# Two Degree of Freedom

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Figure 5.0

*ased on ewtonFs second law of motion 5 m3#or mass m4

m4x45 7 84x49 8"(x"7x4)

m4x49 84x4G 8"x"9 8"x45 ;

m4x49 x4(84 9 8") 5 8"x" 77777 (4)

for mass (")

m"x"5 7 8!x"G 8"(x"G x4)

m"x"9 8!x"9 8"x"G 8"x4

m"x"9 x"(8"9 8!) 5 8"x4 77777 (")

..

..

..

..

..

..

;0;0

;/

m/m/

m0

/ ;/

/;/

0;/

0;/

0;0

0;0

3;03;0

0);0 ;/+

0);0 ;/+

:et ;08 ;/

;/

m0

..

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/et us assume that under steady state conditions the solutions for x4and x"be harmonictherefore, assume x45 34sin t, x"5 3"sin t

x45 7 "34sin t, x"5 7 "3"sin t

-ubstitute these in (4) and (")7 m4"34sin t 9 (849 8") 34sin t 5 8"3"sin t7 m""3"sin t 9 (8"9 8!) 3"sin t5 8" 34 sint.

%emoving sin t through out and re arranging the terms.

34M3"5 8"M(849 8"G m4") 5 A(8"9 8!) G m""CM8"

&ross multiplying

8""5 (849 8"G m4") (8"9 8!G m"")\$n simplification we getm4m"DG Am4(8"9 8!) 9 m"(849 8")C "

9 A848"9 848!9 8"8!C 5 ;

The above e6uation is 6uadratic in " and gives two values of "and therefore the twopositive values of correspond to the two natural fre6uencies n4and n"of the system. Theabove e6uation is called fre6uency e6uation since the roots of the above e6uation give thenatural fre6uencies of the system.

iscussions@/et 845 8!5 8m45 m"5 m

Then the fre6uency e6uation becomesm"DG " m (8 9 8") "9 (8"9 "88") 5 ;

/et@ "5 "5 D,

m""G " m (8 9 8") 9 (8"9 " 88") 5 ;

m""G " m (8 9 8") 9 (8"9 "88") 5 ;

The roots of the above e6uation are as follows@

/et a 5 m", b 5 7" m (8 9 8") c 5 (8"9 "88")

4,"5 A7 b (b"G Dac)CM"a

5 A7 (7"m) (8 9 8") A7"m (898")C"G D (m") (8"9 "88")CM"m"

.. ..

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5 A9 "m (8 98")CM"m"ADm"A(8" 9 :""9 " 88") G (8"9 "88")CMDmD

5 (89 8")Mm (8""Mm")

5 (8 98")Mm 8"Mm

"5 (8 9 "8")Mm

n""5 (8 9 "8")Mm 06 M) K 00+m45 (8 9 8")Mm G 8"M m 5 8Mm

n4"5 8Mm

/6 )m+n4is called the first or fundamental fre6uency or I mode fre6uency, n"is called the secondor II mode fre6uency. Thus the number of natural fre6uencies of a system is e6ual to thenumber of degrees of freedom of system.

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Se##io;II )/9..-+ BS

T(o DOF Sy#tem )cot'.+

*o'e# Sh%pe#1

#rom 34M3"5 8"M(898") 7m"5 (8"9 8) 7 m"M8"

-ubstitute n4in any one of the e6uation.

(34M3")n45 8"M 89 8"G m . 8Mm

(34M3")n45 4

(34M3")n"5 8" M 8 9 8"G m(89 "8"Mm) 5 8"M78"5 74

(34M3")n"5 74

The displacements 34and 3"corresponding to the two natural fre6uency of the system canbe plotted as shown in #igure .!, which describe the mode in which the masses vibrate.-uch a diagram is called principal mode shape of the system. When the system vibrates inprincipal mode the masses oscillate in such a manner that they reach maximumdisplacements simultaneously and pass through their e6uilibrium points simultaneously or allmoving parts of the system oscillate in phase with one fre6uency. -ince the ratio 34M3" isimportant rather than the amplitudes themselves, it is customary to assign a unit value ofamplitude to either 34 or 3". When this is done, the principal mode is referred as normalmode of the system.

Figure 5.3

/

0

3

m/

m0

I *o'e

m/

m/

m0

m0

;/

;/

;0

;0

.No'e

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5.3 Di#cu##io o N%tur%l fre"uecie# %' mo'e #h%pe#1

O,#er\$%tio /@ It can be seen from the figure when the system vibrates in first mode, theamplitude of two masses remain same. The motion of both the masses are in phase i.e., themasses move up or down together. When the system vibrates in II mode the displacement of

two masses have the same magnitude with opposite signs. Thus the motions of m 4and m"are4O;;out of phase.

O,#er\$%tio 0@ When the system vibrates in first mode, the length of the middle springremains constant, this spring (coupling spring) is neither stretched nor compressed. It movesbodily with both the masses and hence totally ineffective as shown in #igure .D. 'ven if thecoupling spring is removed the two masses will vibrate as " -\$# system with n5 (8Mm).Where as when it vibrates in II mode, the midpoint of the middle spring remains stationaryfor all the time. -uch a point which experiences no vibratory motion is called a node, asshown in #igure .H.

O,#er\$%tio 3@ When the two masses are given e6ual initial displacements in the samedirection and released, they will vibrate in I mode. When they are given e6ual initialdisplacements in opposite direction and released they will vibrate in II mode as shown in#igures .D and .H

Figure 5.&

m/

m/

m/

m0

m0

m0

0

0

0

/

/

/

3

3

3

;/

;/

;0

;0

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#rom-/ for mass (4)m434 5 7 84349 8"(3"G 34)

5 7 84349 8"3"G 8"34m4349 34(849 8") 5 8"3" 77777 (4)

#or mass (")m"3" 5 7 8"(3"G 34)

5 7 8"3"9 8"34m"3"9 8"3"5 8"34 77777 (")

/et 345 + -in t 3"5 * -in t345 7 "+ -in t, 3"5 7 "* -in t

-ubstitute these in (4) and (")

7m4"

+ -in t 9 (849 8") + -in t 5 8"* -in t+ (849 8"G m4") 5 8"*+M* 5 8"M A(849 8"G m4")C 77777 (!)7 m""* -int 9 8"* -in t 5 8"+ -in t

(8"G m"") * 5 8"++M* 5 A8"G m""C M 8" 77777 (D)

'6uating (!) and (D)

8"M (849 8"G m4") 5 A8"G m""C M8"

8""5 (849 8"G m4") (8"G m"")

8""5 (849 8") 8"G m4"8"G m""(849 8") 9 m4m"D

m4m"D7 "Am48"9 m"(849 8")C 9 848"5 ;

Put "5

m4m""G Am48"9 m"(849 8")C 9 848"5 ;

\$r

5 AAm48"9 m"(849 8")C [ L Am48"9 m"(8498")"

C7 D m4m"848"CC M "m4m"

#re6uency e6uation of the system

To determine the natural fre6uencies

iven 845 " 8, 8"5 8m45 m, m"5 "m

..

..

....

..

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5 Am8 9 "m ("8 98) [ Am8 9 m8)"G Dm "m8"8CC M "m . "m

5 AN m8 [ A(Nm8)"G D (Dm"8")CC M Dm"

5 ANm8 [ (DBm"8"G 4m"8"C M Dm"

5 ANm8 [ H.NDD m8C MDm"

45 n4"5 AN m8 G H.NDD m8C MDm"5 4."HH m8 MDm"5 ;.!4!O 8Mm

/6 -.5 )m+"5 n""5 ANm8 9 H.NDD m8C MDm"5 !.4O 8 Mm

06 /.79& )m+To determine the mode shapes@I mode shape@ -ubstituting n4"5 ;.!4!O 8Mm

+M* 5 A8"G m" "C M8"5 A8"Gm". n4"CM8"

+M* 5 (8 G "m.;.!4!O 8Mm)M8 5 47"(;.!4!O)C

AB 6 -.370&

If A 6 /! B 6 0.590

II mode@ -ubstituting n""5 !.4O 8Mm

+M* 5 A8"Gm"n""CM8"

5 (8 G "m. !.4O 8Mm) M 8 5 (4 G !.4O ") 5 7 H.!N"

+M* 5 7 H.!N", if + 5 4, * 5 7 ;.4O

.. . .

A 6/A 6/

B 6 0.590

B 6 -./95

I *o'e

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". etermine the natural fre6uency and the corresponding mode shapes for the system shownin figure

iven 845 !8, 8"5 "8, 8!5 8m45 m, m"5 "m

#ree body diagram

/6 )m+ 06 ). m+

/;

/

0;

/

0;

/

3;

0

0;

0

0;

0

m/

m0

;/ ;0

/

;/

3

;0

0);

0;

/+

0

);0

;/

+

/

m/

0 3m

0

;/

;0

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Se##io;III )0-..-+ BS

T(o DOF #y#tem# )cot'.+

!. etermine the atural fre6uencies and ratio of amplitudes of the system shown in #igure.

-olution similar to example o. 4

/6 -./7 )m+ 06 /.@3/ ). m+)AB+/6 -.73/ )AB+06 -.0730

D. -ame as above

iven

m45 4.H :g m"5 ;.O :g

845 8"5 D; Mm

/6 @.3@r%'#ec 06 3.99 r%'#ec)AB+/6 -.75 )AB+06 -.5@5

m

0m

0

m/

m0

/

0

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H. etermine the natural fre6uencies of the system shown in figure. +lso determine the ratioof amplitudes and locate the nodes for each mode of vibration. +ssume that the tension ^TF inthe string remains unchanged, when the masses are displaced normal to the string.

*%##e# i 'i#pl%ce' po#itio

Free ,o'y 'i%gr%m

-/. #or mass (4)

l l l

m4

m"

T T T Tm4

m"

x4 x"

x47 x"

TT T

T

m4

m"

x4

x"T cos T cos

T cos T sin T sin

T sin

T sin

T cos x4

(x47 x")

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mx45 7 T -in 7 T -in -in 5 x4Ml, -in5 x"Ml (x4G x"Ml

5 7 T x4Ml G T (x47x")Ml 5 7 Tx4Ml G Tx4Ml 9 Tx"Ml

mx49 "Tx4Ml 5 Tx"Ml 77777 (a)-/. #or mass (")

mx"5 7 T -in9 T -in

5 7 Tx"Ml 9 T. (x4Gx")Ml

mx"9 "Tx"Ml 5 Tx4Ml 77777 (b)

/et x45 + -in t, x"5 * -in t

x45 7 "

+ -in t x"5 7"

* -in t,-ubstitute in (a) and (b)

7 m4"+ -in t 9 ("TMl) + -in t 5 (TMl) * -in t. %emoving sin t throughout

+ A("TMl) G m4")C 5 *. (TMl)

+M* 5 (TMl)M A("TMl) G m4")C 77777 (a4)

-imilarly

7 m"". * -in t 9 ("T* -in t)Ml 5 (T. + -in t)Ml

* A("TMl) G m"")C 5 +. (TMl)

+M* 5 A("TMl) G m"")CMTMl 77777 (a")

'6uating (a4) and (a") and cross multiplying

(TMl)M A("T7lm4")MlC 5 A("T G lm"")MlCM(TMl)

T"

5 ("T G lm4"

) ("T G lm""

)

T"5 DT"G "Tlm4"G "Tlm""9 l"m4m"D

l0m/m0& 0Tl )m/K m0+ 0K 3T06 - Fre"uecy E"u%tio/et 5 "

..

..

..

.. ..

..

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l"m4m""G "Tl (m49 m") 9 !T"5 ;

4, "5 A"Tl (m49 m") A"T (m49 m")l"G D l"m4m"!T")C M " m4m"l"

/et 5 m45 m" 5 m

4." 5 A"Tl (m 9 m) A"T ("ml)"G Dl"m". !T")C M ". m". l"

5 DmTl A(DmTl)"G 4" m"l"T"C M "m"l"

\$n further simplification/6 /06 Tml /6 )Tml+06 06 3Tml 06 )3Tml+*o'e Sh%pe1

+M* 5 (TMl)MA("TMl) G m4"C

I mo'e1

+M* 5 4 if + 5 4, * 5 94

II mo'e1

+M* 5 74+M* 5 74 if + 5 4, * 5 74

Se!i &efinite S'ste!s or &egenerate S'ste!

m/ m0

;/ ;0

I *o'e

(+M*)n45 4

II *o'e

m/

m0

;/

No'e. (+M*)

n45 74

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'g@ Couple' locomoti\$e

-ystems for which one of the natural fre6uencies is e6ual to Eero are called semi definitesystems.

#*@

ass (4)m4x45 8 (x"Gx4)

m4x49 8x45 8x" 77777 (4)

m"x"5 7 8 (x"G x4)

m"x"9 8x"5 8x4 77777 (")

/et x45 + -in t, x"5 * -in tx45 7 "+ -in t x"5 7 "* -in t,

m/

m0

;/

;0

x"Q x

4

;0

;0

;/

;/

m/

m0

;/ ;0

);0;

/+

);0;

/+

m/

m0

..

..

..

..

.. ..

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-ubstitute in (4) and (")

m4(7 "+ -in t) 9 8 + -in t 5 8 * -in t

+M* 5 (8)M A8 G m4"C 77777 (!)

m"(7 "* -in t) 9 8 * -in t 5 8 + -in t

+M* 5 A8 G m""C M (8) 77777 (D)

&ross multiplying and simplifying further

m/m0& )m/K m0+ 06 - Fre"uecy e"u%tio"Am4m""G 8 (m49 m")C 5 ;

#inding the roots we get the natural fre6uencies

/6 /6 - 06 06 MR)m/K m0+)m/J m0+When one of the roots of the fre6uency e6uation is Eero, one of the natural fre6uencies isEero. -uch systems are referred as semi definite systems. The system will move as a rigidbody without any distortion of spring. The amplitudes of two masses are e6ual. They are alsoreferred as free7free system.

*o'e Sh%pe#@

I mode@

(+M*)n45 (8)M A8 G m4"C

/6 -(+M*)n45 4

II mode@

B 6/

- -

m/

m0

A 6/

I mo'e

- m0

- m/

.No'e

II mo'e

A 6 /

B 6 /

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(+M*)n"5 (8)M A8 G m4"C

06 MR)m/K m0+)m/J m0+

if m45 m"5mThen

)AB+06 /

. etermine the natural fre6uency and mode shapes of the system shown in #igure. ivenm45 4; :gs, m"5 4H :gs, 8 5 !"; Mm

-olution@ It is a free Gfree system#ree body diagram

#re6uencies

/6 -n"5 A8 (m49 m")M(m4 m")C

n" 5 A!";(4; 9 4H)M (4;4H)C6 7.3- r%'#ec

ode -hapes

I mode

m/

m0

;/

;0

x"Q x

4

;0

;0

;/

;/

m/

m0

;/ ;0

);0;

/+

);0;

/+

m/

m0

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(+4M+")n45 4.;, if + 5 4, * 5 4

II mode

(+4M+")n"5 (8)MA8 G m4n""C

5 !"; M A!"; G 4; (N.!;)"C

5 7 4.DBif + 5 4, * 5 7;.N4

N. +n electric train made of two cars each of mass ";;; :gs is connected by couplings ofstiffness e6ual to D; 4;Mm. etermine the natural fre6uency of the system.

Couple' C%r#

-olution@ This is an example similar to problem o. only the answer are given here.

B 6/

- -

m/

m0

A 6/

I mo'e

- m0

- m/

.No'e

II mo'e

A 6 /

B 6 -.57/

m/m0

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iven m45m"5 ";;; :gs. 8 5 D; 4;Mm

n45 ;

n" 5 ("8Mm)

5 ("D;4;) M";;;

(nal'sis of )wo &*F )orsional S'ste!s

#igure above shows a two rotor system which can be represented as follows.

#ree body diagram is as given below.

t/ 0

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4and "in &&W direction loo:ing from left.

-/ for %otor (4)

445 7 8t(47 ")//K t/6 t0 )/+#or rotor (")""5 9 8t(47 ")""5 9 8t4G 8t"

00K t06 t/ )0+/et,

..

..

..

..

..

/ 0

/

8t0

8t/

0

8t0

8t/

/

t)/ 0+

t)/ 0+0

/ t)/ 0+t)/ 0+

0

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45 + sin t, "5 * sin t45 7 "+ sin t, "5 7 *"sin t-ubstituting the above in 4 and " and simplifying we get the amplitude ratios and fre6uencye6uation as follows.

+M* 5 8tMA8tG 4"C 77777 (a4)

+M* 5 A8tG ""C M 8t 77777 (a")

#re6uency e6uation

4"DG (49 ") 8t"5 ;

"A4""G (49 ") 8tC 5 ;

06 -! /6 -and or

4""G (49 ") 8t5 ;

n""5 A(49 ") 8tC M4 "

06 MR)/K 0+ t /J0O. etermine the natural fre6uency of Torsional vibrations of a shaft with two circular dis:s

of uniform thic:ness at its ends. The masses of the discs are m45 H;; :gs and m"5 4;;; :gsand their outer diameter 45 4"H cm and "5 4B; cm. The length of the shaft is ! m and itsdiameter 5 4; cm. odulus of rigidity for shaft material of the shaft 5 ;.O! 4;44Mm"

+lso determine in what proportion the natural fre6uency of the shaft gets changed if alonghalf the length of the shaft the diameter is increased from 4; cm to "; cm.

-olution@Part (4) #or free body diagram and expression for fre6uencies refer previous discussion.

m45 H;; :g m"5 4;;; :g

45 4."H m "5 4.B ml 5 !.;; md 5 ;.4; m 5 ;.O! 4;44Mm"

Two rotor system is a semi definite system whose natural fre6uency is given byn45 ;06MR)/K 0+ t /J0

.. ..

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n5 A8t(49")M4"C

45 m4%4"5 BO :g G m", "5 m"%""5 DH! :gm"

8t5 IpMl 5 A;.O! 4;44

M !.;;C A(dD

)M !"C 5 ".N"H 4;H

06 9./ r%'#ecPart (")@ -ince the diameters are different along the length e6uivalent stiffness is to bedetermined as follows.

'6uivalent -ystem

iven,d45 4; cm, d"5 "; cm, l45 l"5 4.H m

/te6 /t/K /t0

06 7@.@7 r%'#ecGece there i# 37U icre%#e i the %tur%l fre"uecy of the #y#tem.B. etermine the fre6uency e6uation, natural fre6uency and mode shapes for a doublependulum shown in figure.

/ 0

te

/

0

t/

t0

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iven m45 m"l45 l"5 l

Free \$ody diagram

m/

m0

l/

l0

2

m"g

m4

m4g

m"

1

x4

x"

l4

l"

T4

T"

T"

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&onsidering only the oscillation

+pplying -/ for mass (4)

m4x45 7 T4sin 49 T"sin "

but x45l4x45 l4, x"5 l49 l"x"5 l49 l"

m4l45 7 T4sin 49 T"sin " 77777 (a)

+t mass (4) T4cos45 mg 9 T"cos" 77777 (4)

+t mass (") T"cos"5 mg

"being very small cos5 4

T"5 mg 77777 (")

T45 mg 9 mg

T45 "mg 77777 (!)

.... ..

.. ..

..

T4&os 1

T4-in 1

T"-in 2

T4

T"T"

T"&os 2

T"&os 2

m4g

m"g

T"-in 2

1

2

2

..

..

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ml45 7 " mg sin 49 mg sin "

l/K 0g/ g06- ),+-imilarly for mass (")

mx" 5 7 T"sin " T"cos "5 m"g, T"5 mg

5 7 mg sin "5 7 mg"

ml (49 ") 9 mg "5 ;

l/K l0K g06 - )c+'6uations (b) and (c) represent '

/et 45 + sin t "5 * sin t

45 7 "+ sin t, "5 7 * "sin t

-ubstitute in (b) and (c)

7 l"+ sin t 9 " g + sin t G g * sin t 5 ;

+ ("g 7 l") 5 *g

AB 6 gM0gl 0 ),/+7l"+ sin t 7 l"* sin t 5 7 g * sin l

+M* 5 (l"G g)Ml"

AB 6 Mg l0 l 0 )c/+'6uating b4and c4and cross multiplying we get fre6uency e6uation.

l0& &gl0K 0g06 - fre"uecy e"u%tio/et5 "

l""G Dgl9 "g"5 ;

..

.. ..

.. ..

.. ..

..

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The roots are45 ;.HOHN gMl 5 n4"

/6 -.75 )gl+"5 n""5 !.D4D gMl

06 06 /.9&7 )gl+ode shapes

I mode

(+M*)n45 g MA"g 7 ln4"C 5 4M4.D4D!

(/+ , 1/1.-1-3 ( , 1 + , 1.-1-3

II mode

(+M*)n"5 g MA"g 7 ln""C 5 4M74.D4D!

( , 1 + , 01.-1-