ENMEC4060 Vibration and machine dynamics

Equations of motion about P and Q
Assuming θ1 and θ2 are small, the moment equilibrium equations for the two moves about P and
Q.
At point P, the equation of motion will be;
j ̈θ1 + mglsin θ1+ k(x1-x2) d =0
Taking account of horizontal displacement, we have
X1= dsin θ1
X2=d sin θ2
Note θ2 is the displacement for pendulum 2
Substituting j with ml2, x1 with dsin θ1 and x2 with d sin θ2, the simplified equation will be:
Ml2 ̈θ1+ mgl sin θ1+ k (dsin θ1- dsin θ2) d=0 simplified as
Ml2 ̈θ1+ (mgl +kd2) θ1 – kd2 θ2 =0 for the 1st pendulum
At point Q, the equation of motion will be;

j ̈θ2 + mglsin θ2+ k(x2- x1) d =0
Note the acceleration for pendulum 2 will be ̈θ2
Following the same process as point P, the simplified equation is shown below
Ml2 ̈θ2+ (mgl +kd2) θ2 – kd2 θ1 =0 for the 2nd pendulum eqn 2
Natural frequencies and modal shapes
Assuming motion to be S.H.M with
θi (t) = θi cos (ωt- φ) with i= 1,2,3 eqn 3
Where θ1 and θ2 are the amplitudes for pendulum 1 and 2 respectively, φ as the phase angle and
ω as the natural frequency.
With equations 1, 2 and 3 the equation of motion can be expressed in form of a matrix as shown
below
-ω2 ml2
(1 0
0 1)[θ1
θ2]+
[mgl+kd2 −kd2
−kd2 mgl+kd2].
[θ1
θ2] =
[0
0] eqn 4
Taking into account non trivial solutions, the frequency equation will be
|−ω2 ml2+mgl+kd2 −kd2
−kd2 −ω2 ml2+mgl+kd2|= 0
With det (-ω2 m + k) =0 similarly expressed as
-ω4 –ω2 [2g
l+2kd2
ml2 ¿ + g2
l2 +2gkd2
ml3 =0
Substituting ω4 with Y2 and ω2 with Y
The above equation can be rewritten as
M2l4x2- 2ml2(mgl +kd2) x2 + (m2g2l2+ 2mglkd2) =0
The roots of the above are

X1= g
l and x2= g
l+2kd2
ml2
Substituting Y1 for ω12
ω1= (g/l) 1/2
Also ω1 can stand for 2πf1
Therefore, the 1st natural frequency can be written as
f1=1
2π(g/l) 1/2
Following the same steps, the 2nd natural frequency will be
f2=1
2 π(g
l+2kd2
ml2 ) 1/2
Applying into our matrix equation the amplitude vector, the following equations will be
generated.
-ml2 ω2- kd2x2 + (mgl+ kd2) x1 =0
-ml2 ω2+ (mgl+ kd2) x2- kd2x1=0
Applying our natural frequencies and the amplitudes, we get the first and second amplitude ratios
respectively as follows.
Amplitude ratio one= 1
r2=kd2
−ml2 ω22+(mgl+kd2)
Amplitude ratio two= 1
r1=kd2
−ml2ω12+(mgl+kd2)
The generated amplitude vectors will be
λ1=θ11
(1
kd2
−ml2ω12+(mgl+kd2))