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Forced Vibration Under Harmonic Loading Analysis

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FORCED VIBRATION UNDER HARMONIC LOADING
1 Response of Undamped SDOF System to Harmonic Motion
In the forced vibration system, due to periodic force applied to the system, the vibrations continue as long
as the periodic force is present. Periodic force which is harmonic in nature will be considered.
𝑓(𝑑) = π‘“π‘œ π‘ π‘–π‘›πœ”π‘“ 𝑑
Where π‘“π‘œ is the peak amplitude and πœ”π‘“ is the frequency of the force in radians per second.
Thus
π‘š
𝑑2 π‘₯
+ π‘˜π‘₯ = π‘“π‘œ π‘ π‘–π‘›πœ”π‘“ 𝑑
𝑑𝑑 2
So the differential equation of the periodic motions has a solution.
π‘₯𝑐 = π‘₯(𝑑) = 𝑒 𝛼𝑑 (𝐴1 π‘π‘œπ‘ π›½π‘‘ + 𝐴2 𝑠𝑖𝑛𝛽𝑑) = 𝐴1 π‘π‘œπ‘ πœ”π‘‘ + 𝐴2 π‘ π‘–π‘›πœ”π‘‘
Particular solution
π‘₯𝑝 = π‘‹π‘ π‘–π‘›πœ”π‘“ 𝑑
𝑑π‘₯𝑝
= π‘‹πœ”π‘“ π‘π‘œπ‘ πœ”π‘“ 𝑑
𝑑𝑑
𝑑2 π‘₯𝑝
= −π‘‹πœ”π‘“ 2 π‘ π‘–π‘›πœ”π‘“ 𝑑
𝑑𝑑 2
Where 𝑋 is the peak value of a particular solution
Substituting
−π‘šπ‘‹πœ”π‘“ 2 π‘ π‘–π‘›πœ”π‘“ 𝑑 + π‘˜π‘‹π‘ π‘–π‘›πœ”π‘“ 𝑑 = π‘“π‘œ π‘ π‘–π‘›πœ”π‘“ 𝑑
−π‘šπ‘‹πœ”π‘“ 2 + π‘˜π‘‹ = π‘“π‘œ
𝑋=
π‘“π‘œ
π‘˜ − π‘šπœ”π‘“ 2
π‘“π‘œ
π‘“π‘œ
π‘˜
π‘˜
𝑋=
=
π‘šπœ”π‘“ 2
πœ”π‘“ 2
1−
1− 2
π‘˜
πœ”
Where π‘Ÿ =
πœ”π‘“ 2
πœ”2
is the frequency ratio of the applied force frequency to natural frequency of vibration of the
system.
Therefore
π‘“π‘œ
𝑋= π‘˜ 2
1−π‘Ÿ
1
Therefore the final solution will be
π‘“π‘œ
π‘₯(𝑑) = 𝐴1 π‘π‘œπ‘ πœ”π‘‘ + 𝐴2 π‘ π‘–π‘›πœ”π‘‘ + π‘˜ 2 π‘ π‘–π‘›πœ”π‘“ 𝑑
1−π‘Ÿ
Initial conditions are applied to obtain constant𝐴1 π‘Žπ‘›π‘‘π΄2 . (Recall: differential equations)
In the case of initial condition
Considering initial conditions 𝑑 = 0 π‘₯(0) = 0
𝑣=0
π‘“π‘œ
π‘˜
𝐴2 = −
1 − π‘Ÿ2
π‘Ÿ
𝐴1 = 0,
Thus
π‘“π‘œ
π‘“π‘œ
π‘˜ π‘ π‘–π‘›πœ”π‘‘ + π‘˜ π‘ π‘–π‘›πœ” 𝑑
π‘₯(𝑑) = −
𝑓
1 − π‘Ÿ2
1 − π‘Ÿ2
π‘Ÿ
π‘“π‘œ
π‘₯(𝑑) = − π‘˜ 2 [π‘ π‘–π‘›πœ”π‘“ 𝑑 − π‘Ÿπ‘ π‘–π‘›πœ”π‘‘]
1−π‘Ÿ
As damping forces will always be present, the free frequency term will vanish eventually. Therefore the
term (π‘Ÿπ‘ π‘–π‘›πœ”π‘‘) represents transient response of vibration. Transient response of vibration occurs to a system
while it is approaching steady state.
The forcing term is referred to as steady-state response is given by:
π‘“π‘œ
π‘₯(𝑑) = π‘˜ 2 π‘ π‘–π‘›πœ”π‘“ 𝑑, π‘Ÿ ≠ 1
1−π‘Ÿ
When r =1, πœ”π‘“ = πœ” t the amplitude of motions becomes infinitely large. This is resonance. The amplitude
will increase gradually to infinite. The materials used in practice have strength limitation, thus structural
failures occur long before extremely large amplitudes can be attained.
Let
π‘“π‘œ
= 𝑋0
π‘˜
𝑋0 is the static deflection
𝑋=
𝑋0
1 − π‘Ÿ2
𝑋
1
=
= 𝐹(πœ”π‘“ )
𝑋0 1 − π‘Ÿ 2
𝑋
𝐹(πœ”π‘“ ) = 𝑋 is called frequency response function. It gives the magnitude and he sign of steady state motion
0
as a function of frequency ratio “r”
2
The magnitude𝐷𝑠 , is called the steady-state or dynamic magnification factor, or gain as shown in the figure
below
𝐷𝑠 = |𝐹(πœ”π‘“ )|
If r<1, the response is in phase with the excitation, since (1-r2) is positive. If r is more than 1.0, the response
is 180o out of phase with the excitation, that is
π‘“π‘œ
π‘₯(𝑑) = π‘˜ 2 (−π‘ π‘–π‘›πœ”π‘“ 𝑑)
1−π‘Ÿ
π‘₯(𝑑) =
𝑋0
(−π‘ π‘–π‘›πœ”π‘“ 𝑑)
1 − π‘Ÿ2
When r is close to one but not equal to 1, the forced motion and natural motion alternatively reinforce each
other and cancel each other giving the appearance of a beat phenomenon.
2 Response of Viscous-Damped SDOF System to Harmonic Motion
Equation of motion
π‘š
𝑑2 π‘₯
𝑑π‘₯
+𝑐
+ π‘˜π‘₯ = π‘“π‘œ π‘ π‘–π‘›πœ”π‘“ 𝑑
2
𝑑𝑑
𝑑𝑑
Solution
π‘₯(𝑑) = π‘₯𝑐 (𝑑) + π‘₯𝑝 (𝑑)
3
Where π‘₯𝑐 (𝑑) is the complementary solution and π‘₯𝑝 (𝑑) is the particular solution. For underdamped case
where 𝐢 < πΆπ‘π‘Ÿ
π‘₯𝑐 (𝑑) = 𝑒 −πœ‰πœ”π‘‘ (𝐴1 π‘π‘œπ‘ πœ”π· 𝑑 + 𝐴2 π‘ π‘–π‘›πœ”π· 𝑑)
Particular solution is given by
𝑓(𝑑) = π‘ π‘–π‘›πœ”π‘“ 𝑑
𝑑𝑓(𝑑)
𝑑𝑑
𝑑2 𝑓(𝑑)
2
𝑑𝑑
= πœ”π‘“ π‘π‘œπ‘ πœ”π‘“ 𝑑
= −πœ”π‘“ 2 π‘ π‘–π‘›πœ”π‘“ 𝑑
Gives two independent function π‘ π‘–π‘›πœ”π‘“ 𝑑 π‘Žπ‘›π‘‘ π‘π‘œπ‘ πœ”π‘“ 𝑑
π‘₯𝑝 = π‘˜1 π‘ π‘–π‘›πœ”π‘“ 𝑑 + π‘˜2 π‘π‘œπ‘ πœ”π‘“ 𝑑
𝑑π‘₯𝑝
𝑑𝑑
𝑑2 π‘₯𝑝
2
𝑑𝑑
Substituting intoπ‘š
𝑑2 π‘₯
𝑑𝑑 2
+𝑐
𝑑π‘₯
𝑑𝑑
= π‘˜1 πœ”π‘“ π‘π‘œπ‘ πœ”π‘“ 𝑑 − π‘˜2 πœ”π‘“ π‘ π‘–π‘›πœ”π‘“ 𝑑
= −π‘˜1 πœ”π‘“ 2 π‘ π‘–π‘›πœ”π‘“ 𝑑 − π‘˜2 πœ”π‘“ 2 π‘π‘œπ‘ πœ”π‘“ 𝑑
+ π‘˜π‘₯ = π‘“π‘œ π‘ π‘–π‘›πœ”π‘“ 𝑑
π‘š(−π‘˜1 πœ”π‘“ 2 π‘ π‘–π‘›πœ”π‘“ 𝑑 − π‘˜2 πœ”π‘“ 2 π‘π‘œπ‘ πœ”π‘“ 𝑑) + 𝑐(π‘˜1 πœ”π‘“ π‘π‘œπ‘ πœ”π‘“ 𝑑 − π‘˜2 πœ”π‘“ π‘ π‘–π‘›πœ”π‘“ 𝑑) + π‘˜(π‘˜1 π‘ π‘–π‘›πœ”π‘“ 𝑑 + π‘˜2 π‘π‘œπ‘ πœ”π‘“ 𝑑)
= π‘“π‘œ π‘ π‘–π‘›πœ”π‘“ 𝑑
−π‘šπ‘˜1 πœ”π‘“ 2 π‘ π‘–π‘›πœ”π‘“ 𝑑 − π‘šπ‘˜2 πœ”π‘“ 2 π‘π‘œπ‘ πœ”π‘“ 𝑑 + π‘π‘˜1 πœ”π‘“ π‘π‘œπ‘ πœ”π‘“ 𝑑 − π‘π‘˜2 πœ”π‘“ π‘ π‘–π‘›πœ”π‘“ 𝑑 + π‘˜π‘˜1 π‘ π‘–π‘›πœ”π‘“ 𝑑 + kπ‘˜2 π‘π‘œπ‘ πœ”π‘“ 𝑑
= π‘“π‘œ π‘ π‘–π‘›πœ”π‘“ 𝑑
(−π‘šπ‘˜1 πœ”π‘“ 2 − π‘π‘˜2 πœ”π‘“ + π‘˜π‘˜1 )π‘ π‘–π‘›πœ”π‘“ 𝑑 + (−π‘šπ‘˜2πœ”π‘“ 2 + π‘π‘˜1 πœ”π‘“ + kπ‘˜2 )π‘π‘œπ‘ πœ”π‘“ 𝑑 = π‘“π‘œ π‘ π‘–π‘›πœ”π‘“ 𝑑
The equations yields two algebraic equations
−π‘šπ‘˜1 πœ”π‘“ 2 − π‘π‘˜2 πœ”π‘“ + π‘˜π‘˜1 = π‘“π‘œ
(π‘˜ − π‘š πœ”π‘“ 2 )π‘˜1 − π‘πœ”π‘“ π‘˜2 = π‘“π‘œ
−π‘šπ‘˜2 πœ”π‘“ 2 + π‘π‘˜1 πœ”π‘“ + kπ‘˜2 = 0
π‘π‘˜1 πœ”π‘“ + (k − π‘šπœ”π‘“ 2 )π‘˜2 = 0
Divide through by k
(1 −
π‘š 2
𝑐
π‘“π‘œ
πœ”π‘“ )π‘˜1 − πœ”π‘“ π‘˜2 =
π‘˜
π‘˜
π‘˜
4
𝑐
π‘š
π‘˜1 πœ”π‘“ + (1 − πœ”π‘“ 2 )π‘˜2 = 0
π‘˜
π‘˜
(1 − π‘Ÿ 2 )π‘˜1 − 2π‘Ÿπœ‰π‘˜2 = π‘‹π‘œ
2π‘Ÿπœ‰π‘˜1 + (1 − π‘Ÿ 2 )π‘˜2 = 0
𝑐
𝑐
Where πœ‰ = 𝑐 = 2π‘šπœ” π‘‘β„Žπ‘’π‘Ÿπ‘’π‘“π‘œπ‘Ÿπ‘’ 𝑐 = 2πœ‰π‘šπœ”
π‘π‘Ÿ
π‘˜1 =
(1 − π‘Ÿ 2 )π‘‹π‘œ
(1 − π‘Ÿ 2 )2 + (2π‘Ÿπœ‰)2
π‘˜2 = −
(2π‘Ÿπœ‰)π‘‹π‘œ
(1 − π‘Ÿ 2 )2 + (2π‘Ÿπœ‰)2
Therefore
π‘₯𝑝 =
(1 − π‘Ÿ 2 )π‘‹π‘œ
(2π‘Ÿπœ‰)π‘‹π‘œ
π‘ π‘–π‘›πœ”π‘“ 𝑑 −
π‘π‘œπ‘ πœ”π‘“ 𝑑
2
2
2
(1 − π‘Ÿ ) + (2π‘Ÿπœ‰)
(1 − π‘Ÿ 2 )2 + (2π‘Ÿπœ‰)2
π‘₯𝑝 =
(1 −
π‘‹π‘œ
[(1 − π‘Ÿ 2 )π‘ π‘–π‘›πœ”π‘“ 𝑑 − (2π‘Ÿπœ‰)π‘π‘œπ‘ πœ”π‘“ 𝑑]
+ (2π‘Ÿπœ‰)2
π‘Ÿ 2 )2
Which can be written as
π‘₯𝑝 =
π‘‹π‘œ
√(1 − π‘Ÿ 2 )2 + (2π‘Ÿπœ‰)2
sin(πœ”π‘“ 𝑑−∝)
Where ∝ is the phase angle defined by
∝= π‘‘π‘Žπ‘›−1 (
2π‘Ÿπœ‰
)
1 − π‘Ÿ2
Can be written in compact form as
π‘₯𝑝 = π‘‹π‘œ 𝛽 sin(πœ”π‘“ 𝑑−∝)
Where 𝛽 is the magnification factor in the case of damped system.
𝛽=
π‘₯𝑝
1
=
π‘‹π‘œ √(1 − π‘Ÿ 2 )2 + (2π‘Ÿπœ‰)2
Total response
π‘₯(𝑑) = 𝑒 −πœ‰πœ”π‘‘ (𝐴1 π‘π‘œπ‘ πœ”π· 𝑑 + 𝐴2 π‘ π‘–π‘›πœ”π· 𝑑) +
π‘‹π‘œ
√(1 −
π‘Ÿ 2 )2
+ (2π‘Ÿπœ‰)2
sin(πœ”π‘“ 𝑑−∝)
The constants 𝐴1 and 𝐴2 are determined from the initial conditions.
The figure below gives the steady-state magnification factor ‘𝛽’ against frequency ratio ‘r’ and phase angle
∝ against frequency ratio “r” for various amounts of damping ratios 𝝃
5
FORCE
3 Transmission
It is clear that by increasing the spring stiffness k and the damping coefficient c, the amplitude of vibration
decreases. The increase in the stiffness and damping coefficients, however, may have an adverse effect on
the force transmitted to the support. In order to reduce the force transmitted to the support, the stiffness
and damping coefficients must be properly selected.
Given the equation of motion
π‘š
𝑑2 π‘₯
𝑑π‘₯
+𝑐
+ π‘˜π‘₯ = π‘“π‘œ π‘ π‘–π‘›πœ”π‘“ 𝑑
2
𝑑𝑑
𝑑𝑑
With particular solution
π‘₯𝑝 = π‘‹π‘œ 𝛽 sin(πœ”π‘“ 𝑑−∝) (π‘Ÿπ‘’π‘π‘Žπ‘™π‘™? )
6
𝑑π‘₯
The force transmitted to the support through the spring is kx and through the damping element𝑐 𝑑𝑑 . Hence
the total force transmitted to foundation
𝑓𝑑 = π‘˜π‘₯ + 𝑐
𝑑π‘₯
𝑑𝑑
Therefore
π‘₯𝑝 = π‘‹π‘œ 𝛽 sin(πœ”π‘“ 𝑑−∝)
𝑑π‘₯𝑝
𝑑𝑑
= π‘‹π‘œ π›½πœ”π‘“ cos(πœ”π‘“ 𝑑−∝)
Therefore
𝑓𝑑 = π‘˜(π‘‹π‘œ 𝛽 sin(πœ”π‘“ 𝑑−∝)) + 𝑐(π‘‹π‘œ π›½πœ”π‘“ cos(πœ”π‘“ 𝑑−∝))
𝑓𝑑 = π‘‹π‘œ 𝛽[π‘˜ sin(πœ”π‘“ 𝑑−∝) + 𝑐 πœ”π‘“ cos(πœ”π‘“ 𝑑−∝)]
Given that
π‘πœ”π‘“
)
π‘˜
πœ“ = π‘‘π‘Žπ‘›−1 (
Then
𝑓𝑑 = π‘‹π‘œ 𝛽√π‘˜2 + (𝑐 πœ”π‘“ )2 sin(πœ”π‘“ 𝑑−∝ +πœ“)
π‘‹π‘œ =
𝑓𝑑 =
π‘“π‘œ
π‘˜
𝛽√π‘˜2 + (𝑐 πœ”π‘“ )2 sin(πœ”π‘“ 𝑑−∝ +πœ“)
𝑓𝑑 = π‘“π‘œ 𝛽√1 + (
𝑐
π‘“π‘œ
π‘˜
𝑐 πœ”π‘“
π‘˜
2
) sin(πœ”π‘“ 𝑑−∝ +πœ“)
𝑐
Where πœ‰ = 𝑐 = 2π‘šπœ” π‘‘β„Žπ‘’π‘Ÿπ‘’π‘“π‘œπ‘Ÿπ‘’ 𝑐 = 2πœ‰π‘šπœ”
π‘π‘Ÿ
2πœ‰π‘šπœ”πœ”π‘“
𝑓𝑑 = π‘“π‘œ 𝛽√1 + (
𝑓𝑑 = π‘“π‘œ 𝛽√1 + (
π‘˜
2πœ‰πœ”π‘“
πœ”
2
) sin(πœ”π‘“ 𝑑−∝ +πœ“)
2
) sin(πœ”π‘“ 𝑑−∝ +πœ“)
𝑓𝑑 = π‘“π‘œ 𝛽√1 + (2πœ‰π‘Ÿ)2 sin(πœ”π‘“ 𝑑−∝ +πœ“)
7
Maximum force
𝐹𝑑 = π‘“π‘œ 𝛽√1 + (2πœ‰π‘Ÿ)2
𝐹𝑑 = π‘“π‘œ
√1 + (2πœ‰π‘Ÿ)2
√(1 − π‘Ÿ2 )2 + (2π‘Ÿπœ‰)2
1 + (2πœ‰π‘Ÿ)2
√
𝐹𝑑 = π‘“π‘œ
(1 − π‘Ÿ2 )2 + (2π‘Ÿπœ‰)2
Transmissibility Tr is defined as the ratio between the amplitude of the force transmitted to the foundation
and the amplitude of the force. Therefore
π‘‡π‘Ÿ =
𝐹𝑑
π‘“π‘œ
1 + (2πœ‰π‘Ÿ)2
=√
(1 − π‘Ÿ2 )2 + (2π‘Ÿπœ‰)2
The transmissibility of motion from the foundation to the structure and the transmissibility of force from
the structure to the foundation is given by the same expression.
4 Response to Support Motion
There are cases where the foundation or support of a structure is subjected to time varying motion. Ground
motion of earthquakes or other excitations such as explosion or dynamic action of machinery causes motion
of structures.
Harmonic motion
𝑧(𝑑) = π‘§π‘œ π‘ π‘–π‘›πœ”π‘“ 𝑑
Where π‘§π‘œ us the maximum amplitude and πœ”π‘“ is the frequency of the support motion.
Given the equation of motion
8
π‘š
𝑑2 π‘₯
𝑑π‘₯
𝑑𝑧
+𝑐
+ π‘˜π‘₯ = 𝑐 + π‘˜π‘§
2
𝑑𝑑
𝑑𝑑
𝑑𝑑
𝑧(𝑑) = π‘§π‘œ π‘ π‘–π‘›πœ”π‘“ 𝑑
𝑑𝑧
= π‘§π‘œ πœ”π‘“ π‘π‘œπ‘ πœ”π‘“ 𝑑
𝑑𝑑
Substitution
π‘š
𝑑2 π‘₯
𝑑π‘₯
+𝑐
+ π‘˜π‘₯ = π‘§π‘œ (π‘˜π‘ π‘–π‘›πœ”π‘“ 𝑑 + π‘πœ”π‘“ π‘π‘œπ‘ πœ”π‘“ 𝑑)
2
𝑑𝑑
𝑑𝑑
π‘š
𝑑2 π‘₯
𝑑π‘₯
+𝑐
+ π‘˜π‘₯ = π‘§π‘œ √π‘˜ 2 + (π‘πœ”π‘“ )2 sin(πœ”π‘“ 𝑑 +∝𝑧 )
2
𝑑𝑑
𝑑𝑑
π‘š
π‘πœ”π‘“ 2
𝑑2 π‘₯
𝑑π‘₯
+𝑐
+ π‘˜π‘₯ = π‘§π‘œ π‘˜√1 + (
) sin(πœ”π‘“ 𝑑 +∝𝑧 )
2
𝑑𝑑
𝑑𝑑
π‘˜
π‘š
𝑑2 π‘₯
𝑑π‘₯
+𝑐
+ π‘˜π‘₯ = π‘§π‘œ π‘˜√1 + (2π‘Ÿπœ‰)2 sin(πœ”π‘“ 𝑑 +∝𝑧 )
2
𝑑𝑑
𝑑𝑑
π‘πœ”π‘“
∝𝑧 = π‘‘π‘Žπ‘›−1 (
) = π‘‘π‘Žπ‘›−1 (2π‘Ÿπœ‰)
π‘˜
𝑑2 π‘₯
𝑑π‘₯
This is same as π‘š 𝑑𝑑 2 + 𝑐 𝑑𝑑 + π‘˜π‘₯ = π‘“π‘œ π‘ π‘–π‘›πœ”π‘“ 𝑑 except for the addition of angle ∝ in the sine function
Therefore steady state response
π‘₯𝑝 = π‘§π‘œ √1 + (2π‘Ÿπœ‰)2 𝛽 sin(πœ”π‘“ 𝑑 +∝𝑧 −∝)
Magnification factor/Transmissibility
π‘₯𝑝
√1 + (2πœ‰π‘Ÿ)2
=
π‘§π‘œ √(1 − π‘Ÿ 2 )2 + (2π‘Ÿπœ‰)2
Thus the expression for the relative transmission of the support motion to the oscillator is given by
π‘₯𝑝
√1 + (2πœ‰π‘Ÿ)2
=
× sin(πœ”π‘“ 𝑑 +∝𝑧 −∝)
π‘§π‘œ √(1 − π‘Ÿ 2 )2 + (2π‘Ÿπœ‰)2
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If r=0, Tr=1.0
If π‘Ÿ = √2, Tr=1
Transmitted force
In the case of the support motion, the force carried by the support can be obtained
𝑑π‘₯ 𝑑𝑧
𝐹𝑑 = 𝑐 ( − ) + π‘˜ (π‘₯ − 𝑧)
𝑑𝑑 𝑑𝑑
𝐹𝑑 = −π‘š
𝑑2 π‘₯
𝑑𝑑 2
π‘₯𝑝 = π‘§π‘œ √1 + (2π‘Ÿπœ‰)2 𝛽 sin(πœ”π‘“ 𝑑 +∝𝑧 −∝)
𝑑π‘₯
= π‘§π‘œ πœ”π‘“ √1 + (2π‘Ÿπœ‰)2 𝛽 cos(πœ”π‘“ 𝑑 +∝𝑧 −∝)
𝑑𝑑
𝑑2 π‘₯
= −π‘§π‘œ πœ”π‘“ 2 √1 + (2π‘Ÿπœ‰)2 𝛽 sin(πœ”π‘“ 𝑑 +∝𝑧 −∝)
𝑑𝑑 2
Therefore
𝐹𝑑 = π‘š. π‘§π‘œ πœ”π‘“ 2 √1 + (2π‘Ÿπœ‰)2 𝛽 sin(πœ”π‘“ 𝑑 +∝𝑧 −∝)
π‘˜
𝐹𝑑 = ( 2 ). π‘§π‘œ πœ”π‘“ 2 √1 + (2π‘Ÿπœ‰)2 𝛽 sin(πœ”π‘“ 𝑑 +∝𝑧 −∝)
πœ”
𝐹𝑑 = π‘˜. π‘§π‘œ π‘Ÿ 2 √1 + (2π‘Ÿπœ‰)2 𝛽 sin(πœ”π‘“ 𝑑 +∝𝑧 −∝)
10
Mass motion relative to the support.
The relative displacement denoted as z can be written as z=x-y
𝑧=π‘₯−𝑦
𝑑𝑧 𝑑π‘₯ 𝑑𝑦
=
−
𝑑𝑑 𝑑𝑑 𝑑𝑑
𝑑2 𝑧 𝑑2 π‘₯ 𝑑2 𝑦
=
−
𝑑𝑑 2 𝑑𝑑 2 𝑑𝑑 2
𝑑2 π‘₯
𝑑π‘₯
Equation π‘š 𝑑𝑑 2 + 𝑐 𝑑𝑑 + π‘˜π‘₯ = π‘§π‘œ (π‘˜π‘ π‘–π‘›πœ”π‘“ 𝑑 + π‘πœ”π‘“ π‘π‘œπ‘ πœ”π‘“ 𝑑) can be written as
π‘š(
𝑑2 𝑧 𝑑2 𝑦
𝑑𝑧
+ 2 ) = −π‘˜π‘§ − 𝑐
2
𝑑𝑑
𝑑𝑑
𝑑𝑑
Which can be written as
π‘š
𝑑2 𝑧
𝑑𝑧
𝑑2 𝑦
+
𝑐
+
π‘˜π‘§
=
−π‘š
= π‘šπ‘§π‘œ πœ”π‘“ 2 π‘ π‘–π‘›πœ”π‘“ 𝑑
𝑑𝑑 2
𝑑𝑑
𝑑𝑑 2
Therefore steady state response (particular solution)
π‘₯𝑝 =
π‘₯𝑝 =
π‘šπ‘§π‘œ πœ”π‘“ 2 /π‘˜
√(1 − π‘Ÿ 2 )2 + (2π‘Ÿπœ‰)2
π‘§π‘œ π‘Ÿ 2
√(1 − π‘Ÿ 2 )2 + (2π‘Ÿπœ‰)2
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× sin(πœ”π‘“ 𝑑−∝)
× sin(πœ”π‘“ 𝑑−∝)
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