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MECH 375 MECHANICAL VIBRATIONS FINAL CRASH PART 2-Harmonically Excited Vibrations Concordia University CA$17.15   Add to cart
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MECH 375 MECHANICAL VIBRATIONS FINAL CRASH PART 2-Harmonically Excited Vibrations Concordia University

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MECH 375 MECHANICAL VIBRATIONS FINAL CRASH PART 2-Harmonically Excited Vibrations Concordia University

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  • November 21, 2023
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  • 2023/2024
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  • mech 375 mechanical vibrations final crash part 2
  • mech 375 mechanical vibrations

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MECH 375 MECHANICAL VIBRATIONS FINAL CRASH PART 2-
Harmonically Excited Vibrations Concordia University

,Harmonically Excited Vibrations
General Systems
A system with an applied force is system with a
forced vibration, or harmonically excited vibration.

The general differential equation describing this
system is:

𝑚 + 𝑐 + 𝑘 = 𝐹 sin(𝜔𝑡)

Note that ω is the input frequency.

For undamped systems, if the input frequency
matches the natural frequency of the system, the
system’s oscillations will continue to grow. This
would result in catastrophic failure of the system.

The system does not require initial conditions in
order to be moving. If there were no initial
conditions, the system would start at the steady-
state solution.



Parts of the Solution:

The total response of the solution has two parts:

 Homogenous solution
 Particular/steady-state solution




2

,Method of Solving:

1. Solve for the particular solution of the system by doing a substitution of = Xeie𝜔𝑡 and ƒ(𝑡) =
𝐹eie𝜔𝑡.
2. For the homogeneous solution yH, it takes the form that f(t) has, ie, if ƒ(𝑡) = 𝐹 sin 𝜔𝑡, then
𝑦𝐻 = 𝐴 sin(𝜔𝑡 − 𝜑).
3. The most general form is: 𝑦𝐻 = 𝐴e−𝛼𝑡 sin(𝜔𝑡 − 𝜑). Use the Laplace transform to partially solve
the characteristic equation to find α and ω.
4. Find the total solution as the sum of the homogeneous and particular solutions. Apply the initial
conditions to solve for A and φ.

Note: The Laplace transform can be used to find the total solution.

For complex roots: If the solution takes the form:

𝐴 + 𝐵i
X= sin(𝜔𝑡) , o𝑟 X = 𝐴 sin(𝜔𝑡) + 𝐵 cos(𝜔𝑡)
𝐶 + 𝐷i

The amplitude, Xo, and phase angle, φ, can be found by:

√𝐴2 + 𝐵2
Xo = , o𝑟 Xo = √𝐴2 + 𝐵2
√𝐶2 + 𝐷2
𝐵 𝐷 𝐵
𝜑 = t a n−1 ( ) − t a n ( ) , o𝑟 𝜑 = t a n ( )
−1 −1

𝐴 𝐶 𝐴


Response of a Damped System Under the Harmonic Motion of the Base
Sometimes the system is attached to a frame that moves under harmonic
motion.

Force Transmissibility:

The force experienced between the base and the body is considered the
force transmissibility, FT.


Response of a Damped System Under Rotating Unbalance
Due to the rotating unbalanced mass, m, at an eccentric distance, e from the center,
the system will experience a force:

ƒ(𝑡) = 𝑚e𝜔2sin(𝜔𝑡)




3

, 1. A spring-mass system, resting on an inclined plane, is subjected to a harmonic force as shown. Find
the response of the system by assuming zero initial conditions.

Given:
m = 100 kg
k = 4000 N/m
xo = 0
ẋo = 0

Equation of Motion:

𝑚 + 𝑘 = 𝐹o cos 𝜔𝑡
100 + 4000 = 𝐹o cos 𝜔𝑡

Method 1:
Homogeneous Solution: Particular Solution:

100 + 4000 = 0 100 + 4000 = 𝐹o cos 𝜔𝑡

For an undamped system, the solution is: Let:

(𝑡) = 𝐴 cos 𝜔𝑛𝑡 + 𝐴 sin 𝜔𝑛𝑡 = X cos 𝜔𝑡
= 𝜔X sin 𝜔𝑡
= −𝜔2X cos 𝜔𝑡
𝑘
𝜔𝑛 = √
𝑚 −100𝜔2X cos 𝜔𝑡 + 4000X cos 𝜔𝑡 = 𝐹o cos 𝜔𝑡
(4000 − 100𝜔2)X = 𝐹o
𝜔𝑛 = √ 4000 𝐹o
100 X=
4000 − 100𝜔2
𝜔𝑛 = √40 𝐹
o
𝑝 (𝑡) = cos 𝜔𝑡
4000 − 100𝜔2
ℎ (𝑡) = 𝐴 cos √40𝑡 + 𝐵 sin √40𝑡

Total Solution:

(𝑡) = ℎ(𝑡) + 𝑝(𝑡)
𝐹o
(𝑡) = 𝐴 cos √40𝑡 + 𝐵 sin √40𝑡 + cos 𝜔𝑡
4000 − 100𝜔2
𝐹o𝜔
(𝑡) = −√40𝐴 sin √40𝑡 + √40𝐵 cos √40𝑡 − sin 𝜔𝑡
4000 − 100𝜔2




4

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