THE UNIVERSITY OF ZAMBIA
SCHOOL OF ENGINEERING
2021 ACADEMIC YEAR
TERM TEST
MEC 4702 VIBRATION AND CONTROL Test May 2022
Clearly show the working for your answers.
Useful formulas
�=
�
��
��
�
��+�
�=
�
�=
�
� ��
��� + ��
Question 1
a) Derive the expression for the equivalent spring constant that relates the applied force F to
the resulting displacement x of the system shown in Fig. Q1a (Assume the displacement of the
link to be small) and fig Q1b
Fig Q1a Rigid bar connected by springs
Ans--
FigQ1b springs in parallel and series
10 Marks
Let the link OABC undergo a small angular displacement θ as shown in above figure. The spring
reaction forces are also indicated in the figure. Equilibrium of moments about the pivot point O gives:
or
−�1 �
�
2�
− �3 � � − �2 �
+� � =0
3
3
k1 2
+ k + k3 x
(1)
3 3 2
If ��� denotes the equivalent spring constant of the link along the direction of F at point C, we have
F=
Equations (1) and (2) give
F = keq x
�1 2�2
� 2
+
+ �3 = + 2� + 3�
3
3
3 3
14
∴ keq =
k
3
��� =
(2)
(3)
1
1
1
1
=
+ +
��� 2�1 �2 2�3
��� =
��� =
�5 ��� + �4
�5 + �4 + ���
2�1 �2 �3
�2 �3 + 2�1 �3 + �1 �2
1
1
1
=
+
��� ��� + �4 �5
=
�2 �3 �4 �5 + 2�1 �3 �4 �5 + �1 �2 �4 �5 + 2�1 �2 �3 �5
�2 �3 �4 + �2 �3 �5 + 2�1 �3 �4 + 2�1 �3 �5 + �1 �2 �4 + �1 �2 �5 + 2�1 �2 �3
b) What frequency of the system of mass M in fig Q1b when initially loaded gives a
displacement of 4 mm? The deflection of a fixed free beam at its free end due to a unit
��
concentrated load at its free end is ���
M1
Fig Q1b
Ans-The deflection of a fixed-free beam at its free end due to a unit concentrated load at its free end is
�3 /(3��). Thus the equivalent stiffness of the cantilever beam is
�
−5 4
9
3�� 3 210 × 10 �2 1.5 × 10 �
�
=
=
= 6.05 × 105
��
�
2.5� 3
�3
The beam and the upper spring act in parallel. This parallel combination is in series with the spring
placed between the beam and the block. This series combination is in parallel with the spring
between the block and the fixed surface. Thus using the formulas for parallel and series combinations,
the equivalent stiffness is calculated as
��� =
1
�
�
+ 3 × 105 = 4.69 × 105
1
1
�
�
�
�+
�
6.05 × 105 + 5 × 105
2 × 105
�
�
�
The displacement is given as
x=
W
= 0.004 m
keq
W = 0.004 keq = 0.004 m × 4.69 × 105
From which the mass is obtained as
The natural frequency is
m=
N
= 1876 N
m
1876 N
W
=
= 191.2 Kg
g
9.81 m/s2
�� =
�
� = 49.5 ���
191.2 ��
�
4.69 × 105
���
=
�
The frequency, or linear frequency is therefore
ωn 49.5
=
= 7.9 Hz
2π
2π
�=
Question 2
a) A single-degree-of-freedom system has m = 1, k = 2, and c = 0.5:
Determine the following
i. Natural frequency,
Ans-�
=
�
�� =
ii. Linear frequency,
Ans--
2
= 1.4142 ���/�
1
ωn
�=
2π
1.4142
= 0.2251Hz
�=
2π
iii. Natural time period,
Ans--
�� =
iv. Damped frequency,
Ans-ζ=
c
2 mk
2π
= 4.4429 s
��
=
2
0.5
1 2
= 0.1768
�� = �� 1 − �2
�� = 1.4142 1 − 0.17682
�� = 1.3919 ���/�
v. Critical damping constant,
Ans--
vi. Damping ratio,
Ans-vii. Logarithmic decrement,
Ans--
�� = 2�
�
= 2 �� = 2��� = 2.8284
�
ζ=
c
0.5
=
= 0.1768
Cc 2.8284
δ=
2π c
2π
0.5
∙
=
∙
= 1.1285
ωd 2m 1.3919 2(1)
b) A 500-kg vehicle is mounted on springs such that its static deflection is 1.5 mm. What is the
damping coefficient of a viscous damper to be added to the system in parallel with the springs, such
that the system is critically damped?
[15 + 10 = 25 Marks]
Ans-The static deflection is related to the natural frequency by
�� =
�
=
∆��
�
9.81 2
� = 80.9 ���
0.0015�
�
The addition of a viscous damper of damping coefficient c leads to a damping ratio of
ζ=
c
→ c = 2ζmwn
2mwn
The system is critically damped when the damping ratio is 1, requiring a damping coefficient of
c = 2(1)(500kg) 80.9
rad
N∙s
= 8.09 × 104
s
m
Question 3
a) For what value of K will resonance occur for the system of figure Q3a?
150kg
[12 Marks]
Ans-The springs attached to the block act in parallel, leading to an equivalent stiffness of Keq = K + 1 ×
105
�
�
. Resonance occurs when the excitation frequency of 80 rad/s is equal to the natural frequency,
Which leads to
80
rad
= wn =
s
��� = ��2 = (150��) 80
keq
m
��� 2
�
= 960000
�
�
K = 960000 − 100000 = 860000
N
m
b) Determine keq and meq for the system of figure Q3b when x, the displacement of the center of the
disk measured from equilibrium, is used as the generalized coordinate, Assume the disk is thin and
rolls without slip.
Fig Q3b
[12 + 13 = 25 Marks]
Ans-If the disk rolls without slip, then the angular rotation of the pulley θ and the downward displacement
of the block y are
x
θ=
y = 2rp θ = 2x
rp
Noting that the effects of gravity and static deflection cancel, potential energy calculations lead to
1
1
1
1
1
V = kx2 + (2k)y2 = kx2 + 2k(2x)2 = 9kx2 → keq = 9k
2
2
2
2
2
The kinetic energy of the system is
If the disk is thin �� =
��2
2
T=
1 2 1
1
1
mx + Id w2d + Ip θ2 + (2m)y2
2
2
2
2
�
and if it rolls without slip, �� = �. Thus
T=
2
x
1 2 1 1 2 x 2 1
1
mx +
mr
+ Ip
+ (2m)(2x)2
2
2 2
r
2 rp
2
Ip 2
Ip
1 19
19
m + 2 x → meq =
=
m+ 2
2 2
2
rp
rp
QUESTION 4
a) The displacement of a vibrating spring-mass-damper system is recorded on an x-y plotter and
reproduced in Figure Q4a. The y coordinate is the displacement in cm and the x coordinate is
time in seconds. From the plot determine the natural frequency, the damping ratio, and the
damped natural frequency.
Fig Q4a. A plot of displacement versus time for a vibrating system
Ans--
12 Marks
Logarithmic decrement, δ is
x1
0.725
δ = ln
= ln
= ln (3.625) = 1.2879 =
0.2
x2
Making ζ subject of the formula
1.2879
δ
ζ=
=
= 0.2
4π2 + 1.28792
4π2 + δ2
The damped period of vibration is
2�
2�
�� =
=
=4
�� �� 1 − �2
Making �� subject of the formula
2�
= 1.571
�� =
��
Also, making �� subject of the formula
2�
2�
�� =
=
= 1.6032
2
4 1 − 0.22
4 1−�
∴ Natural frequency, �� = 1.6032 ���/�
Damping ratio, ζ = 0.2
Damped natural frequency, �� = 1.571 ���/�
2πζ
1 − ζ2
b) Calculate the natural frequency and damping ratio for the system in Figure Q4b. Assume that
no friction acts on the rollers, is the system overdamped, critically damped, or underdamped?
Fig Q4b. A mass spring damper vibration system.
The End
13 Marks
Ans-The rules for calculating the equivalent stiffness of parallel and series connection of springs.
Springs in series
��� =
1
�1 �2
=
1 �1 + 1 �2 �1 + �2
Springs in parallel
��� = �1 + �2
From Figure Q2b, the equivalent spring constant is:
��� = �1 + �2 + �3 +
�4 �5
2×3
= 10 + 1 + 4 +
��/� = 16.2��/�
�4 + �5
2+3
Then using the standard formulas for frequency and damping ratio:
���
=
�
16.2 × 103
= 40.25 ���/�
10
c
1
ζ=
=
= 0.001242 ≈ 0.001
2mωn 2(10)(40.25)
�� =
Thus the system is underdamped, in fact very lightly damped.