Recommended problems from the book Engineering vibration
(Inman) in the course Mechanical Vibrations, M0015T
1:st ed.
2:nd ed.
3:rd ed.
4:th ed.
1.2, 5, 6,
12, 18, 19, 20, 27 1, 37, 41, 44, 48, 57
1.2, 5, 6, 11, 23, 28, 29, 30, 481, 58, 66, 69, 73, 82
1.2, 7, 8, 13, 26, 32, 33, 34, 531, 67, 75, 78, 82, 65
1.7, 12 2, 13, 18, 36 3, 44, 45, 46, 66 4, 86 5, 95, 98 6, 102, 80
1:st ed.
2:nd ed.
3:rd ed.
4:th ed.
2.
6, 9, 12 7, 14 8, 17 9, 23, 24 10, 28 11, 18
2.7, 11, 14, 177, 318, 349, 40, 4110, 4611, 47
2.7, 16, 19, 227, 378, 409, 46, 4710, 5411, 55
2.8 12, 20, 24 13, 287, 478, 509, 56, 5710, 6411, 65
1:st &
2:nd ed.
3:rd ed.
4:th ed.
4.1-7, 10, 11, 13, 40, 41, 43 14, 50, 71 (nn) not mandatory problems
1:st, 2:nd
& 3:rd ed.
5.8, 9, 17, 25,61
4.1-7, 11, 12, 14, 43, 44, 4614, 53, 79
4.1, 215, 3, 4, 516 , 6, 12, 1317, 1518, 4719, 4914, 56 87
1
Replace the total energy with the kinetic and potential energy i.e. E (t ) = T + U
Use frequency 10Hz and amplitude 1mm
3
Use unloaded automobile mass of 1000 kg, loaded automobile mass of 1300 kg and spring stiffness of
400000 N/m.
4
Replace the total energy with the kinetic and potential energy i.e. E (t ) = T + U . Use frequency=2
rad/s , ζ=0.01 and m=10 kg.
5
Use m=100 kg, l=1 m and an ocillation frequency of 3 Hz.
6
Use m=1000 kg and k=400000 N/m
7
Gravity is included in the problem !!!. Use l1=1.5 m, l2=0.5 m, l=1.0 m and a 10N force to match
the answer of 1:st ed.
8
Calculate the amplitude of the transmitted force. For the 4th ed use m=2000 N and a support deflection
of 5 cm.
9
In order to get correct answer for 2:nd, 3:rd and 4th ed the answer for 1:st ed shall be multiplied by -1
(the coordinate system has been changed compared to 1st ed)
10
Use r=2.5, ζ=0.05
11
Use r=1
12
Use x(0)=0.05
13
Use m=100 kg, k=30000 N/m, c=1000 Ns/m, Harmonic force amplitude=80 N
2
14
Use E = 6.9 ⋅ 109 , l = 2, m = 3000, I = 5.2 ⋅ 10 −6 (SI-enheter)
15
Use m1 = 9kg , m2 = 1kg
16
17
18
19
Use m11 = 9kg , k11 = 27 N / m, k12 = k 21 = −3N / m, k 22 = 3N / m
Use m = 2000kg , k = 280000 N / m
Use m2 = 50kg , m1 = 2000kg
Use Hub stiffness = 10000 N / m
Problem number is associated with the 1:st ed !!
1.2
1.5
1.6
1.12
1.18
1.27
1.37
1.44
1.48
1.57
2.6
x(t)=1.5sin(2t+0.7297)
vmax=62.8 mm/s
amax=3948 mm/s2
B=Acos(Φ)
C=Asin(Φ)
ω1max=20 rad/sec
ω2max=17.54 rad/sec
v1max=2 m/s
v2max=1.754 m/s
a1max=40 m/s2
a2max=30.77 m/s2
x(t)=e-tsin(t)
E(t)=0.05e-0.04t(-0.02 sin(2t+1.56)+2cos(2t+1.56))2 +0.2e-0.04t(sin(2t+1.56))2
EI=11843 Nm2
C = 1.3 ⋅ 104 Ns/m
C=87.2 Ns/m
a + r 4k
ω=
rad/s
r
3m

( x0 − A0 cos(φ ))ω d

 v0 + ( x0 − A0 cos(φ ))ζω − A0ω dr sin(φ ) 

θ = tan −1 
X 0 − A0 cos(φ )
sin(θ )
no φorcing φunction :
A=
x 0ω d 

 v0 + x0ζω 

θ = tan −1 
A=
x0
sin(θ )
2.9
A0=0.0041 m
Φ=-1.29 rad
2.12 c= 1228 Ns/m
A0=0.0023 rad
2.14 Ft=4001 N
2.17
2kζωωbY
Ft (t ) = −
(ω 2 − ωb2 ) 2 + (2ζωωb ) 2
2.18 k=2.21·106 N/m
2.23 Speed ωb (rad/s)
20
5.817
80
23.271
100
29.088
150
43.633
r1
0.32
1.28
1.60
2.40
r2
0.39
1.55
1.94
2.91
cos(ωbt − φ )
x1 (cm)
1.10
1.16
0.83
0.46
x2 (cm)
1.16
0.84
0.56
0.31
2.24 k=53.650 N/m
c=922.2 Ns/m
at v=50 km/h
xmax=0.00196 m
at v=100 km/h
xmax=0.000466 m
2.28 e=0.1 m
4.1
[M ]{x}+ [K ]{x} = {0}
where
0
k + k
, [K ] =  1 2

 0 m2 
 − k2
ω1=1.642 rad/s
ω2=2.511 rad/s
m
[M ] =  1
4.2
4.3
4.4
4.5
4.6
− k2 
x 
, and {x} =  1 

k 2 + k3 
 x2 
1 
− 0.101
, {u}2 = 

0.909
 1 
{u}1 = 
x1 (t ) = 0.916 cos(1.642t ) + 0.0841 cos(2.511t )
x 2 (t ) = 0.833(cos(1.642t ) − cos(2.511t ))
x1 (t ) = 0.25( 2 sin( 2t ) + sin( 2t ))
x 2 (t ) = 0.75( 2 sin( 2t ) − sin( 2t ))
[M ]{x} + [K ]{x} = {0}
where
− k2 
0
m
x 
k
, [K ] =  2
, and {x} =  1 


 x2 
− k 2 k 2 
 0 m2 
4.7 ω1=0 rad/s
ω2=3.333 rad/s
4.10
3 0 θ1 
 2 − 1 θ 1  0
J2 
   + k 

  =  
0 1 θ 2 
− 1 1  θ 2  0
[M ] =  1
4.11
0   x1   280000 − 280000  x1  0
2000
 = 
 +
 0
2000  x2  − 280000 280000   x2  0

ω1 = 0 rad / s, ω 2 = 280 rad / s
0.7071
− 0.7071
, {u}2 = 

0.7071
 0.7071 
{u}1 = 
4.13
4.40
2000 0   x1   1000 − 1000  x1  0
 +
 = 
 0
50  x 2  − 1000 11000   x 2  0

ω1 = 0.674 rad / s, ω 2 = 14.8 rad / s
x1 (t ) = 2.277 ⋅ 10 −5 cos(0.674t ) − 2.277 ⋅ 10 −5 cos(14.834t )
x 2 (t ) = 2.074 ⋅ 10 −6 cos(0.674t ) + 9.998 ⋅ 10 −3 cos(14.834t )
4.41 ω1=0 rad/s
ω2=8.829 rad/s
ω3=19.028 rad/s
0.5774
− 0.9229
− 0.5036




{u}1 = 0.5774, {u}2 = − 0.3833, {u}3 =  0.8638 
0.5774
 0.0359 
− 0.0162






4.43 ω1=0 rad/s
ω2=4.4850 rad/s
ω3=6.7275 rad/s
0.5774
− 0.7071
 0.6667 




{u}1 = 0.5774, {u}2 =  0 , {u}3 = − 0.3333
0.5774
 0.7071 
 0.6667 






4.50 ω1=0 rad/s
ω2=8.9311 rad/s
ω3=19.0675 rad/s
0.5774
 0.9266 
 0.5010 





{u}1 = 0.5774, {u}2 =  0.3723 , {u}3 = − 0.8651
0.5774
− 0.0534
 0.0245 






4.71
5.8
5.9
5.17
5.25
5.61
FT=132.2 N
k=765 N/m, c=2.47 kg/s
X=0.0062 mm, FT=86 N
ζ = 0.29
k2=9512.8 N/m, c=9.51 kg/s
Additional problem number associated with
3:rd ed.
1.13
2.7
k1 + k2
m
k +k +k
b) ω = 1 2 3
m
a) ω =
a) x(t ) = 2.795 ⋅10 −3 sin( 20t ) − 1.25 ⋅10 −3 sin(10t )
b) x(t ) = 2.795 ⋅10 −3 sin( 20t ) + 0.05 cos( 20t ) − 1.25 ⋅10 −3 sin(10t )
Additional problem number associated with 4:th
ed.
1.102 c=87.2 kg/s