Mechanical Vibrations: 4600-431
December 20, 2006
Example Problems
Contents
1 Free Vibration of Single Degree-of-freedom Systems
1
2 Frictionally Damped Systems
33
3 Forced Single Degree-of-freedom Systems
42
4 Multi Degree-of-freedom Systems
69
1
Free Vibration of Single Degree-of-freedom Systems
Problem 1:
In the figure, the disk and the block have
mass m and the radius of the disk is r.
a) Find the equations of motion for this
system.
c
b) What are the natural frequency and
damping ratio of the system in terms
of m, c, and k?
k
c) If the block is displaced 18 cm to the
right and released from rest find the resulting angular displacement of the disk
with
m = 3 kg,
k = 21 N/m,
(m, r)
m
k
r = 9 cm,
c = 63 N · s/m,
1
4k
Problem 2:
For the system shown to the right, the disk of
mass m rolls without slip and x measures the
displacement of the disk from the unstretched
position of the spring.
x
k
a) Find the equations of motion.
c
(m, r)
b) With
c = 16 N/(m/s),
m = 2 kg,
r = 0.10 m
for what value of the spring stiffness k
is the damping ratio of the system onehalf of the critically damped value, so
that ζ = ζcr /2?
c) With these parameter values, find the
displacement of the disk if it is rolled
20 cm to the right (from static equilibrium) and released from rest.
Problem 3:
From the figure shown to the right
k2
a) find the equations of motion in terms of
the angular rotation of the disk;
r
b) what are the damping ratio and natural
frequency of the system in terms of the
parameters m, b, k1 , and k2 ;
r
2
m
c) can you draw an equivalent springmass-damper system?
m
b
2
k1
Problem 4:
For the mechanical system shown to the
right, the uniform rigid bar has mass m and
pinned at point O. For this system:
̂
k
a) find the equations of motion;
ı̂
b) Identify the damping ratio and natural
frequency in terms of the parameters
m, c, k, and ℓ.
m
ℓ
2
c) For:
y
ℓ
2
θ
m = 1.50 kg,
c = 0.125 N/(m/s),
ℓ = 45 cm,
O
k = 250 N/m,
2m
x
find the angular displacement of the bar
θ(t) for the following initial conditions:
c
k
θ(0) = 0,
θ̇(0) = 10 rad/s.
Assume that in the horizontal position the
system is in static equilibrium and that all
angles remain small.
Solution:
a) In addition to the coordinate θ identified in
the original figure, we also define x and y as
the displacment of the block and end of the
bar respecively. The directions ı̂ and ̂ are
defined as shown in the figure.
A free body diagram for this system is
shown to the right. Note that the tension in
the cable between the bar and the block is
unknown and represented with T while the
reaction force FR is included, although both
its magnitude and direction are unspecified.
In terms of the identifed coordinates, the
angular acceleration of the bar αβ/F and the
F
linear acceleration of the block aG are
αβ/F = θ̈ k̂,
F
−k y ̂
−T ̂
FR
T ̂
−k x ̂
−c ẋ ̂
aG = ẍ ̂.
We can also relate the identified coordinates as
ℓ
x = θ,
y = ℓ θ.
2
The equations of motion for this system can be obtained with linear momentum balance
applied to the block and angular momentum balance aout O on the bar. These can be
written as
³
´
X
F
F = m aG −→ T − k x − c ẋ ̂ = 2 m ẍ ̂,
¶
µ
X
m ℓ2
ℓ
θ̈ k̂.
M O = I O αβ/F −→ − T − k y ℓ k̂ =
2
3
3
Solving the first equation for T and substituting into the second equation yields
− (2 m ẍ + k x + c ẋ)
ℓ
m ℓ2
−kyℓ =
θ̈.
2
3
Using the coordinate relations we can obtain the equation of motion as
c ℓ2
5 k ℓ2
5 m ℓ2
θ̈ +
θ̇ +
θ = 0.
6
4
4
b) In the above equation the equivalent mass, damping, and stiffness are
meq =
5 m ℓ2
,
6
beq =
c ℓ2
,
4
keq =
5 k ℓ2
.
4
From these the damping ratio and natural frequency are
ζ=
√
c ℓ2
3c
beq
,
= q 4
= √
5 k ℓ 2 5 m ℓ2
2 keq meq
2
50
km
2
4
6
s
s
r
2
5kℓ
keq
3k
4
=
ωn =
=
5 m ℓ2
meq
2
m
6
p
c) Evaluating the damping ratio and natural frequency we find that for the given values
of the parameters
ωn = 15.8 rad/s,
ζ = 7.91 × 10−4 .
Therefore the system is underdamped and the general solution can be written as
³
³ p
´
³ p
´´
θ(t) = e−ζ ωn t a sin ωn 1 − ζ 2 t + b cos ωn 1 − ζ 2 t ,
where a and b are arbitrary constants used to fit the initial conditions. Evaluating θ(t)
and θ̇(t) at t = 0 yields
p
θ(0) = b = 0,
θ̇(0) = −ζ ωn b + ωn 1 − ζ 2 a = 10 rad/s,
so that the general solution becomes
³
´
θ(t) = 0.632 e−0.0125 t sin 15.8 t .
4
Problem 5:
In the figure, the disk has mass m, radius r,
2
and moment of inertia IG = mr
about the
2
mass center G and is assumed to roll without
slip. The identified coordiante θ measures the
rotation of the disk with respect to the equilibrium position.
k
g
c
a) Find the equations of motion for this
system.
G
(m, r)
k
b) If the disk is released from rest with
θ(0) = − π2 rad, find the resulting angular displacement θ(t) for
m = 3 kg,
k = 36 N/m,
r = 15 cm,
c = 3 N · s/m,
c) What is the force in the upper spring
during this motion?
Problem 6:
For the mechanical system shown to the
right, the uniform rigid bar has mass m and
pinned at point O. For this system:
ℓ
2
k
a) find the equations of motion;
θ
m
b) Identify the damping ratio and natural
frequency in terms of the parameters
m, c, k, and ℓ.
z
ℓ
2
c) For:
m
̂
x
m = 2 kg, ℓ = 25 cm,
c = 0.25 N/(m/s), k = 50 N/m,
k
c
ı̂
find the angular displacement of the bar
θ(t) for the following initial conditions:
θ(0) = 0,
θ̇(0) = 10 rad/s.
d) for this motion, find the tension in the
cable connecting the rod and the block
as a function of time.
Assume that the system is in static equilibrium at θ = 0, and that all angles remain
small.
Solution:
a) We identify the coordinates x and z as shown above, which are related to the angular
5
displacement θ as:
x=
ℓ
θ,
2
z=
An appropriate free-body diagram is
shown to the right. Applying linear momentum balance on the block yields
X
F
F = m aG ,
(T − k x − c ẋ) ̂
ℓ
θ.
2
k z ̂
−T ̂
T ̂
= m ẍ ̂.
FR
Likewise, angular momentum balance on
the bar provides
X
MO = I O αβ/F ,
µ
¶
ℓ
m ℓ2
ℓ
−T − k z
k̂ =
θ̈ k̂.
2
2
12
−k x ̂
−c ẋ ̂
Combining these equations and eliminating the tension, the equation of motion can be
written as
7m
θ̈ + c θ̇ + 2 k θ = 0.
6
b) For the above equation the equivalent mass, damping, and stiffness are
meq =
7m
,
6
beq = b,
keq = 2 k,
and the natural frequency and damping ratio are
s
r
√
keq
12 k
3b
beq
=
,
ζ= p
.
=√
ωn =
meq
7m
2 keq meq
28k m
Problem 7:
The block shown to the right rests on a frictionless surface. Find the response of the system if the block is displaced from its static
equilibrium position 15 cm to the right and
released from rest.
m = 4.0 kg,
k1 = 1.5 N/m,
̂
x
ı̂
k1
k2
m
b = 0.25 N/(m/s),
k2 = 0.50 N/(m/s).
b
Solution:
An appropriate free-body diagram is shown to the
right. Notice that the two springs are effectively in
parallel, as the displacement across each spring is
identical. Linear momentum balance on this block
provides
X
F
F = m aG ,
(−k1 x − k2 x − b ẋ) ı̂ = m ẍ ı̂,
6
−k1 x ı̂
−k2 x ı̂
−b ẋ ı̂
or, writing this in standard form
m ẍ + b ẋ + (k1 + k2 ) x = 0.
Further, the system is released from rest so that the initial conditions are
x(0) = x0 = 15 cm,
ẋ(0) = 0 cm/s.
Problem 8:
For the system shown to the right, the disk of
mass m rolls without slip and x measures the
displacement of the disk from the unstretched
position of the spring. The surface is inclined
at an angle of φ with respect to vertical.
x
k
z
a) find the equations of motion. Do not
neglect gravity;
c
b) if the system is underdamped, what is
the frequency of the free vibrations of
this system in terms of the parameters
k, c, and m;
φ
θ
(m, r)
̂
C
ê2
c) for what value of the damping constant
c is the system critically damped;
ı̂
d) what is the static equilibrium displacement of the disk?
ê1
Solution:
a) In addition to x, the displacement of the center of the disk, we identify the coordinates
z and θ, the displacement across the spring and the rotation of the disk respectively.
These additional coordinates are related to x as
z = 2 x,
x = −r θ.
An appropriate free-body diagram is shown to the
right. We note that (ı̂, ̂) are related to the directions (ê1 , ê2 ) as
ı̂ = cos φ ê1 + sin φ ê2 ,
̂ = − sin φ ê1 + cos φ ê2 .
−k z ê1
−m g ̂
−c ẋ ê1
fr ê1
N ê2
The moment produced by gravity about point C is
Mgravity
= rGC × (−m g ̂),
=
(r ê2 ) × (−m g ̂) = −m g r sin φ k̂.
Angular momentum balance about the contact point C yields
X
MC = I C αD/F ,
µ
¶
³
´
3 m r2
(2 r) k z + r c ẋ − m g r sin φ k̂ =
θ̈ k̂.
2
7
Eliminating the coordinates z and θ, we can write the equation of motion in terms of x
as
3m
ẍ + c ẋ + 4 k x = m g sin φ.
2
Since the gravitational force has been included in the development of this equation of
motion, the coordinates are measured with respect to the unstretched position of the
spring.
b) Assuming
the system is underdamped, the frequency of the free vibrations is ωd =
p
ωn 1 − ζ 2 , where
s
r
c
beq
keq
8k
= √
ωn =
=
,
ζ= p
,
meq
3m
2 keq meq
2 6km
so that
ωd =
r
8k
3m
r
1−
c2
,
24 k m
c) The system is critically damped when ζ = 1, which corresponds to a damping coefficient
of
√
ccr = 2 6 k m.
d) The system is stationary in static equilibrium, so that x ≡ x0 = constant—both ẋ and
ẍ vanish, and the equation of motion reduces to
4 k x0 = m g sin φ.
Solving for x0 , the equilibrium displacement is
x0 =
8
m g sin φ
.
4k
Problem 9:
In the figure shown to the right, in the absence of gravity the springs are unstretched
in the equilibrium position.
z2
k2
r2
a) Determine the deflection of each spring
from its unstretched length when the
system shown is in equilibrium.
θ
̂
b) If the system is released from the unstretched position of the springs, what
is the maximum angular velocity of the
disk during the resulting motion?
r1
I
ı̂
z1
m
x
k1
Solution:
a) We define the coordinates x, θ, z1 , and z2
as shown in the figure, which are related
as
x = −r1 θ,
z1 = r1 θ,
−k2 z2 ı̂
z2 = −r2 θ.
Notice that because of these coordinate
definitions, a rotation with positive θ
gives rise to a negative value in both x
and z2 . Likewise, we see that x = −z1 .
Using the free-body diagram shown to
the right, linear momentum balance on
the block provides
X
F
F = m aG ,
(T − m g) ̂
FR
−T ̂
k1 z1 ̂
T ̂
= m ẍ ̂,
−m g ̂
while angular momentum balance on the
disk yields
X
MO
(T r1 − k1 z1 r1 + k2 z2 r2 ) k̂
= I O αD/F ,
= I θ̈ k̂.
Eliminating the unknown tension T from these equations and using the coordinate
relations, the equation of motion becomes
´
´
³
³
I + m r12 θ̈ + k1 r12 + k2 r22 θ = m g r1 .
9
The equilibrium rotation of the disk thus is found to be
θeq =
m g r1
.
k1 r12 + k2 r22
With this, the equilibrium deflection of each spring is found to be
m g r12
,
k1 r12 + k2 r22
m g r1 r2
=−
.
k1 r12 + k2 r22
z1,eq = r1 θeq =
z2,eq = −r2 θeq
b) The general free response of the disk can be expressed as
θ(t) = θeq + A sin(ωn t) + B cos(ωn t),
where θeq is given above, A and B are arbitrary constants, and
s
k1 r12 + k2 r22
.
ωn =
I + m r12
The system is released with the initial conditions:
θ(0) = 0,
θ̇(0) = 0,
so that solving for the arbitrary constants
A = 0,
B = −θeq .
Therefore the solution is
m g r1
θ(t) = θeq (1 − cos(ωn t)) =
k1 r12 + k2 r22
Ã
1 − cos
Ãs
k1 r12 + k2 r22
t
I + m r12
The angular velocity of the disk becomes
´
³
θ̇(t) = θeq ωn sin(ωn t),
which has amplitude
Ω = θeq ωn = p
10
m g r1
(k1 r12
+ k2 r22 ) (I + m r12 )
!!
.
Problem 10: (Spring 2003)
For the spring-mass-damper system shown to
the right, x is measured from the static equilibrium position. If the surface is assumed to
be frictionless:
x
a) determine the governing equations of
motion;
m
6k
b
b) what is the frequency of oscillation of
the system;
c) what value of the damping coefficient b
corresponds to critical damping?
d) if k = 2 N/m, b = 4 N/(m/s), and m =
3 kg, find the displacement of the mass
x(t) when the system is started with the
initial conditions:
x(0) = 0.10 m,
ẋ(0) = 0 m/s.
Problem 11: (Spring 2003)
For the mechanical system shown to the
right, the uniform rigid bar has mass m and
pinned at point O. For this system:
k
a) find the equations of motion;
z2
b) what value of the damping constant c
gives rise to a critically damped system?
m
ℓ
2
k
θ
m
z1
ℓ
2
̂
c
ı̂
Solution:
a) In addition to θ, we define two additional coordinates, z1 and z2 , to measure the deflection at the left and right ends of the bar.
11
These coordinates are related as:
ℓ
z1 = θ,
2
−k z2 ̂
ℓ
z 2 = θ = z1 .
2
A free-body diagram for this system is
shown to the right. Applying angular momentum balance on the bar eliminates the
appearance of the reaction force and leads
to:
X
MG = I G θ̈ k̂,
´ℓ
³
m ℓ2
k̂ =
θ̈ k̂.
T − k z1 − cż1
2
12
−T ̂
T ̂
k z1 ̂
G
FR
c ż1 ̂
Likewise, application of linear momentum balance on the block yields:
X
F
F = m aG ,
³
´
− k z2 − T ̂ = mz̈2 ̂
Eliminating the unknown tension T and solving for z1 in terms of z2 , the equation of
motion becomes:
c ℓ2
k ℓ2
m ℓ2
θ̈ +
θ̇ +
θ = 0.
3
4
2
b) A critically damped system occurs when ζ = 1. For this system:
√
3c
.
ζ=√
32 k m
Solving for ccr yields:
ccr =
Problem 12: (Spring 2003)
Find the response of the system shown to the
right if the block is pulled down by 15 cm and
released form rest.
m = 2.0 kg,
k1 = 0.5 N/m,
r
32 k m
.
3
̂
k1
b
b = 0.5 N/(m/s),
k2 = 0.25 N/(m/s).
k2
ı̂
x
m
Solution:
For this system, the two springs in series may be replaced by an equivalent spring, with
constant:
1
k1 k2
keq = 1
1 = k +k .
+
1
2
k1
k2
12
Therefore, the free-body diagram is shown
to the right. Applying linear momentum
balance to the block yields:
X
F
F = m aG ,
³
´
keq x + b ẋ ̂ = −m ẍ ̂,
keq x ̂
b ẋ ̂
which can finally be written as:
m ẍ + b ẋ + keq x = 0.
With the numerical values given above, this becomes:
¶
µ
¶
µ
3
1 N
1 N
ẋ +
x = 0,
x(0) =
m,
(2 kg) ẍ +
2 m/s
6 m
20
ẋ(0) = 0 m/s.
With this, the damping ratio and natural frequency are:
r
√
1 −1
3
s ,
ζ=
.
ωn =
12
4
Therefore, the system is underdamped and the general response can be written as:
³
´
x(t) = e−ζ ωn t A cos(ωd t) + B sin(ωd t) .
Using the initial conditions to solve for A and B, we find:
! r
!!
Ã√
Ã
Ã√
13
3 −t/8
3
13
√ t +
√ t
x(t) =
.
e
cos
cos
20
13
8 3
8 3
Problem 13: (Spring 2003)
In the figure shown to the right, in the absence of gravity the springs are unstretched
in the equilibrium position.
x2
k2
a) Determine the deflection of each spring
from its unstretched length when the
system shown is in equilibrium.
r2
θ
b) If the system is released from the unstretched position of the springs, what
is the maximum angular velocity of the
disk during the resulting motion?
r1
̂
ı̂
m
x1
k1
13
Solution:
a) We define θ, x1 , and x2 as indicated in the above figure. In particular, x1 and x2
represent the displacement in their respective springs as measured from their unstretched
position. These coordinates are related through the following transformations:
x1 = −r1 θ,
x2 = −r2 θ.
An appropriate free-body diagram for this
system is shown to the right. Notice that
the gravitational force must be included to
determine the equilibrium deflection in the
system. To eliminate the reaction force on
the disk, angular momentum balance is applied about the center, yielding:
X
MG = I G θ̈ k̂,
´
³
T r1 + k2 r2 x2 k̂ = I G θ̈ k̂.
−k2 x2 ı̂
G
FR
−T ̂
T ̂
Also, applying linear momentum balance to
the block yields:
X
F
F = m aG ,
³
´
T − k1 x1 − m g ̂ = m ẍ1 ̂.
−m g ̂
−k1 x1 ̂
Finally, eliminating the unknown tension from these equations and using the above
coordinate transformations, this single-degree-of-freedom system can be modeled with
the equation:
´
´
³
³
I G + m r12 θ̈ + k1 r12 + k2 r22 θ = (m g r1 ).
This equation of motion determines the equilibrium position θeq (with θ̈eq = 0) to be:
θeq =
m g r1
.
k1 r12 + k2 r22
Therefore, the equilibrium displacements in each spring are:
x1,eq = −r1 θeq =
m g r12
,
k1 r12 + k2 r22
x2,eq = −r2 θeq =
m g r1 r2
.
k1 r12 + k2 r22
b) Define new coordinates z1 and z2 , which measure the displacement in springs 1 and 2
with respect to the static equilibrium position, that is:
z1 = x1 − x1,eq ,
z2 = x2 − x2,eq .
Likewise, let φ represent the angular displacement of the disk from the static equilibrium
position:
φ = θ − θeq .
14
Therefore, the potential energy of this system can be written as:
V
=
=
1
1
k1 z12 + k2 z22 ,
2
2
´
1³
k1 r12 + k2 r22 φ2 .
2
Also, the kinetic energy becomes:
T
1 G 2 1
I φ̇ + m ż12 ,
2
2
´
1³ G
I + m r12 φ̇2 .
2
=
=
If the system is released from rest at the unstretched position of the springs, then:
φ(0) = −θeq = −
m g r1
,
k1 r12 + k2 r22
φ̇(0) = 0.
At this initial state, the potential and kinetic energies become:
T0 = 0,
V0 =
(m g r1 )2
.
2(k1 r12 + k2 r22 )
Because this system is conservative, the total energy, E = T + V remains constant.
Therefore, when the kinetic energy is maximal, the potential energy is minimal, that is:
V1 = 0,
T1 =
´
1³ G
I + m r12 φ̇2max .
2
Finally, conservation of energy implies that V0 = T1 , and solving for φ̇max we find that:
s
(m g r1 )2
.
φ̇max =
G
(I + m r12 )(k1 r12 + k2 r22 )
Problem 14: (Spring 2003)
For the system shown to the right, the disk of
mass m rolls without slip and x measures the
displacement of the disk from the unstretched
position of the spring.
̂
x
3k
z
ı̂
a) find the equations of motion;
θ
b) if the system is underdamped, what is
the frequency of the free vibrations of
this system in terms of the parameters
k, c, and m;
k
m
Solution:
a) We define the three coordinates as shown as the figure, related as:
x = −r θ,
z = −2 r θ,
15
z = 2 x.
c
A free-body diagram for this system is
shown to the right. Notice that the force in
the upper spring depends on z, rather than
x, while the friction force has an unknown
magnitude f . Because the disk is assumed
to roll without slip, we are unable to specify
the value of f , but instead can relate the displacement and rotation of the disk through
the coordinate relations above.
−3k z ı̂
G
−k x ı̂
−c ẋ ı̂
C
f ı̂
The equations of motions can be developed directly with angular momentum balance
about the contact point, so that:
X
MC = I C θ̈ k̂,
³
´
3 m r2
(3k z) 2r + (k x) r + (c ẋ) r k̂ =
θ̈ k̂.
2
Finally, writing this equation in terms of a single coordinate, we obtain:
¶
µ
¡
¢
3 m r2
θ̈ + (c r2 ) θ̇ + 13 k r2 θ = 0.
2
b) For an underdamped response, the frequency of oscillation is ωd = ωn
this system, we find that:
r
c
26 k
,
ζ=√
,
ωn =
3m
78 k m
so that:
ωd =
r
p
1 − ζ 2 . With
26 k
2 c2
.
−
3m
9 m2
Problem 15:
In the system shown to the right, the pulley
has mass m and radius r, so that the moment
2
of inertia about the mass center is IG = mr
2 .
r
a) Find the governing equations of motion;
r
2
b) Find the frequency of oscillation for free
vibrations of the system;
m
c) For what value of the damping constant
is the system critically damped?
m
k
k
Solution:
16
c
a) We choose coordinates (x1 , x2 , θ), where x1 measures the displacement of the first mass
in the −̂ direction, x2 measures the displacement of the second mass in the ̂ direction,
and θ measures the angular rotation of the wheel in the k̂ direction. The kinetic and
potential energies for this system are:
T
=
V
=
1
1
1 mr2 2
θ̇ ,
m ẋ21 + m ẋ22 +
2
2
2 2
1
1
k x21 + k x22 .
2
2
However, these three coordinates are dependent through the transformations:
r
x2 = r θ.
x1 = θ,
2
Therefore, the Lagrangian reduces to:
µ
¶
µ
¶
1 7mr2
1 5kr2
L=T −V =
θ2 .
θ̇2 −
2
4
2
4
Further, the generalized force resulting from the viscous damper becomes:
Qθ = −
cr2
θ̇,
4
so that the equation of motion for this system can be reduced to:
(7m) θ̈ + c θ̇ + (5k) θ = 0.
b) For this system the natural frequency and damping ratio are:
r
c
5k
.
,
ζ= √
ωn =
7m
2 35 km
Therefore, the damped natural frequency becomes:
r
r
p
c2
5k
2
1−
ωd = ωn 1 − ζ =
.
7m
140 km
c) For a critically damped system, ζ = 1, so that we may solve for c = ccr to yield:
√
ccr = 2 35 km.
Problem 16:
In the figure, the disk has mass m, radius r,
2
and moment of inertia IG = mr
about the
2
mass center G, and is attached to a block of
mass m which rolls across the surface. If the
disk rolls without slip (µ is sufficiently large),
while the block moves without friction:
b
G
k
a) find the equations of motion for this
system;
b) for m = 2 kg, b = 0.5 (N · s)/m, and
k = 8 N/m, find the frequency of oscillation for the system.
17
m
Solution:
a) Let x denote the translational displacement of G in the ı̂ direction, while θ denotes the
angular displacement of the disk in the k̂ direction. If T is the tension between the disk
and the mass and f is the frictional force acting on the disk, the equations of motion
on the disk and the mass are:
(−kx − bẋ + T + f ) ı̂ = mẍ ı̂,
mr2
θ̈ k̂,
(rf ) k̂ =
2
(−T ) ı̂ = mẍ ı̂.
In addition, x and θ are related through the kinematic constraint:
x = −rθ.
Eliminating (θ, T, f ) we obtain a single degree-of-freedom system on x of the form:
5m
ẍ + bẋ + kx = 0.
2
b) From the above equation we identify the undamped natural frequency and damping
ratio as:
r
2k
b
ωn =
,
ξ=√
.
5m
10km
Thus, the damped natural frequency is:
ωd = ωn
p
1−
ξ2
=
r
10km − b2
.
25m2
For the given values of the parameters, this reduces to ωd = 1.264 rad/s.
Problem 17:
We model a nonuniform beam as a singledegree-of-freedom system in the form:
mẍ + bẋ + kx = 0,
x1 = 1.00 m
bc
and experimentally measure the mass as m =
kg. In free vibration we experimentally determine the equivalent spring constant to be
k = 4 N/m, and we measure the response as
shown.
x2 = 0.75 m
bc
a) Determine the equivalent damping constant.
b) What is the exponential decay rate of
the transient solution?
Solution:
18
a) We use the logarithmic decrement so that:
µ ¶
¶
µ
x1
δ1
0.75 m
p
,
δ1 = − ln
ζ=
= 0.288,
= − ln
x0
1.00 m
4π 2 + δ12
and we find that ζ = 0.0457. With m = 1 kg, the damping constant b is given as:
√
b = 2ζ km = 0.183 N/(m/s).
We note that with m given, we can also find the period of the oscillations to be:
T =
2π
ωd
=
=
2π
p
,
ωn 1 − ζ 2
√ p 2
m 4π + δ12
√
= 3.145 s.
k
b) The exponential decay rate is σ = ζωn , which is found to be σ = 0.0914 s−1 .
Problem 18:
The rigid beam (mass m = 2 kg, length l =
1.5 m) is supported by an elastic spring (k =
4 N/m) and damper (b = 2 N/(m/s)), and is
pinned to the ground.
k
a) Find the linearized equations of motion
in terms of z, the relative displacement
between the end of the beam and the
ground;
(m, l)
z
b
b) what is the frequency of the resulting
motion;
c) if the mass of the spring is taken to be
mspring = 1 kg, find the new frequency
of the oscillations.
Solution:
a) We will define the inclination of the bar from the horizontal position as θ, so that the
angular acceleration is θ̈ k̂. Therefore, using angular momentum balance about the
point of rotation, we find:
mbar l2
θ̈ + bl ż + kl z = 0.
3
We assume that the coordinates θ and z are related by z = l sin θ, which for small
rotations reduces to z = lθ. Using this constraint to eliminate θ, the equation of motion
reduces to:
mbar
z̈ + b ż + k z = 0.
3
For the parameter values given above, this becomes:
¶
µ
³
´
³
´
2
kg z̈ + 2 N/(m/s) ż + 4 N/m z = 0.
3
19
For this system, the damping ratio and natural frequency can be expressed as:
r
√
3k
3b
ωn =
,
ζ = √
mbar
2 kmbar
r
√
3
.
= 6 rad/s
=
8
p
b) The frequency of oscillation, ωd = ωn 1 − ζ 2 , reduces to:
s
√
µ
¶2
15
3k
3b
−
rad/s = 1.94 rad/s.
=
ωd =
mbar
2mbar
2
c) If the mass of the spring is considered, it is treated as an additional equivalent mass
meq = mspring /3 located at the end of the bar. Therefore the new moment of inertia of
the bar about the point of rotation is:
IO
=
=
=
mbar l2
+ meq l2 ,
3
mbar l2 ³ mspring ´ 2
l ,
+
3
3
(mbar + mspring ) l2
.
3
Therefore, the new frequency of oscillation is:
s
¶2
µ
√
3b
3k
= 3 rad/s = 1.73 rad/s.
−
ωd =
(mbar + mspring )
2(mbar + mspring )
Problem 19:
The non-uniform beam supports an end-mass
of m = 10 kg. If the response of the system
is such that:
x1 = 0.25 m
t1 = 1.00 s
m
x2 = 0.20 m,
t2 = 4.00 s,
as shown in the figure, find the equivalent
stiffness and equivalent damping of the beam
(assume that the beam is massless).
(t1 , x1 )
bc
(t2 , x2 )
bc
Solution:
20
a) The logarithmic decrement is defined as δ = ln |x2 /x1 |. In terms of this quantity and
the period of oscillation T , the damping ratio and natural frequency are defined as:
p
(2π)2 + δ 2
δ
.
ωn =
,
ζ=p
T
(2π)2 + δ 2
Therefore, the stiffness and damping constants reduce to:
k = m ωn2 = m
(2π)2 + δ 2
,
T2
b = m 2ζωn = m
2δ
.
T
For this system, we find that T = 4 s and δ = 0.223, and the stiffness and damping
constant reduce to:
k = 24.7 N/m,
Problem 20:
For the single-degree-of-freedom mechanical
system shown in the figure:
b = 1.12 N/(m/s).
2r
k
a) determine the governing equations of
motion;
r
b) what are the damping ratio and undamped natural frequency of this system;
m
c) find the response of the system x(t)
subject to the initial conditions x(0) =
x0 , ẋ(0) = 0,
c
when m = 1 kg, b = 12 (N · s)/m, and k =
9 N/m. Assume the pulley is massless and
neglect the effects of gravity.
Solution:
a) Let x represent the displacement of mass m in the vertical direction and θ measure the
angular displacement of the pulley, both measured from the static equilibrium position.
If T represents the tension in the cable supporting mass m, then T = kr
2 θ, where θ is
the angular displacement of the massless pulley from static equilibrium. In addition,
x
we find that θ is related to the linear displacement of mass m as θ = 2r
. As a result, in
the ̂ direction, the equation governing the motion of the mass is:
mẍ + bẋ +
k
x = 0.
4
As a result, we find that the the equivalent spring constant is keq = k4 .
b) The damping ratio and natural frequency are simply:
r
√
b
b
keq
k
=√
= √ .
,
ωn =
ζ= p
m
2
m
2 keq m
km
21
In terms of the given parameters, we find ζ = 4.0 and ωn = 3 rad/s. We note that the
damping ratio has no units.
c) For an overdamped system, the general solution is:
³
³
´
³
´´
p
p
x(t) = exp(−ζωn t) c1 exp (ωn ζ 2 − 1)t + c2 exp −(ωn ζ 2 − 1)t ,
and with these initial conditions this reduces to:
Ã
p
´
³
p
ζ + ζ2 − 1
p
exp (ωn ζ 2 − 1)t +
x(t) = x0 exp (−ζωn t)
2 ζ2 − 1
!
p
³
´
p
ζ − ζ2 − 1
2
p
exp −(ωn ζ − 1)t
,
2 ζ2 − 1
Ã
!
√
√
4 + 15 3 √15t 4 − 15 − 3 √15t
−6t
√
+ √
e2
e 2
= x0 e
,
2 15
2 15
¡
¢
= x0 e−6.0t 1.016e5.81t + 0.01640e−5.81t .
Problem 21:
For the spring-mass-damper system shown to
the right, x is measured from the static equilibrium position, and the surface is frictionless.
x
a) Determine the governing equations of
motion.
m
6k
b) What is the period of each oscillation in terms of the system parameters
(m, k, c)?
c) For what value of c is the system critically damped?
d) If the system is released with the initial
conditions:
x(0) = 0.0 m,
ẋ(0) = 5.0 m/s,
find the resulting solution x(t) if m =
2 kg, k = 48 N/m and c = 4 N/(m/s).
Solution:
a) In terms of x, the spring and damping forces can be written as:
Fspring = −6k x ı̂,
Fdamper = −2c ẋ ı̂.
Using linear momentum balance on the block, we find that:
X
F
F = (−6k x − 2c ẋ) ı̂ = mẍ ı̂ = m aG ,
22
2c
and the equation of motion can be written in standard form as:
ẍ + (2ζωn ) ẋ +
(ωn2 )
x = 0,
with
ωn =
r
6k
,
m
c
ζ=√
.
6km
b) In p
terms of ζ and ωn , the period of oscillation is simply T = (2π)/ωd , where ωd =
ωn 1 − ζ 2 . Therefore, for this system, the period reduces to:
T =r
6k
m
s
2π
m
µ
¶2 = 2π √6km − c2 ,
c
1− √
6km
provided c2 < 6km.
c) For a critically damped system, ζ = 1, and solving for c, yields:
√
ccritical = 6km.
d) With these parameter values, the equation of motion reduces to:
ẍ + 4 ẋ + 144 x = 0,
so that ωn = 12 rad/s, and ζ = 1/6. For these values, the general solution can be
written as:
³ √
´
x(t) = A e−2 t sin 2 35 t + φ ,
where A and φ are arbitrary constants used to fit the initial conditions. Solving for A
and φ, we find:
r
5
A=
m,
φ = 0 rad,
28
so that the general solution can be written as:
r
³ √
´
5 −2 t
e
sin 2 35 t m.
x(t) =
28
23
Problem 22:
For the system shown to the right, x is measured from the unstretched position of the
spring. Each block has mass m and the disk
has moment of inertia I and radius r. If the
gravitational constant is g:
x
(I, r)
k
m
a) find the equations of motion which determine x(t);
b) what is the period of the free oscillations?
m
Problem 23:
For the single-degree-of-freedom mechanical
system shown in the figure:
a) determine the governing equations of
motion;
b) what are the damping ratio and undamped natural frequency of this system;
k
2k
b
g
c) what is the stretch in the spring when
the system is in equilibrium?
m
Solution:
We assume that the ı̂ and ̂ directions are standard orthonormal basis in the horizontal
and vertical directions respectively.
a) The forces due to the springs in parallel and damping are:
Felastic = 3k · x̂,
Fdamping = b · ẋ̂.
With the inclusion of the gravitational force, the equation of motion for this system can
be written:
X
F = m (−ẍ̂) ,
3k · x̂ + b · ẋ̂ − mg̂ = −mẍ̂.
Taking components in the ̂ direction, we obtain the governing equation of motion:
mẍ + bẋ + 3kx = mg.
24
b) Dividing through by the mass m, we find:
ẍ +
so that:
2ζωn =
b
3k
ẋ +
x = g,
m
m
b
,
m
ωn2 =
3k
,
m
which can be solved to yield:
ωn =
r
b
ζ= √
.
2 3mk
3k
,
m
c) In equilibrium, the system is stationary, so that ẋeq = 0 and ẍeq = 0. Substitution into
the governing equations yields:
3kxeq = mg, → xeq =
mg
.
3k
Problem 24:
For the single-degree-of-freedom mechanical
system shown in the figure:
a) find the linearized governing equations
of motion for small θ;
k
b) find the frequency of oscillation for the
response;
G
c) if the initial velocity is zero, and θ(0) =
θ0 , determine the time response of the
system.
(m, l)
θ
4k
Solution:
a) The displacement of each end of the bar in the ̂ direction is:
x± = ±
l sin θ
.
2
As a result, the total moment produced by the springs about the center of mass G is:
X
MG
l2
(4k sin θ + k sin θ) k̂,
4
5kl2
= −
sin θk̂.
4
= −
Thus angular momentum balance about G provides:
X
MG = I G θ̈k̂,
−
5kl2
sin θk̂
4
25
=
ml2
θ̈k̂.
12
For small angular displacements sin θ ∼ θ, and the governing equations of motion are
therefore:
15k
θ̈ +
θ = 0.
m
b) This system is undamped. Therefore the frequency of the oscillation is equal to the
undamped natural frequency:
r
15k
ω = ωn =
.
m
c) This system possesses the general solution:
θ(t) = c1 sin ωt + c2 cos ωt,
which, for the initial conditions given above, yields the solution:
r
15k
t.
θ(t) = θ0 cos
m
Problem 25:
We obtain the differential equation:
mẍ + bẋ + kx = 0,
as a model for a spring-mass-damper system with:
m = 2,
k = 18.
a) Identify the damping constant b that gives rise to critical damping;
b) If, instead, b = 24, approximately how long will it take for the the amplitude of free vibration
to be reduced to within 2% of zero?
Solution:
a) Written in nondimensional form, the equations of motion are:
ẍ +
k
b
ẋ + x = 0,
m
m
and so we identify the damping ratio and natural frequency as:
r
b
k
ζ= √
,
ωn =
.
m
2 km
ζ = 1 corresponds to critical damping, so that:
√
bcritical = 2 km = 12.
b) For b = 24 we find:
b
ζ= √
= 2,
2 km
26
ωn =
r
k
= 3.
m
Therefore the eigenvalues of this system are:
³
´
p
√
λ1,2 = ωn ζ ± ζ 2 − 1 = 3(−2 ± 3).
√
The dominant eigenvalue is λ = −6 + 3 3, and so the time t = τ required for the
amplitude of the free vibration to be reduced to within 2% of zero is:
τ∼
−4
√ ∼ 5.0
−6 + 3 3
Problem 26:
For the single-degree-of-freedom mechanical
system shown in the figure, the bar has mass
2
m and length l (so that IO = ml3 ). If the
spring is unstretched when θ = 0:
k
a) find the linearized governing equations
of motion for small θ;
ℓ
2
ℓ
2
θ
(m, l)
b) find the frequency of oscillation for the
free response;
c
Neglect gravity.
Solution:
a) Using angular momentum balance about the fixed point O, we find:
X
MO = IO αβ/F F,
µ
¶
l2
ml2
−kl2 sin θ − c θ̇ cos θ k̂ =
θ̈k̂,
4
3
so that the equation of motion can be written as:
ml2
l2
θ̈ + c θ̇ cos θ + k l2 sin θ = 0.
3
4
Linearizing this equation about θ = 0, we obtain:
θ̈ +
3k
3c
θ̇ +
θ = 0.
4m
m
b) The frequency of oscillation, that is, the damped natural frequency, is given as:
p
ωd = ωn 1 − ζ 2 ,
where ωn is the undamped natural frequency and ζ
we find:
r
3k
,
ζ=
ωn =
m
27
is the damping ratio. For this system
√
3c
√
,
8 km
so that the damped natural frequency is:
s
¶2
µ
3k
3c
ωd =
.
−
m
8m
Problem 27:
In the figure, the disk has mass m, radius r,
2
and moment of inertia IG = mr
about the
2
mass center G.
g
a) Find the equations of motion for this
system assuming that the disk rolls
without slip.
x
c
b) If the disk is released from rest with
initial displacement x(0) = x0 , find the
minimum value of the coefficient of friction for which the disk does not slip.
G
(m, r)
k
Problem 28:
For the spring-mass-damper system shown to
the right, x is measured from the static equilibrium position. If the surface is assumed to
be frictionless:
x
a) determine the governing equations of
motion;
m
6k
b
b) what is the period of each oscillation;
c) what value of the damping coefficient b
corresponds to critical damping?
d) if k = 1 N/m and m = 4 kg, find the
displacement of the mass x(t) if the system is critically damped and started
with the initial conditions x(0) = 0,
ẋ(0) = ẋ0 ;
Problem 29:
For the system shown to the right, x is measured from the unstretched position of the
spring. Each block has mass m and the disk
has moment of inertia I. If the gravitational
constant is g:
x
(I, r)
k
m
a) what is the displacement of the spring
at static equilibrium;
b) find the kinetic energy of the system in
terms of the coordinate x (and/or its
velocity).
m
28
Problem 30:
In the system shown to the right, the pulley
has mass m and radius r, so that the moment
2
of inertia about the mass center is IG = mr
2 .
r
a) Find the governing equations of motion;
r
2
b) What is the equivalent mass of the system;
m
c) Find the frequency of oscillation for free
vibrations of the system?
m
k
c
k
Solution:
a) Define θ as the angular displacement of the disk in the −k̂ direction (clockwise), and
x1 and x2 as the displacements of the two blocks so that:
x1 =
r
θ,
2
x2 = rθ.
We define the tension in the left and right cable as T1 and T2 respectively. Thus, linear
momentum balance on the two blocks, and angular momentum balance on the disk
yield:
r
− T2 r
2
T1 − kx1 − cẋ1
T2 + kx2
T1
mr2
θ̈,
2
= mẍ1 ,
= −mẍ2 .
= −
Eliminating the two unknown tensions from these three equations, we find that the
equation of motion (on θ) can be reduced to:
¶
¶
µ 2¶
µ
µ
cr
5kr2
7mr2
θ = 0.
θ̈ +
θ̇ +
4
4
4
2
b) Examination of the above equation shows that meq = 7mr
4 . Note that this answer is
not unique, but depends on the equation of motion. For example, if we had written the
equation of motion as 7m θ̈ + c θ̇ + k θ = 0, the equivalent mass would be meq = 7m.
c) The frequency of oscillation is given by ωd , which reduces to:
r
r
p
c2
5k
1−
ωd = ωn 1 − ζ 2 =
.
7m
140km
29
Problem 31:
In the spring-mass-damper system shown,
the block slides with no friction. With m =
2 kg, k = 18 N/m, b = 13 (N · s)/m:
x
a) Find the resulting solution if the system is released from the unstretched
position with initial velocity ẋ(0) =
−10.0 m/s
b
m
k
b) Identify the damping constant b that
gives rise to critical damping;
Solution:
For this system the differential equation of motion can be written as:
mẍ + bẋ + kx = 0,
k
b
ẍ + ẋ + x = 0,
m
m
subject to specified initial conditions.
a) With the given values for the mass, stiffness, and damping coefficient, the above equation
becomes:
ẍ + (6.5 kg/s)ẋ + (9 kg/s2 )x = 0,
whose characteristic equation reduces to:
λ2 +
13
λ + 9 = 0,
2
which has two real solutions of the form:
λ=
−13 ± 5
.
4
Thus the system has two purely real eigenvalues and the resulting system is overdamped
and decays exponentially with no sustained oscillations. The general solution is given
as:
9
x(t) = c1 e− 2 t + c2 e−2t .
For the initial conditions x(0) = 0 m, ẋ(0) = 10.0 m/s, we find that:
c1 = 4 m,
c2 = −4 m,
and the general solution to this equation, subject to these initial conditions, becomes:
´
³
´ ³
−1
−1
x(t) = 4 m
e−(9/2 s )t − e−(2 s )t .
b) For this system, the damping ratio is:
b
.
ζ= √
2 km
30
Thus, for a critically damped system ζ = 1, and solving for b with m = 2 kg and
k = 18 N/m, we find:
bcr = 12 (N · s)/m.
Problem 32:
In the figure, the disk has mass m, radius r,
2
and moment of inertia IG = mr
about the
2
mass center G.
g
a) Find the equations of motion for this
system assuming that the disk rolls
without slip.
x
b
b) What value of b correspond to critical
damping?
G
(m, r)
k
c) Find the displacement of the center of
the disk when the system is critically
damped and released from rest with
x(0) = x0 .
Solution:
We define θ as the angular displacement of the disk in the k̂ direction as measured
from the equilibrium position of the disk (when the spring is unstretched). Assuming
that the disk rolls without slip, the rotation and translation of the disk can be related
through the constraint equation:
x = −rθ.
a) The frictional force, which is unknown, is defined as f = fı̂, while the forces due to the
spring and damper are:
Fspring = −kx ı̂,
Fdamper = −bẋ ı̂.
Using linear and angular momentum balance on the disk, we find that:
X
F
F = (f − kx − bẋ) ı̂ = mẍ ı̂ = m aG ,
X
mr2
MG = (f r) k̂ =
θ̈ k̂ = IG αβ/F F.
2
Eliminating the unknown frictional force, and using the kinematic constraint, we find
the equation of motion is:
¶
µ
3m
ẍ + bẋ + kx = 0.
2
b) We can write this differential equation in standard form, that is:
r
ẍ + (2ζωn )ẋ + (ωn2 )x = 0,
with
ωn =
ζ=√
31
2k
,
3m
b
.
6km
If the system is critically damped, then this implies that the damping ratio is unity.
Therefore, solving for b when ζ = 1, we find:
√
bcritical = 6km.
c) When the system is critically damped, the general solution takes the form:
¡
¢
x(t) = c1 + c2 t e−ωn t ,
while this can be differentiated with respect to time to obtain the velocity:
¡
¢
ẋ(t) = (c2 − ωn c1 ) − (ωn c2 ) t e−ωn t .
If the system is released from rest when a known initial displacement, then the initial
conditions are (x(0), ẋ) = (x0 , 0). Thus, returning these to the general solution, we find:
x0 = x(0)
0 = ẋ(0)
= c1 ,
= c2 − ωn c1 .
Solving for c1 and c2 , the solution to these initial conditions becomes:
¡
¢
x(t) = x0 1 + ωn t e−ωn t ,
where recall that ωn =
p
(2k)/(3m).
Problem 33:
In the figure, the disk has mass m, radius r,
and moment of inertia IG about the mass center, and the applied moment has a constant
magnitude M k̂. If the disk rolls without slip
(µ is sufficiently large):
g
x
a) determine the governing equations of
motion;
G
b) what are the equivalent mass, stiffness,
and damping of the system;
(m, r)
k
c) what is the stretch in the spring when
the system is in equilibrium?
Solution:
a) The governing equations of motion are:
µ
¶
I
M
m + 2 ẍ + cẋ + kx = − .
r
r
b) The equivalent mass, damping, and stiffness are:
meq = m +
M k̂
c
I
,
r2
32
ceq = c,
keq = k.
c) When the system is in equilibrium, the displacement of the disk is:
xeq = −
2
M
.
kr
Frictionally Damped Systems
Problem 34:
The block shown to the right rests on a rough
surface with coefficient of friction µ and
m = 6 kg,
̂
k = 128 N/m.
ı̂
a) If the block is displaced 3 cm to the
right and released, for what values of µ
will the block remain in that position?
g
x
z
A
b) With µ = 0.50, if the block is displaced
30 cm to the right and released from
rest, how long will it take the block to
come to rest?
B
k
k
m
µ
Solution:
In addition to the variable x identified in the problem statement, we also define z to
be the stretch in the spring parallel with the cable system. As a one degree-of-freedom
system, the variables x and z are directly related. The relative velocity across the spring
can be identified as
F
F
vB − vA
= ż ı̂,
= (−ẋ ı̂) − (ẋ ı̂) ,
so that ż = −2 ẋ. Therefore the kinematic
relationship becomes
z = −2 x.
−m g ̂
T ı̂
−k x ı̂
An appropriate free-body diagram for this
system is shown to the right. Note that the
unknown friction force is denoted as fr ı̂ and
the tension in the cable is T . Finally, examining the spring in the cable, the tension T
and the displacement z are related as
T ı̂
fr ı̂
N ̂
−T ı̂
T = k z = −2 k x.
Applying linear momentum balance to the block yields
X
F
F = (2 T − k x + fr ) ı̂ + (N − m g) ̂ = m ẍ ı̂ = m aG ,
and in terms of x the equation of motion becomes
m ẍ + 5 k x = fr .
33
T ı̂
If the block slips then fr = −µ m g sgn(ẋ) while is sticking occurs |fr | ≤ µ m g.
a) If the block is in static equilibrium at a displacement x = xeq , then ẍeq ≡ 0 and the
equation of motion reduces to
5 k xeq = fr ,
so that equilibrium is maintained provided
|fr | = |5 k xeq | ≤ µ m g.
This inequality is satisfied provided
|xeq | ≤
µmg
.
5k
Problem 35:
For the system shown to the right, the block
slides on a rough surface (coefficient of friction µ) inclined at an angle of φ with respect
to vertical. If the block is subject to a periodic force of the form
g
k
F (t) = F0 sin(ω t),
a) find the equations of motion. Do not
neglect gravity;
φ
b) find the amplitude of the steady-state
response using Mc when
m = 1.25 kg, k = 20 N/m,
φ = 30◦ , µ = 0.125,
F0 = 4 N,
ω = 2.00 rad/s,
34
m
F (t)
Problem 36:
[(Spring 2003)] The system shown in the figure has mass m and rests on a plane inclined
at an angle φ. The coefficient of friction for
the rough surface is µ and the system is released from rest at the unstretched position
of the spring (with stiffness k).
x
k
a) If µ = 0, what is the equilibrium displacement of the mass (as measured
from the unstretched position)?
̂2
µ
ı̂2
m
̂
θ
b) For µ > 0, at what angle θ does the
block begin to slip?
ı̂
c) Find the value of θ so that the system
comes to rest after one full cycle exactly
at the equilibrium position of the system found in part a (so that the friction
force vanishes when the system comes
to rest), with:
m = 2 kg,
k = 32 N/m,
µ = 0.35,
Solution:
The unit directions ı̂2 and ̂2 are defined to be coincident with the inclined plane and
the coordinate x represents the displacement of the mass from the unstretched position
of the spring, as shown in the figure.
A free-body diagram for this system is
shown to the right. Notice that the force in
the upper spring depends on z, rather than
x, while the friction force has an unknown
magnitude f . Because the disk is assumed
to roll without slip, we are unable to specify
the value of f , but instead can relate the displacement and rotation of the disk through
the coordinate relations above.
−k x ı̂2
N ̂2
−m g ̂
f ı̂2
Therefore, linear momentum balance yields the following equations:
X
F
F = m aG ,
³
´
³ ´
³
´
f − k x ı̂2 + N ̂2 − m g ̂ = m ẍ ı̂2 ,
³
´
³
´
f − k x + m g sin θ ı̂2 + N − m g cos θ ̂2 = m ẍ ı̂2 .
Therefore, this leads to the following scalar equations in the ı̂2 and ̂2 directions:
m ẍ + k x = f + m g sin θ,
N = m g cos θ.
a) If µ = 0, then the friction forces vanishes and the first of the above equations reduces
35
to:
m ẍ + k x = m g sin θ.
The equilibrium displacement of the mass, xeq , then can be found to be:
xeq =
mg
sin θ.
k
b) With µ 6= 0, an equilibrium state is maintained provided:
¯ ¯
¯ ¯
¯f ¯ ≤ µ N,
¯
¯
¯
¯
¯k x − m g sin θ¯ ≤ µ m g cos θ.
Therefore, if the system is released from x = 0, the block begins to slide when:
tan θ = µ.
c) Define z to be a new coordinate measuring the displacement of the system from static
equilibrium:
mg
z =x−
sin θ,
k
so that the equations of motion become:
m z̈ + k z = f,
N = m g cos θ.
with initial displacement z(0) = −(m g sin θ)/k. Over one complete cycle of motion,
for a frictionally damped system the amplitude decreases by a value:
∆A=−
4µN
.
k
Therefore, if the system comes to rest at exactly the equilibrium position, then this
decrease in amplitude must exactly match the initial displacement. That is:
¯
¯
¯
¯ 4 µ N ¯ ¯¯ m g
¯
¯ = ¯−
|∆ A| = ¯¯−
sin
θ
¯ = |z(0)|.
k ¯
k
Solving for θ:
tan θ = 4 µ.
Problem 37:
The spring mass system rests on a surface
with coefficient of friction µ and x is measured from the unstretched position of the
spring.. If the initial conditions of the system are chosen to be x(0) = 0 and ẋ(0) = ẋ0 ,
find the range of ẋ0 so that the system comes
to rest after exactly one cycle of motion.
x
g
m
k
µ
36
Problem 38:
For the spring-mass system with Coulomb
damping:
x
a) determine the governing equations of
motion;
m
b) what is the period of each oscillation.
k
k
µ
Solution:
a) We measure the displacement of the mass from the static equilibrium of the frictionless
F
system, i.e., µ = 0, so that the acceleration of the block is aG = ẍı̂. Thus linear
momentum balance yields:
mẍı̂ = Fspring ı̂ + fµ ı̂ + (N − mg)̂.
The spring force is Fspring = −2kx, while the force due to sliding friction opposes the
velocity and is simply:
ẋ
fµ = −µmg ,
|ẋ|
since the normal force balances the gravitational force, i.e., N = mg. If the block
is stationary the magnitude of the frictional force is less than µmg. Therefore, the
governing equations of motion are:
mẍ + 2kx = fµ ,
with
fµ
|fµ |
= −µmg |ẋẋ| |ẋ| =
6 0,
≤ µmg
ẋ = 0.
b) Coulombic damping does not effect the frequency of oscillation, which is simply:
r
2k
.
ω=
m
Therefore the period of the oscillation is:
T =
37
√
2π
2π m
.
= √
ω
2k
Problem 39:
For the spring-mass-damper system shown to
the right, x is measured from the static equilibrium position. If the coefficient of friction
is µ:
x
a) determine the governing equations of
motion;
m
6k
2k
b) what is the period of each oscillation;
µ
c) if the system is released from rest with
x(0) = x0 > 0, what is the minimum
value of x0 so that the block slips;
d) find the range of initial displacements
so that the system comes to rest after
one complete cycle.
Problem 40:
For the spring-mass system with Coulombic
damping, x is measured from the unstretched
position of the spring. If the coefficient of
friction is µ and the gravitational constant is
g:
x
I=0
k
m
a) determine the governing equations of
motion;
µ
b) if the system is released from rest in the
unstretched position (x(0) = 0, ẋ(0) =
0), for what values of µ will the system
move;
m
c) what is the displacement (from the unstretched position) of the upper block
when it first comes to rest?
Solution:
Notice that x describes the displacement of both masses and, since the pulley is massless,
the tension in the string connecting the masses is constant, say T . Also, notice that in
part c we ask for the displacement from the unstretched position of the spring, rather
than from equilibrium. Therefore we include the gravitational force which will influence
this result.
a) With the frictional force defined as F = f ı̂, linear momentum balance on the upper
and lower block yields:
mẍ + kx = f + T,
mẍ = mg − T,
where the frictional force is defined as:
ẋ
,
f = −µmg |ẋ|
|f | ≤ µmg,
38
ẋ 6= 0,
ẋ = 0.
Eliminating the unknown tension T , the equation of motion is given as:
2m ẍ + kx = f + mg,
where f is defined as above and depends on the motion of the system, that is, the value
of ẋ.
b) If the system is released from rest in the unstretched position, it will remain there
provided the magnitude of the frictional force is less than µmg—the transition to movement occurs when |f | = µmg. Thus the system does not move is |f | ≤ µmg and ẍ = 0.
Substituting these conditions into the equations of motion we find:
|f | = |kx − mg| ≤ µmg,
which, solving for µ with x(0) = 0, implies that the system does not move if µ ≥ 1.
Therefore, the system does move when:
µ < 1.
c) The displacement of the upper block when it first comes to rest is:
x1 =
2(1 − µ)mg
.
k
This can be found by either solving the equations of motion explicitly, or through a
work-energy analysis. Since the initial and final kinetic energy is zero, the work done
by the frictional force balances out the change in potential energy from the spring and
gravity.
Problem 41:
For the spring-mass system with Coulombic
damping, x is measured from the unstretched
position of the spring. If the coefficient of
friction is µ and the gravitational constant is
g:
x
g
m
a) determine the governing equations of
motion;
k
b) if the system is released from rest, so
that ẋ(0) = 0, for what range of initial
displacements (from the unstretched
position) will the block come to rest
when the block first comes again to rest
(ẋ(t1 ) = 0 for t1 > 0)?
µ
Solution:
a) The equations of motion can be written as:
mẍ + kx = f,
39
where f is the force due to friction, modeled by Coulomb’s law of friction as:
f
= −µmg
|f |
≤ µmg,
ẋ
,
|ẋ|
ẋ 6= 0,
ẋ = 0.
b) If the system is released from rest, the initial displacement must be sufficiently large so
that the block slides, rather than remaining at rest. Sliding does not occur if the force
due to friction is sufficient to balance the elastic force, that is, µmg ≥ f = kx(0). Thus,
solving for x(0) we find, that for sliding to occur:
|x(0)| >
µmg
.
k
However, if |x(0)| is too large, the system will undergo multiple reversals as the amplitude of the motion decays. Consider the block sliding to the left (ẋ < 0), released from
rest with initial displacement x(0) > µmg
k . Thus the equation of motion becomes:
mẍ + kx = µmg,
which has the general solution:


µ ¶
µmg
k
µmg


x(t) = x(0) −
cos
t +
.
k
m
k
Therefore, when the block comes again to rest at time t1 (unknown), the mass is at the
position:
µmg
x(t1 ) = 2
− x(0).
k
At this point, the block sticks if and only if |x(t1 )| ≤
we find that:
µmg
x(0) ≤ 3
.
k
µmg
k .
Therefore, solving for x(0),
< x(0) ≤ 3 µmg
So for a block with x(0) > 0, the allowable range for x(0) is µmg
k
k .
Together with an identical argument for x(0) < 0 yields the total allowable range as:
µmg
µmg
< |x(0)| ≤ 3
.
k
k
40
Problem 42:
For the system shown to the right, x is measured from the unstretched position of the
spring. Each block has mass m and the disk
has moment of inertia I and radius r. The
coefficient of friction between the upper block
and the table is µ. If the gravitational constant is g:
x
(I, r)
k
m
µ
a) find the equations of motion which determine x(t);
b) what is the minimum value of µ so that
the system slips when release from rest
with x(0) = 0;
m
c) what is the period of the free oscillations?
d) if the system is released from rest, what
is the range of initial displacements
x(0) so that the systems comes to rest
after exactly one complete cycle?
Solution:
We begin by defining two additional coordinates, θ, which describes the rotation of
the disk in the −k̂ direction (clockwise), and y which measures the displacement of
the hanging mass in the −̂ direction. These additional coordinates are related to the
displacement of the upper mass by the constraint equations:
y = x,
θ=
x
.
r
a) On each mass the equations of motion can be written as:
³
´
³
´
X
F
F = − k x + T1 + f ı̂ + N − mg ̂ = mẍ ı̂ = m aG1 ,
³
´
X
MO = T1 r − T2 r k̂ = −I O θ̈k̂ = I O αD/F F,
³
´
X
F
F = T2 − mg ̂ = −mÿ ı̂ = m aG2 ,
Notice that I O = I 6= 0, so that provided θ̈ 6= 0 the tensions T1 and T2 are not equal.
Taking the components of these equations and eliminating the unknowns (T1 , T2 ), while
using the constraint equations, we find that the equation of motion for this system
reduces to:
¶
µ
I
2m + 2 ẍ + k x = f + mg,
r
where f the value of the frictional force in the ı̂ direction, can be written as:
½
ẋ
−µmg |ẋ|
,
ẋ 6= 0
f=
f0 , |f0 | ≤ µmg, ẋ = 0.
41
b) The minimum value for slip is simply µmin = 0. If we would like to find the range of
µ for which slip occurs, we resort to the value of f at static equilibrium. Assuming
(ẋ, ẍ) = (0, 0), the equations of motion reduce to:
kx = fstatic + mg,
where fstatic represents the force required to maintain static equilibrium. Solving for
this quantity and using the frictional inequality, we find:
|fstatic | = |kx0 − mg| ≤ µmg.
Therefore, solving for µ yields:
¯
¯
¯ ¯
¯ kx0
¯ ¯
kx0 ¯¯
µ ≥ ¯¯
,
− 1¯¯ = ¯¯1 −
mg
mg ¯
which provides a necessary condition for sticking at x = x0 . So for sliding to occur for
x0 = 0, this implies that µ < 1.
c) The period of oscillation for a frictionally damped system is identical to that of an
undamped system. Therefore:
s
2m + rI2
2π
T =
= 2π
.
ωn
k
d) Let δ describe the displacement of the system from equilibrium. The amplitude of
oscillation will decay by a value of ∆ = 4µmg/k over one cycle of motion. Therefore,
Since the system will come to rest within the range:
−
µmg
µmg
< |δfinal | ≤
,
k
k
the initial displacement δ0 from the equilibrium in the absence of friction must be in
the range:
µmg
3µmg
5µmg
µmg
−
+∆=
< |δ0 | ≤
=
+ ∆.
k
k
k
k
However, the equilibrium position corresponds to xeq = mg
k , and so the allowable range
of x0 is:
mg ¯¯ 5µmg
3µmg ¯¯
< ¯x0 −
.
¯≤
k
k
k
3
Forced Single Degree-of-freedom Systems
42
Problem 43:
For the system shown to the right the bar
of length ℓ has mass m and is subject to a
time-dependent moment of the form
M (t) = M0 sin(ω t) k̂.
M (t)
k
G
a) Find the equations of motion for this
rotation of the bar.
(m, ℓ)
b) What is the steady-state amplitude of
the forced response, with
k
m = 3 kg,
b = 16 N/(m/s),
1
k = 8 N/m, ℓ = m,
4
1
ω = 4 rad/s, M0 =
N·m
16
Problem 44:
The block of mass m = 20 kg shown to the
right rests on a rigid foundation and is subject to a time-dependent load
f (t) ̂
F (t) = f0 sin(ω t) ̂.
a) Design an undamped foundation to
achieve isolation ≥ 33% for all forcing
frequencies ω > 3 π rad/s;
m
b) If, for the isolator that you designed,
the damping ratio was measured to be
ζ = 0.125 (rather than ζ = 0 as assumed above), what is the minimum
isolation achieved over this frequency
range?
43
b
Problem 45:
In the figure shown to the right, the disk is
subject to a time dependent moment of the
form
M (t) = M0 sin(ω t).
k
M (t) k̂
r
a) Find the equations of motion for the
angular displacement of the disk.
r/2
I
b) With
k = 280 N/m, b = 12 N/(m/s),
m = 4 kg,
I = 0.40 kg · m2 , r = 0.10 m
M0 = 3 N · m, ω = 5 rad/s,
m
Determine the steady-state response of
disk as a function of time.
k
Problem 46:
The unbalanced rotor shown in the figure is
pinned to a frame and supported by a spring
and damper. The total mass of the structure
is m, with a small rotating component (10%
of the total mass) offset by a distance r from
the center of rotation C.
k
a) Find the distance between the center
of mass of the system and the center of
rotation;
r
C
b) what is the damped natural frequency;
m
c) determine the steady-state amplitude
of vibration when the rotor spins at an
angular speed of ω = 5 rad/s with:
k = 256 N/m,
m = 5 kg,
b
b = 12 N/(m/s),
r = 10 cm
44
b
Problem 47:
In the figure, the disk has mass m, radius r,
2
and moment of inertia IG = mr
about the
2
mass center G and is assumed to roll without
slip. The attached plate undergoes harmonic
motion of the form
g
u(t)
c
u(t) = u0 sin(ω t).
G
a) Find the equation of motion in terms of
the displacement between the moving
plate and the center of the disk;
(m, r)
k
b) What are the damping ratio and natural frequency for this system?
c) If the system is critically damped, find
the amplitude of the relative displacement of the disk for
m = 3 kg,
k = 36 N/m,
r = 0.10 m
c = 3 N/(m/s)
u0 = 0.05 m,
ω = 5 rad/s
Problem 48:
For the mechanical system shown to the
right, the uniform rigid bar is massless and
pinned at point O while a force is applied at
A of the form
ℓ
3
k
O
f (t) = t e−σ t .
θ
A
For this system:
2ℓ
3
a) find the equations of motion;
b) Identify the damping ratio and natural
frequency in terms of the parameters
m, c, k, and ℓ.
f (t)
c) With
m = 2 kg,
ℓ = 30 cm,
c = 0.25 N/(m/s),
σ = 2.00 s
k = 50 N/m,
−1
B
,
find the convolution integral for the response of the system. You need not
evaluate the integral.
Assume that the system is in static equilibrium at θ = 0, and that all angles remain
small.
45
m
k
c
Problem 49:
The block shown to the right rests on a rough
surface with coefficient of friction µ (assume
that any cables can support compression and
tension). Find the amplitude of the vibrations of the block if f (t) = f0 cos(ω t), with
m = 4.0 kg,
x
k
m
k = 64 N/m,
µ = 0.05,
f0 = 40 N,
ω = 4 rad/s.
m
−f (t) ̂
Problem 50:
The unbalanced rotor shown in the figure is
pinned to a frame and supported by a spring
and damper. The total mass of the structure
is m, with a small rotating component (10%
of the total mass) offset by a distance r from
the center of rotation C.
k
b
a) find the damped natural frequency;
r
b) what is the steady-state amplitude of
vibration when the rotor spins at an angular speed of ω = 5 rad/s with:
C
k = 108 N/m,
m = 3 kg,
m
b = 9 N/(m/s),
r = 7.5 cm
46
Problem 51:
For the mechanical system shown to the
right, the uniform rigid bar has mass m and
length ℓ, and is pinned at point O. A harmonic force is applied at A. For this system:
O
k
a) find the equations of motion;
θ
B
b) Identify the damping ratio and natural
frequency in terms of the parameters
m, c, k, and ℓ.
ℓ
2
A
c) For:
4m
m = 6 kg, ℓ = 25 cm,
c = 0.50 N/(m/s), k = 80 N/m,
f0 = 2.00 N, ω = 10 rad/s,
f0 sin(ω t)
c
find the steady-state amplitude of the
displacement of the block.
Assume that the system is in static equilibrium at θ = 0, and that all angles remain
small.
Problem 52:
In the figure, the disk has mass m, radius r,
2
and moment of inertia IG = mr
about the
2
mass center G and is assumed to roll without
slip. The attached plate undergoes harmonic
motion of the form
g
u(t)
c
u(t) = u0 sin(ω t).
G
a) Find the equation of motion in terms of
the angular rotation of the disk;
k
b) What are the damping ratio and natural frequency for this system?
c) If the system is critically damped, find
the amplitude of the rotation of the disk
for
m = 3 kg,
u0 = 10 cm,
k = 36 N/m,
ω = 3 rad/s
47
(m, r)
Problem 53:
The unbalanced rotor shown in the figure is
pinned to a frame and supported by a spring
and damper. The total mass is measured as
m = 200 kg. When the system is operated at
ω = 25 rad/s the phase φ of the response with
respect to the rotation of the unbalanced disk
is measured to be π/2 rad and the steadystate vibration amplitude is X = 2.00 cm.
When the rotation rate of the disk is much
larger than this value the amplitude reduces
to X = 0.50 cm.
Find the stiffness and damping constant for
the foundation and the distance between the
center of rotation C and the mass center G.
k
b
ε
G
C
m
Problem 54:
For the mechanical system shown to the
right, the uniform rigid bar is massless and
pinned at point O while a harmonic force is
applied at A. For this system:
ℓ
3
k
a) find the equations of motion;
b) Identify the damping ratio and natural
frequency in terms of the parameters
m, c, k, and ℓ.
A
B
O
θ
2ℓ
3
m
c) For:
f0 sin(ω t)
m = 2 kg, ℓ = 30 cm,
c = 0.25 N/(m/s), k = 50 N/m,
f0 = 2.00 N,
ω = 10 rad/s,
find the steady-state displacement of
the block.
d) What is the magnitude of the force
transmitted to the ground through the
sping and damper attached to the
block? (Do not include the spring attached at A.)
Assume that the system is in static equilibrium at θ = 0, and that all angles remain
small.
48
k
c
Problem 55:
The block shown to the right rests on a rough
surface with coefficient of friction µ and the
block is subject to a compressive force of N =
20 N (do not include gravity, just this normal
load and assume that any cables can support
compression and tension).
x
k
a) If f (t) = f0 = constant, find the range
of initial displacements for which the
block will remain stationary if released
from rest (it will stick).
m
(m, r)
−f (t) ı̂
b) Find the amplitude of the vibrations of
the block if f (t) = f0 cos(ω t), with
m = 4.0 kg,
r = 12.5 cm,
f0 = 40 N,
k = 64 N/m,
µ = 0.50,
ω = 4 rad/s.
Problem 56:
For the system shown to the right, the disk of
mass m rolls without slip and x measures the
displacement of the disk from the unstretched
position of the spring. The surface is inclined
at an angle of φ with respect to vertical.
k
x
a) find the equations of motion. Do not
neglect gravity;
b) what is the static equilibrium displacement of the disk?
φ
c) if the disk is subject to a periodic moment
M (t) = M0 sin(ω t)k̂,
find the displacement of the center of
the disk if it is released from rest at the
unstretched position of the spring.
49
(m, r)
M (t)
Problem 57: (Spring 2003)
For the system shown to the right, the disk of
mass m rolls without slip and the inner hub
has radius ρ/2.
x
̂
z
2k
a) Find the equations of motion (in terms
of the given parameters—do not substitute in numerical values yet);
ı̂
θ
G
b
b) If the applied moment takes the form:
k
M (t)
(m, ρ)
M (t) = (2 N · m) sin(4 t),
C
find the steady-state amplitude of the
translation of the center of the disk
when:
k = 16 N/m,
m = 2 kg,
b = 2 N/(m/s),
ρ = 0.125 m
c) Determine the steady state amplitude
of the friction force.
Solution:
a) We identify the three coordinates x, z, and
θ as shown in the figure above. These are
related as:
x = −ρ θ,
3
z = x.
2
µ
ρ (k x + b ẋ) +
M (t) k̂
−2 k z ı̂
An appropriate free-body diagram for this
system is shown to the right. Since the disk
is assumed to roll without slip, the equation
of motion can be directly obtained with angular momentum balance about the contact
point C
X
MC = I C αD/F ,
which yields
−m g ̂
−b ẋ ı̂
−k x ı̂
fr ı̂
N ̂
¶
3 m ρ2
3ρ
θ̈ k̂.
(2 k z) + M (t) k̂ =
2
2
Using the above coordinate transformations this equation can be written as
3m
11 k
M (t)
ẍ + b ẋ +
x=−
.
2
2
ρ
b) For the numerical values given above (with consistent units), this equation reduces to
3 ẍ + 2 ẋ + 88 x = −8 sin(4 t),
50
from which we can identify the appropriate parameters as:
r
r
88
6
1
ωn =
,
r=
,
ζ=√
.
3
11
264
Therefore, the amplitude of the translational oscillations becomes
X=
8
F
r
M(r, ζ) =
k
88 ¡
1
1
=√ .
q
´
³
2
¢
26
6 2
6
1
1 − 11
+ 2 √264
11
Likewise, the phase shift of the response is:
tan φ =
2ζ r
=
1 − r2
1
2 √264
1−
q
6
11
6
11
=
1
,
5
so that φ = 0.20 rad = 11.3◦ .
c) In the development of the equation of motion, the friction force was eliminated by
summing moments about C. Using linear momentum balance we can reintroduce the
friction force as
³
´
³
´
X
F
F = fr − k x − 2 k z − b ẋ ı̂ + N − m g = m ẍ ı̂ = m aG .
Therefore, solving for fr we find that
fr = m ẍ + b ẋ + 4 k x.
With x(t) represented as x(t) = X sin(ω t − φ), where X and φ are found above, the
friction force becomes
³
´
³ ´
fr (t) = 4 k − m ω 2 X sin(ω t − φ) + b ω X cos(ω t − φ).
The magnitude of the friction force is then found to be
r³
° °
´2 ³ ´2
° °
4 k − m ω2 + b ω .
°f ° = X
For these parameter values kf k = 6.47 N.
51
Problem 58: (Spring 2003)
The unbalanced rotor shown in the figure is
pinned to a frame and supported by a spring
and damper. If the total mass is m while the
mass center G is located at an eccentricity of
ε from the the center of rotation O,
k
b
a) find the damped natural frequency;
b) what is the steady-state amplitude of
vibration when the rotor spins at an angular speed of ω = 8 π rad with:
k = 32 N/m,
m = 4 kg,
ε
G
O
b = 16 N/(m/s),
ε = 2.5 cm
m
c) If the system is undamped (i.e., b =
0 N/(m/s)) for what range of operating speeds (ω) will the amplitude of the
force transmitted to the ground FT be
less than 1 N.
Problem 59: (Spring 2003)
For the mechanical system shown to the
right, the uniform rigid bar has mass m and
pinned at point O. For this system:
a) find the equations of motion (in terms
of the given parameters—do not substitute in numerical values yet);
f (t) = sin(ω t)
ℓ
2
b) if c = 0.25 N/(m/s), k = 32 N/m,
m = 2 kg, and ℓ = 0.25 m, find the
amplitude of the force transmitted to
the ground through combination of the
spring and damper when ω = 4 rad/s.
k
θ
m
z
ℓ
2
̂
c
c) if c = 0, m = 2 kg, and ℓ = 0.25 m,
find the value of the stiffness k so that
the bar’s amplitude of oscillation is less
than π/6 rad for all forcing frequencies
greater than 20 rad/s.
ı̂
Solution:
a) In addition to θ, we define the additional coordinate z, which measures the deflection
at the left end of the bar, with θ and z related as:
52
ℓ
θ
2
A free-body diagram for this system is
shown to the right. Applying angular momentum balance on the bar eliminates the
appearance of the reaction force and leads
to:
X
MG = I G θ̈ k̂,
³
´ℓ
m ℓ2
k̂ =
θ̈ k̂.
f (t) − k z − cż
2
12
z=
f (t) ̂
k z1 ̂
G
FR
c ż1 ̂
Solving for z in terms of θ, the equation of motion becomes:
c ℓ2
k ℓ2
ℓ
m ℓ2
θ̈ +
θ̇ +
θ = f (t).
12
4
4
2
b) In standard form, this equation of motion can be written as:
¶
¶
µ
¶
µ
µ
6
3k
3c
θ=
sin(ω t),
θ̇ +
θ̈ +
m
m
mℓ
so that:
ωn =
r
3k
,
m
√
3c
,
ζ= √
2 km
M0 =
6
.
mℓ
The amplitude of the moment transmitted to the ground can be written as:
p
1 + (2 ζ r)2
,
MT = (meq M0 ) p
(1 − r2 )2 + (2 ζ r)2
q
¡ ¢2
p
1 + c kω
k 2 + (c ω)2
ℓ
ℓ
q¡
q¡
=
=
.
¢
¡
¢
2
2 2
2 ¢2
2
2
2
1 − m3 ωk
k − m 3ω
+ c kω
+ (c ω)
The amplitude of the force transmitted to the ground is then FT = MT /(ℓ/2), or:
p
k 2 + (c ω)2
.
F T = q¡
2 ¢2
2
+ (c ω)
k − m 3ω
Substituting in the numerical values given in the problem statement, we find that:
FT = 1.50 N
c) The amplitude of the steady-state vibrations can be written as:
Θ =
=
M0
1
p
,
2
2
ωn2
(1 − r ) + (2 ζ r)2
1
2
2
q¡
q¡
=
¢
¡
¢
2
2
2
kℓ
ℓ
c ω
m ω
+ k
1− 3 k
k−
53
1
¢
m ω2 2
3
2
+ (c ω)
.
Substituting in the numerical values given in the problem statement, we find that:
Θ = ¯¡
¯ k−
8
2 ω2
3
¢¯
¯
Therefore, if the amplitude of vibration is less than π/6:
8
¯
¯k −
2 ω2 ¯
3
¯
48
π
This inequality has two solutions:
k≥
π
,
6
¯
¯
¯
2 ω 2 ¯¯
.
≤ ¯¯k −
3 ¯
≤
48 2 ω 2
+
,
π
3
k≤
2 ω2
48
− .
3
π
Since this condition must be satisfied for all ω ≥ 20 rad/s, we take the second inequality
and find that:
48
2 (20)2
−
= 251.
k≤
3
π
Problem 60: (Spring 2003)
For the system shown to the right, the disk of
mass m rolls without slip and x measures the
displacement of the disk from the unstretched
position of the spring.
̂
x
3k z
ı̂
a) find the equations of motion;
θ
b) if the forcing takes the form:
½
f0 ,
0 ≤ t < t0 ,
f (t) =
f0 /2, t0 ≤ t,
k
f (t)
m
find the response of the system with
zero initial conditions.
Solution:
a) We define the three coordinates as shown as the figure, related as:
x = −r θ,
z = −2 r θ,
A free-body diagram for this system is
shown to the right. Notice that the force
in the upper spring depends on z, rather
than x, while the friction force has an unknown magnitude fr . Because the disk is
assumed to roll without slip, we are unable
to specify the value of fr , but instead can
relate the displacement and rotation of the
disk through the coordinate relations above.
54
z = 2 x.
−3k z ı̂
G
−k x ı̂
C
fr ı̂
f (t) ı̂
The equations of motions can be developed directly with angular momentum balance
about the contact point, so that:
X
MC = I C θ̈ k̂,
³
´
3 m r2
(3k z) 2r + (k x) r − f (t) r k̂ =
θ̈ k̂.
2
Finally, writing this equation in terms of a single coordinate, we obtain:
µ
¶
¡
¢
3 m r2
θ̈ + 13 k r2 θ = r f (t).
2
In standard form:
θ̈ +
µ
26 k
3m
¶
θ=
2 f (t)
.
3mr
b) We use the convolution integral to determine the response, so that:
θ(t) =
Z
0
t
F (τ ) h(t − τ ) dτ,
and for this system:
f (t)
2 f (t)
F (t) =
=
,
meq
3mr
1
sin(ωn t) =
h(t) =
ωn
r
3m
sin
26 k
Ãr
!
26 k
t .
3m
Because the forcing function changes abruptly at t = t0 , the solution must be written
separately for 0 < t ≤ t0 , and t > t0 :
Z
2 f0 t sin(ωn (t − τ ))
dτ,
0 < t ≤ t0 ,
x(t) =
3m 0
ωn
Z
Z t
f0
sin(ωn (t − τ ))
2 f0 t0 sin(ωn (t − τ ))
dτ +
dτ,
t > t0 .
x(t) =
3m 0
ωn
3 m t0
ωn
Evaluating these integrals, the solution becomes:
x(t) =
x(t) =
2 f0
26 k
f0
26 k
³
³
´
1 − cos(ωn t) ,
´
1 + cos(ωn (t − t0 )) − 2 cos(ωn t) ,
55
0 < t ≤ t0 ,
t > t0 .
Problem 61: (Spring 2003)
The system shown to the right is subject to
base excitation. Find the steady-state response of the system in terms of z, with:
m = 2.0 kg,
̂
m
b = 4.0 N/(m/s),
k1 = 3.00 N/m,
k1
k2 = 12.00 N/m.
ı̂
k1
z
b
x
k2
u(t) = 0.50 sin(2 t) m
Solution:
a) We define the addition coordinate x which measures the absolute displacement of the
mass with respect to the ground, so that:
x = z + u(t).
Notice that the collection of springs can be
replaced by a single equivalent spring, with:
keq =
1
2 k1
1
+
1
k2
=
m
2 k1 k2
= 4 N/m.
2 k1 + k2
keq
The new equivalent system is shown to the
right.
An appropriate free-body diagram is shown
to the right. In terms of the identified coordinates, the acceleration of the mass center
is:
F
aG = ẍ ̂ = (z̈ + ü) ̂,
z
u(t)
−keq z ̂
with ü(t) = −(u0 ω 2 ) sin(ω t).
Therefore, linear momentum balance on the mass yields:
X
F
F = m aG ,
³
´
− keq z − b ż ̂ = m ẍ ̂
Writing this in terms of z, the equation of motion is:
m z̈ + b ż + keq z = −m ü(t),
and in standard form:
z̈ + 2 ζ ωn ż + ωn2 z = u0 ω 2 sin(ω t),
56
b
−b ż ̂
with:
ωn =
r
b
ζ= p
.
2 keq m
keq
,
m
Therefore the steady state response of this system becomes:
z(t) = Z sin(ω t − ψ),
with Z = u0 Λ(r, ζ), and:
r2
Λ= p
,
(1 − r2 )2 + (2 ζ r)2
tan ψ =
2ζ r
,
1 − r2
and r =
ω
ωn
For the numerical values of this problem:
ωn =
so that:
√
1
ζ=√ ,
2
2,
2
Λ= √ ,
5
tan ψ =
r=
√
2,
2
−1
Recall that the phase shift ψ must be positive, so that is tan ψ is negative, then ψ is in
the second quadrant, so that ψ = 3.03 rad. Finally:
1
z(t) = √ sin(2 t − 2.03).
5
Problem 62:
In the figure, the disk has mass m, radius r,
2
about the
and moment of inertia IG = mr
2
mass center G, subject to an applied moment
of the form:
g
x
M (t) = M0 sin(ωt)k̂.
M (t)
c
a) If the coefficient of friction is µ, find the
condition which determines if the disk
rolls with or without slip, and find the
governing equations of motion when the
disk rolls with and without slip;
G
k
b) If µ is sufficiently large so that the
disk rolls without slipping, what are the
equivalent mass, stiffness, and damping
of the system;
c) If again the disk rolls without slipping,
find the amplitude of the steady-state
motion of G, that is, x(t), when the system is critically damped, with M0 =
4 N · m, m = 1 kg, r = 0.1 m, and
k = 2 N/m.
57
(m, r)
Problem 63:
In the figure, the disk has mass m, radius r,
and moment of inertia IG = α m about the
mass center G. The disk is subject to a timevarying force f (t) = 4 sin(ω t).
g
x
a) Find the equations of motion for this
system assuming that the disk rolls
without slip.
c
f (t)
G
b) After the transient solution decays, find
the amplitude of the force transmitted to the ground through the springdamper element.
(m, r)
k
c) For what value of the damping ratio is
this transmitted force less than twice
the applied load for all values of the
forcing frequency?
Problem 64:
The disk shown in the figure rolls without
slip and is subject to a time-varying moment
M (t) = sin(t).
g
a) Find the governing equations of motion;
x
M (t) k̂
b
b) Find the frequency of oscillation for the
unforced response, i.e. M (t) = 0;
G
c) What is the steady-state amplitude of
the forced response?
k
(m, r)
d) Determine the amplitude of the frictional force during the steady-state motion.
Solution:
a) We assume that (ı̂, ̂, k̂) represent the standard orthonormal basis, while the translational displacement of G is xı̂ and the angular displacement is θk̂. Linear and angular
momentum balance on the disk yield:
mẍ = −kx − bẋ − fµ ,
mr2
θ̈ = −fµ r − M (t),
2
where fµ is the unknown frictional force. If the disk rolls without slipping, we find the
kinematical relation x = −rθ, and, eliminating fµ from the above balance laws, we find
the governing equation of motion can be expressed as:
M (t)
3m
ẍ + bẋ + kx =
.
2
r
b) For M (t) = 0, the frequency of oscillation is the damped natural frequency ωd =
58
ωn
p
1 − ζ 2 , where:
ωn =
r
2k
=
3m
r
2
,
3
ζ=√
so that the damped natural frequency is ωd =
√1
3
b
1
=√ ,
2
6km
= 0.577.
c) In nondimensional form, the system is represented as:
ẍ + 2ζωn ẋ + ωn2 x = F0 sin(t),
where ζ and ωn are given above, and:
F0 =
2
2
= .
3m
3
Recall that the magnification factor for a harmonically driven system is:
M= p
1
(1 −
r2 )2
+ (2ζr)2
,
where r = ωωn is the frequency ratio. For this system we find r =
As a result, the steady-state amplitude is:
A=
q
3
2,
so that M =
√2 .
13
2
F0
M = √ = 0.555.
ωn2
13
d) The response of the disk is x(t) = A sin(t + φ), where A is given above and:
√
− 3
−2ζr
=
.
tan φ =
1 − r2
− 21
From the expression for angular momentum balance, we find:
fµ
where cos φ =
√1
13
M (t)
m
ẍ −
,
2
r
1
= − √ sin(t + φ) − sin(t),
13
¶
µ
¶
¸
·µ
1
1
√ sin φ cos(t) + √ cos φ + 1 sin(t) ,
= −
13
13
=
and sin φ =
√
2 3
√
,
13
so that the amplitude of the frictional force is:
4
|fµ | = √ .
13
59
Problem 65:
In the figure, the disk has mass m, radius r,
2
and moment of inertia IG = mr
about the
2
mass center G, subject to an applied moment
of the form:
g
x
M (t) = M0 sin(ωt)k̂.
M (t)
b
If the disk rolls without slip (µ is sufficiently
large):
G
a) find the equations of motion for this
system;
(m, r)
k
b) for m = 2 kg, b = 0.5 (N · s)/m, and
k = 8 N/m, find the steady-state amplitude of the rotation of the disk?
Solution:
In addition to x, defined in the figure, we define θ as the angular displacement of the
disk from the unstretched position in the k̂ direction. If the disk rolls without slip, x
and θ are related as:
x = −rθ.
a) The frictional force, which is unknown, is defined as f = fı̂, while the forces due to the
spring and damper are:
Fspring = −kx ı̂,
Fdamper = −bẋ ı̂.
Using linear and angular momentum balance on the disk, we find that:
X
F
F = (f − kx − bẋ) ı̂ = mẍ ı̂ = m aG ,
X
mr2
θ̈ k̂ = IG αβ/F F.
MG = (M0 sin(ωt) + f r) k̂ =
2
Eliminating the unknown frictional force, and using the kinematic constraint, we find
the equation of motion is:
¶
µ
M0
3m
ẍ + bẋ + kx = −
sin(ωt).
2
r
b) For the given values of the parameters, we find that:
ωn2 =
8
,
3
1
ζ= √ ,
8 6
F0 = −
Thus, the steady-state amplitude may be easily found as:
X
=
=
F0
M,
ωn2
q¡
60
8
3
0
−M
3r
¢2 ¡ ω ¢ 2 .
− ω2 + 6
M0
.
3r
Problem 66:
The mass m = 2 kg is supported by an elastic cantilever beam attached to a foundation which undergoes harmonic motion of the
form:
u(t) = 4 sin(ωt) m,
̂
m
ı̂
x
If the beam has length l = 20 cm, while
AE = 16 N:
z
a) find the equations of motion in terms
of z, the relative displacement between
the mass and the foundation (assume
the beam has zero mass);
u(t) = 4 sin(ω t) m
b) what is the amplitude of the resulting motion in terms of the forcing frequency ω?
Solution:
a) The equivalent spring for this cantilever beam is:
keq =
16 N
AE
=
= 80 N/m.
l
0.2 m
F
The acceleration of the block with respect to the ground is aG = (ü + z̈)̂, so that, in
terms of z, the equations of motion become:
mz̈ + keq z
z̈ + 40z
= −mü,
= 4ω 2 sin(ωt).
b) For this undamped system, the amplitude of the resulting steady-state motion is:
X
4ω 2
q¡
40
=
1
1−
4ω 2
.
|40 − ω 2 |
=
61
ω2
40
¢2 ,
Problem 67:
The unbalanced rotor shown in the figure is
pinned to a frame and supported by a spring
and damper. If the total mass is m while the
mass center G is located at an eccentricity of
ε from the the center of rotation O,
̂
k
b
ı̂
a) find the damped natural frequency;
b) what is the steady-state amplitude of
vibration when the rotor spins at this
angular speed.
ε
G
O
x
m
Solution:
We define x(t) as the vertical displacement of the geometric center of the rotor as
measured from static equilibrium. As a result, the mass center G is described by the
position z(t) = x(t) + ε sin(ωt). Note that ε measures the eccentricity of the mass
center, not the location of the mass imbalance. Consequently, the governing equations
of motion can be written:
mz̈
mẍ − εω sin(ωt)
2
= −kx − bẋ,
= −kx − bẋ,
or in more standard form:
ẍ +
b
k
ẋ + x = εω 2 sin(ωt).
m
m
a) In the above system we find ωn =
frequency can be written:
ωd = ωn
q
p
k
m
1−
and ζ =
ζ2
=
r
√b ,
2 km
so that the damped natural
k
b2
−
.
m 4m2
b) For an arbitrary forcing frequency the amplitude of oscillation is A = εΛ, where:
Λ= p
ω2
(ωn2 − ω 2 )2 + (2ζωωn )2
which, with ω = ωd , reduces the amplitude to:
1 − ζ2
A=ε p
,
ζ 4 − 3ζ 2
with ζ defined above.
62
,
Problem 68:
The system shown in the figure is supported
by a foundation that undergoes harmonic motion of the form:
̂
g
m
ı̂
u(t) = 4 sin(ωt) m,
If the mass and stiffness are m = 2 kg, and
k = 8 N/m:
z
k
x
a) find the equations of motion in terms
of z, the relative displacement between
the mass and the base (assume the
spring has zero unstretched length);
u(t) = 4 sin(ω t) m
b) what is the amplitude of the resulting
motion in terms of the frequency ratio
r;
c) for what forcing frequencies is the resulting amplitude of the steady-state
motion Z ≤ 8?
Solution:
a) We construct a free-body diagram as shown. The only forces acting on the mass arise
from the gravitational force and the spring force.
However, the acceleration of the mass with respect to
the ground is written as:
³
´
F
aG = ü + z̈ ̂.
Therefore, linear momentum balance takes the form:
X
F
F = m aG ,
³
´
³
´
− kz − mg ̂ = m z̈ + ü ̂.
−m g ̂
−k z ̂
Substituting in the expression for u(t), and taking the component in the ̂ direction, we
obtain the governing equation of motion:
mz̈ + kz
z̈ + (4 s−2 )z
= −mg − mü,
= −9.81 m/s2 + 4ω 2 sin(ωt) m,
Notice that each term has units of acceleration, that is, m/s2 . In what follows we will
dispense with the explicit inclusion of the units. For this system ωn = 2, ζ = 0, and the
amplitude of the forcing is F = 4ω 2 . Thus we have frequency-squared excitation.
b) Using the above values for the natural frequency and the damping ratio, we find that
63
the forced response can be written as z(t) = Z sin(ωt + φ), where the amplitude Z is:
Z = U · Λ(r, ζ)
= Up
=
r2
(1 − r2 )2 + (2ζr)2
,
4r2
.
|1 − r2 |
c) If Z < 8 then this implies that:
Z<p
4r2
(1 − r2 )2
Defining Zcr as:
Zcr = p
We may solve for r to yield:
r
.
4r2
(1 − r2 )2
,
s
2 ± 4Z
Zcr
cr
,
2 − 16
Zcr
)
(r
r
Zcr
Zcr
.
,
=
Zcr − 4
Zcr + 4
=
Thus for Zcr = 8, we find that Z < 8 in the range:
r
√
2
0<r<
, or r > 2.
3
Problem 69:
The single-degree-of-freedom system shown is
subject to a harmonic force. If the natural
frequency is ωn = 4 rad/s and m = 1 kg:
x
a) determine the spring and damping constants when the system is critically
damped;
F (t) = sin(ω t) N
m
k
b
b) determine the amplitude of the total
force transmitted to the ground under steady-state oscillations when ω =
1 rad/s.
Solution:
The equation of motion for this system takes the form:
ẍ +
b
k
F (t)
ẋ + x =
.
m
m
m
a) In terms of the system parameters, the damping ratio and natural frequency can be
written as:
r
b
k
.
,
ωn =
ζ= √
m
2 km
64
So, for a critically damped system ζ = 1, and solving for k and b we find:
k = 16 N/m,
b = 8 N · s/m.
b) The amplitude
p of the total force transmitted to the ground, which is defined as Ft , is
Ft = Xωn2 1 + (2ζr)2 , where X is the amplitude of the response and r = ωωn = 14 is
the frequency ratio. Thus, for this system:
!
Ã
p
1
F0
p
· ωn2 1 + (2ζr)2 ,
Ft =
2
ωn (1 − r2 )2 + (2ζr)2
p
√
F0 1 + (2ζr)2
8 5
= p
= 1.05.
=
17
(1 − r2 )2 + (2ζr)2
Problem 70:
A constant force is applied to the undamped
single degree-of-freedom system for a duration of t1 , at which point it is removed, that
is:
½
0 ≤ t < t1 ,
F0 ı̂,
F (t) =
0,
t ≥ t1 .
x
F (t)
m
4k
If the system starts with zero initial conditions, determine the resulting displacement
of the mass x(t). You may either use the
convolution integral or you may try to solve
this explicitly.
Solution:
The equation of motion for this system is:
ẍ +
F (t)
4k
x=
,
m
m
which has an impulse response of the form:
sin ωn t
h(t) =
,
mωn
r
ωn = 2
k
.
m
Therefore, using the convolution integral, the response of the system is:
Z t
F (ξ)h(t − τ ) dτ,
x(t) =
0
 Z t
F0 sin(ωn (t − τ ))


dτ,
0 ≤ t < t1 ,

 0 m
ωn
=
 Z τ F sin(ω (t − τ ))

n
0


dτ,
t ≥ t1 .
m
ωn
0

´
F0 ³


1
−
cos(ω
t)
,
0 ≤ t < t1 ,
n


mωn2
=
´

F0 ³



cos(ω
(t
−
t
))
−
cos(ω
t)
,
t ≥ t1 .
n
1
n
mωn2
65
Problem 71:
A mass m = 2 kg is rigidly connected to a
rigid massless bar of length ℓ = 40 cm, which
is pinned to a wall. The system is supported
by a combination of springs and damper as
shown in the figure, and subjected to a timedependent moment M (t) = 12 sin(t). If k =
12 N/m and b = 6 N/(m/s):
b
̂
M (t)
m
ı̂
ℓ
2
a) find the undamped natural frequency
and the damping ratio of the system;
z2
z1
θ
b) what are the steady-state amplitude
and phase of the forced response?
6k
2k
k
Problem 72:
The system shown in the figure is supported
by a foundation that undergoes an exponentially decaying motion of the form:
g
m
u(t) = 16e−t/4 .
If the mass and stiffness are m = 5 kg, and
k = 45 N/m:
z
k
a) find the equations of motion in terms
of z, the relative displacement between
the mass and the base (assume the
spring has zero unstretched length);
u(t)
b) find the resulting solution z(t) if the
mass is started from rest and the spring
is initially unstretched (assume g =
10 m/s2 and use the convolution integral);
c) what is the amplitude of the oscillations
as t increases (i.e. u(t) → 0)?
66
Problem 73:
In the system shown at right, the disk is assumed to be massless while:
m = 2 kg,
b = 14 N/(m/s),
2r
k
k = 4 N/m,
r = 1 cm.
r
a) Determine the governing equations of
motion;
b) What is the period of oscillation;
m
c) If f (t) = sin t, find the amplitude and
phase of the resulting motion as t → ∞.
f (t)
b
Problem 74:
In the mechanical system shown each spring
is identical, with spring constant k. If m =
1 kg:
ı̂
a) determine the spring constant k and
damping constant b so that the unforced system is critically damped and
the exponential rate of decay is τ =
2 s−1 ;
m
b
b) with the initial conditions:
x(0) = 0,
̂
f (t)
ẋ(0) = 2 m/s,
k
find the resulting solution of x(t) for the
unforced problem, i.e., f (t) = 0;
c) with f (t) = sin(2t), determine the amplitude of the force transmitted to the
supporting structure.
67
x
Problem 75:
The unbalanced rotor shown in the figure is
pinned to a frame and supported by a spring
and damper. If the total mass is m while the
mass center G is located at an eccentricity of
ε from the the center of rotation O,
̂
k
b
ı̂
a) find the damped natural frequency;
b) for:
ε
m = 4 kg,
b = 0.5 (N · s)/m,
ε = 0.1 m,
O
k = 8 N/m
x
m
what is the steady-state amplitude of
vibration when the rotor spins at this
angular speed?
2k
Problem 76:
The disk shown in the figure has mass m =
4 kg, and is subject to a time-dependent moment M (t) = M0 sin(t) k̂. If k = 2 N/m and
r = 0.2 m:
4k
IG
a) Find the steady-state amplitude of the
response when:
b = 1 N/(m/s),
G
b
r
M0 = 12 N · m.
b) What is the amplitude of the oscillations when the system is critically
damped?
r
2
k
2k
68
Problem 77:
For the system shown to the right, the block
slides on a rough surface (coefficient of friction µ) inclined at an angle of φ with respect
to vertical. If the block is subject to a periodic force of the form
g
k
F (t) = F0 sin(ω t),
a) find the equations of motion. Do not
neglect gravity;
m
φ
F (t)
b) find the amplitude of the steady-state
response using Mc when
m = 1.25 kg, k = 20 N/m,
φ = 30◦ , µ = 0.125,
F0 = 4 N,
4
ω = 2.00 rad/s,
Multi Degree-of-freedom Systems
Problem 78:
For the system shown to the right
a) use Lagrange’s equations to determine
the equations of motion;
b) find the mode shapes and natural frequencies of the system;
m
2k
c) determine the modal equations for this
system.
69
4m
k
F (t) ı̂
)
,d
(4
m
Problem 79:
For the system shown to the right
a) find the equations of motion;
b) are your equations statically coupled,
dynamically coupled, both or neither?
k
k
2
(m,
d)
f (t)
m
b
2k
Problem 80:
For the system shown to the right
a) use Lagrange’s equations to determine
the equations of motion;
k
4k
b) find the mode shapes and natural frequencies of the system. Normalize the
mode shapes so that u T M u = 1.
3m
k
k
m
Problem 81:
For the system shown to the right the bar
of length ℓ is massless and the block on the
right is subject to a time-dependent force of
the form
k
k
F (t) = F0 sin(ω t).
G
The intermediate springs are located a distance 2ℓ from the center of the bar.
k
x2
x1
a) Find the equations of motion in terms
of x1 and x2 .
m
m
b) Determine the natural frequencies and
mode shapes of this system.
F (t)
c) Determine the modal equations.
d) What is the steady-state amplitude of
the in-phase motion?
70
Problem 82: (Spring 2003)
For the mechanical system shown to the
right, the uniform rigid bar is supported by
identical springs. For this system:
k
a) find the equations of motion in terms
of x and θ (and in terms of the given
parameters—do not substitute in numerical values yet);
k
x
θ
b) if k = 64 N/m, m = 2 kg, and ℓ =
0.25 m, find the natural frequencies and
mode shapes for this system;
ℓ
3
G
m, ℓ
Problem 83: (Spring 2003)
For the system shown to the right:
F0 sin(ω t)
m
a) find the equations of motion (and in
terms of the given parameters—do not
substitute in numerical values yet);
βm
k
αk
b) for α = β = 2, find the natural frequencies and mode shapes of the system and
normalize the mode shapes by M ;
c) determine the equations of motion that
describe the response of each mode.
Problem 84: (Spring 2003)
In the figure shown to the right, in the absence of gravity the springs are unstretched
in the equilibrium position. Determine the
equations of motion for this system.
x2
k2
IG
m
r2
b
θ
r1
x3
z
k1
̂
m
ı̂
x1
71
k1
Solution:
Because we can, we define five different coordinates to describe the dynamical behavior
of this two degree-of-freedom system, leading to the following transformations:
x2 = −r2 θ,
x3 = −r1 θ,
A free-body diagram for this system is
shown to the right. We develop three equations of motion based on linear momentum
balance on both blocks and angular momentum balance on the disk:
z = x3 − x1 .
−(k2 x2 + b ẋ2 ) ı̂
−T ı̂
T ı̂
Block 1:
³
X
k1 z − k1 x1
Block 2:
³
´
X
T − b ẋ2 − k2 x2
Disk:
F
F
= m a G1 ,
FR
̂ = m ẍ1 ̂
F
´
X
−k1 z ̂
F
= m a G2 ,
k1 z ̂
ı̂ = m ẍ2 ı̂
MG
³
´
T r2 + k1 r1 z k̂
= I G αD/F ,
−k1 x1 ̂
= I G θ̈ k̂
From the equations on block 2 and the disk, we eliminate the unknown tension T from
the system to obtain:
I G θ̈ − m r2 ẍ2 − b r2 ẋ2 − k2 r2 x2 − k1 r1 z = 0.
From this equation, and the equation on block 1, we eliminate the coordinates z and
x2 , and obtain the equations of motion to be:
³
I G + m r22
´
m ẍ1 + 2 k1 x1 + k1 r1 θ
´
³
θ̈ + b r22 θ̇ + k1 r1 x1 + k1 r12 + k2 r22 θ
Problem 85:
For the system shown in the figure:
=
0,
=
0.
x
a) what is the degree-of-freedom for this
system?
k
k
(I, r)
2m
b) using Lagrange’s equations, determine
the differential equations that govern
the motion.
m
72
Solution:
a) This system contains three masses which are each allowed to move in only one direction. The upper mass slides horizontally with displacement x1 , while the disk rotates
through an angle θ. Finally, the suspended mass moves vertically and its position can
be described by the coordinate x2 . However, because the disk and the suspended mass
are connected by an inextensible string, their motion can be related by:
x2 = rθ.
So this system has only two independent coordinates and therefore it is a two-degreeof-freedom system.
b) We utilize the coordinates x1 , x2 , and θ, as measured from the unstretched position of
the two springs. Therefore, the kinetic and potential energies are written as:
T
=
V
=
1
1
1
(2m) ẋ21 + m ẋ22 + I θ̇2 ,
2
2
2
1
1
k x21 + k (rθ − x1 )2 − m g x2 .
2
2
However, x2 and θ are related by the above relationship. Thus eliminating θ, the energies
become:
1
1
1 I 2
T =
(2 m) ẋ21 + m ẋ22 +
ẋ ,
2
2
2 r2 2
1
1
k x21 + k (x2 − x1 )2 − m g x2 ,
V =
2
2
which, using Lagrange’s equations, yields the equations of motion:
2m ẍ1 + 2 k x1 − k x2
¶
µ
I
m + 2 ẍ2 − k x1 + k x2
r
Problem 86:
For the system shown in the figure:
=
0,
= m g.
k2
a) find the mass and the stiffness matrix;
k1
b) is your system of equations dynamically
coupled, statically coupled, or both?
m2
m1
k3
m3
k3
Solution:
We choose the coordinates (x1 , x2 , x3 ), which represent the positions of the three masses.
a) To determine the stiffness matrix, we use the stiffness influence coefficients. Maintaining
a unit displacement of each mass in turn requires forces of the form:
(x1 , x2 , x3 ) =
(x1 , x2 , x3 ) =
(x1 , x2 , x3 ) =
(1, 0, 0)
(0, 1, 0)
(0, 0, 1)
→
→
→
73
f
f
f
=
=
=
(k1 + k2 , −k2 , 0)T ,
(−k2 , k2 + 2k3 , −2k3 )T ,
(0, −2k3 , 2k3 )T .
Therefore, with these coordinates the stiffness matrix is:


k1 + k2
−k2
0
k2 + 2k3 −2k3 
K =  −k2
0
−2k3
2k3
Alternatively, we can define the potential energy of the system as:
V
=
=
1
1
1
(k1 )(x1 )2 + k2 (x2 − x1 )2 + (2k3 )(x3 − x2 )2 ,
2
2
2
1
1
1
(k1 + k2 )(x1 )2 − (2k2 )(x1 x2 ) + (k2 + 2k3 )(x2 )2
2
2
2
1
1
− (2k3 )(x2 x3 ) + (2k3 )(x3 )2 ,
2
2
which leads to the same stiffness matrix.
To determine the mass matrix, we could use the inertia influence coefficients, but, for
variety, we determine the kinetic energy as:
T =
1
1
1
m1 ẋ21 + m2 ẋ22 + m3 ẋ23 .
2
2
2
Therefore, the mass matrix is:

m1
M = 0
0
0
m2
0

0
0 
m3
b) With this choice of coordinates, the mass matrix is diagonal and the stiffness matrix
contains nonzero off-diagonal terms. Thus, the system is statically coupled but dynamically uncoupled.
Problem 87:
For the system shown at right:
k2
a) determine the governing equations of
motion;
k1
k1
m1
b) is the system statically or dynamically
coupled, or both;
m1
k2
m2
c) find the matrix M −1 K .
74
Problem 88:
In the system shown to the
and stiffness matrix are:
·
m1 0
M =
0 m2
·
k1 + k2
K =
−k2
right, the mass
¸
,
−k2
k2
¸
while the damping is proportional, i.e., C =
βM +αK . If m1 is subject to harmonic forcing F (t) = sin t, with:
m1 = 2.0 kg,
k1 = 6.0 N/m
m2 = 1.0 kg,
k2 = 2.0 N/m,
β = 0.1,
α = 0.1
a) determine the eigenvectors of this system, normalized by the kinetic energy
inner product;
b) what are the damped natural frequencies of this system;
c) find the steady-state amplitude of vibration of the mode of vibration with
the lowest natural frequency.
Problem 89:
The two-degree-of-freedom system shown is
subject to a harmonic force applied to the
block of mass 2m, of the form:
¡
¢
f (t) = (2 sin(t)) N ı̂.
f (t)
2m
k
If the system is subject to proportional damping with α = 0.25, m = 2 kg, and k =
4 N/m, find:
2m
k
k
a) the mass, damping, and stiffness matrices;
b) the forced, damped equation (singledegree-of-freedom) that describes the
motion of each mode;
c) the steady-state response of the system.
Solution:
a) Using influence coefficients, we find that the mass and stiffness matrices are:
¸
¸
·
·
2k −k
2m 0
.
,
K =
M =
−k 2k
0 2m
Therefore, with proportional damping, the damping matrix becomes C = αK , so that
75
we find:
M = (4 kg)
·
1
0
¸
0
1
,
K = (4 N/m)
C = (1 N/(m/s))
·
Problem 90:
In the multi-degree-of-freedom system shown
in the figure, the block with mass 4m slides
on a smooth, frictionless surface. If the pulley
is massless:
2 −1
−1 2
·
¸
2 −1
−1 2
,
.
x1
z
2k
a) using Lagrange’s equations, determine
the differential equations governing the
motion, as measured from static equilibrium;
¸
r
k
4m
b) with m = 1 kg and k = 16 N/m,
find the natural frequencies and mode
shapes for the free vibration of this system. Normalize the mode shapes so
that with respect to the mass matrix
the amplitude of each mode is one;
x2
2m
c) find the general solution to these equations for the above values of m and k.
Solution:
a) We identify the three coordiantes x1 , x2 , and z, with
z = x2 − x1 .
Measuring the response from static equilibria and neglecting the gravitational potential
energy, the kinetic and potential energies for this system can be written as
T
=
V
=
¢
¢
1¡
1¡
4 m ẋ21 +
2 m ẋ22 ,
2
2
1¡ ¢ 2 1¡ ¢ 2
2 k x1 +
k z ,
2
2
In terms of x1 and x2 , the potential energy becomes
V=
¢2
¢
1 ¡ ¢ 2 1 ¡ ¢¡
1¡ ¢ 2 1¡
1¡ ¢ 2
2 k x1 +
k x2 − x1 =
3 k x1 +
− 2 k x1 x2 +
k x2 .
2
2
2
2
2
Therefore the equations of motion become
4 m ẍ1 + 3 k x1 − k x2
2 m ẍ2 − k x1 + k x2
=
=
0,
0,
or in matrix form
m
·
4
0
0
2
¸ ·
ẍ1
ẍ2
¸
76
+k
·
3
−1
−1
1
¸ ·
x1
x2
¸
= 0.
b) The corresponding eigenvalue problem for the above system is
¸
· 3
¡ −1 ¢
¢
k
k ¡
− 14
4
M
K u =λu
−→
=
βu
1
m − 21
m
2
The characteristic equation is
¶µ
¶
µ
1
5
1
1
3
−β
− β − = β 2 − β + = 0,
4
2
8
4
4
with the solution
β=
5±3
8
−→
λ = ω2 =
½
k
k
,
4m m
¾
.
Returning this to the eigenvalue problem, the mode shapes are defined by the equation
3
1
ui1 + ui2 = βi ui1 ,
4
4
so that
ui =
·
1
4 βi − 3
¸
ci
u1 =
·
1
−2
Normalizing ui by the mass matrix implies that
·
£
¤
4m
1 (4 βi − 3)
1 = uiT M ui = c2i
0
¸
c1 ,
u2 =
¸ ·
0
2m
·
¸
1
1
c2 .
1
(4 βi − 3)
¸
.
Solving for ci
ci =
s
1
2 m (2 + (4 βi − 3)2 )
Finally, the normalized eigenpairs are
"
#
r
√ 1
k
12 m
, u1 =
,
ω1 =
2
− √12
4m
m
−→
c1 =
ω2 =
r
r
1
, c2 =
12 m
k
,
m
u2 =
r
"
1
6m
√1
6m
√1
6m
#
.
c) With the above mode shapes and natural frequencies the general solution becomes
q (t) =
·
x1 (t)
x2 (t)
¸
=
2
X
(Ai sin(ωi t) + Bi cos(ωi t)) ui ,
i=1
=
Ã
Ãr
!
Ãr
!! "
#
√ 1
k
k
12
m
A1 sin
t + B1 cos
t
2
− √12
4m
4m
m
#
!
Ãr
!! "
Ã
Ãr
1
√
k
k
6m
.
t + B2 cos
t
+ A2 sin
√1
m
m
6m
77
Problem 91:
For the system shown in the figure:
a) find the mass and the stiffness matrices;
k3
k1
b) is your system of equations dynamically
coupled, statically coupled, or both?
m1
k2
m3
k1
m2
Solution:
a) For coordinates we choose (x1 , x2 , x3 ) as the displacements of each mass with respect
to inertial space. Using influence coefficients, we find that:




m1 0
0
k1 + k2
−k2
0
k1 + 2k2 −k2 
M =  0 m2 0  ,
K =  −k2
0
−k2
k2
0
0 m3
Alternatively, if we choose coordinates (x1 , x2 , z), where z represents the stretch in the
spring connecting m2 and m3 , we can determine the mass and stiffness matrices from
the Lagrangian. The kinetic and potential energies are:
T
=
V
=
1
1
1
m1 ẋ21 + m2 ẋ22 + m3 (ẋ2 − ẋ3 )2 ,
2
2
2
1
1
1
1
2
k1 x1 + k2 (x2 − x1 )2 + k2 z 2 + k1 x22 .
2
2
2
2
Thus, with these coordinates the mass and stiffness matrices become:



k1 + k2
−k2
m1
0
0
k1 + k2
K =  −k2
M =  0 m2 + m3 −m3  ,
0
0
0
−m3
m3

0
0 
k2
b) With the first choice of coordinates, the mass matrix is diagonal while the stiffness
matrix is not, the system is statically coupled but dynamically uncoupled. With the
latter coordinates neither matrix is diagonal so that the system is both statically and
dynamically coupled.
78
Problem 92:
In the system shown to the right, the pulley
has mass m and radius r, so that the moment
2
of inertia about the mass center is IG = mr
2 .
r
a) What is the degree-of-freedom for this
system?
r
2
b) Find the governing equations of motion;
k
c) If m = 1 kg and k = 4 N/m, what
are the frequencies of oscillation for
the motion and the corresponding mode
shapes, normalized by the kinetic energy inner product?
m
m
k
Solution:
a) Let the displacement of the left block, disk, and right block be described as (−x1 ̂),
(θ k̂), and (x2 ̂) respectively. Although x1 and θ are related by the following constraint:
x1 =
r
θ,
2
x2 is independent from the above two coordinates. Therefore, the system has two
degrees-of-freedom.
b) With the above coordinates, the kinetic and potential energies can be written as:
T
=
V
=
1
1 mr2 2
1
mẋ21 + mẋ22 +
θ̇ ,
2
2
2 2
1 2 1
kx + k(x2 − rθ)2 ,
2 1 2
Thus, using the above kinematic constraint to eliminate θ, the Lagrangian becomes:
L
= T − V,
·
¸
¤ 1
1 £ 2
2
2
2
m 3ẋ1 + ẋ2 − k 5x1 − 4x1 x2 + x2 .
=
2
2
Using Lagrange’s equations of motion, the governing equations are:
3m ẍ1 + 5k x1 − 2k x2
m ẍ2 − 2k x1 + k x2
= 0,
= 0.
c) From the above equations, the mass and stiffness matrices can be written as:
¸
¸
·
·
5 −2
3 0
,
K =k
M =m
−2 1
0 1
79
The characteristic matrix, A = M −1 K becomes:
¸
· 5
k
− 32
3
A=
,
m −2 1
and the characteristic equation can be written as:
β2 −
8
1
β + = 0,
3
3
k
β is an eigenvalue of the characteristic
where, if β is a solution to this equation, λ = m
matrix A. This quadratic equations has solutions of the from:
√
4 ± 13
β=
.
3
To determine the eigenvectors, we return to the characteristic matrix A, so that Au =
λu. The elements of u then satisfy the equation:
2
5
u 1 − u 2 = β u1 .
3
3
Thus, if u1 = 1, this yields:
u2 =
1∓
√
13
2
for
β=
4±
√
13
.
3
Normalizing by the kinetic energy inner product, we find that:
û = p
u
(u, u)M
=p
u
3u21 + u22
Problem 93:
For the system shown in the figure, the surface is assumed to be frictionless.
a) Using Lagrange’s equations, determine
the differential equations governing the
motion (measured from static equilibrium);
k
4m
(m, r)
b) Find the mode shapes and natural frequencies of the resulting motion.
k
m
80
Problem 94:
The multi-degree-of-freedom system shown in
the figure is subject to harmonic forcing of
the form f (t) = sin(4 t).
k
a) Find the equations of motion in terms
of the coordinates x1 , x2 , and x3 , and
identify the mass and stiffness matrices.
m
b) If m = 2 kg and k = 4 N/m, the natural frequencies of the system are found
to be:
ω1 = 0.38 rad/s,
2k
2k
ω2 = 2.62 rad/s,
ω3 = 1.41 rad/s.
m
Find the corresponding mode shapes
and normalize them so that (ui , ui )M =
1.
m
f (t)
c) Determine the forced, uncoupled equations of motion for the modal amplitudes Qi (t).
Problem 95:
For the system shown in the figure, the surface is assumed to be frictionless. If each
block is displaced by a distance d (down and
to the right), find the resulting motion of the
system.
x1
(m, r)
k
m
θ
z
k
m
Solution:
We define the coordinates x1 , x2 , θ, and z as shown in the figure, so that
x1 = −r θ,
81
z = r θ − x2 .
x2
With these coordinates, the kinetic and potential energies can be written as
µ
¶
1 ¡ ¢ 2 1 m r2
1¡ ¢ 2
T =
θ̇2 +
m ẋ1 +
m ẋ2 ,
2
2
2
2
1¡ ¢ 2 1¡ ¢ 2
k x1 +
k z .
V=
2
2
Expressing these only in terms of the coordinates x1 and x2 , we obtain
µ
¶
1¡ ¢ 2
1 3m
ẋ21 +
m ẋ2 ,
T =
2
2
2
1¡ ¢ 2 1¡ ¢
1¡ ¢ 2 1¡ ¢
1¡ ¢ 2
2
V=
k x1 +
k (−x1 − x2 ) =
2 k x1 +
2 k x1 x2 +
k x2 .
2
2
2
2
2
Therefore, the mass and stiffness matrix can be identifed as
· 3
¸
·
¸
·
¸
2 1
x1 (t)
0
M =m 2
,
K =k
with q (t) =
1 1
x2 (t)
0 1
and the equations of motion are:
¸ ·
¸
·
· 3
ẍ1
2
0
2
+k
m
ẍ2
1
0 1
1
1
¸ ·
x1
x2
¸
= 0.
The solution to this equation requires the solution of an eigenvalue problem of the
form
· 4 2 ¸
k
3
3
u = λ u,
(M −1 K ) u =
m 1 1
which is determined from the characteristic equation
·µ
¸
¶
³
´
4
2
k
−1
= 0,
− β (1 − β) −
det M K − λ I =
m
3
3
with λ =
k
m
β. This quadratic equation has the solution
)
(r
r
½
¾
1
7±5
k
2k
=
, 2
−→ ω =
,
β=
6
3
3m
m
With this, the eigenvectors are determined by returning to the original eigenvalue equation
· 4 2 ¸ ·
·
¸
¸
3 βi − 4
ui1
ui1
3
3
, −→ ui2 =
ui1
= βi
ui2
ui2
1 1
2
In addition, normalizing the eigenvectors by the mass matrix
·
¶
¸ ·
¸
µ
£
¤ 3m 0
6 + (3 βi − 4)2
ui1
2
1 = uiT M ui = ui1 ui2
.
= m u2i1
ui2
0 m
4
Solving for ui1 the normalized eigenvectors are
#
"
u1 =
√ 2
15 m
3
− √15
m
82
,
u2 =
"
√ 2
10 m
√ 2
10 m
#
.
One can easily verify that both u1T M u2 = 0 and u2T M u1 = 0.
The general solution is written as
q (t) =
2 ³
X
i=1
´
Ai sin(ωi t) + Bi cos(ωi t) ui
subject to the initial conditions
q (0) =
·
d
−d
Premultiplying by uiT M yields
¡
¢
uiT M q (0) = uiT M ui Bi ,
¸
,
q̇ (0) =
·
0
0
¸
.
¡
¢
uiT M q̇ (0) = uiT M ui (Ai ωi ).
Since uiT M ui = 1 from our normalization, the constants are directly solved to be
r
r
m
m
A1 = 0, B1 = 6 d
,
A2 = 0, B2 = d
15
10
Finally, the solution to the specific initial conditions becomes
(
Ãr
!·
!·
Ãr
¸
¸)
·
¸
k
2k
d
4
1
x1 (t)
cos
t
t
+ cos
.
q (t) =
=
−6
1
x2 (t)
5
3m
m
Problem 96:
The two-degree-of-freedom system shown is
subject to a harmonic force applied to the
block of mass 2m, of the form:
f (t)
m
2m
f (t) = (2 sin(t) N) ı̂.
k
If the mass and stiffness of the system are
assumed to be m = 2 kg, and k = 4 N/m,
find:
x1
k
2k
z
x2
a) the equations of motion;
b) the forced, damped equation (singledegree-of-freedom) that describes the
motion of each mode;
Solution:
a) We define the coordinates x1 , x2 , and z, which are related as
z = x2 − x1 .
The equations of motion become
·
¸
·
¸ ·
¸
2 0
2 −1
f (t)
m
+k
=
.
0 1
−1 3
0
83
b) The corresponding eigenvalue problem can be written as
·
¶
¸
µ
¡ −1 ¢
k
k
1 − 12
M K =
β u
u=
−1 3
m
m
and the characteristic equation becomes
(1 − β) (3 − β) −
5
1
= β 2 − 4 β + = 0.
2
2
This quadratic equation has the solutions
( √
)
r
√
√
√
( 8 − 3) k ( 8 + 3) k
3
2
√
√
,
−→
ω =
β =2±
.
,
2
2m
2m
Returning to the eigenvalue equation, the eigenvectors satisfy the equation
ui1 −
1
ui2 = βi ui1
2
−→
ui2 = 2 (1 − βi ) ui1
so that
s √
√
·
¸
( 8 − 3) k
1
√
√
, u1 =
ω1 =
,
6−2
2m
s √
√
¸
·
( 8 + 3) k
√1
√
.
ω2 =
, u2 =
−( 6 + 2)
2m
For each eigenvector the kinetic energy inner product is
√
√
u1T M u1 = 14 − 4 6,
u2T M u2 = 14 + 4 6,
Finally, the modal equation for the first mode can be written as
¡
¢
¡ T
¢
u1 M u1 Q̈1 + u1T K u1 Q1 = u1T f (t),
Q̈1 +
³
(
√
Q̈1 + ω12 Q1 =
´
√
8− 3) k
√
Q1 =
2m
u1T f (t)
,
u1T M u1
f (t)
√ ,
14−4 6
while the response of the second mode is governed by
¡ T
¢
¡
¢
u2 M u2 Q̈2 + u2T K u2 Q2 = u2T f (t),
Q̈2 + ω22 Q2 =
´
√ √
3) k
√
Q2 =
Q̈2 + ( 8+
2m
³
84
u2T f (t)
,
u2T M u2
f (t)
√ ,
14+4 6