Double Pendulum with Spring Analysis

This assignment delves into the dynamics of a double pendulum system coupled with a spring. It starts by deriving the equations of motion for the system and then proceeds to analyze its characteristics. The assignment focuses on determining the natural frequencies and mode shapes of the system using harmonic motion assumptions. Further, it explores how stiffness variations and operating speeds influence the system's behavior.

Answer: (1) _A
Double pendulum with spring:
System equation of motion are:
¨θ1 + ( g
l + k
m1 )θ1 −
( k
m1 )θ2=0
¨θ2 + ( g
l + k
m2 ) θ2−
( k
m2 ) θ1=0
Characteristic equation
θ1= X 1 cos ( ωt−∅ )
θ2= X 2cos ( ωt−∅ )
¨θ1=−X 1ω2 cos ( ωt−∅ )
¨θ2=−X 2 ω2 cos ( ωt−∅ )
Equation of motion
M ¨x + Kx=0
(−ω2 0
0 −ω2 ) (X 1
X 2 )cos ( ωt −∅ )+
[ ( g
l + k
m1 ) −k
m
−k
m ( g
l + k
m1 ) ] ( X 1
X 2 )cos ( ωt−∅ ) =0
h2 −2h ω2+ω4− ( k
m )2
=0
where h=
( g
l + k
m1 )
Natural frequency and natural modes:
ω2=h ± k
m =( g
l + k
m1 )± k
m
ω1= √ g
l , ω2= √ g
l +2 k
m
(−ω2 + g
l + k
m1 ) X 1− k
m X 2=0
For natural mode formula,
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X 2
X 1 =−mω2+2 k
k
Put ω1= √ g
l
X 2
X 1 =−mω2+2 k
k
X 2
X 1 =1
ω2= √ g
l +2 k
m
X 2
X 1 =−1
Answer:1_ ( C)
sin θ1=θ1= X1
l … … ..(1)
sin θ2=θ2= X2− X1
l … … … … ..(2)
T 2 cosθ2=mg
And T 1 cos θ1=mg+T 2 cos θ2
¿ the value of θ1∧θ2 above relations reduces ¿
T 2=mg … … … … … ..(3)
And T 1=mg+T 2 … … … … .(4)
Write differential equation of two masses for motion in horizontal direction,
We have m ¨x1=T 2 sinθ2−T 1 sin θ1
m ¨x2=−T2 sin θ2
From the equation of (1), (2), (3) and (4) following relation establish.
m ¨x❑+ [ 3 m
l ] g x1= m
l g x2 … … …..(5)
m ¨x❑+ [ m
l ] g x2= m
l g x1 … … … … … … .. ( 6 )
Consider harmonic motion.
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x1=X1 sin ω❑ t … … … … … … … … ..(7)
x2=X2 sin ω❑ t … … … … … … … …..( 8)
Value of Equation (7) and (8) put into equation (5) and (6)
Cancel common sinωt
[ -m ω2 + [ 3 m
l ] g ¿ X1= m
l g X2
[-m ω2 + m
l g ¿ X2= m
l g X1
Re arrange above both equation
X1
X2
= ( g
l )
3 g
l −ω2
X1
X2
= ( g
l )−ω2
g
l
( g
l )
3 g
l −ω2
= ( g
l )−ω2
g
l
ω2
g2 − 4 ω4
lg + 2
l2 =0
ωn 1= √ ( g
l ) ( 2− √2 )
ωn 1= √ ( g
l ) ( 2+ √2 )
Corresponding mode shape
( X1
X2 )1
=−1+ √2=+0.41
( X1
X2 )2
=−1− √ 2=−2.41
Answer: 1_ (E)
Equation of motion using x (t) and θ ( t )
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Force equilibrium equation in vertical direction can be written as:
m ¨x =−k1 ( x−l1 θ ) −k 1 ( x +l2 θ ) … … … …… … ( 1 )
Moment of equation about the C.G. can be expressed as
J0 ¨θ=k1 ( x −l1 θ ) −k2 ( x+l2 θ ) l2 … … … … … ( 2 )
Equation rearranging and written in matrix form as
[m 0
0 J 0 ] { ¨x
¨θ }+ [ k1 +k2 − ( k1 l1−k2 l2 )
− ( k1 l1−k2 l2 ) ( k1 l1
2 +k 2 l2
2 ) ] {x
θ }=0
Mass and stiffness matrix as following:
M = [ m 0
0 IC ]
And K= [ k1+ k2 − (k1 l1−k2 l2 )
– ( k1 l1−k2 l2 ) (k 1 l1
2+k2 l2
2 ) ]
x=Xsinωt
θ=α sinωt
ωn 1= √ k 1+ k2
M
And ωn 2= √ k1 l1
2 +k2 l2
2
J
x=Xsinωt
θ=α sinωt
Therefore mode shape:
X
α =
k1 l1−k2 l2
M
k 1+ k2
M + k1 +k 2
M
= k1 l1−k2 l2
2 k1 +2 k2
If k1 = K2
X
α =(l1−l2 )
4
Answer (1)_F:
Equation of motion are given below
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m ¨x1 +2 kx 1−kx 2=0
m ¨x2−kx 1+2 kx 2=0
Assume simple harmonic motion,
x (t )= Xi cos ( ωt+ ∅ )
Frequency equation can be obtained by:
[ −m ω2 +2 k −k
−k −mω2+2 k ] =0
m2 ω4 −4 kmω2+ 3 k2=0
Solution to above equation gives the natural frequencies:
ω 1={ 4 km− ( 16 k2 m2−12 k2 m2 )
1
2
2m2 }
1/ 2
= √ k
m
ω 1={ 4 km+ ( 16 k2 m2−12 k2 m2 )
1
2
2 m2 }
1 /2
= √ 3 k
m
r1= X2
( 1 )
X1
( 1 ) =−mω1
2+ k1+ k2
k2
= k2
−mω1
2+ k2+ k3
r2= X2
( 1 )
X1
( 1 ) =−mω2
2+ k1+ k2
k2
= k2
−mω2
2+ k2 +k3
Amplitude ratios are given as following:
For k1 = k2 = k3 = k
r1= X2
( 1 )
X1
( 1 ) =−mω1
2+2 k❑
k❑
x k❑
−mω1
2+2 k❑
=1
r2= X2
( 2 )
X1
( 2 ) =−mω2
2+ 2k❑
k❑
x k❑
mω2
2+2 k❑
=−1
Answer: 2
k =3 EI
L3 =3 ( 210 x 109 ) (1.3 x 10−7)
2 =4.095 x 104
And
ω11= √ k
m = √ 4.095 x 104
35 =34.20 rad / sec
Operating speed
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ω=180 ( 2 π ) ( 1
60 )=18.84 rad / sec
k 2=m2 ω2=5 x (18.84)2=1774.728=1.774 x 103 N /m
(B)
ωn= √ k
m= √ 4.095 x 104
5 =90.49 rad / sec
r1= ω
ω1
= 18.84
34.20 =0.55
Answer: 3
K= 48 EI
L3 = 48 x 200 x 109 x 1. 8 x 10−4
27 =640 x 105 N /m
ω= √ k
m= √ 640 x 105
200 =565.6854 rad
sec
Operating speed is ω11=3000 x 2 π x ( 1
60 )
1256.6370 rad
sec
Frequency ratio:
r = ω
ω11
= 565.6854
1256.6370 =0.4501 rad /sec
Steady state amplitude
X1 =( 200
1.41 ) ( 0.9452
1−0.9452 ) =0.01790 m
Since the ratio of the absorber stiffness to absorber mass is fixed, the absorber with the
Minimum mass is also the absorber with the minimum stiffness. The amplitude of the
Absorber is to be limited to 40 cm,
k 2= F0
X2
= 1.41 x ( 1256.6370 ) 2
0.4 =5566456.3379=5.566 x 106 N
m
Minimum steady state amplitude of the absorber mass is
X2 = F0
k2
=1.41 x 1256.63702
640 x 105 =0.034 m
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Answer: (4)
Given Data:
m=500 kg
equivalent stiffness k=5 x 106 N
m
Operating speed, N = 800 rpm
Unbalance harmonic force of magnitude ¿ 50000 N
E = 210 x 109 N/m2
I = 4 x 10-6 m4
m=10 kg mass
Natural Frequency ωn= √ k
m= √ 5 x 106
500 =100 rad /sec
Operating speed
ω=800 x 2 π x 1
60 =83.77 radsec
q= 1
1+μ
mass ratio μ= 10
500 =0.02
q= 1
1+ μ = 1
1+0.02 =0.98
And ζ = √ 3 μ
8 ( 1+ μ ) = √ 3 x 0.02
8 ( 1+ 0.02 ) =0.085
r = √ 1+ ( 1+ μ ) q2 ± √ 1−2q2+ ( 1+ μ ) 2 q4
2+ μ
r =1.0246
r =0.91
Answer: (5)
[ m1 ¨x1 + (k 1+ k2 ) x1 ] −k2 ( x2 ) =F sinωt …… …(1)
[−m2 ¨x2 + ( k2 ) x1 ]−k2 ( x2 ) =0 … … … …(2)
Let assume, for steady state, the solution as
x1=X1 sin ω❑ t … … … … … … … … ..(3)
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x2=X2 sin ω❑ t … … … … … … … …..( 4)
Above both equation substitute into equation (1) and (2),
[−m1 ω2 + ( k1+ k2 ) x1 ] X1 −k2 ( X2 )=F0 … … … ( 5 )
k1 X1− [−m2 ω2+ k2 ] X2=0 … … … … … …..(6)
Solving above two equation,
X1 =¿ ¿
X2 = k2 F0
[ m1 m2 ω4 − (m1 k2+m2 ( k 1+k2 ) ) ω2 +k1 k 2 ] … … … … .. ( 8 )
To obtain above equation in dimensionless forms, let us divide their numerators and
denominators by k1 k2 and introduce following equation.
X st= F0
k1
= 200
25 =8=Zero frequrncy deflection of first mass
μ= m2
m1
=ratio of abosorber mass¿ mainmass
ω1= √ k
m = √ 5 x 104
25 =44.72rad / sec
ω2= √ k
m = √ 4 x 104
31 =35.92 rad /sec
μ= m2
m1
= 31
25 =1.24
X2 = X st
μ ω2
1
ω1
2
= 8
1.24 x 35.922
44.722
=10
X1
Xst
=
1− ω2
ω2
2
ω4
ω4
2
− ( 2+ μ ) ω2
ω2
2
+1
=
1− 672
35.922
674
35.924 − ( 2+1.24 ) 672
35.922 +1
= 1−1.92
0.17 =−0.92
0.17 =5.411
X1
Xst
= 1
ω4
ω4
2
− ( 2+μ ) ω2
ω2
2
+1
= 1
674
35.924 − ( 2+1.24 ) 672
35.922 +1
= 1
−0.17 =−5.88
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ω2
ω2
2
=(1+ μ
2 )+ √μ+ μ2
4
¿ ( 1+ 1.24
2 ) + √ 1.24+ 1.242
4
¿ ( 1+0.62 ) +1.27
ω2
ω2
2
=2.89
ω❑
ω❑
2
= 1.70 …..Answer
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