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NUS / Mechanical Engineering / ME 3112 / How to determine the original mass or spring stiffness?

How to determine the original mass or spring stiffness? Description

Description: A very detailed and structrual summary of the second part of this course. Handwritten. with marking and corrections. If the prof cannot give you the clear thoughts in lecture notes, let this help you to study everything within 2 weeks. Don't miss
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How to determine the original mass or spring stiffness?

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for si Td = period ot damped vibration S= ln leren en de mense p We also discuss several other topics like What is the organic solvent required in halogenation?

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What is damped free vibration?

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c=29 km = 25 km - agte

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Base Excitation: (K=1 for Base Excitation)

malze = - klx-y)-cdr mx + cx+ kx= ky let y = Im sinut Don't forget about the age old question of It is a good that violates the law of demand , what is it?

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mä + cx + k*= kym sinut

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me - kvx-y)-cd met een

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mä + ex+kx= ky + c d en mä + cx+ kx=cwym Cosat + kym sinut

Cet K=1 Xo= xm Yo=ym

Xp (t) = Xo sinlwt+o) studyso Xo = KYOlt (29w/Wn)? We also discuss several other topics like What is an interaction between two animals or species where one benefits and one is unaffected?

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4= 8/8422+82 7.2. Damped Force Vibration:

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ndx = kx- c d et

m

äte & + kx=F

F=Fosinat Wis the frequency of external

excitation.

mä+cŇ + kx = Fo sinut

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Xut)= Xhlt) + Xplt) where Xplt)= Xosincut + 0) where

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Xo= - KFO

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(sime on 12 ) csime

c=251 km

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solution

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and Xhlt) depends on s is stated in 7.1.

and initial condition of the homogenous obtained by Xo = xo - Xp (0)= Xo - Xo sine Vor = vo- dx = Vo-Xow cost

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eavy damping (negative roots) nonvio

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Damping Factor S = &- anwn = 2 km

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X— Се +Сеза. critical damping: 4-1 c=ce Critically damped system

e

henvirony. X = +CD) e-Wut campaan sy SOM light damping: <1 ccce under-damped system

x= è limit (asin wat + Calos wat)

Wd = Wn He

damped frequency Sl>l: x= cient + C2 e At Ako, a

system returns to its equielbrium in finite time t=0, x= xo and ds = vo define wa = ( 2 ) ( h) = WA? (E)= 1) = will 1921) XXL = e sunt | Votle watwad) xo . e wat vot (Swo-wa) Ko e wat ) x= 1 Gt Czt) e whit (21= 22=- Ce m-Wn)

system return to its equillibrium position in the shortest time without oscillation

t=0, x=0 and data = Vo Xet)= é wat ( Xo+ (Not Wn Xo st) x= e lint (a sin wat +Ca caswat) ca de are complex conjungate) wh 1 ) - ( 2 ) khawn wat = win 1-(82] - wer (1-5) x= xme tämt sin ( wd t+) gives the vibratory motion

with diminishing amplitude. of t=0, x=xo, dx = Vo then 9wnt) Vote WnXe sinwatt Noo

Act) =

I I Wal

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Pa = 22

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Period of damed vibration -

► logarithmic devenent 8

= en A = 20 21

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x=asinwatt (2 coswnt + Xm) sincet

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depend on We Magnification factor =

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when we=wn ragnification factors

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\$7 Damped Vibration : 7.1 Damped free vibration

viscous damping friction force is directly proporcional and opposite to the velocity of the moving body. k damping force = c dx

metan u celine tkx=0

mx + cx + kx=0 x= e At

mai+2+k=0> characterstic equation

2=-2m /)=1k) Critical damping coeffcient radical

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cosom = (1-sinan)

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TO Vmage mg L(1-cosom) At O studs v=o Tmax = 1 m ( L Wn Om Vmax = Tmax

žmg Lam = 1 mL we are

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Wm

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\$6. Fored Vibration of Spring Mass System.

P= Pm sinust

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max

md=p-kx

(Let weight balance out the Xo)

mX+kX= (on sinugt

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case

1

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S= Sm Sin wat

(x-S) = m de ä tkx=(k Sm, sinust

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case 2.

To solve the above non-homogeneous Solution Xn= G sinunt + coswnt wark xp = xm sincuft xm = Pm

or with what {Km m

7- 7 COLD

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How

to determine

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the origional mass or spring stiffness?

Original: Tu = (22)? (W) R\$

Soul Ta'= (22) (when)

Tn. Ta can be measured from experiments

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Am is know, we can sowe two unknown

R.

m.

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Due to the sinousidal property:

x= A sin (Wat+0) A=Xm ant = v= WnA cos Cunt +°) o

Vm= Km. Wn a= te rester - wn? A sin Umt+0)

Am=Xm. we?

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85. Application of Principle of Conservation

soup of Energy. for the same mass and spring system

Ve= h k (X+Xo) 2 equillibrium

Vg = -mgx

our T = 4 m (dx 2 x= xm sin (Wat+0)

dx = Wn Km Cos Clint+O) At X=Xm, T,=0 V = 7 kl Xm txo) - mg Xu At x=0, T2 = m ( x Won? V2 = kx 2

From Vitti = Ver Tzudy soup

3 k (X m + Xo } - mg xom = K x 2 + m (Xm Won) 2 Ikkunt ko + k XmKong Kn= kx 2 + I'm de che

7 kNm = mxu con

wn the same as before

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(ab x26) Wn=1

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In- im ftare+13+5)

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O = 0o cos wat two sin wat

tudo

i Io - Ig + mb2

where to Wo is at t=o.

initial condition.

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Let o be the hinged point

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I= Io + md

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Swinging Rod Wall

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consider taking moment about point o without conside kinetic force at G and kinetic torque is applicable

only if o is fixed

up

SMS

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Io = 7 ML2 1. Ia tm 122

dusan mg () sino = 20

Tootmg (2) sindo

sino

o

2

To ah = one

ö + Cyriezo wie 39 and with singo is small lőt who=o.

solution = to cos wat + Wo sin wat

39

NL

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A What if hinged at other point A rather than o? Whaq

TA O + nad sine = 0

o is small őt mad o=0 IA = Ist md2

1A - mb2 [112) + d2

mgd " mt / +18)

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Planar Rigid Body Prob.

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Kinetic torque

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Aroment Equillibrium mg = sino = (mo) Z Za o ng / sino = (a + m (2)

2 ő

- Parallel Axix Theorem

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if

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A is a fixed point IMA = IA Ő v

EM - IG

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Pendulum Ww=lg

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mLo + ing sino=0

öt Zsino0 na & 6+ wnsindo

O is small öt no=0

T= 2t - 22 ay soup

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solution O= A cos wat + B sin wat

with initial conditions: even when amplitude? t-o, O=0. due to energy loss.

to, W = We period remains the

ame (good time keeper

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energy

translational, rotational

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Energy

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T= m va + 2 Ig 2

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T = 1 m / 2 2 2 2 + (1 0

translational rotational

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Axis Theorem

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consider this

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Slider - Crank

Mechanism

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Normal Set up:

ostu x = L cosp + Rcoso o w R L ß Rsino = L siny 3 -one ---eo - derivative of them will give v

V=-BL sing-WR sino ®

WR COJO = Bucose 1 x=/ 42 - (Rsing)2 + Rcoso

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dv

of Imax is required, just calculate y = 0

4. sintecoste

cos W (42_R2 sinze /

(2 - Resino sinza +Rw

w a Resin Rw coso = O

get w and substitute into & get Vmax 2. Off-set Slider - Crank Mechanism

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X= 4 cos Q + REDSO H4 Rsine = L sing

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v=

-

CH+Rsine) Roosow - Rsinow

22-1H+R sino) Lays B2costo

cos (42-B2) 3/2 - R 22-B?

pe sine

- - Riw

olo

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a 22-82 - Rew cose where B= HAR sine

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83. Free Vibration of Spring Mass sys No Static displacement due to the weight of mass os

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Static equilibrium - kXoting=0

- kx-kXoting=ma

- kx=ma matkxo

(radis) (Hg) mdx + kx=0 Wn-2afn

dt2 de + anexo, com

m TEPE dax+ winx=0

Cet solution be x = (Acos Wat +13 sinant another expression of solution or x= xm sin(Watto)

j(watto)

x= xm caso sin(Wit) + (xm singlos (wit) = xm cos (Wutto)+j Xm sin (witte) B

either real or the

Xm2 = A + B2 tande imaginary part is

need initial conditions to sowe ODE the solution.

t=0, x=Xo oup to dx = Vo

So A=Xo B V

* = Xocos Wnt + no sinant with xm = x2 + ( )

tand= Xown

x=xml

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• Complex spring systems

extension sane

kon=k, tkz

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k }

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I

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Ktot = kitka

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yo-yo

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rer

Problem:

a=ra To=lae To Id = I a mg-T=ma mg. Ia ama Tas ng

so if targer the r (with constant m. I)

larger the acceleration, a hence it will drop faster

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from Energy principle of view= way soul initial KE=0

final KE = 4 mv 2 + 3 Iw2

v=rw final KE = 4 mv 2 + 1 - 2 if youyo falls by a distance of

migh = {mv2 + 1 x Luz mgh

3 žm+mt

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