4.1 Solved Problems
Problem 91:
In the figure shown to the right, in the ab-sence of gravity the springs are unstretched in the equilibrium position. Determine the equations of motion for this system.
r2
r1
IG
k1
k1
k2
b
m m
x1
z x3
x2
θ
ˆ
ˆı
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= −r2θ, x3= −r1θ, z = x3− x1. A free-body diagram for this system is
shown to the right. We develop three equa-tions of motion based on linear momentum balance on both blocks and angular momen-tum balance on the disk:
Block 1:
XF = m aG1,
k1z − k1x1
ˆ = m ¨x1ˆ
Block 2:
XF = m aG2,
T − b ˙x2− k2x2
ˆı = m ¨x2ˆı Disk:
XMG = IGαD,
T r2+ k1r1z ˆk = IGθ ˆ¨k
FR
−k1 x1 ˆ
k1z ˆ
−k1 z ˆ
−T ˆı T ˆı
−(k2 x2+ b ˙x2) ˆı
From the equations on block 2 and the disk, we eliminate the unknown tension T from the system to obtain:
IG θ − m r¨ 2x¨2− b r2 ˙x2− k2 r2 x2− k1 r1z = 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:
m ¨x1+ 2 k1x1+ k1r1θ = 0,
IG+ m r22 ¨θ + b r22 ˙θ + k1r1x1+
k1r12+ k2r22
θ = 0.
Problem 92:
For the system shown in the figure:
a) what is the degree-of-freedom for this system?
b) using Lagrange’s equations, determine the differential equations that govern the motion.
x
2 m
m (I, r)
k k
Solution:
a) This system contains three masses which are each allowed to move in only one direc-tion. 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-degree-of-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 = 1
2(2m) ˙x21+1
2m ˙x22+1 2I ˙θ2,
V = 1
2k x21+1
2k (rθ − x1)2− m g x2.
However, x2and θ are related by the above relationship. Thus eliminating θ, the energies become:
T = 1
2(2 m) ˙x21+1
2m ˙x22+1 2
I r2 ˙x22,
V = 1
2k x21+1
2k (x2− x1)2− m g x2,
which, using Lagrange’s equations, yields the equations of motion:
2m ¨x1+ 2 k x1− k x2 = 0,
m + I
r2
¨
x2− k x1+ k x2 = m g.
Problem 93:
For the system shown in the figure:
a) find the mass and the stiffness matrix;
b) is your system of equations dynamically
coupled, statically coupled, or both? m1
k1 m2
m3
k2
k3 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) = (1, 0, 0) → f = (k1+ k2, −k2, 0)T, (x1, x2, x3) = (0, 1, 0) → f = (−k2, k2+ 2k3, −2k3)T, (x1, x2, x3) = (0, 0, 1) → f = (0, −2k3, 2k3)T.
Therefore, with these coordinates the stiffness matrix is:
K=
k1+ k2 −k2 0
−k2 k2+ 2k3 −2k3
0 −2k3 2k3
Alternatively, we can define the potential energy of the system as:
V = 1
2(k1)(x1)2+1
2k2(x2− x1)2+1
2(2k3)(x3− x2)2,
= 1
2(k1+ k2)(x1)2−1
2(2k2)(x1x2) +1
2(k2+ 2k3)(x2)2
−1
2(2k3)(x2x3) +1
2(2k3)(x3)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
2m1˙x21+1
2m2˙x22+1 2m3˙x23. Therefore, the mass matrix is:
M =
m1 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 dynam-ically uncoupled.
Problem 94:
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 ˆı.
If the system is subject to proportional damp-ing with α = 0.25, m = 2kg, and k = 4N/m, find:
a) the mass, damping, and stiffness matri-ces;
b) the forced, damped equation (single-degree-of-freedom) that describes the motion of each mode;
c) the steady-state response of the system.
k k k
2 m 2 m
f(t)
Solution:
a) Using influence coefficients, we find that the mass and stiffness matrices are:
M =
2m 0
0 2m
, K=
2k −k
−k 2k
.
Therefore, with proportional damping, the damping matrix becomes C = αK, so that we find:
M = (4kg)
1 0 0 1
, K= (4N/m)
2 −1
−1 2
,
C= (1N/(m/s))
2 −1
−1 2
.
Problem 95:
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:
a) using Lagrange’s equations, determine the differential equations governing the motion, as measured from static equi-librium;
b) with m = 1kg and k = 16N/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 am-plitude of each mode is one;
c) find the general solution to these equa-tions for the above values of m and k.
4 m
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 = 1
In terms of x1 and x2, the potential energy becomes V = 1
Therefore the equations of motion become
4 m ¨x1+ 3 k x1− k x2 = 0, 2 m ¨x2− k x1+ k x2 = 0,
or in matrix form m
b) The corresponding eigenvalue problem for the above system is M−1 K
The characteristic equation is with the solution
β = 5 ± 3
Returning this to the eigenvalue problem, the mode shapes are defined by the equation 3
Normalizing ui by the mass matrix implies that 1 = uTi M ui = c2i
Solving for ci
ci=
Finally, the normalized eigenpairs are ω1=
c) With the above mode shapes and natural frequencies the general solution becomes q(t) =
For the system shown in the figure:
a) find the mass and the stiffness matrices;
b) is your system of equations dynamically
coupled, statically coupled, or both? m1
k1 m3
m2
k2 k1
k3
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:
M =
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 = 1 Thus, with these coordinates the mass and stiffness matrices become:
M =
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.
Problem 97:
In the system shown to the right, the pulley has mass m and radius r, so that the moment of inertia about the mass center is IG=mr22. a) What is the degree-of-freedom for this
system?
b) Find the governing equations of mo-tion;
c) If m = 1kg and k = 4N/m, what are the frequencies of oscillation for the motion and the corresponding mode shapes, normalized by the kinetic en-ergy inner product?
r
a) Let the displacement of the left block, disk, and right block be described as (−x1 ˆ), (θ ˆk), and (x2ˆ) respectively. Although x1and θ are related by the following constraint:
x1= r 2θ,
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 = 1
2m ˙x21+1
2m ˙x22+1 2
mr2 2 ˙θ2,
V = 1
2kx21+1
2k(x2− rθ)2,
Thus, using the above kinematic constraint to eliminate θ, the Lagrangian becomes:
L = T − V,
= 1
2m3 ˙x21+ ˙x22 − 1 2k
5x21− 4x1x2+ x22
.
Using Lagrange’s equations of motion, the governing equations are:
3m ¨x1+ 5k x1− 2k x2 = 0, m ¨x2− 2k x1+ k x2 = 0.
c) From the above equations, the mass and stiffness matrices can be written as:
M = m
3 0 0 1
, K= k
5 −2
−2 1
The characteristic matrix, A = M−1K becomes:
A= k m
5
3 −23
−2 1
,
and the characteristic equation can be written as:
β2−8 3 β +1
3 = 0,
where, if β is a solution to this equation, λ = mkβ is an eigenvalue of the characteristic matrix A. This quadratic equations has solutions of the from:
β = 4 ±√ 13
3 .
To determine the eigenvectors, we return to the characteristic matrix A, so that Au = λu. The elements of u then satisfy the equation:
5 3 u1−2
3 u2= β u1. 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:
For the system shown in the figure, the sur-face 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.
With these coordinates, the kinetic and potential energies can be written as T =1
Expressing these only in terms of the coordinates x1 and x2, we obtain T = 1 Therefore, the mass and stiffness matrix can be identifed as
M = m
and the equations of motion are:
m
The solution to this equation requires the solution of an eigenvalue problem of the
which is determined from the characteristic equation det
with λ =mk β. This quadratic equation has the solution
β = 7 ± 5
With this, the eigenvectors are determined by returning to the original eigenvalue
equa-tion 4
In addition, normalizing the eigenvectors by the mass matrix 1 = uTi M ui=
Solving for ui1 the normalized eigenvectors are
u1=
The general solution is written as
q(t) =
subject to the initial conditions q(0) =
Premultiplying by uTi M yields uTi M q(0) = uTi M ui
Bi, uTi M ˙q(0) = uTi M ui
(Aiωi).
Since uTi M ui= 1 from our normalization, the constants are directly solved to be
A1= 0, B1= 6 dr m
15, A2= 0, B2= dr m 10
Finally, the solution to the specific initial conditions becomes
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) ˆı.
If the mass and stiffness of the system are assumed to be m = 2kg, and k = 4N/m, find:
a) the equations of motion;
b) the forced, damped equation (single-degree-of-freedom) that describes the motion of each mode;
k k 2 k
The equations of motion become m
b) The corresponding eigenvalue problem can be written as M−1K = k
and the characteristic equation becomes (1 − β) (3 − β) −1
2 = β2− 4 β +5 2 = 0.
This quadratic equation has the solutions
β = 2 ±r 3
Returning to the eigenvalue equation, the eigenvectors satisfy the equation ui1−1
2ui2= βiui1 −→ ui2= 2 (1 − βi) ui1
so that
For each eigenvector the kinetic energy inner product is uT1 M u1= 14 − 4√
6, uT2 M u2= 14 + 4√ 6, Finally, the modal equation for the first mode can be written as
uT1 M u1Q¨1+ uT1 K u1 Q1= uT1 f(t), while the response of the second mode is governed by
uT2 M u2Q¨2+ uT2 K u2 Q2= uT2 f(t),
4.2 Unsolved Problems
Problem 100:
For the system shown to the right the block on the left is wrapped around an inner hub.
a) Find the equations of motion using La-grange’s equations (you may neglect gravity or equivalently assume that all coordinates are measured from static equilibrium); find the mode shapes and natural fre-quencies for this system;
c) Determine the modal equations for this undamped system when
f (t) = (8N) sin(2 t).
You do not need to solve these modal equations.
Problem 101:
For the mechanical system shown to the right, the uniform rigid bar with mass m and length ℓ is supported by identical springs while a block of mass 2 m is suspended from the bar. For this system find the equations of motion in terms of x, z, and θ.
m, ℓ G
ℓ 4
k
k
k
m θ z
x
Problem 102:
For the mechanical system shown to the right, the uniform rigid bar has mass m, length ℓ and is pinned at point O while the block is suspended with a spring from B.
Find the equations of motion for this system in terms of θ and x. Assume that the system is in static equilibrium at θ = 0, and that all angles remain small.
O
ℓ 3
2 ℓ 3
m k
A
x
m k
k B
θ
Problem 103:
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 fre-quencies of the system;
c) determine the modal equations for this system.
m 4 m
2 k k
F (t) ˆı
Problem 104:
For the system shown to the right a) find the equations of motion;
b) are your equations statically coupled, dynamically coupled, both or neither?
(4m, d)
f (t) (m,2 d) k
2 k k
b m
Problem 105:
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 fre-quencies of the system. Normalize the mode shapes so that uTM u= 1.
m
3 m
k k
4 k k
Problem 106:
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
F (t) = F0 sin(ω t).
The intermediate springs are located a dis-tance ℓ2 from the center of the bar.
a) Find the equations of motion in terms of x1 and x2.
b) Determine the natural frequencies and mode shapes of this system.
c) Determine the modal equations.
d) What is the steady-state amplitude of the in-phase motion?
G
k
m
F (t) x1
m x2
k
k
Problem 107: (Spring 2003)
For the mechanical system shown to the right, the uniform rigid bar is supported by identical springs. For this system:
a) find the equations of motion in terms of x and θ (and in terms of the given parameters—do not substitute in nu-merical values yet);
b) if k = 64N/m, m = 2kg, and ℓ = 0.25m, find the natural frequencies and mode shapes for this system;
m, ℓ
G
ℓ 3
k
k
θ x
Problem 108: (Spring 2003) For the system shown to the right:
a) find the equations of motion (and in terms of the given parameters—do not substitute in numerical values yet);
b) for α = β = 2, find the natural frequen-cies 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.
k α k
m β m
F0 sin(ω t)
Problem 109:
For the system shown at right:
a) determine the governing equations of motion;
b) is the system statically or dynamically coupled, or both;
c) find the matrix M−1K.
m1
m2
k1
k2 m1
k2
k1
Problem 110:
In the system shown to the right, the mass and stiffness matrix are:
M =
m1 0 0 m2
,
K =
k1+ k2 −k2
−k2 k2
while the damping is proportional, i.e., C = βM +αK. If m1is subject to harmonic forc-ing F (t) = sin t, with:
m1= 2.0kg, m2= 1.0kg, β = 0.1, k1= 6.0N/m k2= 2.0N/m, α = 0.1 a) determine the eigenvectors of this
sys-tem, normalized by the kinetic energy inner product;
b) what are the damped natural frequen-cies of this system;
c) find the steady-state amplitude of vi-bration of the mode of vivi-bration with the lowest natural frequency.
Problem 111:
For the system shown in the figure, the sur-face is assumed to be frictionless.
a) Using Lagrange’s equations, determine the differential equations governing the motion (measured from static equilib-rium);
b) Find the mode shapes and natural fre-quencies of the resulting motion.
4 m
m (m, r) k
k
Problem 112:
The multi-degree-of-freedom system shown in the figure is subject to harmonic forcing of the form f (t) = sin(4 t).
a) Find the equations of motion in terms of the coordinates x1, x2, and x3, and identify the mass and stiffness matrices.
b) If m = 2kg and k = 4N/m, the natural frequencies of the system are found to be:
ω1= 0.38rad/s, ω2= 2.62rad/s, ω3= 1.41rad/s.
Find the corresponding mode shapes and normalize them so that (ui, ui)M = 1.
c) Determine the forced, uncoupled equa-tions of motion for the modal ampli-tudes Qi(t).
m k
m 2 k
m 2 k
f (t)
Problem 113:
For the mechanical system shown to the right, the uniform rigid bar with mass m and length ℓ is supported by identical springs while a block of mass 2 m is suspended from the bar. For this system
a) find the equations of motion in terms of x, z, and θ.
b) if k = 6400N/m, m = 8kg, and ℓ = 0.50m, and the natural frequencies of the system are
ω1= 15.65rad/s, ω2= 48.48rad/s, ω3= 77.50rad/s,
find the corresponding mode shapes for this system;
m, ℓ
G ℓ 4
k
k
k
m θ z
x