f

nk

dy y

q

y y

n y

y yn

a

c

f

k dy

y

q

y y

n

a

c

f

k dy

y

q

ody

y y

n

a

c

f

k

dy y

q

π

π

π

π π

π

π

π

π


38/60

( ) ( )

( ) ( )

2232

3(2!A2/2;32ln'

3:!

!/;322=;2(2!A2/2;32ln'

3:2

A22

A:CA

3:DC3;3:=

!

;

!

;

!!

×

+

××−

=∂∂

π

π

x

dy y

qo

o

2%

2 y0" y

0=6.8887 x103

5$14$4$ 6ncertaint o# n

( )

( ) ( )

2232

3((!/!ln'

!'(!/!ln'

=

2232

3(!/!ln'

!

(!/!ln'!(!('

!

3'

22323(!/!ln'

!

3((!!ln'

!'

!

;

!;

!!

!

;!

3!!

!

!

3

!!

!

!

×

+××

=

∂

∂

×

+

×××

×−−×

=

∂∂

×

+

××

=

∂∂

+

=∂∂

−

−−

−

no

nno

N

o

o

no

o

no

N

o

o

no N

o

o

o

no

N

o

o

y y

n y y y

a

c

f

nk

dy

y

q

y y

n y

y yn

a

c

f

k dy

y

q

y y

n

a

c

f

k dy

y

q

ody

y y

n

a

c

f

k

dy y

q

π

π

π

π π

π

π

π

π


39/60

( ) ( )

( )

2232

3(2!A2/2;32ln'

3:!

(!/;32A!'(2!A2/2;32ln'

3:2

A22

A:CA

3:=C3;3:=

!

;

!

;

!!

×

+

××−

=∂∂

π

π

x

dy y

qo

o

2%

2 yn" y n=1.0883

5$14$5$ 6ncertaint o# 7 f n 8

2 q

2 f N " f N =|

− k

π f N 2 ( ca )

2

√(2 (n ) π

¿( y0 yn ))2

.1|

|−1315.79

π 6.6592 ( 0.60.15 )

2

√(2 (15 ) π

¿( 3.12.6 ) )2

+1

.1|72!B!2;C


40/60

5$14$9$ 6ncertaint o# "k /

1sing Equation !!+ -e can formulateEquation !!+

compliancek dqq

dqq

k dk σ !

!

3±=±=

∂∂

±=

6&ere+

( )!! ∑∑ −

=ii

ycomplience

x xn

nσ σ

And +

( )∑

−

−

−=

!

3

!

3 x

k y

n ii yσ


41/60

.o determine "k + -e need to first find y follo-ed by

compliance

No .otal force+

x

3eflection0

y−[1

k ] x−b b > / (

y−[ 1

k ] x−b)

2

3 ( 2 ( 2! 32 222B 2222=2222; 3222!E02

; !2 223 2222B2222 =223E02

= ;2 22!= 2223!2222C 3==22!E02

: =2 22; 02222;CCCBB 3:CCCE02

9um 322 22B 222!2222; !=222!E02

k =1315.79 N /m


42/60

y=0.89443×10−3

( ) ( ) !322;222:

:

−= ycomplience σ σ

7!B!B=, 10−5


43/60

::CD2

B=3;3

3((2332/2=;2ln'

(;3'!'

3:2

::2

(2::D'

3

!

!

=∂∂

×+

=∂∂

undamped dk k

q

dk k

q

π

π

5$14$0$ 6ncertaint o# q /

∂∂

+∂∂

+∂∂

+∂∂

+∂∂

+∂∂

±= dk k

qdf

f

qdy

y

qdy

y

qdc

c

qda

a

qdq N

N

n

n

o

o

{ }::CD2B2!=2;DD:22CAA22!2A22D::2 +++++±=dq

C;!;3±=dq

9),iscussions

0.1. 2art -) Measuring the stiffness of the spring


44/60

0..2art $) Natural frequency of oscillation :+ith and +ithout lumped mass

correction;

0.#. 2art

0.. 2art 6

.&e value of damping coefficient+ q calculated is 5 .6613(1.9323 Ns/m + -&ere t&e

uncertainty is ;=3;N of t&e damping coefficient .&e result s&o-s t&at t&e uncertainty is

very &ig& and improvement mig&t need to carryout to improve t&e accuracy of t&e result

Ho-ever+ since t&at t&e damping coefficient is a function of ot&er parameters taken from

t&e e,periment+ t&e accuracy of t&e parameters must be improved to determine a more

accurate damping coefficient


45/60

0)onclusion$art A

In conclusion+ t&e value of spring stiffness determined '3!:2 #/m( -it& appro,imate !N erroris consider acceptable Due to t&e difference in t&eoretical and e,perimental data+ t&e t&eoretical

value 3;;;;; #/m is used in t&e follo-ing e,periments

$art


46/60


47/60


48/60

.1. 2art 6)

2$1$1$ rror Analsis

.&e evaluation of data gained from e,periment al-ays carries a certain level of

uncertainties .&e -ill to reduce t&e uncertainties must be balanced -it& t&enecessity to ac&ieve ma,imum efficiency of information collected and

e,perimental yield -it& t&e resources available'1niversity !23!(

.o determine t&e uncertainty of a function t&at is made up of several parameters

t&e formula belo- is used

For function /=f ( w 0 x 0 y 1) + its uncertainty can be defined as+

&quation 1

+∂

∂+∂∂+

∂∂±= dy

ydx

xdw

wd β β β β

In t&is e,periment+ t&e uncertainty of t&e damping coefficient q+ is to be

determined From Equation 3B+

3(!/!ln'

!!

!

+

=

no

N

y y

na

c

f

k

qπ

π

.&erefore+ t&e uncertainty of q for damped spring can be defined as+

&quation

∂∂

+∂∂

+∂∂

+∂∂

+∂∂

+∂∂

±= dk k

qdf

f

qdy

y

qdy

y

qdc

c

qda

a

qdq N

N

n

n

o

o


49/60

.. 2art ) =iscous damping and its effects on natural frequency

.#. 2art 6) &rror -nalysis

∂∂

+∂∂

+∂∂

+∂∂

+∂∂

+∂∂

±= dk k

qdf

f

qdy

y

qdy

y

qdc

c

qda

a

qdq N

N

n

n

o

o

.o determine t&e

uncertainty of t&e damping coefficient q + Equation !! is used

.o solve t&is long equation at ease it -as broken into smaller parts for solving and later

combined toget&er to determine t&e final result

#ote t&e calculation is only conducted for t&e damped system as t&e undamped system is

assumed t&at no damping took place

2$3$1$ 6ncertaint o# 7a8

2!:2=32

2232

3((A!/3;ln'

3:!'

3:2A:CA

A2DC3;3:

!

3((!!ln'

!'

!

!

;

!

!

;

!

=∂∂

×+

××××

−=∂∂

+

−=∂∂

daa

q

daa

q

da

y y

n

a

c

f

k

daa

q

no

N

π

π

π

π


50/60

2$3$2$ 6ncertaint o# 7c8

!;;

10−3

;32!:CA

2232

3((2!A2/2;32ln'

3:!'

3:22::D

A2DC3;3:

!

3((!!ln'

!'

!

!

!

!

!

−=∂∂

×+

×××=

∂∂

+

=∂∂

xdcc

q

x

dcc

q

dc

y y

n

a

c

f

k

dcc

q

no

N

π

π

π

π


51/60

2$3$4$

( )

( ) ( )

2232

3(

(!/!ln'

!'(!/!ln'

=

2232

3(!/!ln'

!

(!/!ln'!(!('

!

3'

22323(!/!ln'

!

3((!!ln'

!'

!

;

!;

!!

!

;!

3!!

!

!

3!!

!

!

×

+××

=∂∂

×

+

×××

×−−×

=

∂∂

×

+

××

=

∂∂

+

=∂∂

−

−−

−

no

ono

N

o

o

no

o

no

N

o

o

no N

o

o

o

no

N

o

o

y y

n y y y

a

c

f

nk

dy y

q

y y

n y

y yn

a

c

f

k dy

y

q

y y

n

a

c

f

k dy

y

q

ody

y y

n

ac

f k

dy y

q

π

π

π

π π

π

π

π

π

6ncertai

nt o# 7o8


52/60

( ) ( )

( )

;32BBDA

2232

3(A!/3;ln'

3:!

!

;323;

(2!A2/2;32ln'

3:2

A2

A:CA

3:DC3;3:=

!

;

!

;

!!

−=∂∂

×

+

××

−

=∂∂

xdy y

q

x

dy y

q

o

o

o

o

π

π

dfd

2$3$5$

( ) ( )

( ) ( )

( ) ( )

;DD=:2

2232

3(2332/2=;2ln'

;3!2332(2332/2=;2ln'

3:2

::2

2::D

;3;;3;;;=

2232

3(!/!ln'

!(!/!ln'

=

3((!!ln'

!'

!

;!

;

!!

!

;!

;

!!

!

!

=∂∂

×

+

××−

=∂∂

×

+

××

=∂∂

+

=∂

∂

n

n

n

n

no

ono

N n

n

o

no

N n

n

dy y

q

dy y

q

y y

n y y y

a

c

f

nk

dy y

q

dy

y yn

a

c

f

k

dy

y

q

π

π

π

π

π

π

6ncertaint o# n


53/60

2$3$9$ 6ncertaint o# 7 f n 8

B2!=2

(3'

3((23322=;2ln'

(;3'!'

3:2

::2

(2::D'

;;3;;;

3((!!ln'

!'

!

!

!

!

!

!

=∂∂

+

−=

∂∂

+

−=∂∂

N

N

N

N

N

no

N N

N

df f

q

df f

q

df

y y

n

ac

f k

df f

q

π

π

π

π

2$3$0$ 6ncertaint o# "k /

1sing Equation !!+ -e can formulateEquation !!+

compliancek dqq

dqq

k dk σ !

!

3±=±=

∂∂

±=


54/60

6&ere+

( )!! ∑∑ −=

ii

ycomplience

x xn

nσ σ

And +

( )∑

−

−

−=

!

3

!

3 x

k y

n ii yσ


55/60

.o determine "k + -e need to first find y follo-ed by

compliance

From page Error >eference source not found+

i .otal suspended

mass+ mi 'kg(

.otal

force+ Fi 7

mig '#(

9cale

>eading+ ?i

Deflection+

?i @ ?o

Increment in

deflection 'mm(

'mm

(

'm( 'mm

(

'm( 'mm( 'm(

2 2 2 2 222 2 2 2

3 323C 32 : 22: : 222

:

: 222:

! !2;C !2 B; 2B;2 3; 223

;

B 222B

; ;2:B ;2 C2 22C2 !2 22!

2

32 2232

= =2 =2 322 2322 ;2 22;

2

32 2232

k =1333.33 N /m


56/60

;2223

! =∑=i

i x

y=2.345×10−3

( ) ( ) ( ) N m ycomplience /32=3AD2;3A!232;=:!

322;222:

: :;!

−− ×=×=−

= σ σ


57/60

::CD2

B=3;3

3((2332/2=;2ln'

(;3'!'

3:2

::2

(2::D'

3

!

!

=∂∂

×+

=∂∂

undamped dk k

q

dk k

q

π

π

2$3$;$ 6ncertaint o# q /

∂∂

+∂∂

+∂∂

+∂∂

+∂∂

+∂∂

±= dk k

qdf

f

qdy

y

qdy

y

qdc

c

qda

a

qdq N

N

n

n

o

o

{ }::CD2B2!=2;DD:22CAA22!2A22D::2 +++++±=dq

C;!;3±=dq


58/60

3$ ,iscussions

#.1. 2art -) Measuring the stiffness of the spring

#.. 2art $) Natural frequency of oscillation :+ith and +ithout lumped masscorrection;

#.#. 2art

#.. 2art 6

.&e value of damping coefficient+ q calculated is 5 .6613(1.9323 Ns/m + -&ere t&e

uncertainty is ;=3;N of t&e damping coefficient .&e result s&o-s t&at t&e uncertainty is

very &ig& and improvement mig&t need to carryout to improve t&e accuracy of t&e result

Ho-ever+ since t&at t&e damping coefficient is a function of ot&er parameters taken from

t&e e,periment+ t&e accuracy of t&e parameters must be improved to determine a more

accurate damping coefficient


59/60

4$ onclusion$art A

In conclusion+ t&e value of spring stiffness determined '3!:2 #/m( -it& appro,imate !N erroris consider acceptable Due to t&e difference in t&eoretical and e,perimental data+ t&e t&eoretical

value 3;;;;; #/m is used in t&e follo-ing e,periments

$art


60/60

damping lab 1

Documents