Description
Vibrion
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Ce=an mk = 2mWn= zout the critical G= @= 2 mwen  2 link =) damping facer was wall E P = walte? } damped
damping Coeffcient
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damped
free
frequency
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woda wa (6) ? = was a damped
frequency If you want to learn more check out Who states “emotional intelligence is the “something” in each of us
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We also discuss several other topics like What are the some strategies to use for students who have difficulty with sentences are?
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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c=29 km = 25 km  agte
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Base Excitation: (K=1 for Base Excitation)
malze =  klxy)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
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mä + cx + k*= kym sinut
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All solutions just replace to by kym
another form
me  kvxy)cd met een
ť cac
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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0= tant
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X and Xnt are quite the same, so we take
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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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Fo
Xo=  KFO
(we + ( 25wj2  K

Jek  kW2+1/29 kw)
Wnet Wnl
DT250/Wn
(sime on 12 ) csime
c=251 km
studysowy
solution
can be
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 = VoXow cost
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eavy damping (negative roots) nonvio
ystemy
(double roots) honvidrar
vibratory
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Damping Factor S = & anwn = 2 km
C> Ce over de
X— Се +Сеза. critical damping: 41 c=ce Critically damped system
e
henvirony. X = +CD) eWut campaan sy SOM light damping: <1 ccce underdamped 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 (Swowa) Ko e wat ) x= 1 Gt Czt) e whit (21= 22= Ce mWn)
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 (15) 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
Wnt
more often used
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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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The
general
solution is
x
+
%
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@
x=asinwatt (2 coswnt + Xm) sincet
transient
ayson Steadystate
depend on We Magnification factor =
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when we=wn ragnification factors
" Resonance
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2
3 un
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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
Ce=2m/k = amwn (radical=0)
Im
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cosom = (1sinan)
Sample
At o O max
TO Vmage mg L(1cosom) At O studs v=o Tmax = 1 m ( L Wn Om Vmax = Tmax
žmg Lam = 1 mL we are
Lilleosoro
Wm
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$6. Fored Vibration of Spring Mass System.
P= Pm sinust
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max
md=pkx
(Let weight balance out the Xo)
mX+kX= (on sinugt
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case
1
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S= Sm Sin wat
(xS) = m de ä tkx=(k Sm, sinust
m
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case 2.
To solve the above nonhomogeneous Solution Xn= G sinunt + coswnt wark xp = xm sincuft xm = Pm
or with what {Km m
7 7 COLD
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7
.
How
to determine
the
the origional mass or spring stiffness?
Original: Tu = (22)? (W) R$
Soul Ta'= (22) (when)
Tn. Ta can be measured from experiments
Δw
12
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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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National University of Singapore
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No.
Swinging Plate
(ab x26) Wn=1
mtb
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Fь
22
In im ftare+13+5)
To
O = 0o cos wat two sin wat
tudo
i Io  Ig + mb2
where to Wo is at t=o.
initial condition.
stus
Other Rigid Body
Let o be the hinged point
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2
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I= Io + md
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Swinging Rod Wall
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4
studysou
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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
10 Taking moments about A.
 madsino = In
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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National University of Stogapore
ar
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kinetic
force
o
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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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A is a fixed point IMA = IA Ő v
EM  IG
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Pendulum Ww=lg
ng sino = meo
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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? to, O=0. due to energy loss.
to, W = We period remains the
ame (good time keeper
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$ 4. Vibration of Rigid Body.
ANSOUP  kinetie.
energy
translational, rotational
National University of Singapore
Energy
Date
No.
& VG
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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
Thear

Paralel
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consider this
pure rotational
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§ 2.
Slider  Crank
Mechanism
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1.
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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 singWR 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. Offset 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
221H+R sino) Lays B2costo
cos (42B2) 3/2  R 22B?
pe sine
  Riw
olo
+
Rw
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a 2282  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
 kxkXoting=ma
 kx=ma matkxo
(radis) (Hg) mdx + kx=0 Wn2afn
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
forces same extension added
kon=k, tkz
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1
Like
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k }
}k,
forces added
I
L
Ktot = kitka
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yoyo
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Problem:
a=ra To=lae To Id = I a mgT=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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gerr
langer v.
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