The document describes the mathematical modeling and analysis of a dynamic vibration absorber system. It presents the differential equations of motion for a primary system attached to a secondary absorber system. It then derives the frequency response equation and shows that tuning the natural frequencies of the two systems eliminates vibration response. MATLAB code is also included to simulate and analyze the time response and frequency response of the tuned dynamic vibration absorber.

1. Dynamic Vibration Absorber
Dynamic Absorber
F(t)
x1(t)
m1
k1 k2
m2
x2(t)
Primary
system
Secondary
system
Figure: Absorber
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2. Dynamic Vibration Absorber
Differential equations of motion
From free body diagrams of the masses m1 and m2
m1¨x1 = −k1x1 − k2 (x1 − x2) + Feqsinωt
m2¨x2 = k2 (x1 − x2)
. (1)
Rearranging the Eq.(1)
m1¨x1 + (k1 + k2) x1 − k2x2 = Feqsinωt
m2¨x2 + k2x2 − k2x1 = 0
. (2)
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3. Dynamic Vibration Absorber
Formulas
For steady state solution, assume solution x1 = X1 sin ωt and
x2 = X2 sin ωt. Substituting for x1 and x2 and their second time
derivatives in Eq.(2), we obtain
k1 + k2 − mω2 X1 − k2X2 sin ωt = Feqsinωt
−k2X1 + k2 − m2ω2 X2 sin ωt = 0 .
(3)
Collecting out common terms and also because sin ωt = 0 at all
time, we get
k1 + k2 − mω2 X1 − k2X2 = Feq
−k2X1 + k2 − m2ω2 X2 = 0 .
(4)
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4. Dynamic Vibration Absorber
Cramer’s rule
Solving for X1 and X2 using Cramer’s rule, we obtain
X1 =
Feq −k2
0 k2 − m2ω2
∆ω
, X2 =
k1 + k2 − mω2 Feq
−k2 0
∆ω
(5)
The frequency equation is given by
∆ω =
k1 + k2 − mω2 −k2
−k2 k2 − m2ω2 (6)
which semplifies to
∆ω = 0 ⇒ m1m2ω4
− [(k1 + k2) m2 + k2m1] ω2
+ k1k2 = 0 (7)
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5. Dynamic Vibration Absorber
Frequency equation
Dividing out Eq. (7) by k1k2
m1m2
k1k2
ω4
−
k1 + k2
k1
m2
k2
+
m1
k1
ω2
+ 1 = 0 . (8)
In Eq.(8) the natural frequency of the
main system is ω11 = k1/m1;
auxiliary system is ω22 = k2/m2.
The Eq.(8) may be rewritten as
ω
ω11
2
ω
ω22
2
− 1 +
k2
k1
ω
ω22
2
+
ω
ω11
2
+ 1 = 0 .
(9)
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6. Dynamic Vibration Absorber
Dynamic Absorber
When the main system and auxiliary system have same natural
frequency ω11 = ω22, let
ω
ω11
= r
ω
ω22
= r . (10)
Letting m2/m1 = µ, Eq.(10) reduces to
r4
− (2 + µ) r2
+ 1 = 0 . (11)
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7. Dynamic Vibration Absorber
Dynamic Absorber
The new resonance frequencies of tuned system depend on
the mass ratio µ. Let us assume that the resonance occurs at
r ≡ r1 =
ω
ω11
r ≡ r2 =
ω
ω22
. (12)
where the two roots are given by
r2
1 , r2
2 =
2 + µ
2
±
1
2
(2 + µ)2
− 4 . (13)
or
r2
1 , r2
2 = 1 +
µ
2
± 1 +
µ
2
2
− 1 . (14)
When ω11 = ω22, the natural frequencies are given by
ω
ω11
2
= 1 +
µ
2
± 1 +
µ
2
2
− 1 . (15)
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8. Dynamic Vibration Absorber
Tuned condition
Solving the determinants in Eq.(5) we have



X1 =
Feq k2 − m2ω2
∆ω
=
Feq k2 − m2ω2
k1 + k2 − m1ω2 k2 − m2ω2 − k2
2
X2 =
Feqk2
∆ω
=
Feqk2
k1 + k2 − m1ω2 k2 − m2ω2 − k2
2
.
(16)
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9. Dynamic Vibration Absorber
Tuned condition
Under tuned condition, substituting k2 = m2ω2, the Eq.(16)
gives



X1 =
Feq · 0
k1 + k2 − m1ω2 · 0 − k2
2
= 0
X2 =
Feqk2
k1 + k2 − m1ω2 · 0 − k2
2
= −
Feq
k2
.
(17)
X2 = −
Feq
k2
⇒ −X2k2 = Feq . (18)
Fspring = Fexternal
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10. Dynamic Vibration Absorber
Application
A machine runs at 5500 rpm. Its forcing frequency is very
near to its natural frequency.
If the nearest frequency of the machine is to be at least 20
per cent from the forced frequency, design a suitable
vibration absorber for the sistem.
Assume the mass of the machine as m1 = 30 kg.
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11. Dynamic Vibration Absorber
Application
The natural frequency of the system at 478 rpm is
ωn = 2π
478
60
≈ 50 rad/s . (19)
For 20 per cent variation, we obtain
r1 =
ω
ωn
= 0.8 r2 =
ω
ωn
= 1.2 . (20)
From Eq. (11) we have
0.84
− (2 + µ) 0.82
+ 1 = 0 ⇒ µ = 0.202 . (21)
Mass of auxiliary system is
m2 = µ ⇒ m1 = 0.202 × 30 = 6.06 kg . (22)
Stiffness values
k1 =ω2
n × m1 = 502
× 30 = 75000 N/m
⇒ k2 = ω2
n × m2 = 502
× 6.06 = 15150 N/m . (23)
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12. Dynamic Vibration Absorber
Matlab: Main
tspan = 0 : 0.05 : 10;
y0 = [0; 0; 0; 0];
[t, y] = ode23( Dabsorber , tspan, y0);
subplot(211);
plot(t, y(:, 1));
xlabel( t(Solid line : x1(t)Dotted line : xd1(t)) )
holdon;
plot(t, y(:, 2), −− );
subplot(212);
plot(t, y(:, 3));
xlabel( t(Solid line : x2(t)Dotted line : xd2(t)) );
hold on;
plot(t, y(:, 4), −− );
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13. Dynamic Vibration Absorber
Matlab: Function
function f = Dabsorber(t, y)
m1 = 30;
m2 = 0.202 ∗ m1;
omegan = 50;
k1 = omegan ∗ omegan ∗ m1;
k2 = omegan ∗ omegan ∗ m2;
omega = 15 ∗ 2 ∗ pi;
Feq = 10;
F = Feq ∗ sin(omega ∗ t);
f = zeros(4, 1);
f(1) = y(2);
f(2) = F/m1 − ((k1 + k2)/m1) ∗ y(1) + (k2/m1) ∗ y(3);
f(3) = y(4);
f(4) = (k2/m2) ∗ y(1) − (k2/m2) ∗ y(3);
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14. Dynamic Vibration Absorber
Matlab: FRF
clear all
clc
m1 = 30;
m2 = 0.202 ∗ m1;
omegan = 50;
k1 = omegan ∗ omegan ∗ m1;
k2 = omegan ∗ omegan ∗ m2;
Feq = 10;
k = 31;
X1 = zeros(k, 1);
X2 = zeros(k, 1);
omega = zeros(k, 1);
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15. Dynamic Vibration Absorber
Matlab: FRF
for m = 1 : k
omega(m) = m;
Deltaomega = m1 ∗ m2 ∗ omega(m)4 − ((k1 + k2) ∗ m2 + k2 ∗
m1) ∗ omega(m)2 + k1 ∗ k2;
X1(m) = Feq ∗ (k2 − m2 ∗ omega(m)2)/Deltaomega;
X2(m) = Feq ∗ k2/Deltaomega;
end
plot(omega, X1, omega, X2); legend( X1, X2); grid;
xlabel( Frequency ); ylabel( Amplityde );
title( Frequency Response Functions );
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16. Dynamic Vibration Absorber
Time history
0 2 4 6 8 10
−10
−5
0
5
10
t(Solidline: x1(t) Dottedline: xd1(t))
0 2 4 6 8 10
−100
−50
0
50
100
t(Solidline: x2(t) Dottedline: xd2(t))
Figure: Time history
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17. Dynamic Vibration Absorber
Time history
0 2 4 6 8 10
−2
−1.5
−1
−0.5
0
0.5
1
1.5
2
primary system
secondary system
Figure: Time history
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18. Dynamic Vibration Absorber
FRF
0 50 100 150 200
−5
0
5
10
x 10
−3
Frequency [rad/s]
Amplityde
Frequency Response Functions
X1
X2
Figure: Frequency Response Function
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