II" FREE VIBRATION OF UNDAMPED

SINGLE DEGREE OF FREEDOM

SYSTEMS

MECHANICAL VIBRATIONS

| 2.1. Equation of Motionl

mX +kx=0This is the Differential Equation Of Motion (D.E.O.M.)

Equation Of Motion (E.O.M.)

x=Acos(r,r-rnt-B),

r h)n = ,m = Circular natural frequency. General position

A and p are determined by initial conditions Ix(0) and X(0) ].

fn = &)n l2r = Natural frequency. Solution is explained in AppendixA.l.

Equilibriunn position

Mshamed T, l{edaya M EOI-IAN IGAI VI ERATION S First Edition, 0,2,2

Effect of Gravitvs-lmx= L Fx

mX=mg-k(6+x)

l*X=mg-k6-kxmX=-kxmX+kx=0 {D.E.O.M.)

Or

m X = Change of I F* from

equilibrium position

mX=-kxmX+kx=0 (D.E.O.M.)

Unstretchedposition

Equilibriumposition

Note: Gravity only changes the equilibriurn position.

mgGeneralposition

Mohamed T. Hedaya MECHANICAL VIBRATIONS First Edition, p.2.3

2.1.2" Energy Method {Fortqtservative systems, K.E. (T) + P.E. {U) = constant

Differe

t=1mx2.2

ax

I

dt dt

kxdx =1k*'2

Spring force = kx

*^rrx)x+ lnrrx)x=0kx

mxX+kxX=0

mi(+kx=0 (D.E.O.M.)

Equation of motion x =A cos(urnt- B),

X

Mohamed T. Hedaya MECHANIGAL VIBRATIONS First Edition, p. 2--

rnfr-* k

Jc=Ac,o ,

",1,t0 n'vv'{-*-"

-1L\

t{CDn

Lfl

ILnLt,IV

,']H

c

:oI!nL

::-

c

cbs

CJo

,I

(a

I-t*I

CV

A

o

lB)t/

ci.

ffect of

T-\iz

u=t1 k(d+x'2\

dt dt

*{ ut -ms x = [kd x + ]n^'r-ffig x = lr^'lSpring force = kAL

k(6+x)f ----)112*(z x) X +

ik(z x) x = 0

m xX + k x X = 0

mX+kx;0 (DE.O.t\l=)

Note: Gravity onty ehanges the equitibrium position,

Mohamed T. Hedaya MECHANICAL VIBRATIONS First Edition, p.2.

2.2. Tortional Svstem

Tortionat stiffness of straft (q) = restlgriry

lgrgu"e- lr sn9{tJr)angular twist of shaft (6)

Stress anatysis gives r = G.Jo

s----' L ,

G = modulus of rigidity of shaft material,

J = polar moment of inertia of shaft cross-section ( = rrda I 32 ),L = shaft length.

This gives c Q = fl r* */radl

Mohamed T. Hedaya MECHANICAL VIBRATIONS First Edition , p.2.6

2.3rPetermination of NaturaI FrequencyEy Energv M-ethod

ln section 2.1, it was proven that free vibration of undamped single degree

of freedom system is harmonic. ln this section, this result is used to

determine the natural frequency of the harmonic motion, by applying the

principle of energy conservation.

Application of the Principle of Energy Conservation

At equilibrium poqJ/:ion

T = T*a, - ! *r' U = 0 (,i,,,1,i,,..n,,,1

At extreme positiae

T=0

T*r*+0=0*Ur"*o T*r* = U*r,

Equilibriumposition

Extremeposition

Mohamed T. Hedaya MECHANICAL VIBRATIONS First Edition , p.2.7

Example 1

Find the natural frequency of the spring-mass system.

x=-uJnAsin(ulnt-p)

m (u,ln A)2.

kA2

x=Acos(urnt-B)1

Trrr=2m(*rur)'

1

Urr*=rk(x,,.*)'

, -0n 1 retn- zn - 2"!*

1

2

=12

Equilibriumposition

Mohamed T. Hedaya MECHANICAL VIBRATIONS First Edition, p.2.8

Examnle 2

Find the,natural frequency of the

system shown

0=Acos(rrr"t-F)

x=L0=LAcos(tl.t-p)6 --tlrnAsin(urnt*p)

x = L S = - [r]n L A sin(tr.rnt- B)

At equiliprium positiar,, T = T*r*

= )t^(orn A)' . ] r, (r,:n A)2 + 1

At extreme position, U = U*r*

=.)x(L A)'

* t^^r= LA

d5 ,,.*.* = (J)n A

/\- ft'a /

, ..1 -

m (un L A)'

Mohamed T. Hedaya MECHANICAL VIBRATIONS First Edition, p. 2.9

1^_17

(ln I la + m L=) (tlrn A)= = if (L A)=

(ln + la + m L'),^,n' - k L2 = 0

t- kTt--_\ln+1.+mL'

Note: Energy method is useful when the system consists of many rigid

bodies, because the internal forces do not appear in the energy equations.

These internal forces must be considered in the application of Newton's

law method.

lo +l* +mL2

5 o, f x;rm yi'

,,1

"t

:iI

ili*y'i 'r

:

rti i.,l'r 1t),.. i ll

i feep the system frequency unchanged

= spring mass

= spring length

ms

L

x = displacement of mass rn

dz = length of spring element

dm = mass of spring element

y = displacement of spring element

= static position of spring element

Mohamed T. Hedaya

litrir,^"'r I \'ir lr

tq- aJ

1,

MECHANICAL VIBRATIONS First Edition, p.2.11

dm = [uO,L'LL

K.E. of equivalent system (T*) = K.E. of mass m (T*) + K.E. of spring (T')

,"= *ffi*

X2

T,n = ).**'m^- -dz -L

1[r") *,21 3 )lindm

= il[i-)'1*^ *.2=1 mxz*1[T.2' 2 2[3

am^-**'t-3

)*'

Mohamed T. Hedaya MECHANICAL VIBRATIONS First Edition, p.2.12

\-, li':,i:r i ;...-\--,..-J -

" A1 .,trI1 ,

,t'1

A -"/i*/

I,L_r : T. *(nc)*A --tl

i',1,t ['7yt1*.41

j.i ,.,,, , i ,

r/ill -'\ ' '

t,

' ,talII \..:

Li..i l'lt_,.

L

t,I l/I I\t"r,-,

li "'itj

t'

I

A, ltr r .t (;', ,i'r, n i

CY-r' -, I ,,. i .:.Ir

, - l't r, ^, ,1. ,t.r l.- \,' \

Ll

I

l^i{*-, tiq, firt4i() r

2.5. Ditferential Equ,ation of Plaqll Motion of a Riqid Eodv

2.5.1. Translation

rmX=I F,

2.5.2. Rotation about a Fixed Center

MECHANICAL VIBRATIONS

2.5.3. Translation of C.G. and Rotation.. s-trmxG= )_, l-x

o166=I M6

First Edition, p. 2.13

Specia/ Case (Rolling)

mX=P-F,ma6=P-F166=Fa(2)+(1)Xo-)(lc+ma')0=Pa

ii=a0-)(1)

(2)

ols6=I M6

P is external force.

F is friction force.

o166=I Mo

Mohamed T. Hedaya MECHANICAL VIBRATIONS First Edition, p. 2.14

Example 2.{

Find un.

Example 2.2

Find tr-tn.

Example 2.3

Find on.

Example 2.4

Find tltn.

Uniform rod

niform disk

Qu, .L

ka' +

-TI

t*

I \^, ^r- l ''i \

i ,-i .-r,.-.r o .,i',

rlr'i

ft-,iVv -u

t- '̂r

lr

/ I '] I /- 'tttq. !r'

(., o

i \.,

,-1I

-t\

s

A\J=

c(s :+ ,l-t i

\n\+ fr12},')

n?/'A - ( lt,, ,l

*t\i '_' I

-rt-O

ffi2

I

{ -,"L 'Ls

-L. +

iL I'

"- t '.:

I

;

Example 2.5

Mass moment of inertia of the pulley about its axis = lo.

Find on.

[xc*gtr 2,{

Mohamed T. Hedaya MECHANICAL VIBRATIONS First Edition, p.2.17

,n- .,) ,, {ion I t vc*,rcph z' € :

'

1I

' -l 4,'n-,-.Jt.

i.. ..1 , i tun\

.AlQ =;1.*S --f'

I e.^O't}.I

i'itrlI

s*r,,1*

nrl T = -*A-f sira $

LL

*f;i er Qi