MEE411 – MECHANICAL VIBRATIONS
LECTURE NOTES
PART 5 – VIBRATION OF SYSTEMS WITH 2DEGREES OF FREEDOM
DEPARTMENT OF MECHANICAL ENGINEERING
FACULTY OF ENGINEERING
UNIVERSITY OF BENIN
COURSE LECTURER – ENGR. DR. E. G. SADJERE
ASSISTANT LECTURER – ENGR. MRS. A. G. OREAVBIERE
2020
MEE411 LECTURE NOTES
2020
Contents
Introduction ...................................................................................... 3
Vibration in Systems with two-degrees of Freedom ............................. 5
Undamped 2-degreee of freedom Vibration ........................................ 9
Two Connected Simple Pendulum .................................................... 14
MEE 411 – Mechanical Vibrations
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INTRODUCTION
The number of degrees of freedom of a system is the number of independent
coordinates required to completely specify the configuration (position) of the
system
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In all these examples, two coordinates are required to specify the positions of m1
and m2. These could be Cartesian coordinates or polar coordinates. These
coordinates has to be independent.
Degrees of Freedom
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VIBRATION IN SYSTEMS WITH TWO-DEGREES OF FREEDOM
Equations of motions are:
And let 𝑥1 > 𝑥2
Let 𝑚1 be displaced in the direction, 𝑥1 , then the forces acting on the bodies are
as follows:
Notes:
1. Force 𝑘1 𝑥1 is directed to the left as restoring force
2. Force 𝑘(𝑥1 − 𝑥2 ) on mass 𝑚1 is also directed to the left as spring 𝑘 is
compressed
3. Force 𝑘(𝑥1 − 𝑥2 ) on mass 𝑚2 is to the right as action and reaction are equal
and opposite
4. Force 𝑘2 𝑥2 is to the left as spring 𝑘2 is compressed.
The equations of motions are;
For 𝑚1
𝑚1 𝑥̈ 1 = −𝑘1 𝑥1 − 𝑘(𝑥1 − 𝑥2 )
For 𝑚2
𝑚2 𝑥̈ 2 = 𝑘(𝑥1 − 𝑥2 ) − 𝑘2 𝑥2
These are second order differential equations which can be solved by assuming a
solution of the form
𝑥 = 𝐴𝑒 𝑠𝑡
OR
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Since we have solved this in free vibration we can assume a generalized solution
of the form
𝑥1 = 𝐴1 sin(𝜔𝑡 + 𝜑)
And
𝑥2 = 𝐴2 sin(𝜔𝑡 + 𝜑)
If we assume that the bodies are oscillating with the same frequency.
Rearranging eqns 1 and 2, we have
𝑚1 𝑥̈ 1 + 𝑘1 𝑥1 + 𝑘(𝑥1 − 𝑥2 ) = 0
𝑚2 𝑥̈ 2 − 𝑘(𝑥1 − 𝑥2 ) + 𝑘2 𝑥2 = 0
Substituting for 𝑥1 and 𝑥2 and differentiating, we have
−𝑚1 𝜔2 𝐴1 sin(𝜔𝑡 + 𝜓) + 𝑘1 𝐴1 sin(𝜔𝑡 + 𝜓) + 𝑘(𝐴1− 𝐴2) sin(𝜔𝑡 + 𝜓) = 0
−𝑚2 𝜔2 𝐴2 sin(𝜔𝑡 + 𝜓) − 𝑘(𝐴1− 𝐴2) sin(𝜔𝑡 + 𝜓) + 𝑘2 𝐴2 sin(𝜔𝑡 + 𝜓) = 0
Simplifying and rearranging by collecting all 𝐴1𝑠 and 𝐴2𝑠 , we have,
−𝑚1 𝜔2 𝐴1 + 𝑘1 𝐴1 + 𝑘(𝐴1 − 𝐴2) = 0
−𝑚2 𝜔2 𝐴2 − 𝑘(𝐴1− 𝐴2) + 𝑘2 𝐴2 = 0
−𝑚1 𝜔2 𝐴1 + 𝑘1 𝐴1 + 𝑘𝐴1 − 𝑘𝐴2 = 0
−𝑚2 𝜔2 𝐴2 − 𝑘𝐴1 − 𝑘𝐴2 + 𝑘2 𝐴2 = 0
𝐴1 (𝑘 + 𝑘1 − 𝑚1 𝜔2 ) + 𝐴2 (−𝑘) = 0
𝐴1 (−𝑘) + 𝐴2 (𝑘 + 𝑘2 − 𝑚2 𝜔2 ) = 0
This can be written as
𝑘 + 𝑘1 − 𝑚1 𝜔2
(
−𝑘
−𝑘
𝐴1
2 ) (𝐴 ) = 0
𝑘 + 𝑘2 − 𝑚2 𝜔
2
And
𝑘 + 𝑘1 − 𝑚1 𝜔2
−𝑘
|=0
−𝑘
𝑘 + 𝑘2 − 𝑚2 𝜔2
This is the frequency equation or characteristic equation. The determinant can be
multiplied out to give
|
(𝑘 + 𝑘1 − 𝑚1 𝜔2 )(𝑘 + 𝑘2 − 𝑚2 𝜔2 ) − 𝑘 2 = 0
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Frequency Equation
Alternatively
𝐴1
𝑘
=
𝐴2 𝑘 + 𝑘1 − 𝑚1 𝜔 2
𝐴1 𝑘 + 𝑘2 − 𝑚2 𝜔2
=
𝐴2
𝑘
Equating both equations,
𝑘
𝑘 + 𝑘2 − 𝑚2 𝜔2
=
𝑘 + 𝑘1 − 𝑚1 𝜔 2
𝑘
Hence again
(𝑘 + 𝑘1 − 𝑚1 𝜔2 )(𝑘 + 𝑘2 − 𝑚2 𝜔2 ) − 𝑘 2 = 0
Or
(𝑘 + 𝑘1 − 𝑚1 𝜔2 )(𝑘 + 𝑘2 − 𝑚2 𝜔2 ) = 𝑘 2
Simplification
Consider the case where
𝑘1 = 𝑘2 = 𝑘
And
𝑚1 = 𝑚2 = 𝑚
The frequency equation becomes
(𝑘 + 𝑘 − 𝑚𝜔2 )(𝑘 + 𝑘 − 𝑚𝜔2 ) − 𝑘 2 = 0
(2𝑘 − 𝑚𝜔2 )(2𝑘 − 𝑚𝜔2 ) − 𝑘 2 = 0
4𝑘 2 − 4𝑘𝑚𝜔2 + 𝑚2 𝜔4 − 𝑘 2 = 0
𝑚2 𝜔4 − 4𝑘𝑚𝜔2 − 3𝑘 2 = 0
Or
(𝑚𝜔2 − 𝑘)(𝑚𝜔2 − 3𝑘) = 0
i.e
𝑚𝜔2 − 𝑘 = 0
Or
𝑚𝜔2 − 3𝑘 = 0
Which gives
𝜔1 = √
𝑘
𝑚
Or
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3𝑘
𝜔2 = √
𝑚
For 𝜔1 = √ ,
𝑘
𝑚
𝐴1
𝐴2
= +1
3𝑘
𝐴1
𝐴2
= −1
For 𝜔2 = √ 𝑚 ,
These give the two modes of vibration for the system.
𝐼
𝐴
In mode 1, (𝐴1 ) = 1, the bodies move in phase with each other and with the same
2
amplitude as if connected by a rigid link
𝐴
𝐼𝐼
In mode 2, (𝐴1 ) = −1, the bodies move exactly out of phase with each other but
2
with the same amplitude.
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UNDAMPED 2-DEGREEE OF FREEDOM VIBRATION
Masses displaced to the right to start vibration. Positions 𝑥1 , 𝑥2 are measured
from equilibrium.
Position of masses
Free body diagram
𝑚1 = 𝑚; 𝑚2 = 2𝑚
𝑚1 is displaced 𝑥1
𝑚2 is displaced 𝑥2
Assume 𝑥1 > 𝑥2
Equations of motion: One for each mass
Hence
𝑚𝑥̈ 1 = −𝑘𝑥1 − 𝑘(𝑥1 − 𝑥2 )
2𝑚𝑥̈ 2 = 𝑘(𝑥1 − 𝑥2 ) − 𝑘𝑥2
𝑚𝑥̈ 1 = −𝑘𝑥1 − 𝑘𝑥1 + 𝑘𝑥2
2𝑚𝑥̈ 2 = 𝑘𝑥1 − 𝑘𝑥2 − 𝑘𝑥2
𝑚𝑥̈ 1 = −2𝑘𝑥1 + 𝑘𝑥2
2𝑚𝑥̈ 2 = 𝑘𝑥1 − 2𝑘𝑥2
𝑚𝑥̈ 1 + 2𝑘𝑥1 − 𝑘𝑥2 = 0
2𝑚𝑥̈ 2 − 𝑘𝑥1 + 2𝑘𝑥2 = 0
Remember we are dealing with forces which are vectors
𝑥1
Let 𝑥⃗ = (𝑥 ) vector
2
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𝑥̈
𝑚 0
0
2𝑘 −𝑘 𝑥1
(
) ( 1) + (
) (𝑥 ) = ( )
𝑥̈
0 2𝑚
0
−𝑘 2𝑘
2
2
Mass matrix
Stiffness matrix
As usual, we assume oscillatory solution
𝑥1 = 𝑅1 𝑒 𝑖(𝜔𝑡−𝜙)
𝑥2 = 𝑅2 𝑒 𝑖(𝜔𝑡−𝜙)
Same frequency
Differentiating and Substituting
𝑥̇ 1 = 𝑖𝜔𝑅1 𝑒 𝑖(𝜔𝑡−𝜙)
𝑥̈ 1 = −𝜔2 𝑅1 𝑒 𝑖(𝜔𝑡−𝜙)
−𝑚𝜔2 𝑅1 𝑒 𝑖(𝜔𝑡−𝜙) + 2𝑘𝑅1 𝑒 𝑖(𝜔𝑡−𝜙) − 𝑘𝑅2 𝑒 𝑖(𝜔𝑡−𝜙) = 0
−𝑚𝜔2 𝑅1 + 2𝑘𝑅1 − 𝑘𝑅2 = 0
(−𝑚𝜔2 + 2𝑘)𝑅1 − 𝑘𝑅2 = 0
Similarly,
𝑥̇ 2 = 𝑖𝜔𝑅2 𝑒 𝑖(𝜔𝑡−𝜙)
𝑥̈ 2 = −𝜔2 𝑅2 𝑒 𝑖(𝜔𝑡−𝜙)
−2𝑚𝜔2 𝑅2 𝑒 𝑖(𝜔𝑡−𝜙) − 𝑘𝑅1 𝑒 𝑖(𝜔𝑡−𝜙) + 2𝑘𝑅2 𝑒 𝑖(𝜔𝑡−𝜙) = 0
−2𝑚𝜔2 𝑅2 − 𝑘𝑅1 + 2𝑘𝑅2 = 0
−𝑘𝑅1 − 2𝑚𝜔2 𝑅2 + 2𝑘𝑅2 = 0
−𝑘𝑅1 + (−2𝑚𝜔2 + 2𝑘)𝑅2 = 0
So we have,
(−𝑚𝜔2 + 2𝑘)𝑅1 − 𝑘𝑅2 = 0
−𝑘𝑅1 + (−2𝑚𝜔2 + 2𝑘)𝑅2 = 0
2
(−𝑚𝜔 + 2𝑘
−𝑘
𝑅
−𝑘
) ( 1)
2
𝑅
−2𝑚𝜔 + 2𝑘
2
Solving for 𝑅1 and 𝑅2
For non-trivial solution, the determinant of the coefficients must vanish
i.e
2
−𝑘
|−𝑚𝜔 + 2𝑘
|=0
−𝑘
−2𝑚𝜔2 + 2𝑘
(−𝑚𝜔2 + 2𝑘)(−2𝑚𝜔2 + 2𝑘) − (−𝑘)(−𝑘) = 0
2𝑚2 𝜔4 − 2𝑘𝑚𝜔2 − 4𝑘𝑚𝜔2 + 4𝑘 2 − 𝑘 2 = 0
2𝑚2 𝜔4 − 6𝑘𝑚𝜔2 + 3𝑘 2 = 0
Divide through by 2𝑚2
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𝑘 2 3 𝑘 2
𝜔 −3 𝜔 + ( ) =0
𝑚
2 𝑚
This is a quadratic equation for 𝜔2 and solving gives
4
2
𝜔1,2
=
𝑘
𝑘 2
3 𝑘 2
3 𝑚 ± √(−3 𝑚) − 4 × 1 × 2 (𝑚)
2×1
𝑘
𝑘 2
𝑘 2
3 𝑚 ± √9 (𝑚) − 6 (𝑚)
2
𝜔1,2
=
2
3
𝑘
1 𝑘
2
𝜔1,2
=
± . √3
2𝑚 2 𝑚
𝑘 3 √3
2
𝜔1,2
= ( ± )
𝑚 2
2
𝜔12 =
𝑘 3 √3
𝑘
( − ) = 0.634
𝑚 2
2
𝑚
𝑘 3 √3
𝑘
( + ) = 2.366
𝑚 2
2
𝑚
There are two frequencies at which the two masses will oscillate together. They
are the two natural frequencies
𝜔22 =
𝑅
Of particular interest to us is the ratio 1⁄𝑅 , now back to our equations
2
(−𝑚𝜔2 + 2𝑘)𝑅1 − 𝑘𝑅2 = 0
−𝑘𝑅1 + (−2𝑚𝜔2 + 2𝑘)𝑅2 = 0
𝑅1
𝑘
=
𝑅2 2𝑘 − 𝑚𝜔 2
Also
𝑅1
𝑘
=
𝑅2 2𝑘 − 2𝑚𝜔 2
Therefore
2𝑘 − 𝑚𝜔2 = 2𝑘 − 2𝑚𝜔2
Let’s define
𝜆=
Then
𝑅1
𝑅2
𝑅1
𝑘
1
𝜆1 = ( )
=
=
= 0.731
2
𝑚
𝑘
𝑅2 𝜔=𝜔
2𝑘 − 𝑚𝜔
2
−
(
)
(0.634
)
1
𝑚
𝑘
𝜆2 = (
𝑅1′
𝑘
1
=
=
= −2.73
′)
2
𝑚
𝑘
𝑅2 𝜔=𝜔
2𝑘 − 𝑚𝜔
2
−
(
)
(2.366
)
2
𝑚
𝑘
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These defines the two normal modes of vibration of the 2 degree of freedom
system.
First Mode
Second Mode
Since the system is linear, a general solution will be the superposition of the two
modes, hence
𝑥1 = 𝑅1 sin(𝜔1 𝑡 − 𝜙1 ) + 𝑅1 ′ sin(𝜔2 𝑡 − 𝜙2 )
1
1 ′
𝑥2 = 𝑅1 sin(𝜔1 𝑡 − 𝜙1 ) + 𝑅1 sin(𝜔2 𝑡 − 𝜙2 )
𝜆1
𝜆2
𝑥2 = 𝑅2 sin(𝜔1 𝑡 − 𝜙1 ) + 𝑅2 ′ sin(𝜔2 𝑡 − 𝜙2 )
There are 4 unknowns, 𝑅1 , 𝑅1 ′ , 𝜙1 , 𝜙2 . These can be determined from the 4 initial
conditions
- Initial displacement of both masses, 𝑥1 0, 𝑥2 0
- Initial velocities of both masses, 𝑥̇ 1 0, 𝑥̇ 2 0
𝑥1
𝑅
(𝑥 ) = ( 1
2
𝑅2
MEE 411 – Mechanical Vibrations
𝑅1 ′ sin(𝜔1 𝑡 − 𝜙1 )
)(
)
𝑅2 ′ sin(𝜔2 𝑡 − 𝜙2 )
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Classwork 1
Determine the two natural frequencies of vibration and the ratio of the amplitudes
of motion of mass m1 and m2 for the system below
If m1 = 1.5kg, m2 = 1.2kg, k1 =k =k2 = 5N/m
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TWO CONNECTED SIMPLE PENDULUM
Consider two simple pendulum of equal mass m and constant length l connected
by a spring of spring constant k at a distance h below the point of support A and
B.
Assume angle of displacement 𝜙1 , 𝜙2 from the vertical. Spring is unstretched
when the pendulums are vertical
For small displacement
Spring extension = ℎ(𝜙2 − 𝜙1 )
Equation of motion for each pendulum for small angles
(moment Equation)
𝐼1 𝜙̈1 = −𝑚𝑔𝑙𝜙1 + 𝑘ℎ(𝜙2 − 𝜙1 ). ℎ
𝐼2 𝜙̈2 = −𝑚𝑔𝑙𝜙2 − 𝑘ℎ(𝜙2 − 𝜙1 ). ℎ
But
𝐼1 = 𝐼2 = 𝑚𝑙 2
Therefore
𝑚𝑙 2 𝜙̈1 + 𝑚𝑔𝑙𝜙1 − 𝑘ℎ2 (𝜙2 − 𝜙1 ) = 0
𝑚𝑙 2 𝜙̈2 + 𝑚𝑔𝑙𝜙2 + 𝑘ℎ2 (𝜙2 − 𝜙1 ) = 0
Rearranging,
𝑚𝑙 2 𝜙̈1 + 𝑚𝑔𝑙𝜙1 − 𝑘ℎ2 𝜙2 + 𝑘ℎ2 𝜙1 = 0
𝑚𝑙 2 𝜙̈1 + (𝑚𝑔𝑙+𝑘ℎ2 )𝜙1 − 𝑘ℎ2 𝜙2 = 0
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𝑚𝑙 2 𝜙̈2 + 𝑚𝑔𝑙𝜙2 + 𝑘ℎ2 𝜙2 − 𝑘ℎ2 𝜙1 = 0
𝑚𝑙 2 𝜙̈2 − 𝑘ℎ2 𝜙1 + (𝑚𝑔𝑙 + 𝑘ℎ2 )𝜙2 = 0
𝑔 𝑘ℎ2
𝑘ℎ2
𝜙̈1 + ( + 2 ) 𝜙1 − 2 𝜙2 = 0
𝑙 𝑚𝑙
𝑚𝑙
𝜙̈2 −
𝑘ℎ2
𝑔 𝑘ℎ2
𝜙
+
+
(
)𝜙 = 0
𝑚𝑙 2 1
𝑙 𝑚𝑙 2 2
To simplify analysis
𝑔
𝑙
Let a =( +
𝑘ℎ 2
),
𝑚𝑙 2
b=
𝑘ℎ 2
𝑚𝑙 2
Therefore,
𝜙̈1 + 𝑎𝜙1 − 𝑏𝜙2 = 0
𝜙̈2 − 𝑏𝜙1 + 𝑎𝜙2 = 0
Writing this as a vector
𝜙
−𝑎
𝜙⃗⃗ = ( 1 ) ; 𝐴 = (
𝜙2
𝑏
𝑏
)
−𝑎
We now have,
̈
(𝜙⃗⃗) = 𝐴𝜙⃗
Or
𝜙̈
−𝑎
( 1) = (
𝑏
𝜙̈2
𝜙
𝑏
) ( 1)
−𝑎 𝜙2
For our general solution, we assume an oscillatory solution of frequency 𝜔𝑛
Let
𝑅
𝜙⃗ = 𝑅⃗ 𝑒 𝑖(𝜔𝑛 𝑡−𝜃) = ( 1 ) 𝑒 𝑖(𝜔𝑛 𝑡−𝜃)
𝑅2
Differentiate twice
̈
(𝜙⃗⃗) = −𝜔𝑛 2 𝑅⃗𝑒 𝑖(𝜔𝑛 𝑡−𝜃)
̈
(𝜙⃗⃗) = −𝜔𝑛 2 𝜙⃗
Substituting in the differential equation of motion
(𝐴 + 𝜔𝑛 2 𝐼)𝜙⃗ = 0
Where I is a unitary matrix
−𝑎 + 𝜔𝑛 2
𝑏
𝑅1 𝑖(𝜔𝑛 𝑡−𝜃)
=0
(
2 ) (𝑅 ) 𝑒
𝑏
−𝑎 + 𝜔𝑛
2
This has a non-trivial solution only if the determinant vanishes
i.e.
|𝐴 + 𝜔𝑛 2 𝐼| = 0
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𝜔𝑛 2
2020
−𝑎 + 𝜔𝑛 2
𝑏
|
|=0
𝑏
−𝑎 + 𝜔𝑛 2
(𝜔𝑛 2 − 𝑎)2 − 𝑏 2 = 0
𝜔𝑛 2 − 𝑎 = ±𝑏
= 𝑎 ± 𝑏 – The two natural frequencies
𝑔 𝑘ℎ2
𝑘ℎ2 𝑔
𝜔𝑛1 2 = 𝑎 − 𝑏 = ( + 2 ) − 2 =
𝑙 𝑚𝑙
𝑚𝑙
𝑙
𝑔 𝑘ℎ2
𝑘ℎ2 𝑔 2𝑘ℎ2
𝜔𝑛2 2 = 𝑎 + 𝑏 = ( + 2 ) + 2 = +
𝑙 𝑚𝑙
𝑚𝑙
𝑙
𝑚𝑙 2
Ratio of amplitudes
(−𝑎 + 𝜔𝑛 2 )𝑅1 + 𝑏𝑅2 = 0
𝑅1
−𝑏
= 2
𝑅2 𝜔𝑛 − 𝑎
𝑅1
−𝑏
−𝑏
−𝑏
( ) =
=
=
= +1 (𝜆1 = +1)
2
𝑅2 1 𝜔𝑛1 − 𝑎 (𝑎 − 𝑏) − 𝑎 −𝑏
𝑅1
−𝑏
−𝑏
−𝑏
( ) =
=
=
= −1 (𝜆2 = −1)
2
𝑅2 2 𝜔𝑛2 − 𝑎 (𝑎 + 𝑏) − 𝑎
𝑏
There are two modes of vibration
1st Mode
Pendulum moves together as simple pendulum with same frequency. No force in
spring
MEE 411 – Mechanical Vibrations
Page 16 of 17
MEE411 LECTURE NOTES
2020
𝜔𝑛1 = √
𝑔
𝑙
2nd Mode
Pendulum directly out of phase
𝑔 2𝑘ℎ2
𝜔𝑛2 = √ +
𝑙
𝑚𝑙 2
General Solution
The General Solution depends on initial conditions
𝜙1 = 𝑅1 sin(𝜔𝑛1 𝑡 − 𝜃1 ) + 𝑅1 ′ sin(𝜔𝑛2 𝑡 − 𝜃2 )
𝜙1 = 𝑅1 sin(𝜔𝑛1 𝑡 − 𝜃1 ) + 𝑅1 ′ sin(𝜔𝑛2 𝑡 − 𝜃2 )
MEE 411 – Mechanical Vibrations
Page 17 of 17