Problems and Solutions Section 1.1 (1.1 through 1.26)
1.1 Consider a simple pendulum (see Example 1.1.1) and compute the magnitude of the
restoring force if the mass of the pendulum is 3 kg and the length of the pendulum is 0.8 m.
Assume the pendulum is at the surface of the earth at sea level.
, which has
Solution: From example 1.1.1, the restoring force of the pendulum is
maximum value
1.2 Compute the period of oscillation of a pendulum of length 1.2 m at the North Pole where the
acceleration due to gravity is measured to be 9.832 m/s2.
Solution: The natural frequency and period can be computed with the following
relationships:
1.3 The spring of Figure 1.2, repeated here as Figure P1.3, is loaded with mass of 10 kg and
the corresponding (static) displacement is 0.012 m. Calculate the spring's stiffness.
Solution:
Free-body diagram:
From the free-body diagram and static
equilibrium:
)
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1.4
The spring of Figure P1.3 is successively loaded with mass and the corresponding (static)
displacement is recorded below. Plot the data and calculate the spring's stiffness. Note
that the data contain some error. Also calculate the standard deviation.
m(kg)
x(m)
10
1.14
11
1.25
12
1.37
13
1.48
14
1.59
15
1.71
16
1.82
Solution:
Free-body diagram:
From the free-body diagram and static
equilibrium:
The sample standard deviation in
computed stiffness is:
Plot of mass in kg versus displacement in m
Computation of slope from mg/x
m(kg)
x(m)
k(N/m)
10
1.14
86.05
11
1.25
86.33
12
1.37
85.93
13
1.48
86.17
14
1.59
86.38
15
1.71
86.05
16
1.82
86.24
1.5
Consider the pendulum of Example 1.1.1 and compute the amplitude of the
restoring force if the mass of the pendulum is 2 kg and the length of the pendulum is
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0.5 m if the pendulum is at the surface of the moon.
Solution: From example 1.1.1, the restoring force of the pendulum is
which has maximum value
1.5
1.6
,
Consider the pendulum of Example 1.1.1 and compute the angular natural frequency
(radians per second) of vibration for the linearized system if the mass of the
pendulum is 3 kg and the length of the pendulum is 0.8 m if the pendulum is at the
surface of the earth. What is the period of oscillation in seconds?
Solution: The natural frequency and period are:
1.7
Derive the solution of
and plot the result for at least two periods for the case with
ω n = 2 rad/s, x 0 = 1 mm, and v 0 =
mm/s.
Solution:
Given:
Assume:
. Then:
and
(1)
. Substitute into equation (1) to get:
Thus there are two solutions:
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The sum of x 1 and x 2 is also a solution so that the total solution is:
Substitute initial conditions: x 0 = 1 mm, v 0 =
mm/s
Therefore the solution is:
Using Mathcad the plot is:
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1.8
Solve
Solution: Here
for k = 4 N/m, m = 1 kg, x 0 = 1 mm, and v 0 = 0. Plot the solution.
. Calculating the initial conditions:
x(t)= cos (2t )
The following plot is from Mathcad:
Alternately students may use equation (1.10) directly to get
1.9
The amplitude of vibration of an undamped system is measured to be 1.5 mm. The phase
shift from t = 0 is measured to be 2 rad and the frequency is found to be 10 rad/s.
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Calculate the initial conditions that caused this vibration to occur. Assume the response
is of the form
=
x(t ) A sin(ωnt + φ ).
Solution:
Given:
,
Setting t = 0 in the above expressions yields:
1.10
Determine the stiffness of a single-degree-freedom, spring-mass system with a mass of
80 kg such that the natural frequency is 12 Hz.
Solution: Given: Single-degree-freedom spring-mass system with
But,
First change Hertz to radians and then use the formula for natural frequency:
Solution:
1.11
Find the equation of motion for the system of Figure P1.11, and find the natural
frequency. In particular, using static equilibrium along with Newton’s law, determine
what effect gravity has on the equation of motion and the system’s natural frequency.
Assume the block slides without friction.
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Figure P1.11
Solution:
Choosing a coordinate system along the plane with positive down the plane, the freebody diagram of the system for the static case is given and (a) and for the dynamic case
in (b):
In the figures, N is the normal force and the components of gravity are determined by the
angle θ as indicated. From the static equilibrium:
. Summing forces
in (b) yields:
1.12
An undamped system vibrates with a frequency of 8 Hz and amplitude 1.5 mm.
Calculate the maximum amplitude of the system's velocity and acceleration.
Solution:
Given: First convert Hertz to rad/s:
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For an undamped system:
and differentiating yields the velocity:
Realizing that both the sin and cos functions have maximum values of 1 yields:
Likewise for the acceleration:
1.13
Show by calculation that A sin (ωn t + φ) can be represented as A 1 sin ωn t + A 2 cosωn t and
calculate A 1 and A 2 in terms of A and φ.
Solution:
This trig identity is useful:
Given:
1.14
Using the solution of equation (1.2) in the form
calculate the values of A 1 and A 2 in terms of the initial conditions x 0 and v 0 .
Solution:
Using the solution of equation (1.2) in the form
and differentiate to get:
Now substitute the initial conditions into these expressions for the position and velocity
to get:
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Solving for A 1 and A 2 yields:
Thus
1.15
Using the drawing in Figure 1.7, verify that equation (1.10) satisfies the initial velocity
condition.
Solution: Following the lead given in Example 1.1.2, write down the general expression
of the velocity by differentiating equation (1.10):
From the figure:
Figure 1.7
Substitution of these values into the expression for v(0) yields
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verifying the agreement between the figure and the initial velocity condition.
1.16
A 5 kg mass is attached to a linear spring of stiffness 0.1 N/m. a) Determine the natural
frequency of the system in hertz. b) Repeat this calculation for a mass of 50 kg and a
stiffness of 10 N/m. Compare your result to that of part a.
Solution: From the definition of frequency and equation (1.12)
(a)
(b)
1.17
,
Derive the solution of the single degree of freedom system of Figure 1.4 by writing
Newton’s law, ma = -kx, in differential form using adx = vdv and integrating twice.
Solution: Substitute a = vdv/dx into the equation of motion ma = -kx, to get mvdv = kxdx. Integrating yields:
Here c 2 is a second constant of integration that is convenient to write as c 2 = -φ/ωn .
Rearranging yields
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in agreement with equation (1.19).
1.18
Determine the natural frequency of the two systems illustrated.
(a)
(b)
Figure P1.18
Solution:
(a) Summing forces from the free-body diagram in the x direction yields:
Free-body diagram for part a
Examining the coefficient of x
yields:
(b) Summing forces from the free-body diagram in the x direction yields:
Free-body diagram for part b
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1.19* Plot the solution given by equation (1.10) for the case k = 1000 N/m and m = 10 kg for
two complete periods for each of the following sets of initial conditions: a) x 0 = 0 m, v 0 =
1 m/s, b) x 0 = 0.01 m, v 0 = 0 m/s, and c) x 0 = 0.01 m, v 0 = 1 m/s.
Solution: Here we use Mathcad:
a) all units in m, kg, s
parts b and c are plotted in the above by simply changing the initial conditions as
appropriate
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1.20* Make a three dimensional surface plot of the amplitude A of an undamped oscillator
given by equation (1.9) versus x 0 and v 0 for the range of initial conditions given by –0.1
< x 0 < 0.1 m and -1 < v 0 < 1 m/s, for a system with natural frequency of 10 rad/s.
Solution: Working in Mathcad the solution is generated as follows:
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1.21
A machine part is modeled as a pendulum connected to a spring as illustrated in Figure
P1.21. Ignore the mass of pendulum’s rod and derive the equation of motion. Then
following the procedure used in Example 1.1.1, linearize the equation of motion and
compute the formula for the natural frequency. Assume that the rotation is small enough
so that the spring only deflects horizontally.
Figure P1.21
Solution: Consider the free body diagram of the mass displaced from equilibrium:
There are two forces acting on the system to consider, if we take moments about point O
(then we can ignore any forces at O). This yields
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Next consider the small θ approximations to that
. Then the
linearized equation of motion becomes:
Thus the natural frequency is
1.22
A pendulum has length of 300 mm. What is the system’s natural frequency in Hertz?
Solution:
Given:
Assumptions: Small angle approximation of sinθ.
From Window 1.1, the equation of motion for the pendulum is:
The coefficient of θ yields the natural frequency as:
1.23
The pendulum in Example 1.1.1 is required to oscillate once every second. What length
should it be?
Solution:
Given: f = 1 Hz (one cycle per second)
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1.24
The approximation of sin θ = θ, is reasonable for θ less than 10°. If a pendulum of length
0.5 m, has an initial position of θ(0) = 0, what is the maximum value of the initial angular
velocity that can be given to the pendulum with out violating this small angle
approximation? (be sure to work in radians)
Solution: From Window 1.1, the linear equation of the pendulum is
For zero initial position, the solution is given in equation (1.10) by
since sin is always less then one. Thus if we need θ < 10°= 0.175 rad, then we need to
solve:
for v 0 which yields:
v 0 < 0.773 rad/s.
1.25
A machine, modeled as a simple spring-mass system, oscillates in simple harmonic motion. Its
acceleration is measured to have an amplitude of 5,000 mm/s2 with a frequency of 10 Hz.
Compute the maximum displacement the machine undergoes during this oscillation.
Solution: The equations of motion for position and acceleration are
and
Since sin is max at 1, the maximum acceleration is
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Solving for A yields:
1.26
Derive the relationships given in Window 1.4 for the constants a 1 and a 2 used in the
exponential form of the solution in terms of the constants A 1 and A 2 used in sum of sine
and cosine form of the solution. Use the Euler relationships for sine and cosine in terms
of exponentials as given following equation (1.18).
Solution: Let θ = ωt for ease of notation. Then:
Adding these to in order to form x(t) yields:
Comparing this last expression to
yields:
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Problems and Solutions for Section 1.2 and Section 1.3 (1.27 to 1.64)
Problems and Solutions Section 1.2 (Numbers 1.27 through 1.40)
1.27
The acceleration of a machine part modeled as a spring mass system is measured
and recorded in Figure P 1.27. Compute the amplitude of the displacement of the
mass.
Figure P1.27
Solution: From Window 1.3 the maximum amplitude of the acceleration versus
time plot is just
where A is the maximum amplitude of the displacement and
the quantity to be determined here. Looking at P1.27, not that the plot repeats
itself twice after 2.5 s so that T = 2.5/2 = 1.25 s. Also the plot has 1 m/s2 as its
maximum value. Thus
and
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1.28
A vibrating spring and mass system has a measured acceleration amplitude of 12
mm/s2 and measured displacement amplitude of 1.5 mm. Calculate the system’s
natural frequency.
Solution: Given: The amplitude of displacement is 𝐴𝐴 = 1.5𝑚𝑚𝑚𝑚, and that of
acceleration is 𝜔𝜔𝑛𝑛2 𝐴𝐴 = 12 𝑚𝑚𝑚𝑚⁄𝑠𝑠 2 .
12
Therefore: 𝜔𝜔𝑛𝑛 = �1.5 = 2.828 𝑟𝑟𝑟𝑟𝑟𝑟⁄𝑠𝑠
1.29
A spring-mass system has measured period of 8 seconds and a known mass of 15
kg. Calculate the spring stiffness.
Solution: Given: 𝑇𝑇 = 8𝑠𝑠𝑠𝑠 = 15𝑘𝑘𝑘𝑘
𝑇𝑇 =
2𝜋𝜋
𝜋𝜋
⇒ 𝜔𝜔𝑛𝑛 = 𝑟𝑟𝑟𝑟𝑟𝑟⁄𝑠𝑠
𝜔𝜔𝑛𝑛
4
Using the basic formula for period and frequency
𝜋𝜋 2
𝜋𝜋 2
Solution:𝑘𝑘 = 𝜔𝜔𝑛𝑛2 × 𝑚𝑚 = �4 � × 𝑚𝑚 = �4 � × 15
𝑘𝑘𝑘𝑘
𝑠𝑠2
= 9.26 𝑁𝑁⁄𝑚𝑚
1.30* Plot the solution of a linear, spring and mass system with frequency 𝜔𝜔𝑛𝑛 = 3 rad/s,
x 0 = 1.2 mm and v 0 = 2.34 mm/s, for at least two periods.
Solution: Given initial data
𝑥𝑥0 = 1.2𝑚𝑚𝑚𝑚, 𝑣𝑣0 = 2.34 𝑚𝑚𝑚𝑚⁄𝑠𝑠 ; 𝜔𝜔𝑛𝑛 = 3 𝑟𝑟𝑟𝑟𝑟𝑟⁄𝑠𝑠
1
𝐴𝐴 = � � �((𝑤𝑤𝑛𝑛 𝑥𝑥0 )2 + 𝑣𝑣𝑜𝑜2 ) = 1.4312 𝑚𝑚𝑚𝑚,
𝜔𝜔𝑛𝑛
𝜙𝜙 = 𝑡𝑡𝑡𝑡𝑡𝑡−1 (𝜔𝜔𝑛𝑛 𝑥𝑥0 ⁄𝑣𝑣0 ) = 0.99442 radians.
𝑥𝑥(𝑡𝑡) = 𝐴𝐴𝐴𝐴𝐴𝐴𝐴𝐴(𝜔𝜔𝑛𝑛 𝑡𝑡 + 𝜙𝜙) = 1.4312sin(3t+0.9442)
1.31* Compute the natural frequency and plot the solution of a spring-mass system with
mass of 2 kg and stiffness of 4 N/m, and initial conditions of x 0 = 1 mm and v 0 =
0 mm/s, for at least two periods.
Solution: Given initial data
𝑚𝑚 = 2𝑘𝑘𝑘𝑘, 𝐾𝐾 = 4 𝑁𝑁⁄𝑚𝑚 , 𝑥𝑥0 = 1.0𝑚𝑚𝑚𝑚, 𝑣𝑣0 = 0 𝑚𝑚𝑚𝑚⁄𝑠𝑠 ;
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𝜔𝜔𝑛𝑛 = �(𝐾𝐾 ⁄𝑚𝑚) = �(4⁄2) = √2 𝑟𝑟𝑟𝑟𝑟𝑟⁄𝑠𝑠
𝐴𝐴 = (1⁄𝜔𝜔𝑛𝑛 )�((𝑤𝑤𝑛𝑛 𝑥𝑥0 )2 + 𝑣𝑣𝑜𝑜2 ) = (1.0⁄2)�(22 + 02 )
= 1.0
𝜙𝜙 = 𝑡𝑡𝑡𝑡𝑡𝑡−1 (𝜔𝜔𝑛𝑛 𝑥𝑥0 ⁄𝑣𝑣0 ) = 𝜋𝜋⁄2
𝑥𝑥(𝑡𝑡)=Asin(𝜔𝜔𝑛𝑛 t+ϕ)
𝜋𝜋
𝑥𝑥(𝑡𝑡) = 1.0sin �√2t+ �
2
1.32
When designing a linear spring-mass system it is often a matter of choosing a
spring constant such that the resulting natural frequency has a specified value.
Suppose that the mass of a system is 5 kg and the stiffness is 100 N/m. How
much must the spring stiffness be changed in order to increase the natural
frequency by 20%?
Solution: Given 𝑚𝑚 = 5𝑘𝑘𝑘𝑘and 𝑘𝑘 = 100 𝑁𝑁⁄𝑚𝑚, the natural frequenccy is
100
5
𝜔𝜔𝑛𝑛 = �
= √20 𝑟𝑟𝑟𝑟𝑟𝑟⁄𝑠𝑠
Increasing this value by 10% requires the new
frequency to be √20 x 1.2 = 5.366 rad/s
Solving for k given m and𝜔𝜔𝑛𝑛 yields
𝑘𝑘
𝑁𝑁
5.366 = � ⇒ 𝑘𝑘 = (5.366)2 (5) ≅ 144
5
𝑚𝑚
Thus the stiffness k must be increased by about
44%.
1.33
The pendulum in the Chicago Museum of Science and Industry has a length of 20
m and the acceleration due to gravity at that location is known to be 9.803 m/s2.
Calculate the period of this pendulum.
Solution:Following along through Example 1.2.2:
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1.34
Calculate the RMS values of displacement, velocity and acceleration for the
undamped single degree of freedom system of equation (1.19) with zero phase.
Solution: Calculate RMS values
Let
Mean Square Value:
Therefore,
1.35
A foot pedal mechanism for a machine is crudely modeled as a pendulum
connected to a spring as illustrated in Figure P1.35. The purpose of the spring is
provide a return force for the pedal action. Compute the spring stiffness needed
to keep the pendulum at 1° from the horizontal and then compute the
corresponding natural frequency. Assume that the angular deflections are small,
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such that the spring deflection can be approximated by the arc length, that the
pedal may be treated as a point mass and that pendulum rod has negligible mass.
The pedal is horizontal when the spring is at its free length. The values in the
figure are m = 0.8 kg, g = 9.8 m/s2, l 1 = 0.2 m and l 2 = 0.5 m.
Solution: You may want to note to your students that many systems with springs are
often designed based on static deflections to hold parts in specific positions as in this
case, and yet allow some motion. The free-body diagram for the system is given in
the figure.
For static equilibrium the sum of moments about point O yields (𝜃𝜃 1 is the static
deflection):
R
� 𝑀𝑀𝑂𝑂 = −𝑙𝑙1 𝜃𝜃1 (𝑙𝑙1 )𝑘𝑘 + 𝑚𝑚𝑚𝑚𝑚𝑚2 = 0
⇒ 𝑙𝑙12𝑘𝑘 𝜃𝜃1 = 𝑚𝑚𝑚𝑚𝑚𝑚2
𝑚𝑚𝑚𝑚𝑚𝑚2 0.8 × 9.81 × 0.5
⇒ 𝑘𝑘 = 2
=
= 5620.72 𝑁𝑁⁄𝑚𝑚
𝜋𝜋
𝑙𝑙1 𝜃𝜃1
0.22
180
(1)
Again taking moments abut point O, to get the dynamic equation of motion
� 𝑀𝑀𝑜𝑜 = 𝐽𝐽𝜃𝜃¨ = −𝑙𝑙12𝑘𝑘 (𝜃𝜃 + 𝜃𝜃1 ) + 𝑚𝑚𝑚𝑚𝑚𝑚2 = −𝑙𝑙12 𝑘𝑘𝑘𝑘 + 𝑙𝑙12 𝑘𝑘𝜃𝜃1 − 𝑚𝑚𝑚𝑚𝑚𝑚2 𝜃𝜃
Next using equation (1) above for the static deflection yields:
𝑚𝑚𝑚𝑚22 𝜃𝜃¨ + 𝑙𝑙12 𝑘𝑘𝑘𝑘 = 0
⇒ 𝜃𝜃¨ + �
𝑙𝑙12 𝑘𝑘
� 𝜃𝜃 = 0
𝑚𝑚𝑚𝑚22
⇒ 𝜔𝜔𝑛𝑛 =
𝑙𝑙1
0.2 5620.72
𝑟𝑟𝑟𝑟𝑟𝑟
𝑘𝑘
�� � =
��
� = 37.49
𝑙𝑙2 𝑚𝑚
0.5
𝑠𝑠
0.8
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1.36
An automobile is modeled as a 1200-kg mass supported by a spring of stiffness k
= 480,000 N/m. When it oscillates it does so with a maximum deflection of 10
cm. When loaded with passengers, the mass increases to as much as 1000 kg.
Calculate the change in frequency, velocity amplitude, and acceleration amplitude
if the maximum deflection remains 10 cm.
Solution:
Given: m 1 = 1200 kg,
m 2 =1000 kg,
k = 480,000 N/m,
x max = A = 10 cm
𝑘𝑘
480000
𝜔𝜔𝑛𝑛1 = ��𝑚𝑚 � = ��
𝑘𝑘
1
480000
𝜔𝜔𝑛𝑛2 = ��𝑚𝑚 � = ��
2
1200
1000
� = 20
𝛥𝛥𝛥𝛥 = 2𝜋𝜋 =
1.91
2𝜋𝜋
𝑠𝑠
� = 21.91
𝛥𝛥𝛥𝛥 = 21.91 − 20 = 1.91
𝛥𝛥
𝑟𝑟𝑟𝑟𝑟𝑟
𝑟𝑟𝑟𝑟𝑟𝑟
𝑣𝑣1 = 𝐴𝐴𝜔𝜔𝑛𝑛1 = 10 × 20 = 200
𝑐𝑐𝑐𝑐
𝑠𝑠
𝑣𝑣2 = 𝐴𝐴𝜔𝜔𝑛𝑛2 = 10 × 21.19 = 211.9
𝑎𝑎1 =
𝑎𝑎2 =
𝜟𝜟𝜟𝜟 = 𝑉𝑉2 − 𝑣𝑣1 = 𝟏𝟏𝟏𝟏. 𝟗𝟗
2
𝐴𝐴𝜔𝜔𝑛𝑛1
2
𝐴𝐴𝜔𝜔𝑛𝑛2
2
𝒔𝒔
= 10 × 20 = 4000
2
𝑠𝑠
𝑠𝑠
= 0.304𝐻𝐻𝐻𝐻
𝒄𝒄𝒄𝒄
𝑟𝑟𝑟𝑟𝑟𝑟
𝑐𝑐𝑐𝑐
𝑠𝑠
𝑐𝑐𝑐𝑐2
𝑠𝑠
= 10 × 21.19 = 4490.161
𝜟𝜟𝜟𝜟 = 4490.161 − 4000 = 𝟒𝟒𝟒𝟒𝟒𝟒. 𝟏𝟏𝟏𝟏𝟏𝟏
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𝑐𝑐𝑐𝑐2
𝑠𝑠
𝒄𝒄𝒄𝒄𝟐𝟐
𝒔𝒔
1.37
The front suspension of some cars contains a torsion rod as illustrated in Figure
P1.37 to improve the car’s handling. (a) Compute the frequency of vibration of
the wheel assembly given that the torsional stiffness is 2500 N m/rad and the
wheel assembly has a mass of 40 kg. Take the distance x = 0.26 m. (b)
Sometimes owners put different wheels and tires on a car to enhance the
appearance or performance. Suppose a thinner tire is put on with a larger wheel
raising the mass to 45 kg. What effect does this have on the frequency?
Solution: (a) Ignoring the moment of inertia of the rod and computing the
moment of inertia of the wheel as 𝑚𝑚𝑚𝑚 2 , the frequency of the shaft mass system is
𝜔𝜔𝑛𝑛 = �
𝑘𝑘
2500
𝑟𝑟𝑟𝑟𝑟𝑟
= ��
� = 30.41
2
2
𝑚𝑚𝑚𝑚
40 × 0.26
𝑠𝑠
𝜔𝜔𝑛𝑛 = �
𝑘𝑘
2500
𝑟𝑟𝑟𝑟𝑟𝑟
= ��
� = 28.67
2
2
𝑚𝑚𝑚𝑚
45 × 0.26
𝑠𝑠
(b) The same calculation with 45 kg will reduce the frequency to
1.38
A machine oscillates in simple harmonic motion and appears to be well modeled
by an undamped single-degree-of-freedom oscillation. Its acceleration is
measured to have an amplitude of 12,000 mm/s2 at 8 Hz. What is the machine's
maximum displacement?
Solution:
Given 𝑎𝑎𝑚𝑚𝑚𝑚𝑚𝑚 = 12000 𝑚𝑚𝑚𝑚⁄𝑠𝑠 2 @ 8𝐻𝐻𝐻𝐻
The equations of motion for position and acceleration are:
The amplitude of acceleration is 𝐴𝐴𝜔𝜔𝑛𝑛2 = 12,000
16𝜋𝜋 𝑟𝑟𝑟𝑟𝑟𝑟⁄𝑠𝑠, from equation (1.12).
The machine’s displacement is A =
𝑨𝑨 = 𝟒𝟒. 𝟕𝟕𝟕𝟕𝟕𝟕 𝒎𝒎𝒎𝒎
12000
𝜔𝜔𝑛𝑛 2
𝑚𝑚𝑚𝑚
𝑠𝑠2
and 𝜔𝜔𝑛𝑛 = 2𝜋𝜋𝜋𝜋 = 2𝜋𝜋(8) =
12000
= (16𝜋𝜋)2
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1.39
A simple undamped spring-mass system is set into motion from rest by giving it
an initial velocity of 80 mm/s. It oscillates with a maximum amplitude of 10 mm.
What is its natural frequency?
Solution:
Given: 𝑥𝑥0 = 0, 𝑣𝑣0 = 80 𝑚𝑚𝑚𝑚/𝑠𝑠, 𝐴𝐴 = 10 𝑚𝑚𝑚𝑚
1.40
From equation (1.9), 𝐴𝐴 =
𝑣𝑣0
𝜔𝜔𝑛𝑛
or𝜔𝜔𝑛𝑛 =
𝑣𝑣0
𝐴𝐴
=
80
, so that: 𝝎𝝎𝒏𝒏 = 𝟖𝟖 rad/s
10
An automobile exhibits a vertical oscillating displacement of maximum amplitude
5 cm and a measured maximum acceleration of 2000 cm/s2. Assuming that the
automobile can be modeled as a single-degree-of-freedom system in the vertical
direction, calculate the natural frequency of the automobile.
Solution:
Given: A = 5 cm. From equation (1.15)
Solving for ωn yields:
Problems Section 1.3 (Numbers 1.41 through 1.64)
1.41
Consider a spring mass damper system, like the one in Figure 1.9, with the
following values: m =12 kg, c = 4 N/s and k = 1200 N/m. a) Is the system
overdamped, underdamped or critically damped? b) Compute the solution if the
system is given initial conditions x 0 = 0.01 m and v 0 = 0.
Solution:a) Using equation 1.30 the damping ratio is
𝜁𝜁 =
𝑐𝑐
2√𝑘𝑘𝑘𝑘
=
4
2√1200 ∙ 12
Thus the system is underdamped.
= 0.016 < 1
b) Using equations (1.38) the amplitude and phase can be calculated from the
initial conditions:
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𝑘𝑘
𝜔𝜔𝑛𝑛 = �� �
𝑚𝑚
𝜔𝜔𝑑𝑑 = ((1 − 𝜁𝜁 2 )0.5 )
𝜔𝜔𝑛𝑛 𝐴𝐴 = 𝑥𝑥0 𝐵𝐵 =
𝑥𝑥(𝑡𝑡) = 𝑒𝑒
𝜁𝜁𝜔𝜔𝑛𝑛 𝑥𝑥0 − 𝑣𝑣0
𝑥𝑥 = �𝑒𝑒 (−𝜁𝜁𝜔𝜔𝑛𝑛𝑡𝑡) ��𝐴𝐴𝐴𝐴𝐴𝐴𝐴𝐴[𝜔𝜔𝑑𝑑 𝑡𝑡] + 𝐵𝐵(𝑠𝑠𝑠𝑠𝑠𝑠[𝜔𝜔𝑑𝑑 𝑡𝑡])�
𝜔𝜔𝑑𝑑
So the solution is
+ 0.00016669𝑠𝑠𝑠𝑠𝑠𝑠[9.99861𝑡𝑡])
−0.166667∗𝑡𝑡 (0.01𝑐𝑐𝑐𝑐𝑐𝑐[9.99861𝑡𝑡]
Note that for any system with v 0 = 0 the phase is strictly a function of the
damping ratio.
1.42
Consider a spring-mass-damper system with equation of motion given by
. Compute the damping ratio and determine if the system is
overdamped, underdamped or critically damped.
Solution: The parameter values are m = 1, k = 2 and c = 2. From equation (1.30)
the damping ratio is
Hence the system is underdamped.
1.43
Consider the system
for x 0 = 1.2 mm, v 0 = 0 mm/s. Is this system
overdamped, underdamped or critically damped? Compute the solution and
determine which root dominates as time goes on (that is, one root will die out
quickly and the other will persist).
Solution:Using equation 1.30 the damping ratio is
𝜁𝜁 =
Hence the system is overdamped.
𝑐𝑐
2√𝑘𝑘𝑘𝑘
=2
Given x + 4x + x=0 where x0 = 1.2mm; v0 = 0
Let 𝑥𝑥=aert ⇒ 𝑥𝑥˙ =arert ⇒ 𝑥𝑥¨ =ar2 𝑒𝑒 rt
Substitute these into the equation of motion to get:
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ar2 𝑒𝑒 rt + 4arert +aert = 0
⇒ 𝑟𝑟 2 + 4r+1 = 0 ⇒ 𝑟𝑟1,2 = −2 ± �(3)
So
x = a1 𝑒𝑒
�−2+�(3)�𝑡𝑡
+ a2 𝑒𝑒
𝑥𝑥˙ = �−2 + �(3)� 𝑎𝑎1 𝑒𝑒
�−2−�(3)�𝑡𝑡
�−2+�(3)�𝑡𝑡
+ �−2 − �(3)� 𝑎𝑎2 𝑒𝑒
Applying initial conditions yields,
x 0 = a1 + a 2 ⇒ x 0 − a 2 = a1
(
)
(
)
(1)
v 0 = − 2 + 3 a1 + − 2 − 3 a 2
Substitute equation (1) into (2)
(
)
�−2−�(3)�𝑡𝑡
(2)
(
)
v0 = − 2 + 3 ( x0 − a 2 ) + − 2 − 3 a 2
(
)
v0 = − 2 + 3 x0 − 2 3a 2
Solve for a 2
𝑎𝑎2 =
−𝑣𝑣0 + ��−2 + �(3)� 𝑥𝑥0 �
2�(3)
Substituting the value of a 2 into equation (1), and solving for a 1 yields,
𝑎𝑎1 =
𝑣𝑣0 + �2 + �(3)� 𝑥𝑥0
𝑥𝑥(𝑡𝑡) =
2�(3)
𝑣𝑣0 + �2 + �(3)� 𝑥𝑥0
2�(3)
𝑒𝑒
�−2+�(3)�𝑡𝑡
+
−𝑣𝑣0 + �−2 + �(3)� 𝑥𝑥0
2�(3)
𝑒𝑒 �−2−√3�𝑡𝑡
The response is dominated by the root �−2 + √3�𝑡𝑡 as the other root dies off very fast.
1.44
Compute the solution to
for x 0 = 0 mm, v 0 = 1 mm/s and write
down the closed form expression for the response.
Solution:
The parameter values are m = 1, k = 2 and c = 2. From equation (1.30) the
damping ratio is
Hence the system is underdamped. The natural frequency is
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Thus equations (1.36) and (1.38) can be used directly or one can follow the last
expression in Example 1.3.3:
The damped natural frequency is
Thus the solution is
Alternately use equations (1.36) and (1.38). The plot is similar to figure 1.11.
1.45
Derive the form of λ 1 and λ 2 given by equation (1.31) from equation (1.28) and
the definition of the damping ratio.
Solution:
Equation (1.28):
Rewrite,
Rearrange,
Substitute:
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1.46
Use the Euler formulas to derive equation (1.36) from equation (1.35) and to
determine the relationships listed in Window 1.4.
Solution:
Equation (1.35):
From Euler,
Let: A 1 =
, A2=
, then this last expression becomes
Next use the trig identity:
to get:
1.47
Using equation (1.35) as the form of the solution of the underdamped system,
calculate the values for the constants a 1 and a 2 in terms of the initial conditions x 0
and v 0 .
Solution:
Equation (1.35):
Initial conditions
(1)
(2)
Substitute equation (1) into equation (2) and solve for a 2
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Solve for a 2
Substitute the value for a 2 into equation (1), and solve for a 1
1.48
Calculate the constants A and φ in terms of the initial conditions and thus verify
equation (1.38) for the underdamped case.
Solution:
From Equation (1.36),
Applying initial conditions (t = 0) yields,
(1)
(2)
Next solve these two simultaneous equations for the two unknowns A and φ.
From (1),
(3)
Substituting (3) into (1) yields
.
Hence,
(4)
From (3),
(5)
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and From (4),
(6)
Substituting (5) and (6) into (2) yields,
which are the same as equation (1.38)
1.49
Calculate the constants a 1 and a 2 in terms of the initial conditions and thus verify
equations (1.42) and (1.43) for the overdamped case.
Solution: From Equation (1.41)
taking the time derivative yields:
Applying initial conditions yields,
Substitute equation (1) into equation (2) and solve for a 2
Solve for a 2
Substitute the value for a 2 into equation (1), and solve for a 1
1.50
Calculate the constants a 1 and a 2 in terms of the initial conditions and thus verify
Copyright © 2015 Pearson Education Ltd.
equation (1.46) for the critically damped case.
Solution:
From Equation (1.45),
Applying the initial conditions yields:
(1)
and
(2)
solving these two simultaneous equations for the two unknowns a 1 and a 2 .
Substituting (1) into (2) yields,
which are the same as equation (1.46).
1.51
Using the definition of the damping ratio and the undamped natural frequency,
derive equitation (1.48) from (1.47).
Solution:
thus,
thus,
Therefore,
becomes,
1.52
For a damped system, m, c, and k are known to be m = 1.5 kg, c = 2.2 kg/s, k = 12
N/m. Calculate the value of ζ and ωn . Is the system overdamped, underdamped,
or critically damped?
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Solution:
𝑁𝑁
Given: 𝑚𝑚 = 1.5𝑘𝑘𝑘𝑘, 𝑘𝑘 = 12 𝑚𝑚 , 𝑐𝑐 = 2.2
𝑘𝑘
𝑘𝑘𝑘𝑘
𝑠𝑠
12𝑁𝑁⁄𝑚𝑚
Natural Frequency: 𝜔𝜔𝑛𝑛 = � = � 1.5𝑘𝑘𝑘𝑘 = 2.82843𝑟𝑟𝑟𝑟𝑟𝑟⁄𝑠𝑠
𝑀𝑀
Damping ratio: 𝜁𝜁 =
𝑐𝑐
𝑐𝑐
2.2 𝑘𝑘𝑘𝑘⁄𝑠𝑠
⇒ 𝜁𝜁 =
=
= 0.25927
𝑐𝑐𝑐𝑐
2𝑀𝑀𝜔𝜔𝑛𝑛 2 × 1.5𝑘𝑘𝑘𝑘 × 2.82843 𝑟𝑟𝑟𝑟𝑟𝑟⁄𝑠𝑠
Damped natural frequency: 𝜔𝜔𝑑𝑑 = 𝜔𝜔𝑛𝑛 �1 − 𝜁𝜁 2 = 2.7317 rad/s
As 0 < 𝜁𝜁 < 1, the system is underdamped.
1.53
Plot x(t) for a damped system of natural frequency ωn = 2 rad/s and initial
conditions x 0 = 1 mm, v 0 = 1 mm, for the following values of the damping ratio:
ζ = 0.01, ζ = 0.2, ζ = 0.1, ζ = 0.4, and ζ = 0.8.
Solution:
Given: ωn = 2 rad/s, x 0 = 1 mm, v 0 = 1 mm, ζ i = [0.01; 0.2; 0.1; 0.4; 0.8]
Underdamped cases:
From equation 1.38,
The response is plotted for each value of the damping ratio in the following using
Matlab:
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1.54
Plot the response x(t) of an underdamped system with ωn = 2 rad/s, ζ = 0.1, and
v 0 = 0 for the following initial displacements: x 0 = 10 mm and x 0 = 100 mm.
Solution:
Given: ωn = 2 rad/s, ζ = 0.1, v 0 = 0, x 0 = 10 mm and x 0 = 100 mm.
Underdamped case:
where
The following is a plot from Matlab.
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1.55
Calculate the solution to
with x 0 = 1 and v 0 =0 for x(t) and sketch the
response.
Solution: This is a problem with negative damping which can be used to tie into
Section 1.8 on stability, or can be used to practice the method for deriving the
solution using the method suggested following equation (1.13) and eluded to at
the start of the section on damping. To this end let
the equation of
motion to get:
This yields the characteristic equation:
There are thus two solutions as expected and these combine to form
Using the Euler relationship for the term in parenthesis as given in Window 1.4,
this can be written as
Next apply the initial conditions to determine the two constants of integration:
Differentiate the solution to get the velocity and then apply the initial velocity
condition to get
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This function oscillates with increasing amplitude as shown in the following plot
which shows the increasing amplitude. This type of response is referred to as a
flutter instability. This plot is from Mathcad.
1.56
A spring-mass-damper system has mass of 120 kg, stiffness of 3600 N/m and
damping coefficient of 330 kg/s. Calculate the undamped natural frequency, the
damping ratio and the damped natural frequency. Does the solution oscillate?
Solution:Working straight from the definitions:
𝜁𝜁 =
𝑘𝑘
3600
𝑟𝑟𝑟𝑟𝑟𝑟
𝜔𝜔𝑛𝑛 = �� � = ��
� = 5.48
𝑚𝑚
120
𝑠𝑠
𝑐𝑐
330
330
=
=
= 0.251
𝑐𝑐𝑐𝑐𝑐𝑐 2�(𝑘𝑘𝑘𝑘) 2�(120)(3600)
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Since 𝜁𝜁 is less than 1, the solution is under damped and will oscillate. The damped
natural frequency is 𝜔𝜔𝑑𝑑 = 𝜔𝜔𝑛𝑛 �(1 − 𝜁𝜁 2 ) = 5.48 ��(1 − 0.2512 )� = 5.305
1.57
𝑟𝑟𝑟𝑟𝑟𝑟
.
𝑠𝑠
A sketch of a valve and rocker arm system for an internal combustion engine is
give in Figure P1.57. Model the system as a pendulum attached to a spring and a
mass and assume the oil provides viscous damping in the range of ζ = 0.01.
Determine the equations of motion and calculate an expression for the natural
frequency and the damped natural frequency. Here J is the rotational inertia of
the rocker arm about its pivot point, k is the stiffness of the valve spring and m is
the mass of the valve and stem. Ignore the mass of the spring.
Figure P1.57
Solution: The model is of the form given in the figure. You may wish to give this
figure as a hint as it may not be obvious to all students.
Taking moments about the pivot point yields:
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Next divide by the leading coefficient to get;
From the coefficient of q, the undamped natural frequency is
From equation (1.37), the damped natural frequency becomes
This is effectively the same as the undamped frequency for any reasonable
accuracy. However, it is important to point out that the resulting response will
still decay, even though the frequency of oscillation is unchanged. So even
though the numerical value seems to have a negligible effect on the frequency of
oscillation, the small value of damping still makes a substantial difference in the
response.
1.58
A spring-mass-damper system has mass of 160 kg, stiffness of 2000 N/m and
damping coefficient of 250 kg/s. Calculate the undamped natural frequency, the
damping ratio and the damped natural frequency. Is the system overdamped,
underdamped or critically damped? Does the solution oscillate?
Solution:Working straight from the definitions:
𝑘𝑘
2000 𝑁𝑁⁄𝑚𝑚
𝜔𝜔𝑛𝑛 = � = �
= 3.535 𝑟𝑟𝑟𝑟𝑟𝑟⁄𝑠𝑠
𝑚𝑚
160𝑘𝑘𝑘𝑘
𝜁𝜁 =
𝑐𝑐
250
250
=
=
= 0.221
𝑐𝑐𝑐𝑐𝑐𝑐 2√𝑘𝑘𝑘𝑘 2�(160)(2000)
This last expression follows from the equation following equation (1.29). Since 𝜁𝜁
is less than 1, the solution is underdamped and will oscillate. The damped natural
frequency is 𝜔𝜔𝑑𝑑 = ��1 − 𝜁𝜁 2 � = 3.447 𝑟𝑟𝑟𝑟𝑟𝑟⁄𝑠𝑠, which follows from equation
(1.37).
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1.59* The spring mass system of 120 kg mass, stiffness of 3200 N/m and damping
coefficient of 350 Ns/m is given a zero initial velocity and an initial displacement
of 0.12 m. Calculate the form of the response and plot it for as long as it takes to
die out.
Solution: Working straight from the definitions:
𝑘𝑘
3200 𝑁𝑁⁄𝑚𝑚
𝜔𝜔𝑛𝑛 = � ⟹ 𝜔𝜔𝑛𝑛 = �
= 5.164 𝑟𝑟𝑟𝑟𝑟𝑟⁄𝑠𝑠
𝑚𝑚
120𝑘𝑘𝑘𝑘
𝜁𝜁 =
𝑐𝑐
350
350
=
=
= 0.2824
𝑐𝑐𝑐𝑐𝑐𝑐 2√𝑘𝑘𝑘𝑘 2�(120)(3200)
𝜔𝜔𝑑𝑑 = ��1 − 𝜁𝜁 2 � = ��1 − (0.2824)2 � 5.164 𝑟𝑟𝑟𝑟𝑟𝑟⁄𝑠𝑠 = 4.9538 𝑟𝑟𝑟𝑟𝑟𝑟⁄𝑠𝑠
Since 𝜁𝜁 is less than 1, the solution is underdamped and will oscillate.
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1.60*The spring mass system of 180 kg mass, stiffness of 1800 N/m and damping
coefficient of 250 Ns/m is given an initial velocity of 12 mm/s and an initial
displacement of -6 mm. Calculate the form of the response and plot it for as long as it
takes to die out. How long does it take to die out?
Solution:Working straight from the definitions:
𝑘𝑘
1800 𝑁𝑁⁄𝑚𝑚
𝜔𝜔𝑛𝑛 = � ⟹ 𝜔𝜔𝑛𝑛 = �
= 3.162277 𝑟𝑟𝑟𝑟𝑟𝑟⁄𝑠𝑠
𝑀𝑀
180𝑘𝑘𝑘𝑘
𝜁𝜁 =
𝑐𝑐
= 0.219
𝑐𝑐𝑐𝑐𝑐𝑐
𝜔𝜔𝑑𝑑 = ��1 − 𝜁𝜁 2 �𝜔𝜔𝑛𝑛 as 𝜁𝜁 < 1 ⟹ 𝜔𝜔𝑑𝑑 = ��1 − (0.219)2 � 3.162277 𝑟𝑟𝑟𝑟𝑟𝑟⁄𝑠𝑠 =
3.08508 𝑟𝑟𝑟𝑟𝑟𝑟⁄𝑠𝑠
Since 𝜁𝜁 is less than 1, the solution is underdamped and will oscillate.
1.61* Choose the damping coefficient of a spring-mass-damper system with mass of
160 kg and stiffness of 2500 N/m such that it’s response will die out after about 2
s, given a zero initial position and an initial velocity of 12 mm/s.
Solution:Working straight from the definitions:
𝑘𝑘
2500 𝑁𝑁⁄𝑚𝑚
𝜔𝜔𝑛𝑛 = � ⟹ 𝜔𝜔𝑛𝑛 = �
= 3.9528 𝑟𝑟𝑟𝑟𝑟𝑟⁄𝑠𝑠
160𝑘𝑘𝑘𝑘
𝑀𝑀
𝜁𝜁 =
𝑐𝑐
= 0.219
𝑐𝑐𝑐𝑐𝑐𝑐
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𝜔𝜔𝑑𝑑 = ��1 − 𝜁𝜁 2 � 𝜔𝜔𝑛𝑛 = ��1 − (0.219)2 � 3.162277 𝑟𝑟𝑟𝑟𝑟𝑟⁄𝑠𝑠 = 3.08508 𝑟𝑟𝑟𝑟𝑟𝑟⁄𝑠𝑠
Since ζ is less than 1, the solution is underdamped and will oscillate.
1.62
Derive the equation of motion of the system in Figure P1.62 and discuss the effect
of gravity on the natural frequency and the damping ratio.
Solution: This requires two free body diagrams. One for the dynamic case and
one to show static equilibrium.
(a)
(b)
From the free-body diagram of static equilibrium (b) we have that mg = k∆x,
where ∆x represents the static deflection. From the free-body diagram of the
dynamic case given in (a) the equation of motion is:
From the diagram, y(t) = x(t) + ∆x. Since ∆x is a constant, differentiating and
substitution into the equation of motion yields:
where the last term is zero from the relation resulting from static equilibrium.
Dividing by the mass yields the standard form
It is clear that gravity has no effect on the damping ratio ζ or the natural
frequency ωn . Not that the damping force is not present in the static case because
the velocity is zero.
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1.63
Derive the equation of motion of the system in Figure P1.63 and discuss the effect
of gravity on the natural frequency and the damping ratio. You may have to make
some approximations of the cosine.
Assume the bearings provide a viscous
damping force only in the vertical direction. (From the A. Diaz-Jimenez, South
African Mechanical Engineer, Vol. 26, pp. 65-69, 1976)
Solution: First consider a free-body diagram of the system:
Let α be the angel between the damping and stiffness force. The equation of
motion becomes
From static equilibrium, the free-body diagram (above with c = 0 and stiffness
. Thus the equation of motion
force kδ s ) yields:
becomes
(1)
Next consider the geometry of the dynamic state:
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From the definition of cosine applied to the two different triangles:
Next assume small deflections so that the angles are nearly the same cos α = cos
θ, so that
For small motion, then this last expression can be substituted into the equation of
motion (1) above to yield:
, α and x small
Thus the frequency and damping ratio have the standard values and are not
effected by gravity. If the small angle assumption is not made, the frequency can
be approximated as
as detailed in the reference above. For a small angle these reduce to the normal
values of
as derived here.
1.64
Consider the response of an underdamped system given by
whereA and φ are given in terms of the initial conditions x 0 = 0, and v 0 ≠ 0.
Determine the maximum value that the acceleration will experience in terms of v 0 .
Solution: From equation (1.XX) for underdamped systems the phase and
amplitude for x 0 = 0 become:
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Problems and Solutions Section 1.4 (problems 1.65 through 1.81)
1.65
Calculate the frequency of the compound pendulum of Figure P1.65 if a mass m T
is added to the tip, by using the energy method. Assume the mass of the pendulum
is evenly distributed so that its center of gravity is in the middle of the pendulum
of length l.
Figure P1.65 A compound pendulum with a tip mass.
Solution Adding a tip mass adds both kinetic and potential energy to the system.
If the mass of the pendulum bar is m, and it is lumped at the center of mass the
energies become:
Potential Energy:
Kinetic Energy:
Conservation of energy (Equation 1.51) requires T + U = constant:
Differentiating with respect to time yields:
Rearranging and approximating using the small angle formula sin θ ~ θ, yields:
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Note that this solution makes sense because if m t = 0 it reduces to the frequency
of the pendulum equation for a bar, and if m = 0 it reduces to the frequency of a
massless pendulum with only a tip mass.
1.66
Calculate the total energy in a damped system with frequency 2.5 rad/s and
damping ratio ζ = 0.02 with mass 12 kg for the case x 0 = 0.1 m and v 0 = 0. Plot
the total energy versus time.
Solution: Given
Calculate the stiffness and damped natural frequency:
The total energy of the damped system is
Where
Applying the initial conditions to evaluate the constants of integration yields:
Substitution of these values into E(t) yields:
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1.67
Use the energy method to calculate the equation of motion and natural frequency
of an airplane's steering mechanism for the nose wheel of its landing gear. The
mechanism is modeled as the single-degree-of-freedom system illustrated in
Figure P1.54.
The steering wheel and tire assembly are modeled as being fixed at ground for
this calculation. The steering rod gear system is modeled as a linear spring and
mass system (m, k 2 ) oscillating in the x direction. The shaft-gear mechanism is
modeled as the disk of inertia J and torsional stiffness k 2 . The gear J turns
through the angle θ such that the disk does not slip on the mass. Obtain an
equation in the linear motion x.
Solution: From kinematics:
Kinetic energy:
Potential energy:
Substitute
:
Derivative:
Equation of motion:
Natural frequency:
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1.68 Consider the pendulum and spring system of Figure P1.68. Here the mass of the
pendulum rod is negligible. Derive the equation of motion using the energy
method. Then linearize the system for small angles and determine the natural
frequency. The length of the pendulum is l, the tip mass is m, and the spring
stiffness is k.
Figure P1.68 A simple pendulum connected to a spring
Solution: Writing down the kinetic and potential energy yields:
Here the soring deflects a distance lsin θ, and the mass drops a distance l(1 –cosθ).
Adding up the total energy and taking its time derivative yields:
For small θ, this becomes
1.69
A control pedal of an aircraft can be modeled as the single-degree-of-freedom
system of Figure P1.69. Consider the lever as a massless shaft and the pedal as a
lumped mass at the end of the shaft. Use the energy method to determine the
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equation of motion in θ and calculate the natural frequency of the system. Assume
the spring to be unstretched at θ = 0.
Figure P1.69
Solution: In the figure let the mass at θ = 0 be the lowest point for potential
energy. Then, the height of the mass m is (1-cosθ) 2 .
Kinematic relation: x =  1 θ
Kinetic Energy:
Potential Energy:
Taking the derivative of the total energy yields:
Rearranging, dividing by dθ/dt and approximating sinθ with θ yields:
1.70
To save space, two large pipes are shipped one stacked inside the other as
indicated in Figure P1.70. Calculate the natural frequency of vibration of the
smaller pipe (of radius R 1 ) rolling back and forth inside the larger pipe (of radius
R). Use the energy method and assume that the inside pipe rolls without slipping
and has a mass m.
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Figure P1.70
Solution: Let θ be the angle that the line between the centers of the large pipe and
the small pipe make with the vertical and let α be the angle that the small pipe
rotates through. Let R be the radius of the large pipe and R 1 the radius of the
smaller pipe. Then the kinetic energy of the system is the translational plus
rotational of the small pipe. The potential energy is that of the rise in height of
the center of mass of the small pipe.
From the drawing:
Likewise examination of the value of x yields:
Let β denote the angle of rotation that the small pipe experiences as viewed in the
inertial frame of reference (taken to be the truck bed in this case). Then the total
kinetic energy can be written as:
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The total potential energy becomes just:
Now it remains to evaluate the angle β. Let α be the angle that the small pipe
rotates in the frame of the big pipe as it rolls (say) up the inside of the larger pipe.
Then
β=θ–α
were α is the angle “rolled” out as the small pipe rolls from a to b in figure
P1.56. The rolling with out slipping condition implies that arc length a’b must
equal arc length ab. Using the arc length relation this yields that Rθ =R 1 α.
Substitution of the expression β = θ – α yields:
which is the relationship between angular motion of the small pipe relative to the
ground (β) and the position of the pipe (θ). Substitution of this last expression into
the kinetic energy term yields:
Taking the derivative of T + V yields
Using the small angle approximation for sine this becomes
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1.71
Consider the example of a simple pendulum given in Example 1.4.2.
The
pendulum motion is observed to decay with a damping ratio of ζ = 0.005.
Determine a damping coefficient and add a viscous damping term to the
pendulum equation.
Solution: From example 1.4.2, the equation of motion for a simple pendulum is
So
. With viscous damping the equation of motion in normalized form
becomes:
or with as given:
The coefficient of velocity terms is
1.72
Determine a damping coefficient for the disk-rod system of Example 1.4.3.
Assuming that the damping is due to the material properties of the rod, determine
c for the rod if it is observed to have a damping ratio of ζ = 0.008.
Solution: The equation of motion for a disc/rod in torsional vibration is
or
Add viscous damping:
From the velocity term, the damping coefficient must be
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1.73
The rod and disk of Window 1.1 are in torsional vibration. Calculate the damped
natural frequency if J = 1200 m2 ⋅ kg, c = 25 N⋅ m⋅ s/rad, and k = 500 N⋅m/rad.
Solution: The equation of motion is
The damped natural frequency is
Where
and
Thus the damped natural frequency is
1.74
Consider the system of P1.74, which represents a simple model of an aircraft
landing system. Assume, x = rθ. What is the damped natural frequency?
Figure P1.74
Solution: Ignoring the damping and using the energy method the equation of
motion is
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Thus the undamped equation of motion is:
From examining the equation of motion the natural frequency is:
An add hoc way do to this is to add the damping force to get the damped equation
of motion:
The value of ζ is determined by examining the velocity term:
Thus the damped natural frequency is
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1.75
Consider Problem 1.74 with k = 500,000 N/m, m = 2000 kg, J = 200 m2⋅kg/rad, r
= 30 cm, and c = 8000 kg/s. Calculate the damping ratio and the damped natural
frequency. How much effect does the rotational inertia have on the undamped
natural frequency?
Solution: From problem 1.74:
Given :
N/m
and
Inserting the given values yields:
For the undamped natural frequency,
(1)
To find the Differential equation:
Rate of change of energy in the system
(2)
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(3)
(4)
as
To find
when no rotational components are present
That is when
Thus the rotating component reduces the undamped natural frequency by
With rotational inertia,
Without rotational inertia,
The effect of the rotational inertia is that it reduces the natural frequency by 45.3 %.
1.76
Use Lagrange’s formulation to calculate the equation of motion and the natural
frequency of the system of Figure P1.76. Model each of the brackets as a spring
of stiffness k, and assume the inertia of the pulleys is negligible.
Figure P1.76
Solution: Let x denote the distance mass m moves, then each spring will deflects
a distance x/4. Thus the potential energy of the springs is
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The kinetic energy of the mass is
Using the Lagrange formulation in the form of Equation (1.64):
1.77
Use Lagrange’s formulation to calculate the equation of motion and the natural
frequency of the system of Figure P1.77. This figure represents a simplified
model of a jet engine mounted to a wing through a mechanism which acts as a
spring of stiffness k and mass m s . Assume the engine has inertia J and mass m and
that the rotation of the engine is related to the vertical displacement of the engine,
x(t) by the “radius” r 0 (i.e.
).
Figure P1.77
Solution: This combines Examples 1.4.1 and 1.4.4. The kinetic energy is
The kinetic energy in the spring (see example 1.4.4) is
Thus the total kinetic energy is
The potential energy is just
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Using the Lagrange formulation of Equation (1.64) the equation of motion results
from:
1.78
Consider the inverted simple pendulum connected to a spring of Figure P1.68.
Use Lagrange’s formulation to derive the equation of motion.
Solution: The energies are (see the solution to 1.68):
Choosing θ as the generalized coordinate, the spring compresses a distance x = l
sin θ and the mass moves a distance h = l cos θ from the reference position. So
the Lagrangian becomes:
The terms in Lagrange’s equation are
Thus from the Lagrangian the equation of motion is
Where the last expression is the linearized version for small θ.
1.79
Lagrange’s formulation can also be used for non-conservative systems by adding
the applied non-conservative term to the right side of equation (1.63) to get
Here R i is the Rayleigh dissipation function defined in the case of a viscous
damper attached to ground by
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Use this extended Lagrange formulation to derive the equation of motion of the
damped automobile suspension driven by a dynamometer illustrated in Figure
P1.79. Assume here that the dynamometer drives the system such that x =rθ.
Figure P1.79
Solution: The kinetic energy is
The potential energy is:
The Rayleigh dissipation function is
The Lagrange formulation with damping becomes
1.80
Consider the disk of Figure P1.80 connected to two springs. Use the energy
method to calculate the system's natural frequency of oscillation for small angles
θ(t).
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Figure P1.80
Solution:
Known:
and
Kinetic energy:
Potential energy:
Conservation of energy:
Natural frequency:
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1.81
A pendulum of negligible mass is connected to a spring of stiffness k at halfway
along its length, l, as illustrated in Figure P1.81. The pendulum has two masses
fixed to it, one at the connection point with the spring and one at the top. Derive
the equation of motion using the Lagrange formulation, linearize the equation and
compute the systems natural frequency. Assume that the angle remains small
enough so that the spring only stretches significantly in the horizontal direction.
Figure P1.81
Solution: Using the Lagrange formulation the relevant energies are:
From the trigonometry of the drawing:
So the potential energy writing in terms of θ is:
Setting L = T-U and taking the derivatives required for the Lagrangian yields:
Thus the equation of motion becomes
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Linearizing for small θ this becomes
So the natural frequency is
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Problems and Solutions Section 1.5 (1.82 through 1.93)
1.82
A bar of negligible mass fixed with a tip mass forms part of a machine used to
punch holes in a sheet of metal as it passes past the fixture as illustrated in Figure
P1.82. The impact to the mass and bar fixture causes the bar to vibrate and the
speed of the process demands that frequency of vibration not interfere with the
process. The static design yields a mass of 60 kg and that the bar be made of steel
of length 0.30 m with a cross sectional area of 0.02 m2. Compute the system’s
natural frequency.
Solution: From equation (1.63)
This is equivalent to 2000 Hz. This frequency is very high and might result in
noise.
1.83
Consider the punch fixture of Figure P1.82. If the system is giving an initial
velocity of 8 m/s, what is the maximum displacement of the mass at the tip if the
mass is 1200 kg and the bar is made of steel of length 0.40 m with a cross
sectional area of 0.012 m2?
Solution: First compute the frequency:
From equation (1.9) the maximum amplitude is
or about 0.36 mm, not much.
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1.84
Consider the punch fixture of Figure P1.82. If the punch strikes the mass off
center it is possible that the steel bar may vibrate in torsion. The mass is 1200 kg
and the bar 0.20 m-long, with a square cross section of 0.12 m on a side. The mass
polar moment of inertia of the tip mass is 12 kg/m2. The polar moment of inertia
for a square bar is b4/6, where b is the length of the side of the square. Compute
both the torsion and longitudinal frequencies. Which is larger?
Solution: First compute the longitudinal frequency of the bar
rad/s
Next compute the torsional frequency of the bar (square cross section):
rad/s
In this case the torsional frequency is lower and should be considered in any
design.
1.85
A helicopter landing gear consists of a metal framework rather than the coil
spring based suspension system used in a fixed-wing aircraft. The vibration of the
frame in the vertical direction can be modeled by a spring made of a slender bar
as illustrated in Figure 1.23, where the helicopter is modeled as ground. Here l =
0.42 m, E = 20 × 1010 N/m2, and m = 120 kg. Calculate the cross-sectional area
that should be used if the natural frequency is to be f n = 520 Hz.
Solution: Given:
From equation (1.63)
and
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1.86
The frequency of oscillation of a person on a diving board can be modeled as the
transverse vibration of a beam as indicated in Figure 1.26. Let m be the mass of
the diver (m = 80 kg) and l = 1.5 m. If the diver wishes to oscillate at 4 Hz, what
value of EI should the diving board material have?
Solution: From equation (1.67),
ω n2 =
3EI
ml 3
and
Solving for EI
1.87
Consider the spring system of Figure 1.32. Let k 1 = k 5 = k 2 =80 N/m, k 3 = 40
N/m, and k 4 = 10 N/m. What is the equivalent stiffness?
Solution: Given:
From Example 1.5.4
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1.88
Springs are available in stiffness values of 10, 100, and 1000 N/m. Design a
spring system using these values only, so that a 120-kg mass is connected to
ground with frequency of about 1.5 rad/s.
Solution: Using the definition of natural frequency with m = 120 kg, and
, the equivalent stiffness must be
There can be many configurations of the springs with this equivalent stiffness.
One particular configuration is to choose two springs of 100 N/m in parallel to get
200 N/m and use 2 springs of 100 N/m in series to get 50 N/m and use 10 N/m in
parallel to get 20 N/m.
Thus using 6 springs with the following configuration gives the equivalent 270
N/m stiffness.
1.89
Calculate the natural frequency of the system in Figure 1.32(a) if k 1 = k 2 = 0.
Choose m and nonzero values of k 3 , k 4 , and k 5 so that the natural frequency is 80
Hz.
and
Solution: Given
from the figure 1.29
the natural frequency is
Equating the given value of frequency to the analytical value yields
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Any values of k 3 , k 4 , k 5 , and m must satisfy this equation. The answer is not
unique. One solution is taking:
1.90* Example 1.4.4 examines the effect of the mass of a spring on the natural
frequency of a simple spring-mass system. Use the relationship derived there and
plot the natural frequency (normalized by the natural frequency, ωn , for a
massless spring) versus the percent that the spring mass is of the oscillating mass.
Determine from the plot (or by algebra) the percentage where the natural
frequency changes by 1% and therefore the situation when the mass of the spring
should not be neglected.
Solution: The solution here depends on the value of the stiffness and mass ratio
and hence the frequency. Almost any logical discussion is acceptable as long as
the solution indicates that for smaller values of m s , the approximation produces a
reasonable frequency. Here is one possible answer. For
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From this plot, for these values of m and k, a 10 % spring mass causes less then a
1 % error in the frequency.
1.91
Calculate the natural frequency and damping ratio for the system in Figure P1.91
given the values m = 20 kg, c = 120 kg/s, k 1 = 4800 N/m, k 2 = 400 N/m and k 3 =
800 N/m. Assume that no friction acts on the rollers. Is the system overdamped,
critically damped or underdamped?
Solution: Following the procedure of Example 1.5.4, the equivalent spring
constant is:
Then using the standard formulas for frequency and damping ratio:
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Thus the system is underdamped.
1.92
Calculate the natural frequency and damping ratio for the system in Figure P1.92.
Assume that no friction acts on the rollers. Is the system overdamped, critically
damped or underdamped?.
Figure P1.92
Solution: Again using the procedure of Example 1.5.4, the equivalent spring
constant is:
Then using the standard formulas for frequency and damping ratio:
Thus the system is underdamped, in fact very lightly damped.
1.93
A manufacturer makes a cantilevered leaf spring from steel (E = 2.1 x 1011 N/m2)
and sizes the spring so that the device has a specific frequency. Later, to save
weight, the spring is made of aluminum (E = 7.1 x 1010 N/m2). Assuming that the
mass of the spring is much smaller than that of the device the spring is attached
to, determine if the frequency increases or decreases and by how much.
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Solution: Use equation (1.67) to write the expression for the frequency twice:
rad/s
Dividing yields:
Thus the frequency is decreased by about 40% by using aluminum.
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Problems and Solutions Section 1.6 (1.94 through 1.101)
1.94
The displacement of a vibrating spring-mass-damper system is recorded on an x y plotter and reproduced in Figure P1.94. The y coordinate is the displacement in
cm and the x coordinate is time in seconds. From the plot determine the natural
frequency, the damping ratio and the damped natural frequency.
P1.94 A plot of displacement versus time for a vibrating system.
Solution: From the plot, the period is T =4 seconds. Thus the damped natural
frequency is
Using the formula for log decrement and noting from the plot that the first peak is
about y(t 1 ) = 0.74 cm and the second peak is about y(t 2 ) = 0.2 cm yields
Using equation (1.83) the damping ration is then:
The undamped natural frequency is
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1.95
Show that the logarithmic decrement is equal to
where x n is the amplitude of vibration after n cycles have elapsed.
Solution:
(1)
Since
Hence, Eq. (1) becomes
Since
Then
Therefore,
Here x 0 = x(0).
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1.96
Derive the equation (1.78) for the trifalar suspension system.
Solution: Using the notation given for Figure 1.32, and the following geometry:
Write the kinetic and potential energy to obtain the frequency:
Kinetic energy:
From geometry,
and
Potential Energy:
Two term Taylor Series Expansion of cos φ
For geometry, sin
, and for small φ, sin φ = φ so that φ
Conservation of energy requires that:
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At maximum energy, x = A and
Substitute
were T is the period of oscillation of the suspension.
1.97
A prototype composite material is formed and hence has an unknown modulus.
An experiment is performed consisting of forming it into a cantilevered beam of
length 1.2 m and I = 10-9 m4 with a 6.2-kg mass attached at its end. The system is
given an initial displacement and found to oscillate with a period of 0.5 s.
Calculate the modulus E.
Solution: Using equation (1.65) for a cantilevered beam,
Solving for E and substituting the given values yields
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1.98
The free response of a 1200-kg car with stiffness of k = 500,000 N/m is observed
to be of the form given in Figure 1.35. Modeling the car as a single-degree-offreedom oscillation in the vertical direction, determine the damping coefficient if
the displacement at t 1 is measured to be 2 cm and 0.22 cm at t 2 .
Solution: Given: x 1 = 2 cm and x 2 = 0.22 cm where t 2 = T + t 1
Logarithmic Decrement:
Damping ratio:
Damping Coefficient:
1.99
A pendulum decays from 8 cm to 2 cm over one period. Determine its damping
ratio.
Solution: Using Figure 1.34:
Logarithmic decrement:
Damping Ratio:
1.100 The relationship between the log decrement δ and the damping ratio ζ is often
approximated as δ =2πζ. For what values of ζ would you consider this a good
approximation to equation (1.82)?
Solution: From equation (1.82),
For small ζ,
A plot of these two equations is shown:
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The lower curve represents the approximation for small ζ, while the upper curve
is equation (1.82). The approximation appears to be valid to about ζ = 0.3.
1.101 A damped system is modeled as illustrated in Figure 1.9. The mass of the system
is measured to be 6 kg and its spring constant is measured to be 6000 N/m. It is
observed that during free vibration the amplitude decays to 0.25 of its initial value
after five cycles. Calculate the viscous damping coefficient, c.
Solution:
Note that for any two consecutive peak amplitudes,
by definition
So,
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and
Solving for c,
N-s/m
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Problems and Solutions Section 1.7 (1.102 through 1.110)
1.102 Consider the system of Example 1.7.2 consisting of a helical spring of stiffness
103 N/m attached to a 12-kg mass Place a dashpot parallel to the spring and
choose its viscous damping value so that the resulting damped natural frequency
is reduced to 9 rad/s.
Solution:
The frequency of oscillation is
From example 1.7.2:
So,
Then
1.103 For an underdamped system, x 0 = 0 mm and v 0 = 8 mm/s. Determine m, c, and k
such that the amplitude is less than 1 mm..
Solution: Note there are multiple correct solutions. The expression for the
amplitude is
So,
1)
2)
Choose
3)
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1.104 Repeat problem 1.103 if the mass is restricted to be between 10 kg < m < 15 kg.
Solution: Referring to the above problem, the relationship between m and k is
k >1.01x10-4 m
after converting to meters from mm. Choose m =10 kg and repeat the calculation
at the end of Problem 1.102 to get ωn (again taking ζ = 0.01). Then k = 1000 N/m
and:
1.105 Use the formula for the torsional stiffness of a shaft from Table 1.1 to design a 1m shaft with torsional stiffness of 105 N⋅m/rad.
Solution: Referring to equation (1.64) the torsional stiffness is
Assuming a solid shaft, the value of the shaft polar moment is given by
Substituting this last expression into the stiffness yields:
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Solving for the diameter d yields
Thus we are left with the design variable of the material modulus (G). Choose
10
2
steel, then solve for d. For steel G = 8 × 10 N/m . From the last expression the
numerical answer is
1.106 Consider designing a helical spring made of aluminum, such that when it is
attached to a 12-kg mass the resulting spring-mass system has a natural frequency
of 10 rad/s. Thus repeat Example 1.7.2 which uses steel for the spring and note
any difference.
Solution:
Given:
For Aluminum G = 25 x 109 N /m2
For example 1.7.2, the stiffness is
and d = 0.1
So,
Solving for nR3 yields
=
m3
Choose R =10 cm = 0.1m, so that
Thus, aluminum requires 1/3 fewer turns than steel.
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1.107 Try to design a bar that has the same stiffness as the helical spring of Example
1.7.2 (i.e., k = 103 N/m). This amounts to computing the length of the bar with its’
cross sectional area taking up about the same space at the helical spring (R = 8
cm). Note that the bar must remain at least 12 times as long as it is wide in order
to be modeled by the stiffness formula given for the bar in Figure 1.23.
Solution:
Given:
The length is not practical at all. Sometimes in the course of design, the
requirements cannot be met.
1.108 Repeat Problem 1.107 using plastic (E = 1.40 × 109 N/m2) and rubber (E = 7 × 106
N/m2). Are any of these feasible?
Solution:
From problem 1.53,
For plastic,
So,
For rubber,
So,
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Rubber may be feasible, plastic would not.
1.109 Consider the diving board of Figure P1.109. For divers, a certain level of static
deflection is desirable, denoted by ∆.
Compute a design formula for the
dimensions of the board (b, h and ) in terms of the static deflection, the average
diver’s mass, m, and the modulus of the board.
Figure P1.109
Solution: From Figure 1.15 (b),
holds for the static deflection. The
period is:
(1)
From Figure 1.27, we also have that
(2)
Equating (1) and (2) and replacing I with the value from the figure yields:
Alternately just use the static deflection expression and the expression for the
stiffness of the beam from Figure 1.27 to get
1.110 In designing a vehicle suspension system using a “quarter car model” consisting
of a spring, mass and damper system, studies show that a desirable damping ratio
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is ζ = 0.2. If the model has a mass of 800 kg and a frequency of 16 Hz, what
should the damping coefficient be?
Solution First convert the 16 Hz to rad/sec:
Next use the formula for the damping ratio
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Problems and Solutions Section 1.8 (1.111 through 1.115)
1.111 Consider the system of Figure P1.111. (a) Write the equations of motion in terms
of the angle, θ, the bar makes with the vertical. Assume linear deflections of the
springs and linearize the equations of motion. (b) Discuss the stability of the
linear system’s solutions in terms of the physical constants, m, k, and . Assume
the mass of the rod acts at the center as indicated in the figure.
Figure P1.111
Note that from the geometry, the springs deflect a distance
and the cg is a distance
up from the horizontal line
through point 0 taken as zero gravitational potential. Thus the total potential
energy is
Solution:
Using the inertia for a thin rod of length l rotating about its end, the total kinetic
energy is
The Lagrange equation (1.64) becomes
Using the linear, small angle approximations
yields
Since the leading coefficient is positive the sign of the coefficient of θ determines
the stability.
b)
Note that physically these results state that the system’s response is stable as long
as the spring stiffness is large enough to overcome the force of gravity.
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1.112 Consider the inverted pendulum of Figure 1.40 as discussed in Example 1.8.1 and
repeated here as Figure P1.112. Assume that a dashpot (of damping rate c) also
acts on the pendulum parallel to the two springs. How does this affect the
stability properties of the pendulum?
Figure P1.112 The inverted pendulum of Example 1.8.1
Solution: The equation of motion is found from the following FBD:
Moment about O:
When θ is small, sinθ ≈ θ and cosθ ≈ 1
For stability,
and c > 0.
The result of adding a dashpot is to make the system asymptotically stable.
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1.113 Replace the massless rod of the inverted pendulum of Figure 1.40 with a solid
object compound pendulum of Figure 1.20(b).
Calculate the equations of
vibration and discuss values of the parameter relations for which the system is
stable.
Solution:
Moment about O:
When θ is small, sinθ ≈ θ and cosθ ≈ 1.
For stability,
1.114 A simple model of a control tab for an airplane is sketched in Figure P1.114. The
equation of motion for the tab about the hinge point is written in terms of the
angle θ from the centerline to be
.
Here J is the moment of inertia of the tab, k is the rotational stiffness of the hinge,
c is the rotational damping in the hinge and
is the negative damping
provided by the aerodynamic forces (indicated by arrows in the figure). Discuss
the stability of the solution in terms of the parameters c and f d .
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Figure P1.114 A simple model of an airplane control tab
Solution: The stability of the system is determined by the coefficient of
the inertia and stiffness terms are both positive. There are three cases
Case 1 c - f d > 0 and the system’s solution is of the form
and the solution is asymptotically stable.
Case 2 c - f d < 0 and the system’s solution is of the form
and the solution oscillates and grows without bound, and exhibits flutter
instability as illustrated in Figure 1.36.
Case 3 c = f d and the system’s solution is of the form
the solution is stable as illustrated in Figure 1.37.
since
and
1.115* In order to understand the effect of damping in design, develop some sense of
how the response changes with the damping ratio by plotting the response of a
single degree of freedom system for a fixed amplitude, frequency and phase as
ζ changes through the following set of values ζ = 0.01, 0.05, 0.1, 0.2, 0.3 and 0.4.
That is plot the response
for each value of ζ.
Solution: Any plotting program can be used (Grapher, Excel, Matlab, etc.). Here
is a Matcad version. If assigned it’s good to discuss the differences as the
damping increases.
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Copyright © 2015 Pearson Education Ltd.
Problems and Solutions Section 1.9 (1.116 through 1.123)
1.116* Compute and plot the response to
using Euler’s method for time
steps of 0.1 and 0.5. Also plot the exact solution and hence reproduce Figure 1.41.
Solution: The code is given here in Mathcad, which can be run repeatedly with different
∆t to see the importance of step size. Matlab and Mathematica can also be used to show
this.
Copyright © 2015 Pearson Education Ltd.
1.117* Use numerical integration to solve the system of Example 1.7.3 with m = 1361
kg, k = 2.688 x 105 N/m, c = 3.81 x 103 kg/s subject to the initial conditions x(0) = 0 and
v(0) = 0.01 mm/s. Compare your result using numerical integration to just plotting the
analytical solution (using the appropriate formula from Section 1.3) by plotting both on
the same graph.
Solution: The solution is shown here in Mathcad using an Euler integration. This can
also been done in the other codes or the Toolbox:
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1.118* Consider again the damped system of Problem 1.117 and design a damper such
that the oscillation dies out after 2 seconds. There are at least two ways to do this. Here it
is intended to solve for the response numerically, following Examples 1.9.2, 1.9.3 or
1.9.4, using different values of the damping parameter c until the desired response is
achieved.
Solution: Working directly in Mathcad (or use one of the other codes). Changing c until
the response dies out within about 2 sec yields c =6500 kg/s or ζ = 0.17.
Copyright © 2015 Pearson Education Ltd.
1.119* Consider again the damped system of Example 1.9.2 and design a damper such
that the oscillation dies out after 25 seconds. There are at least two ways to do this. Here
it is intended to solve for the response numerically, following Examples 1.9.2, 1.9.3 or
1.9.4, using different values of the damping parameter c until the desired response is
achieved. Is your result overdamped, underdamped or critically damped?
Solution: The following Mathcad program is used to change c until the desired response
results. This yields a value of c = 1.1 kg/s or ζ = 0.225, an underdamped solution.
Copyright © 2015 Pearson Education Ltd.
Copyright © 2015 Pearson Education Ltd.
1.120* Repeat Problem 1.119 for the initial conditions x(0) = 0.1 m and v(0) = 0.01
mm/s.
Solution: Using the code in 1.96 and changing the initial conditions does not change the
settling time, which is just a function of ζ and ωn . Hence the value of c = 6.5x103 kg/s (ζ
= 0.17) as determined in problem 1.96 will still reduce the response within 2 seconds.
Copyright © 2015 Pearson Education Ltd.
1.121* A spring and damper are attached to a mass of 80 kg in the arrangement given in
Figure 1.9. The system is given the initial conditions x(0) = 0.1 m and v(0) = 1 mm/s.
Design the spring and damper ( i.e. choose k and c) such that the system will come to rest
in 2 s and not oscillate more than two complete cycles. Try to keep c as small as
possible. Also compute ζ.
Solution: In performing this numerical search on two parameters, several underdamped
solutions are possible. Students will note that increasing k will decrease ζ. But increasing
k also increases the number of cycles which is limited to two.
Using the code below, the computed values of the coefficients are:
Octave code for calculating the function.
function v = func_p1121(t,x)
m = 80;
k=2000;
c= 350;
Fo= 100;
omega_n = sqrt(k/m)
zeta = c/(2*sqrt(k*m))
omega_d = omega_n*sqrt(1- zeta**2)
T = 2*pi/omega_d;
omega = 8.162;
v=[x(2);x(1).*-k/m - 2.0*x(2).*zeta*omega_n ];
end
clear all
clc
xo =[0.1;1];
ts=[0,2];
[t,x] = ode45(@func_p1121,ts,xo);
figure(1)
plot(t,x(:,1),'r') ;
xlabel("t")
ylabel("x(t)")
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Copyright © 2015 Pearson Education Ltd.
1.122* Repeat Example 1.7.1 by using the numerical approach of the previous 5
problems.
Solution: The following Mathcad session can be used to solve this problem by varying
the damping for the fixed parameters given in Example 1.7.1.
The other codes or the toolbox may also be used to do this.
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1.123* Repeat Example 1.7.1 for the initial conditions x(0) = 0.01 m and v(0) = 1 mm/s.
Solution: The above Mathcad session can be used to solve this problem by varying the
damping for the fixed parameters given in Example 1.7.1. For the given values of initial
conditions, the solution to Problem 1.100 also works in this case. Note that if x(0) gets
too large, this problem will not have a solution.
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Problems and Solutions Section 1.10 (1.124 through 1.136)
1.124 A 4-kg mass connected to a spring of stiffness 103 N/m has a dry sliding friction
force (F c ) of 3 N. As the mass oscillates, its amplitude decreases 10 cm. How
long does this take?
Solution: Given:
From equation (1.101): slope
Solving the last inequality for
yields
1.125 Consider the system of Figure 1.41 with m = 6 kg and k = 8 × 103 N/m with a
friction force of magnitude 6 N. If the initial amplitude is 4 cm, determine the
amplitude one cycle later as well as the damped frequency.
Solution: Given
and the amplitude
after one cycle is
It should be noted that the damped natural frequency is the same as the natural
frequency in the case of coulomb damping, hence
1.126* Compute and plot the response of the system of Figure P1.104 for the case where
x 0 = 0.1 m, v 0 = 0.1 m/s, µ κ = 0.05, m = 250 kg, θ = 20° and k =3000 N/m. How long
does it take for the vibration to die out?
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Figure P1.126
Solution: Choose the x y coordinate system to be along the incline and perpendicular to
it. Let µ s denote the static friction coefficient, µ k the coefficient of kinetic friction and ∆
the static deflection of the spring. A drawing indicating the angles and a free-body
diagram is given in the figure:
For the static case
For the dynamic case
Combining these three equations yields
Note that as the angle θ goes to zero the equation of motion becomes that of a spring
mass system with Coulomb friction on a flat surface as it should.
Answer: The oscillation dies out after about 0.9 s. This is illustrated in the following
Mathcad code and plot.
Copyright © 2015 Pearson Education Ltd.
Alternate Solution (Courtesy of Prof. Chin An Tan of Wayne State University):
Static Analysis:
In this problem,
is defined as the displacement of the mass
from the equilibrium position of the spring-mass system under
friction. Thus, the first issue to address is how to determine this
equilibrium position, or what is this equilibrium position. In
reality, the mass is attached onto an initially unstretched spring on
the incline. The free body diagram of the system is as shown. The
governing equation of motion is:
where
is defined as the displacement measured from the unstretched position of the
spring. Note that since the spring is initially unstretched, the spring force
is zero
initially. If the coefficient of static friction
is sufficiently large, i.e.,
, then
the mass remains stationary and the spring is unstretched with the mass-spring-friction in
equilibrium. Also, in that case, the friction force
, not necessarily equal
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y
x
to the maximum static friction. In other words, these situations may hold at equilibrium:
(1) the maximum static friction may not be achieved; and (2) there may be no
and one would expect
displacement in the spring at all. In this example,
that
(not given) should be smaller than 0.364 since
(very small). Thus, one
would expect the mass to move downward initially (due to weight overcoming the
maximum static friction). The mass will then likely oscillate and eventually settle into an
equilibrium position with the spring stretched.
Dynamic Analysis:
The equation of motion for this system is:
where
is the displacement measured from the equilibrium position. Define
and
. Employing the state-space formulation, we transform the
original second-order ODE into a set of two first-order ODEs. The state-space equations
(for MATLAB code) are:
MATLAB Code:
x0=[0.1, 0.1];
ts=[0, 5];
[t,x]=ode45('f1_93',ts,x0);
plot(t,x(:,1), t,x(:,2))
title('problem 1.93'); grid on;
xlabel('time (s)');ylabel('displacement (m), velocity (m/s)');
%--------------------------------------------function xdot = f1_93(t,x)
% computes derivatives for the state-space ODEs
m=250; k=3000; mu=0.05; g=9.81;
angle = 20*pi/180;
xdot(1) = x(2);
xdot(2) = -k/m*x(1) - mu*g*cos(angle)*sign(x(2));
% use the sign function to improve computation time
xdot = [xdot(1); xdot(2)];
Plots for
and
cases are shown. From the
simulation results,
the oscillation dies out after about 0.96 seconds (using ginput(1) command to
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estimate). Note that the acceleration may be discontinuous at
the friction force.
due to the nature of
Effects of µ:
Comparing the figures, we see that reducing µ leads to more oscillations (takes longer
time to dissipate the energy). Note that since there is a positive initial velocity, the mass
is bounded to move down the incline initially. However, if µ is sufficiently large, there
may be no oscillation at all and the mass will just come to a stop (as in the case of
). This is analogous to an overdamped mass-damper-spring system. On the
other hand, when µ is very small (say, close to zero), the mass will oscillate for a long
time before it comes to a stop.
Discussion on the ceasing of motion:
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Note that when motion ceases, the mass reaches another state of equilibrium. In both
simulation cases, this occurs while the mass is moving upward (negative velocity). Note
that the steady-state value of
is very small, suggesting that this is indeed the true
equilibrium position, which represents a balance of the spring force, weight component
along the incline, and the static friction.
1.127* Compute and plot the response of a system with Coulomb damping of equation
(1.90) for the case where x 0 = 0.5 m, v 0 = 0, µ = 0.1, m = 80 kg and k =2000 N/m. How
long does it take for the vibration to die out?
Solution: Here the solution is computed using the following code. Any of the codes may
be used.
clear all
clc
xo =[0.5;1e-6];
ts=[0,6];
[t,x] = ode45(@func_p1127,ts,xo);
figure(1)
plot(t,x(:,1),'r') ;
xlabel("t")
ylabel("x(t)")
grid on
function v = func_p1127(t,x)
m = 80;
mu = 0.1;
k=2000;
g = 9.81;
a = x(2)/abs(x(2));
v=[x(2);(x(1).*(-k/m) -cos(20*pi/180)*(mu*g.*a))];
end
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1.128* A mass moves in a fluid against sliding friction as illustrated in Figure P1.106.
Model the damping force as a slow fluid (i.e., linear viscous damping) plus Coulomb
friction because of the sliding, with the following parameters: m = 240 kg, µ =0.01, c =
25 kg/s and k =3200 N/m . a) Compute and plot the response to the initial conditions: x 0
= 0.1 m, v 0 = 0.1 m/s. b) Compute and plot the response to the initial conditions: x 0 = 0.1
m, v 0 = 1 m/s. How long does it take for the vibration to die out in each case?
Figure P1.128
Solution: A free-body diagram yields the equation of motion.
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where the vertical sum of forces gives
the magnitude µN = µmg for the
Coulomb force as in figure 1.41.
The equation of motion can be solved by using any of the codes mentioned or by using
the toolbox. While the next problem shows that the viscous damping can be changed to
reduce the settling time, this example shows how dependent the response is on the value
of the initial conditions. In a linear system the settling time, or time it takes to die out is
only dependent on the system parameters, not the initial conditions. This makes design
much more difficult for nonlinear systems.
clear all
clc
xo =[0.1;0.1];
ts=[0,6];
[t,x] = ode45(@func_p1128,ts,xo);
figure(1)
plot(t,x(:,1),'r') ;
xlabel("t")
ylabel("x(t)")
grid on
hold on
xo =[0.1;1];
ts=[0,6];
[t,x] = ode45(@func_p1128,ts,xo);
figure(1)
plot(t,x(:,1),'b') ;
Function to evaluate
function v = func_p1128(t,x)
m = 240;
mu = 0.01;
c = 25;
k=3200;
g = 9.81;
a = x(2)/abs(x(2));
v=[x(2);(x(1).*(-k/m) -(mu*g.*a)- (c/m).*x(2))];
end
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1.129* Consider the system of Problem 1.128 part (a), and compute a new damping
coefficient, c, that will cause the vibration to die out after one oscillation.
Solution: Working in any of the codes, use the simulation from the last problem and
change the damping coefficient c until the desired response is obtained. A Mathcad
solution is given which requires an order of magnitude higher damping coefficient,
c = 275 kg/s
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Ý+ ω 2n x + βx 2 = 0. How many are there?
1.130 Compute the equilibrium positions of xÝ
Solution: The equation of motion in state space form is
The equilibrium points are computed from:
Solving yields the two equilibrium points:
1.131 Compute the equilibrium positions of Ý
xÝ+ ω n2 x − β 2 x 3 + γx 5 = 0. How many are
there?
Solution: The equation of motion in state space form is
The equilibrium points are computed from:
Solving yields the five equilibrium points (one for each root of the previous
equation). The first equilibrium (the linear case) is:
Next divide
by x 1 to obtain:
which is quadratic in x 1 2 and has the following roots which define the remaining
four equilibrium points: x 2 = 0 and
Thus there are 5 equilibrium. Of course some disappear for certain combinations
of the coefficients.
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1.132* Consider the pendulum example 1.10.3 with length of 1 m an initial conditions of
θ =π/10 rad and θÝ = 0 . Compare the difference between the response of the linear
0
0
version of the pendulum equation (i.e. with sin(θ) = θ) and the response of the nonlinear
version of the pendulum equation by plotting the response of both for four periods.
Solution: First consider the linear solution. Using the formula’s given in the text
the solution of the linear system is just:
. The
following Mathcad code, plots the linear solution on the same plot as a numerical
solution of the nonlinear system.
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Note how the amplitude of the nonlinear system is growing. The difference
between the linear and the nonlinear plots are a function of the ration of the linear
spring stiffness and the nonlinear coefficient, and of course the size of the initial
condition. It is work it to investigate the various possibilities, to learn just when
the linear approximation completely fails.
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1.133* Repeat Problem 1.132 if the initial displacement is θ 0 = π/2 rad.
Solution: The solution in Mathcad is:
Here both solutions oscillate around the “stable” equilibrium, but the nonlinear
solution is not oscillating at the natural frequency and is increasing in amplitude.
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1.134 If the pendulum of Example 1.10.3 is given an initial condition near the
equilibrium position of θ 0 = π rad and θÝ0 = 0 , does it oscillate around this
equilibrium?
Solution
The pendulum will not oscillate around this equilibrium as it is
unstable. Rather it will “wind” around the equilibrium as indicated in the solution
to Example 1.10.4.
1.135* Calculate the response of the system of Problem 1.121 for the initial conditions
of x 0 = 0.02 m, v 0 = 0, and a natural frequency of 4 rad/s and for β = 100, γ = 0.
Solution: The solution of the problem using simple Euler integration is given using the
following Octave code:
clear
clc
delta_t = 0.01;
x =zeros(1,1000);
theta = zeros(1,1000);
v= zeros(1,1000);
x(1) = 0.02;
v(1) = 0.0;
omega = 4
A = (1 / omega)*sqrt(omega**2 *x(1)**2)
beta = 100
for i = 1 : 1000
x(i+1) = x(i) + v(i)*delta_t;
v(i+1) = v(i) - delta_t*(omega**2 * x(i) - beta**2*x(i)**3);
theta(i) = A*sin(3*delta_t*i + pi/2);
end
for i = 1 : 1001
theta(i) = A*sin(3*delta_t*i + pi/2);
end
t = linspace(0,10,1001);
plot(t,x,'r')
hold on
plot (t,theta,'b')
hold off
grid on
ylabel("x_i/theta_i")
xlabel("t")
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The other codes may be used to compute this solution as well.
1.136* Repeat problem 1.135 and plot the response of the linear version of the system (β
=0) on the same plot to compare the difference between the linear and nonlinear versions
of this equation of motion.
Solution: The solution is computed and plotted in the solution of Problem 1.113. Note
that the linear solution starts out very close to the nonlinear solution. The two solutions
however diverge. They look similar, but the nonlinear solution is growing in amplitude
and period.
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Problems and Solutions Section 2.1 (2.1 through 2.19)
2.1
The forced response of a single-degree-of-freedom, spring-mass system is
modeled by (assume the units are Newtons)
6 
x(t )  24 x(t )  6 cos t
Compute the magnitude of the forced response for the two cases  =2.1 rad/s and
 = 2.5 rad/sec. Comment on why one value is larger than the other.
Solution:
Given:
6 
x(t )  24 x(t )  6 cos t
We have to find the magnitude of the forced response xmax for ω1 = 2.1rad / s
and ω2 = 2.5rad / s
First divide through by mass 6 kg to reveal the natural frequency
xt  + 4 xt  = cosωt
Thus f 0 = 1and ωn = 2
ω = 2.1
X=
rad
. Hence the magnitude of the forced response for
s
rad
and
s
fo
1
=
= 2.4390m
2
ω ω
4  2.12
2
N
Forced response for ω = 2.5
X=
rad
is given by
s
fo
1
=
= 0.4444 m
2
ω ω
4  2.5 2
2
N
The value of the magnitude of the response at the driving frequency of 2.5 rad/s is
much smaller than at 2.1 rad/s because it is much farther away from the resonance
condition.
2.2
Consider the forced response of a single-degree-of-freedom, spring-mass system
that is modeled by (assume the units are Newtons)
3(t) + 12x(t) = 3 cosωt
Compute the total response of the system if the driving frequency is 5.0 rad/s and
the initial position and velocity are both zero.
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Solution:
3(t) + 12x(t) = 3 cosωt
We have to find the total response of the system if the driving frequency is
ω = 5.0rad / s and the initial position x 0  = 0 and velocity x́ (0 )= 0
The total response is given by the equation
xt  =

vo
f
sin ωn t  +  x0  2 o 2
ωn
ωn  ω




f
cosωn t  +  2 o 2 cosωt 

ω ω 

 n

For zero initial velocity and position becomes
ωn =
x t  =
 k 
 =
m
12 / 3  =
2
rad
s
 f0
f
cos ω n t  + 2 0 2 cos ωt 
2
ω ω
ωn  ω
2
n
3
3
cos 4 t  +
cos 2 t 
4  25
4  25
x t  = 0.4287 4 t   0.4287cos 2 t 
x t  =
The value of the magnitude of the response at a driving frequency 2.5 Hz:
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2.3
Compute the response of a spring–mass system modeled by equation (2.2) to a
force of magnitude 25 N, driving frequency of twice the natural frequency, and
initial conditions given by x0 = 0 m and v0 = 0.2 m/s. The mass of the system is 10
kg, the spring stiffness is 1000 N/m, and the mass of the spring is considered and
known to be 1 kg. What percent does the natural frequency change if the mass of
the spring is not taken into consideration?
Solution:Given:
The force of magnitude F = 25 N, the driving frequency is twice the natural
frequency, and initial conditions are x0 = 0 m and v0 = 0.2 m/s. The mass of the
system is 10 kg, the spring stiffness is 1000 N/m, and the mass of the spring is
considered and known to be 1 kg.
We have to find the response of a spring–mass system and what percent does the
natural frequency change if the mass of the spring is not taken into consideration.
Given msp = 1kg , Example 1.4.4 yields that the effective mass is me = m +
m sp
3
.
Thus the natural frequency, X and the coefficients in the equation 2.11 for the
system now become
1

me = 10 + 
3





1000 
rad

ωn =
= 9.837

1
s
 10 +  
3

rad v 0
0.2
ω = 2ωn = 19.675
,
=
= 0.020331
s ωn 9.837
F0
f
me
2.4194
X= 2 0 2=
=
= 0.008334
2
ωN  ω
9.837  19.675
9.8372  19.6752
xt  = 0.02033sin 9.837t  + 8.3  103 cos 9.837t   cos 19.675t m
The frequency without the spring mass considered is given in example 2.1.1 as 10
rad/s.
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The percentage change in natural frequency is
 1000 


 10 
0.5




1000 



1
 10 +  
3

 1000 


 10  
2.4
0.5
0.5
= 0.0162  1.62
Compute the response of a spring–mass system modeled by equation (2.2) to a
force of magnitude 23 N, driving frequency of twice the natural frequency, and
initial conditions given by x0 = 0 m and v0 = 0.2 m s.The mass of the system is
10 kg, the spring stiffness is 1000 N m and the mass of the spring is considered
and known to be 1 kg. What percent does the natural frequency change if the
mass of the spring is not taken into consideration?
Solution: Given: m sp = 1 kg, Example 1.4.4 yields that the effective mass is
me = m +
msp
1
= 10 + = 10.333 kg.
3
3
Thus the natural frequency, X and the coefficients in equation (2.11) for the
system now become
1000
= 9.837 rad/s, w = 2w n = 19.675 rad/s
10 + 1
3
f
F /m
v
2.2258
X = 2 0 2 = 20 e2 =
= -0.0077 m, 0 = 0.02033 m
2
2
wn
wn - w
wn - w
9.837 -19.675
Thus the response as given by equation (2.11) is
wn =
x(t) = 0.02033sin 9.837t + 7.7 ´10 -3 (cos9.837t - cos19.675t) m
The frequency without the spring’s mass consider is given in example 2.1.1 as 10
rad/s. So the percent change by including the spring’s mass is
9.837 - 10
×100 = -1.63%
10
So the mass of the spring reduced the natural frequency by 1.63 %.
2.5
A spring-mass system is driven from rest harmonically such that the displacement
response exhibits a beat of period of 0.4π s. The period of oscillation is measured
to be 0.04π s. Calculate the natural frequency and the driving frequency of the
system.
Copyright © 2015 Pearson Education Ltd.
Solution: Given that the displacement response exhibits a beat of period of 0.4 π s.
The period of oscillation is measured to be 0.04 π s.
The beat frequency is
Tb = 2
π
π
rad
 ωb = 2 = 50
ωb
Tb
s
The period of oscillation is such that
To = 2
ω +ω
π
2π
 ωo =
= 50 = n
ωo
To
2
ωn + ω = 100
rad
s
Solving these two equations we get ωn = 75
2.6
rad
rad
and ω = 25
s
s
An airplane wing modeled as a spring-mass system with natural frequency 50 Hz
is driven harmonically by the rotation of its engines at 49.9 Hz. Calculate the
period of the resulting beat.
Solution: ωn = 2π (50)rad / s ω = 2π ( 49.9) rad / s
BeatPeriod =
ωn  ω
= 50  49.9 = 0.1Hz
2π
 T = 10 s
2.7
Compute the total response of a spring-mass system with the following values:
k=1500 N/m, m = 15 kg, subject to a harmonic force of magnitude F0 = 150 N and
frequency of 8.162 rad/s, and initial conditions given by x0 = 0.01 m and v0 = 0.01
m/s. Plot the response.
Solution:Given: k = 1500 N / m , m = 15kg , F0 = 150N , ω = 8.162 rad / s ,
x0 = 0.01m and v0 = 0.01m / s .
Solution:
Differential equation :
15xt + 1500 xt  = 150sin8.162t 
x 0 = 0.01 and x0= 0.01
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Solving the above differential equation for the given conditions we have:
From equation 2.11
xt  =

vo
f
sin ωn t  +  x0  2 o 2
ωn
ωn  ω




f
cosωn t  +  2 o 2 cosωt 

ω ω 

 n

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2.8
Consider the system in Figure P2.8, write the equation of motion and calculate the
response assuming a) that the system is initially at rest, and b) that the system has
an initial displacement of 0.04 m.
Solution: The equation of motion is m ẍ +k x= 10 sin (10 t )
Let us first determine the general solution for
x + ω 2 x = f 0 sin ωt 
Replacing the cosine function with a sine function in eq 2.4 and following the
same argument, the general solution is
xt  = A1sin ωn t + A2 cosωn t +
f0
sin ωt 
ω  ω2
2
n
Using the initial conditions x0  = x0, x 0  = v0 a general expression for the
response of spring mass system to a harmonic (sine) excitation is
v
f0
ω
xt  =  0 
2
2
 ωn ωn ωn  ω
Given k = 2000

f
sin ωn t  + x0 cosωn t  + 2 0 2 sin ωt 

ωn  ω

N
rad
, m = 100kg,ω = 10
m
s
a) Using the general expression obtained above
x0 = 0m, v0 = 0
xt  = 0 
10
20
0.1
20
2
 102
sin

m
s
20t + 0 +
0.1
202  102
sin 10t 
xt  = 0.002975sin 4.472t  + 0.04cos4.472t   1.25103 sin10t 
b) x0 = 0.04m, v0 = 0
xt  = 0 
10
0.1
20  202
m
s
sin

20t   0.04 cos 20t + 0 +
10
20 
 10 2 



3
xt  = 0.002975sin 4.472t + 0.04cos4.472t   1.25  10 sin 10t 
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0.1
202
 10 2 

sin 10t 
2.9
Consider the system in Figure P2.9, write the equation of motion and calculate the
response assuming that the system is initially at rest for the values k1= 200 N/m,
k2 = 1000 N/m and m = 178 kg.
Solution: The equation of motion is m ẍ +k x= 10 sin (10 t )
The general expression obtained for the response of an under-damped spring –
mass system to a harmonic excitation input in problem 2.8 was
v
f0 
f
ω
sin ωn t  + x0 cosωn t  + 2 0 2 sin ωt 
xt  =  0 
2
2 
ωn  ω
 ωn ωn ωn  ω 
Substituting the following values
k=
1
1 
 1
+


 200 1000 
= 166.67
N
,
m
m = 178kg,
rad
k
 166.67 
ωn =   = 
,
 = 0.96764
s
m
 178 
f0 =
F0 10
N
=
= 0.05618
M 178
kg
and initial conditions x0 = 0 and v0 = 0
The system response is evaluated as
− 04
x (t )= 0.0058607 sin (0.0562t )− 5.6711× 10
2.10
sin(10 t )
Consider the system in Figure P2.10, write the equation of motion and calculate
the response assuming that the system is initially at rest for the values Ө = 30°, k
= 1500 N/m and m = 60 kg.
Solution: Assuming x = 0 to be at equilibrium
∑ F x = m ẍ= − k ( x +Δ)+m g sin (θ)+90 sin ( 25t )
where Δ is the static deflection of the spring.
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From the static equilibrium in the x direction − k Δ+mgsin (θ)
The equation of motion becomes m ẍ (t )+ k x = 90 sin(2.5 t)
The general expression for the response of a spring- mass system to a harmonic
(sine) excitation (See problem 2.7) is
v
f0 
f
ω
sin ωn t  + x0 cosωn t  + 2 0 2 sin ωt 
xt  =  0 
2
2 
ωn  ω
 ωn ωn ωn  ω 
Given v0 = 0 , x0 = 0 and ω = 2.5
rad
s
rad
F
90
N
k
 1500 
ωn =   = 
and f 0 =
=
= 1.5
=5
s
M 60
kg
m
 60 
So the response is x t  = 0.04sin 5t + 0.08sin 2.5t 
2.11
Consider the system in Figure P2.11, write the equation of motion and calculate
the response assuming that the system is initially at rest for the values q = 30°, k
= 1000 N/m and m = 50 kg.
Figure P2.11

x
m
Fs
mg sin 
F=90 sin 2.5 t
(Forces that are normal
to the x direction are
neglected)
Solution: Free body diagram:
Assuming x = 0 to be at equilibrium:
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åF
x
= m
x = -k(x + D) + mgsin q + 90sin 25t
(1)
where Δ is the static deflection of the spring. From static equilibrium in the x
direction yields
-kD + mgsin q
(2)
Substitution of (2) onto (1), the equation of motion becomes
m 
x + k x = 90sin 2.5t
The general expression for the response of a mass-spring system to a harmonic
(sine) excitation (see Problem 2.7) is:
x(t) = (
v0 w
f
f
× 2 0 2 )sin w n t + x0 cosw n t + 2 0 2 sin w t
wn wn wn - w
wn - w
Given: v0= 0, x0= 0, w = 2.5 rad/s
wn =
k
1000
=
= 20 = 4.472 rad/s ,
m
50
f0 =
F0 90 9
=
= N/kg
m 50 5
So the response is:
x(t) = -0.0732sin 4.472t + 0.1309sin 2.5t
2.12
The natural frequency of a 75-kg person illustrated in Figure P2.12 is measured
along vertical, or longitudinal direction to be 4.5 Hz. a) What is the effective
stiffness of this person in the longitudinal direction? b) If the person, 1.86 m in
length and 0.60 m2 in cross sectional area, is modeled as a thin bar, what is the
modulus of elasticity for this system?
Solution: First change the frequency in Hz to rad/s:
cycles
rad
rad
4.5
2π
= 9π
s
cycles
s
Then, from the definition of natural frequency:
k = ωn2 m = 2π  4.5  75 = 59957.846N / m
2
From section 1.4 the value of stiffness for the longitudinal vibration of a beam is
E=
kl 59957.846  1.86
=
= 1858693.22 6 N / m 2
A
0.06
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2.13
If the person in Problem 2.12 is standing on a floor, vibrating at 3.5 Hz with an
amplitude of 1 N (very small), what longitudinal displacement would the person
“feel”? Assume that the initial conditions are zero.
Solution Using equation 2.12 for a cosine excitation and zero initial conditions
F  1  1

1
= 
= 4.2217e -05 m
yields  X  = 0  2
2
2
2
m ωn  ω  759π   3.5  2π  
2.14
Vibration of body parts is a significant problem in designing machines and
structures. A jackhammer provides a harmonic input to the operator’s arm. To
model this situation, treat the forearm as a compound pendulum subject to a
harmonic excitation (say of mass 5 kg and length 30 cm) as illustrated in Figure
P2.14. Consider point O as a fixed pivot. Compute the maximum deflection of the
hand end of the arm if the jackhammer applies a force of 10 N at 2 Hz.
Solution: Using equation 2.12 for a cosine excitation and zero initial conditions
yields (Converting the frequency from Hertz to rad/s using the value of k
calculated in 2.11)

  

F
10
  





 = 0.066874m
m
5
Maximum Amplitude 2
=2
  mgR


 3  9.81
2
2
 


 16π  
 16π  

I
0.3
    
 
 
2.15
An airfoil is mounted in a wind tunnel for the purpose of studying the
aerodynamic properties of the airfoil’s shape. A simple model of this is illustrated
in Figure P2.15 as a rigid inertial body mounted on a rotational spring, fixed to
the floor with a rigid support. Find a design relationship for the spring stiffness k
in terms of the rotational inertia, J, the magnitude of the applied moment, M0, and
the driving frequency, w, that will keep the magnitude of the angular deflection
less than 6°. Assume that the initial conditions are zero and that the driving
frequency is such that  n2   2  0 ..
Solution: Given
From the figure we have,
q
Jθ + qθ = M 0 cosωt , ωn =  
J
0.5
θ0 = 0 and θ 0  = 0
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 M0

Max Amplitude 2 2 J 2
 ωn  ω



 ωn2 = ω 2 +
2.16



 = Amax



2M 0
2M 0
2M 0
k
   = ω2 +
 k = Jω 2 +
JAmax
JAmax
Amax
J
The spar of an airplane wing is a relatively rigid beam extending along the length
of the wing inside the wing to provide strength. It is typical to model a spar as a
cantilever beam with the fixed end at the body of the aircraft. An example is
given in Figure P2.16. Using the modeling methods given in Section 1.5determine
a single-degree-of-freedom model for the spar and compute its natural frequency.
The spar here is modeled as a cantilever beam of dimensions length 560 mm,
width 38mm and thickness 3.175mm, and has a mass of 13.975 grams. The
beam’s Young's modulus is 10.29 GPa and its shear modulus: 1.65 GPa.
MAKE THIS A BEAM MODELD AS A SDOF SM SYSTEM
Figure P2.16A small, unmanned air vehicle with a rigid spar, modeled as a beam
Solution: This problem requires students to think back to Section 1.5 where an
SDOF model of a beam is given. The stiffness of the beam is given by equation
(1.76) and in Figure 1.26 as
k=
3EI
l3
Where E is the beams elastic modulus, l is the length and I is the area moment of
inertia of a beam in the bending direction given by
(thickness)(width)3 hb 3
I=
=
12
12
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This can be found on google or a strength of materials text. The formula for
frequency now requires the mass which will causes some students confusion.
Looking at Figure 1.28, the effective mass of a system with a tip mass and
substantial beam mass is results in Equation (1.76) for the frequency:
k
w n = 133
M +m
140
Combining these three expressions and letting the tip mass, m, be zero yields the
following expression for the frequency
wn =
2.17
1.65 ´ 10 9 N/m 2 (0.003175)(0.038)3 4
3E hb3
×
m
×
(0.560m)3
4
l 3 12 =
= 352.459 rad/s » 56.1 Hz
33
33
M
×13.975 ´ 10 -3 kg
140
140
Compute the response of a shaft and disc system to an applied moment of
M = 12 sin 312 t
as indicated in Figure P2.17. Assume that the shaft is initially at rest (zero initial
conditions) and J = 0.5 kg m2, the shear modulus is G = 8.2× 1010 N/m2, the shaft
is 1.2 m long, of diameter 6 cm and made of steel.
Solution: Given:
2
J = 0.5 kgm G = 8.2  1010 N / m 2 L = 1.2m d = 0.06m a =
 GJ 
M = 12sin312t   Jθ + 
θ = 12sin312t 
 L 
θ0 = 0 and θ 0  = 0
Solution: Summing moments, the equation of motion is
 M  
   


ω
J
θ t  =  2  2  sinωt    sinωnt 
 ωn  ω 
 ωn 





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πd 2
= 0.0028274 m 2
4





24
 sin312t    312 sinωn t 
 θ t  =  2
2 



 ωn 

 ωn  312  
The natural frequency (see also example 1.5.1) is
Where ωn =
G
=
L
8.2  10  = 261406rad / s
10
1.2
 θ t  = 3.512 1010 sin312t   0.0012sin261406.452t 
2.18
Consider a spring-mass system with zero initial conditions described by

x(t )  4 x(t )  10 cos 2t , x(0)  0, x (0)  0
and compute the form of the response of the system.
Solution: Given:
xt  + 4 xt  = 10cos2t  x 0  = 0 x 0  = 0
Solution: Note that here ω = 2
rad
, so the system is in resonance and the solution
s
is given by equation 2.17.
The solution is of the form:
xt  =
v0
f
sin ωt  + x0 cosωt  + 0 tsin ωt 
ω
2ω
rad
ω = 4 = 2
s
10
xt  = tsin 2t 
4
So, the system is in resonance.
2.19
Consider a spring mass system with zero initial conditions described by

x(t )  4 x(t )  10sin 4t , x(0)  0, x (0)  0
and compute the form of the response of the system.
Solution: Given:
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xt  + 4 xt  = 10cos4t  , x 0  = 0 x 0  = 0
Solution:
For equation 2.25 the response is of the form
v
f0 
f
ω
sin ωn t  + x0 cosωn t  + 2 0 2 sin ωt 
xt  =  0 
2
2 
ωn  ω
 ωn ωn ωn  ω 
ω
f0 
f0



xt  = 
sin
ω
t
+
sin ωt 
n
2
2 
2
2
ω
ω

ω
ω

ω
n
 n n

rad
ω = 4  = 2
,ω = 4
s
 4 10
10
xt  =
sin 2t  + 2
sin 4t 
2
2
2 2 4
2  42
xt  = 1.6667sin 2t   0.8333sin 4t 
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Problems and Solutions Section 2.2 (2.20 through 2.38)
2.20
Calculate the constants A and f for arbitrary initial conditions, x 0 and v 0 , in the
case of the forced response given by
()
- zw n t
x t = Ae
(
)
(
sin w d t + f + X cos wt - q
)
Compare this solution to the transient response obtained in the case of no forcing
function (i.e. F 0 = 0).
Solution: From equation (2.37)
x(t) = Ae-zwn t sin(w d t + f ) + X cos(wt - q) Þ
x˙ (t) = - zw n Ae -zw n t sin(w d t + f ) + Awd e -zwn t cos(w d t + f ) - X w sin(wt - q )
Next apply the initial conditions to these general expressions for position and
velocity to get:
x(0) = A sin f + X cos q
x˙ (0) = -zwn Asin f + A w d cos f + X w sin q
Solving this system of two equations in two unknowns yields:
æ
ö
(x0 - X cos q )w d
f = tan -1 ç
÷
è v0 + (x0 - X cosq )zw n - X w sin q ø
x0 - X cosq
sin f
Recall that X has the form
A=
æ
F0 / m
-1 ç 2zw nw ö
÷
and
q
=
tan
2
2
è wn - w ø
(w n2 - w 2 )2 + (2zw nw )2
Now if F0 = 0, then X = 0 and A and  from above reduce to:
X=
æ x0 w d ö
÷
f = tan -1 ç
è v0 + x 0zw n ø
x0
(v0 + zw n x0 )2 + (x 0w d )2
=
sin f
w d2
These are identical to the values given in equation (1.38).
A=
2.21
Consider the spring-mass-damper system defined by (use basic SI units)

x  t   6 x  t   25 x  t   4 cos 5t
First determine if the system is underdamped, critically damped or overdamped.
Then compute the magnitude and phase of the steady state response.
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Solution:Dividing by the mass yields
xt  + 6 x t  + 25 xt  = 4cos5t 
From examining the coefficients and comparing these definitions of the frequency
and damping ratio we get
c
6
ζ=
=
= 0.6
0.5
0.5
2km 
225
The system is underdamped.
Note that f 0 = 4 . From equation 2.36 magnitude of the steady state response is
X=
F0
k  mω  + cω
2 2
2
=
4
25  252 + 6  52
= 0.1333m
Note that the system is in resonance. The phase is computed from equation 2.36
cω
π
θ = tan 1
= rad (Resonance) or 90 o as it should for resonance.
2
2
k  mω
2.22
Show that the following two expressions are equivalent:
x p( t ) = X cos ( wt - q) and x p ( t ) = As cos wt + Bs sin wt
Solution: From equation (2.28) and expanding the trig relation yields
x p = X cos(w t - q ) = X [ cos w t cosq + sin w t sin q ]
= (X cosq ) cos w t + (X sin q ) sin w t






As
Bs
Now with As and Bs defined as indicated, the magnitude is computed:
X = As2 + Bs2
and
æB ö
Bs X sin q
=
Þ q = tan -1 ç s ÷
As X cosq
è As ø
2.23
Calculate the total solution of
x + 2ωn x + ωn 2 x = f 0 cost
for the case that m = 1.2 kg,  = 0.01, ωn = 2.5 rad/s. f0= 5 N/kg, and ω = 10
rad/s, with initial conditions x0= 1 m and v0 = 1 m/s, and then plot the response
Solution:Given:
xt  + 2ξωn x t  + ωn 2 xt  = f 0 cost
m = 1.2kg , ζ = 0.01 , ωn = 2.5rad / s , f 0 = 5 N / kg , and ω = 10rad / s ,
with initial conditions x0 = 1m and v0 = 1m / s .
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The total solution has the form
xt  = Ae 0.025t sin 2.4999t +   + 0.0533cos10t + 0.0533
xt  = e 0.025t  Asin 2.499t  + Bcos2.499t  + 0.0533cos10t + 0.0533
Differentiating then yields
x t  = 0.025e 0.025t  Asin 2.499t + Bcos2.499t 
 0.533sin10t + 0.0533+ e 0.025t 2.499 Acos2.499t   2.499Bsin 2.499t 
Applying the initial conditions, A = 0.40963 and B = 0.94667
Copyright © 2015 Pearson Education Ltd.
Octave code for plotting is
octave:64> m = 1.2;
octave:65> zeta = 0.01;
octave:66>omega_n = 2.5;
octave:67> f_0=5
octave:68> omega = 10;
octave:69> t1 = f_0/(sqrt((omega_n**2omega**2)**2+(2*zeta*omega_n*omega)**2))
t1 = 0.053333
octave:70> theta = atan((2*zeta*omega_n*omega)/(omega_n**2-omega**2))
theta = -0.0053333
octave:71> zeta*omega_n
ans = 0.025000
octave:72>omega_d = omega_n*sqrt(1-zeta**2)
omega_d= 2.4999
octave:73> cos(30)
ans = 0.15425
octave:74> cos(30*pi/180)
ans = 0.86603
octave:75> cos(60*pi/180)
ans = 0.50000
octave:76> cos(0.05333*pi/180)
ans = 1.00000
octave:77> sin(0.05333*pi/180)
ans = 9.3078e-04
ans = 0.94667
ans = 0.40963
A = 0.40963
B = 0.94667
t = linspace(0,10,1000);
xt = exp(-zeta*omega_n*t).*(A*sin(omega_d*t) +
B*cos(omega_d*t))+t1*cos(10*t+0.0533);
plot(t,xt)
t = linspace(0,100,1000);
xt = exp(-zeta*omega_n*t).*(A*sin(omega_d*t) +
B*cos(omega_d*t))+t1*cos(10*t+0.0533);
plot(t,xt)
2.24
A 120 kg mass is suspended by a spring of stiffness 32 103 N/m with a viscous
damping constant of 1200 Ns/m. The mass is initially at rest and in equilibrium.
Calculate the steady-state displacement amplitude and phase if the mass is excited
by a harmonic force of 100 N at 3 Hz.
Copyright © 2015 Pearson Education Ltd.
Solution: Given: m = 120 kg, k =32000 N/m, c = 1000 Ns/m, F= 100 N
rad
ω = 6π
s
The governing equation is
120xt  + 1200xt  + 32  10 3 xt  = 100cos2π  3t 
F
f o = o = 100 / 120 = 0.8333
M
c
ζ=
= 0.25516
2 km 

X=

F0
k  mω  + cω
2 2
2
=
 100 


 120 
32  10   1206π   + 1200 2π  3
3
2 2
= 0.00003334m
2
Next compute the angle from
1200   2π  3 = 1.1312rad (Resonance)
θ = tan 1
32  10 3   1206π 2
The angle is negative hence must be found in 4th quadrant. To find this we
have to use window 2.3 and then use Octave or find the principle value and add π
to it. Either way the phase is θ = 1.132 rad.
2.25
Plot the total solution of the system of Problem 2.24 including the transient.
Solution:The total response is given in the solution in the problem 2.24. For the
values given in the previous problem, and with zero initial conditions the response
is plotted using octave as given below
Plot:
octave:12> m =120;k=32000;c=1000;F=100;X=0.00003334;theta=-1.1312;
octave:13>omega_n = sqrt(k/m)
omega_n= 16.330
octave:14> zeta = c/(2*sqrt(k*m))
zeta = 0.25516
octave:15>omega_d= omega_n*sqrt(1-zeta**2)
omega_d= 15.789
octave:16> omega=6*pi
omega = 18.850
octave:17> phi = atan((-X*cos(theta)*omega_d)/(-X*cos(theta)*zeta*omega_n X*omega_n*sin(theta)))
phi = -0.47694
octave:18> A = -X*cos(theta)/sin(phi)
A = 3.0908e-05
octave:19> t= linspace(0,2,1000);
Copyright © 2015 Pearson Education Ltd.
octave:20>xt = A*exp(-zeta*omega_n*t).*sin(omega_d*t+phi) + X*cos(6*pi*t theta);
octave:21> plot(t,xt)
octave:22>xlabel("t")
octave:23>ylabel("x(t)")
Copyright © 2015 Pearson Education Ltd.
2.26
A damped spring-mass system modeled by (units are Newtons)
10 
x (t )  x (t )  170 x (t )  100 cos 4t
is also subject to initial conditions: x0 = 1mm and v0 = 20 mm/s. Compute the
total response, x(t), of the system.
.
:
Solution:Given
10xt  + x t  + 170 xt  = 100cos4t  that m = 10kg , ζ = 2cωn m , ωn = 17rad / s ,
f 0= 10 N / kg , and ω= 4 rad / s , with initial conditions x = 10 3 m and
0
3
v 0 = 20  10 m / s
N
m
rad
m = 10kg, k = 170 , c = 1, x0 = 10 3 m, v0 = 20  10 3 , ω = 4
m
s
s
rad
k
 170 
ωn =   = 
 = 4.1231
s
m
 10 
Following equation 2.26 with the values give here, the system to be solved has
damping ratio and thus is under-damped. Next convert the initial conditions to m
and m/s to be consistent given in the equation of motion
c
ζ=
= 1 / 2 170  10  = 0.012127
2 km 
Dividing through the mass yields the vibration properties
rad
ωd = ωn 1  ζ 2 = 4.1228
s
F
f = = 10 ,
m
rad
Since ω = 4.1231
the system is near resonance. Computing the amplitude
s
and phase of the particular solution from the values in window 2.3 yields
f0
X=
= 9.2848
2
2
2 2
ωn  ω + 2ζωn ω
Compute the phase of the transient response:
 2ζω ω 
 = tan 1  2 n 2  = 0.38051
 ωn  ω 
The amplitude of the transient is
x  Xcosθ 
A= 0
= 9.2848
sin  
The total response is then written from the equation 2.37 as
 ζω t
xt  = Ae n sin ωd t + + Xcosωt  q 





xt  = 9.2827e 0.05t sin 4.1228t +1.1906 + 9.2848cos4  0.38051
Copyright © 2015 Pearson Education Ltd.
2.27
Consider the pendulum mechanism of Figure P2.27 which is pivoted at point O.
Calculate both the damped and undamped natural frequency of the system for
small angles. Assume that the mass of the rod, spring, and damper are negligible.
What driving frequency will cause resonance?
k
l1 0.05 m
0
l2 0.07 m
c
l 0.10 m
m
F(t)
Figure P2.27
Solution: Assume the driving frequency to be harmonic of the standard form to
get resonance. The free body diagram is:
Here x1 = 1q and x2 =  2q so that x2 =  2q . To get the equation of motion take
the moments about point O to get:
å M 0 = Jq(t) = m2q(t)
= -k 1q (1 cosq ) - c 2q ( 2 cos q )
- mg(sin q ) + F0 cos w t( cosq )
Rearranging and approximating sin ~  and cos~1 yields:
m 2q˙˙(t) + c22 q˙ (t) + (k 21 + mg)q (t) = F0 cos wt
Dividing through by the coefficient of the inertia term and using the standard
definitions for  and  yields:
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wn =
z=
k21 + mg
which is the resnonant frequency
m2
c 22
2 (k 21 + mg)m2
wd = wn 1 - z 2 =
2.28
ö
k 21 + mg æ
c 2  42
1
2
2
2÷
ç
m
4(k 1 + mg)m ø
è
Consider the pivoted mechanism of Figure P2.27 with k = 4 x 103 N/m,l1 = 0.06
m,l2 = 0.09 m, and l = 0.12 m. and m = 40 kg. The mass of the beam is 40 kg; it
is pivoted at point 0 and assumed to be rigid. Design the dashpot (i.e. calculate c)
so that the damping ratio of the system is 0.2. Also determine the amplitude of
vibration of the steady-state response if a 15-N force is applied to the mass, as
indicated in the figure, at a frequency of 10 rad/s.
Solution:
2.29Compute the response of a shaft and disc system to an applied moment of
M = 10 sin 312 t
as indicated in Figure P2.29. Assume that the shaft is initially at rest (zero initial
conditions) and J = 0.6 kg m2, the shear modulus is G = 8.2× 1010 N/m2, the shaft
is 1 m long, of diameter 5 cm and made of steel. Assume the damping ratio of
steel is ζ = 0.01.
Copyright © 2015 Pearson Education Ltd.
GJ p
Solution: θt  + 2ζωn θ t  +
θ t  = 16.67sin 312t 
0.6l
The natural frequency and damped natural frequency of the system is
 GJ p 
k
 = 289.58
ωn =   = 
lJ
J




ωd = ωn  1.0  0.012 = 289.57
The magnitude and the phase of the steady state response are from window 2.3
m0
X=
= 0.0012249
2
2
ωn2  ω 2 + 2ζωn ω 



 2ζω ω
θ = tan 1  2 n 2
 ωn  ω



 = 0.13319

The phase of the transient is
= tan 1 ωd  x 0  Xcos θ  / v 0 +  x 0  Xcos θ ζωn   ωXsin θ  = 1.4732
The amplitude of the transient is
x  Xcosθ 
A= 0
= 0.0012249
sin  
The Total response is
θ t  = Ae
θ t  = 0.001249e
 0.01289.58t
t
+ Xcosωt  θ 
sin 289.57t  1.4732 + 0.0012249cos312t  0.13319 
 ζωn
sin ωd t +
2.30Compute the forced response of a spring-mass-damper system with the following
values: c = 250 kg/s, k = 2500 N/m, m = 125 kg, subject to a harmonic force of
magnitude F0= 15 N and frequency of 10 rad/s and initial conditions of x0 = 0.01
Copyright © 2015 Pearson Education Ltd.
m and v0 = 0.1 m/s. Plot the response. How long does it take for the transient
part to die off?
Solution:
Given:
125xt  + 250x t  + 2500xt  = 15cos10t 
m = 125kg ,  = 2cωn m , ωn = 20rad / s , F0 = 15 N , and ω = 10rad / s ,
with initial conditions x0 = 0.01m and v0 = 0.1m / s
The total solution is:
k
 2500 
ωn =   = 
=
m
 125 
ωn =
20 rad
s
f = F0/ m
k / m = 2500 / 125 = 4.4721
ζ = c / 2 km   = 0.22361
 
a = ω  ω 
ωd = ωn 1   2 = 4.3589
2
n

2
b = 2ωn ω b = 2ωn ω
t1 = ωn2  ω 2
 + 2ζω ω t1 = ω
2
2
n
2
n
 ω2
 + 2ζω ω 
2
2
n
A = x0  fa / t1
 ζω
B =  n
 ωd
 fa 
v
b
  
+ 0
 t1  ωd t1  ωd
C = f / t1
  ζ ωnt   Acosω t  + Bsin ω t  + C acosωt  + 2ζω ωsin ωt 
xt  = e
d
d
n
The Transient nature lasts for about 3 seconds.
Copyright © 2015 Pearson Education Ltd.
2.31
Compute a value of the damping coefficient c such that the steady state response
amplitude of the system in Figure P2.31 is 0.01 m.
x (t)
c
20 cos 6.3t N
100 kg
Friction
-free
surface
2,000 N/m
Figure P2.31
Solution:
From Eq. (2.39), the amplitude of the steady state response is given by
f0
X=
(w n 2 - w 2 )2 + (2zw nw )2
Then substitute, 2n = c/m, c =
and solve for c:
2
2 2
F0 2
2 (w n - w )
into this equation
m
w2 × X 2
w2
X = 0.01m w = 6.3 rad / s F0 = 20N m = 100kg
w n2 =
2.32
k 2000
=
= 20 (rad/s) 2 Þ c = 55.7 kg/s
m 100
Consider a spring-mass-damper systems like the one in Figure P2.31 with the
following values: m = 80 kg, c = 80 kg/s, k = 2400 N/m, F0 = 20 N, and the
driving frequency  = 4.3 rad/s. Compute the magnitude of the steady-state
response and compare it to the magnitude of the forced response of an undamped
system.
Solution:
The magnitude of the steady state response in the undamped condition is:


F
20






m

80
 = 0.04344m


=2 2
=2
  2400 

 ωn  ω 2 
2


  4.3 



  80 

The magnitude of the steady state response in the damped condition is:
F
20
m
80
=
=
= 0.00025433m
k  mω  + cω
2 2
2.33
2
2400  804.3  + 804.3
2 2
2
Compute the response of the system in Figure P2.33 if the system is initially at
Copyright © 2015 Pearson Education Ltd.
rest for the values k1 = 200 N/m, k2 = 1000 N/m, c = 20 kg/s and m = 175 kg.
Solution:
Given:




1
 xt  = 25cos3t 
175xt  + 20x t  + 
1 
 1
+


 200 1000 
1
1
k=
=
1
1
1
1
+
+
k1 k 2
200 1000
that m = 175kg , ζ = 2cωn m , F0 = 25 N , and ω = 3rad / s ,
With initial conditions x0 = 0.0m and v0 = 0.0m / s
k
 2500 
ωn =   = 
=
m
 125 
Total Solution
20 rad
s
f = F0 / m
ωn =
k / m  = 2500 / 125 = 4.4721
ζ = c / 2 km   = 0.22361
 
a = ω  ω 
ωd = ωn 1   2 = 4.3589
2
n

2
b = 2ζωn ω
t1 = ωn2  ω 2
 + 2ζω ω 
2
2
n
A = x0  fa / t1
 ζω
B =  n
 ωd
xt  = e
2.34
  ζ ωnt  
 fa 
v
b
  
+ 0
 t1  ωd t1  ωd
C = f / t1
Acosωd t  + Bsin ωd t  + C acosωt  + 2ζωn ωsin ωt 
Write the equation of motion for the system given in Figure P2.34 for the case
that F(t) = F cos w t and the surface is friction free. Does the angle  affect the
magnitude of oscillation?
Copyright © 2015 Pearson Education Ltd.
c
k
m
F (t)
Figure P2.34
Solution:
Free body diagram:
q
x
Fs
m
mg sinq
(Forces that are normal
to the x direction are
neglected)
F(t)=F cos wt
Assuming x = 0 to be at the equilibrium:
å F = F + mgsinq - F = mx˙˙
x
s
mg sin 
)
where Fs  k ( x 
k
and
F(t) = F cos w t
Then the equation of motion is:
m ˙x˙ + k x = F cos w t
Note that the equation of motion does not contain  which means that the
magnitude of the response is not affected by the angle of the incline.
2.35
A foot pedal for a musical instrument is modeled by the sketch in Figure P2.35.
With k = 2400 N/m, c = 30 kg/s, m = 32 kg and F(t) = 50 cos tcompute the
steady state response assuming the system starts from rest. Also use the small
angle approximation.
Solution: Given: m = 32 kg, k = 2400 N/m, c= 30 kg/s, F t  = 50cos2πt N , a =
0.05m
Summing the moments about the point O
2
θ o = m3a  = 9a 2 m, Fs = kasin θ , Fc = c2  a  sin ' = 2cacos
M
o
= F 3a   Fc 2a   Fs a  = I
Substituting these equations and simplifying
9a 2 m2 2 t 
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Observing the equation of motion, equivalent mass, damping and stiffness
coefficients are
2.36
Consider the system of Problem 2.15, repeated here as Figure P2.36 with the
effects of damping indicated. The physical constants are J =24 kg m2, k = 2500
Nm/rad, and the applied moment is 5 Nm at 1.432 Hz acting through the distance
r = 0.5 m. Compute the magnitude of the steady state response if the measured
damping ratio of the spring system is  = 0.01. Compare this to the response for
the case where the damping is not modeled ( = 0).
SolutionGiven: From equation 2.39 the magnitude of the steady sate response for
an under-damped system is
J = 24 k = 2500 M = 5 ω = 2π  1.432 = 8.9975rad / s r = 0.5
ζ = 0 . 01
X=
ω
 M0 


 J 
2
n
 ω2
 + 2ζω ω 
2
2
n
For ζ = 0 Magnitude of Steady State response damped Condition
5
 M 






24
 = 0.01795m
= 2 2 J 2  = 2
 ωn  ω 
 2500 / 24   8.99752 








for ζ = 0.01 , the magnitude of Steady State response for an underdamped system
is
Copyright © 2015 Pearson Education Ltd.
=
=
M
J
k  J ω  + cω
2 2
2
5
24
2500  248.9975  + 248.9975
2 2
= 0.00021537 m
2
where X is the vertical displacement of the wing tip. Thus a small amount of
damping can greatly reduce the amplitude of vibration. Strictly speaking, the
undamped case does not have a steady state, and the above answer is really the
magnitude of the particular solution. To get the actual value undamped value
look at equation (2.13).
2.37
A machine, modeled as a linear spring-mass-damper system, is driven at
resonance (n =  =2 rad/s). Design a damper (that is choose a value of c) such
that the maximum deflection at steady state is 0.05 m. The machine is modeled as
having a stiffness of 2000 kg/m and the excitation force has a magnitude of 100
N.
Solution: Since the frequency and stiffness are given, we can find the mass from
w n2 =
k 2000
=
= 4 rad 2 / s2 Þ m = 500 kg
m
m
Thus, f0 = 100/500 = 0.2 N/kg. At resonance the steady state amplitude becomes
f
f0
0.2
X= 02=
=
= 0.05
c
æ c ö 2
2zw
4
2ç
w
(2000)(500)
è 2 km ÷ø
Þ c = (2000)(500) = 1,000 kg/s
2.38
Derive the total response of the system to initial conditions x0 and v0 using the
homogenous solution in the form xh (t) = e-zw n t (A1 sin w d t + A2 cos w d t) and hence
verify equation (2.38) for the forced response of an underdamped system.
Solution:
From Sec. 1.3, the homogeneous solution is:
xh (t) = e-zw n t (A1 sin w d t + A2 cos w d t)
From equations (2.29) and (2.35), the particular solution is:
(w n2 - w 2 ) f 0
2zw nwf0
x p (t) =
sin wt
2 2
2 cos w t +
2
2
(w n - w ) + (2zw nw )
(w n - w 2 )2 + (2zw nw )2
Then the general solution is:
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x(t) = xh (t) + x p (t) = e-z w n t (A1 sin w d t + A2 cos w d t)
+
(w n 2 - w 2 ) f 0
2zw n wf0
cos w t +
sin wt
2
2 2
2
2
2 2
2
(w n - w ) + (2zw n w )
(w n - w ) + (2zw nw )
Using the initial conditions, x(0) = x0 yields:
(w n2 - w 2 ) f0
x(0) = x0 = A2 + 2
(w n - w 2 )2 + (2zw n w )2
(w n2 - w 2 ) f0
Þ A2 = x 0 - 2
(w n - w 2 )2 + (2zw n w )2
x˙ (0) = v0 yields:
 = v0 = zw ne -zw n 0 (A1 sin w d 0 + A2 cos w d 0) + e -zw n 0 ( A1 cos w d 0 - A2 sin w d 0)
x(t)
-
(w n2 - w 2 )w f0
2zw n w f0
sin w 0 + 2
cos w 0
2
2 2
2
(w n - w ) + (2zw n w )
(w n - w 2 )2 + (2zw n w )2
Þ v0 = zw n A2 + A1 +
A1 =
2zw n w f0
(w - w 2 )2 + (2zw n w ) 2
2
n
v0 w
2zw n w f0
wn
(w n2 - w 2 ) f0
+
× 2
+
z
(x
)
w d w d (w n - w 2 )2 + (2zw n w )2
w d 0 (w n2 - w 2 ) 2 + (2zw n w )2
Then, Eq. (2.38) is obtained by substituting the expressions for A1 and A2 into the
general solution and simplifying the resulting equation.
Copyright © 2015 Pearson Education Ltd.
Problems and Solutions Section 2.3 (2.39 through 2.44)
2.39
Referring to Figure 2.11, draw the solution for the magnitude X for the case m = 80 kg, c
= 3200 N s/m, and k = 8,000 N/m. Assume that the system is driven at resonance by a
10-N force.
Solution:
Given: Given m = 80 kg, c = 3200 N s/m, k = 8000 N/m F = 10 N
rad
k
 8000 
ω=   = 
 = 10
s
m
 80 
cω
π
θ = tan −1
= rad
2
(k − mω ) 2
From the figure
F
m
X=
(k - mω ) + (cω)
2 2
=
2
10
80
((8000 - 80 × 100)
2
+ (3200 × 10)
Copyright © 2015 Pearson Education Ltd.
2
)
= 3.125e - 04m
2.40
Use the graphical method to compute the phase shift for the system with m = 100 kg, c =
4000 N s/m, k = 10,000 N/m, and F 0 =10 N, if ω = ωn /2 and again for the case ω = 2ωn .
Solution:
From Problem 2.32
(a)
= 10 rad/s
= 5 rad/s
m
kX = (10,000)(.000468) = 4.68 N
cωX = (4000)(5)(.000468) = 9.36 N
= (100)
(.000468) = 1.17 N
From the figure given in problem 2.32:
rad
(b)
rad/s
m
kX = (10000)(.000117) = 1.17 N
cωX = (4000)(20)(.000117) = 9.36 N
=(100)
(.000117) = 4.68 N
From the figure:
rad
Copyright © 2015 Pearson Education Ltd.
2.41
A body of mass 80 kg is suspended by a spring of stiffness of 25 kN/m and dashpot of
damping constant 800 N s/m. Vibration is excited by a harmonic force of amplitude 60 N
and a frequency of 3 Hz. Calculate the amplitude of the displacement for the vibration
and the phase angle between the displacement and the excitation force using the graphical
method.
Solution:
Given: m = 80 kg, C = 25 kN/m, F 0 =60 N c = 800 Ns / m
ω = 32π = 18.85
Kx = 25000 x
cωω= 800 × 28.274 x = 15079.6447 x
mω 2 x = 80 × 28.274 2 x = 28424.4606 x
X=
F
m
(k − mω ) + (cω)
θ = tan −1
2 2
=
2
60
((25000 − 28424.4606)
2
+ 15079.6447 2
cω
4800π
= tan −1
= −1.3479rad = −77.2056
2
2
k − mω
25 − (80)(6π )
Copyright © 2015 Pearson Education Ltd.
)
= 0.00388m
2.42
Calculate the real part of equation (2.55):
to verify that this is consistent with the equation (2.36):
and hence establish the equivalence of the exponential approach to solving the damped
vibration problem with method of undetermined coefficients.
Solution:
Equation (2.55)
whereθ
Using Euler’s Rule:
The real part is:
Rearranging:
which has the same form as Equation (2.36).
2.43
Referring to equation (2.56):
and a table of Laplace Transforms (see Appendix B), calculate the solution x(t) by using
a table of Laplace transform pairs and show that the solution obtained this way is
equivalent to (2.36).
Solution: Taking the Laplace transform of the equation of motion is given in Equation
(2.56):
Solving this expression algebraically for X yields
Using Laplace Transform pairs from the table, this last expression is transformed into the
time domain to get:
Copyright © 2015 Pearson Education Ltd.
cos (ωt- )
2.44
Solve the following system using the Laplace Transform method and the table in
Appendix B:
Check your solution against equation (2.11) obtained via the method of undetermined
coefficients.
Solution: Frist divide through by the mass to get:
Taking the Laplace Transform of the equation of motion considering the initial conditions
yields (remember the formulas in table B.1 are for zero initial conditions):
Solving this for X(s) yields
Using table B.1 and taking the Inverse Laplace transform of each term yields
which is in agreement with equation (2.11) derived by the method of undetermined
coefficients. Note that the last term used table B.1, entry 21.
Copyright © 2015 Pearson Education Ltd.
2- 27
Problems and Solutions Section 2.4 (2.45 through 2.60)
2.45
For a base motion system described by
m
x + c x + kx = cYw b cos w b t + kYsin w b t
With m = 100 kg, c = 50 kg/s, k = 1000 N/m, Y = 0.03 m and b = 3 rad/s. compute the
magnitude of the particular solution. Last compute the transmissibility ratio.
Solution: First define the usual vibration properties by dividing through by the mass to
get
1000
c
50
wn =
= 3.1623 rad/s, z =
=
= 0.079
100
2 mk 2 100,000
r=
wb
3
=
= 0.94868
w n 3.1623
Then use equation (2.70) to compute the magnitude of the particular solution:
1
é
ù
é 1+ ( 2z r ) 2
ù2
1+ ( 2 × 0.079 × 0.94868 )2
ê
ú = 0.033 m
X =Y ê
ú
=
0.03
2
2 2
2 2
2
ê
ú
êë (1- r ) + ( 2z r ) úû
ë 1- ( 0.94868 ) + ( 2 × 0.079 × 0.94868 ) û
Thus the transmissibility is 1.111> 1 and motion is amplified by the system. This is
because the system is near resonance and there is very little damping in the system.
(
2.46
)
For a base motion system described by
m
x + c x + kx = cYw b cos w b t + kYsin w b t
with m = 100 kg, c = 50 N/m, Y = 0.03 m and b = 3 rad/s, find largest value of the
stiffness k and that makes the transmissibility ratio less than 0.75.
Solution: Using the formula from equation (2.71) students can write a program to
calculate values of X/Y for decreasing values of k. Note that if you increase k the ratio r
increases moving the response to values lower then 1 per figure 2.14. Such a procedure
yields a value of about k = 665 N/m for a transmissibility ratio of 0.747.
2.47
A machine weighing 1800 N rests on a support as illustrated in Figure P2.47. The
support deflects about 4 cm as a result of the weight of the machine. The floor under the
support is somewhat flexible and moves, because of the motion of a nearby machine,
harmonically near resonance (r =1) with an amplitude of 0.2 cm. Model the floor as base
motion, and assume a damping ratio of = 0.01, and calculate the transmitted force and
the amplitude of the transmitted displacement.
Solution:
The differential equation will be of the form:
m x́ +c (x́ − ý )+ k (x − y )= 0
And in this case after substituting the known parameters we have
(183.48 )x́ + (406.37 )(x́ − ý )+ (45000 )(x − y )= 0
k=
1800
= 45000 N / m
0.04
ζ = 0.01
Copyright © 2015 Pearson Education Ltd.
2- 28
 45000  1800 
0.5
c = 2ζ km  = 2  0.01  
 = 406.37 Ns / m
9.81


Transmitted displacement from equation 2.70 is
1
1
2
 1 + 2ζr 2
2
1+ 2  0.01  2
X =Y
= 0.10002m
 = 0.002 
2 
2
2 2
 2  0.01 
 1  r + 2ζr  
0.5


The Transmitted force is given from equation 2.77
F = kr 2 X = 45000  0.10002 = 4500.9 N
2.48
Derive Equation (2.70):
1/ 2
é 1+ (2z r)2
ù
X = Yê
2 2
2ú
ë (1- r ) + (2z r) û
from (2.68):
1/ 2
é
ù
w 2 + (2zw b )2
x p (t) = w nY ê 2 n 2 2
2ú
êë (w n - w b ) + (2zw nw b ) úû
cos( w bt - q1 - q 2 )
to see if the author has done it correctly.
Solution:
Equation (2.68) states:
1/ 2
é
ù
w n2 + (2zw b )2
x p (t) = w nY ê 2
2 2
2ú
êë (w n - w b ) + (2zw nw b ) úû
cos( w bt - q1 - q 2 )
1/ 2
é
ù
w n2 + (2zw b )2
The magnitude is: X = w nY ê 2
2 2
2ú
êë (w n - w b ) + (2zw nw b ) úû
1/ 2
é
ù
(w n-4 )(w n2 + (2zw b )2 )
= w nY ê -4
2
2 2
2 ú
êë (w n )((w n - w b ) + (2zw nw b ) ) úû
1/ 2
é (w n-2 )(1 + (2z r)2 ) ù
= w nY ê
2 2
2 ú
êë (1 - r ) + (2z r) úû
Þ
1/ 2
ù
1 é 1 + (2z r) 2
= w nY
ê
ú
w n ë (1 - r 2 )2 + (2z r)2 û
Þ
1/ 2
é 1+ (2z r)2
ù
X
= Yê
2 2
2ú
ë (1- r ) + (2z r) û
This is equation (2.71).
Copyright © 2015 Pearson Education Ltd.
2- 29
2.49
From the equation describing Figure 2.13, show that the point ( 2 , 1) corresponds to the
value TR > 1 (i.e., for all r < 2 , TR > 1).
Solution:
1/ 2
ù
X é 1+ (2z r)2
Equation (2.71) is
TR =
=ê
ú
Y ë (1- r 2 )2 + (2z r)2 û
Show TR > 1 for r< 2
1/ 2
ù
X é 1 + (2z r)2
TR =
= ê
2 2
2 ú
Y
ë (1- r ) + (2z r) û
1 + (2z r)2
>1
(1 - r 2 )2 + (2z r)2
1 + (2z r)2 > (1 - r 2 )2 + (2z r)2
1 > (1 - r 2 )2
Take the real solution:
>1
1 - r 2 < +1 or 1 - r 2 < -1 Þ
-r 2 > -2 Þ r 2 < 2 Þ r < 2
Copyright © 2015 Pearson Education Ltd.
2- 30
2.50
Consider the base excitation problem for the configuration shown in Figure P2.50. In this
case the base motion is a displacement transmitted through a dashpot or pure damping
element. Derive an expression for the force transmitted to the support in steady state.
Figure P2.50
Solution: The entire force passes through the spring. Thus the support sees the force FT =
kX where X is the magnitude of the displacement. From equation (2.65)
2zw nw b kY
FT = kX =
2
(w n - w b2 )2 + (2zw nw b )2
=
2.51
2z rkY
(1 - r 2 )2 + (2z r)2
A very common example of base motion is the single-degree-of-freedom model of an
automobile driving over a rough road. The road is modeled as providing a base motion
displacement of y(t) = (0.01)sin (5.818t) m. The suspension provides an equivalent
stiffness of k = 3.273 x 104 N/m, a damping coefficient of c = 231 kg/s and a mass of
1007 kg. Determine the amplitude of the absolute displacement of the automobile mass.
Solution:
From the problem statement we have (working in Mathcad)
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2- 31
2.52
A vibrating mass of 250 kg, mounted on a massless support by a spring of stiffness
32,000 N/m and a damper of unknown damping coefficient, is observed to vibrate with a
10-mm amplitude while the support vibration has a maximum amplitude of only 2.5 mm
(at resonance). Calculate the damping constant and the amplitude of the force on the
base.
Solution:
Given:
m= 250 kg k = 32000 X = 0.01 m Y = 0.0025 m r= 1
Solution:
Finding damping constant (Equation 2.71)
1
1 + 2ζ 2  2
We know that X = Y 
at resonance

2

2
ζ



X2
1
2=
+1
Y
4 ξ2
Y
0.0025
ξ=
=
= 0.13
2
2
2
2
2 X Y
2 0.01  0.0025
 c = 2ζ km = 2  0.13  32000 250 = 735.4Ns / m
Amplitude of force on base (Equation 2.76)
FT = kYr
2


1 + 2ζr 2
2

 = kr X
2
2 2
 1  r + 2ζr  


Also F = kr X = 32000  0.01 = 320 N
2
2.53
Referring to Example 2.4.2, at what speed does car 1 experience resonance? At what
speed does car 2 experience resonance? Calculate the maximum deflection of both cars
at resonance.
Solution:
Given: m 1 = 1007 kg, m 2 =1585 kg, k = 4x10 5 N/m; c = 2,000 kg/s, Y = 0.01m
Velocity for resonance: (from Example 2.4.1)
w b = 0.2909v (v in km/h)
Car 1: w 1 =
Car 2: w 2 =
k
4 ´ 10 4
=
= w b = 0.2909v1
m
1007
v 1 = 21.7 km/h
k
4 ´ 10 4
=
= w b = 0.2909v2
m
1585
v 2 = 17.3 km/h
Copyright © 2015 Pearson Education Ltd.
2- 32
Maximum deflection: (Equation 2.71 with r = 1)
1/ 2
é 1 + 4z 2 ù
Þ
X = Yê
2 ú
ë 4z û
c
2000
Car 1: z1 =
=
= 0.158
2 km1 2 (4 ´ 10 5 )(1007)
1/ 2
é 1+ 4(0.158) 2 ù
X 1 = (0.01) ê
2 ú
ë 4(0.158) û
Car 2: z 2 =
= 0.033 m
c
2000
=
= 0.126
2 km2 2 (4 ´ 10 4 )(1585 )
1/ 2
é 1+ 4(0.126) 2 ù
X 2 = (0.01) ê
2 ú
ë 4(0.126) û
2.54
= 0.041 m
For cars of Example 2.4.2, calculate the best choice of the damping coefficient so that the
transmissibility is as small as possible by comparing the magnitude of z = 0.01, z = 0.1
and z = 0.2 for the case r = 2. What happens if the road “frequency” changes?
Solution:
From Equation 2.62, with r = 2, the displacement transmissibility is:
1/ 2
1/ 2
ù
é 1 + 16z 2 ù
X é 1+ (2z r)2
=ê
ú =ê
2 ú
Y ë (1 - r 2 )2 + (2z r)2 û
ë 9 + 16z û
X
For z = 0.01, = 0.334
Y
X
For z = 0.1, = 0.356
Y
X
For z = 0.2, = 0.412
Y
The best choice would be z = 0.01.
If the road frequency increases, the lower damping ratio would still be the best choice.
However, if the frequency decreases, a higher damping ratio would be better because it
would approach resonance.
2.55
A system modeled by Figure 2.13, has a mass of 200 kg with a spring stiffness of 3.0 
104 N/m. Calculate the damping coefficient given that the system has a deflection (X) of
0.7 cm when driven at its natural frequency while the base amplitude (Y) is measured to
be 0.3 cm.
Copyright © 2015 Pearson Education Ltd.
2- 33
Solution:
Given: m = 200 kg , k = 30000 , X = 0.007m , Y = 0.003m , r = 1
Solution:
We know that the base excitation Equation 2.71 gives:
1
2
1 + 2ζ   2
X =Y
 with r =1
2
 2ζ  
2
X
1
=
+1
2
Y
4ζ2
ζ=
Y
2 X Y
2
 c = 2ζ
2.56
2
=
0.003
2 0.007  0.003
2
2
= 0.23717
km = 2  0.23717 30000 200 = 1161.89Ns / m
Consider Example 2.4.2 for car 1 illustrated in Figure P2.56, if three passengers totaling
200 kg are riding in the car. Calculate the effect of the mass of the passengers on the
deflection at 20, 80, 100, and 150 km/h. What is the effect of the added passenger mass
on car 2?
Figure P2.56 Model of a car suspension with the mass of the occupants, mp, included.
Solution:
Add a mass of 200 kg to each car. From Example 2.4.2, the given values are:
m 1 = 1207 kg, m 2 = 1785 kg, k = 4x104 N/m; c = 2,000 kg/s, w b = 0.29v.
Car 1: w 1 =
z1 =
k
=
m
c
2 km1
=
4 ´ 10 4
= 5.76 rad/s
1207
2000
2 (4 ´ 10 5 )(1207)
= 0.144
Copyright © 2015 Pearson Education Ltd.
2- 34
4 ´ 10 4
= 4.73 rad/s
1785
2000
z2 =
=
= 0.118
2 km2 2 (4 ´ 10 5 )(1785)
Car 2: w 2 =
k
=
m
c
1/ 2
é 1 + (2z r)2
ù
Using Equation (2.71): X = Y ê
produces the following:
2 2
2ú
(1
r
)
+
(2
z
r)
ë
û
Speed (km/h)
wb
r1
r2
20
80
100
150
(rad/s)
5.817
23.271
29.088
2.40
1.01
3.871
5.05
7.58
1.23
4.71
6.15
9.23
x1
(cm)
3.57
0.107
0.072
0.042
x2
(cm)
1.77
0.070
0.048
0.028
At lower speeds there is little effect from the passengers weight, but at higher speeds the
added weight reduces the amplitude, particularly in the smaller car.
Copyright © 2015 Pearson Education Ltd.
2- 35
2.57 Consider Example 2.4.2. Choose values of c and k for the suspension system for
car 2 (the sedan) such that the amplitude transmitted to the passenger compartment is as
small as possible for the 1 cm bump at 50 km/h. Also calculate the deflection at 100
km/h for your values of c and k.
Solution:
For car 2, m = 1585 kg.
Also, w b = 0.2909(50) = 14.545 rad/s and Y = 0.01 m.
From equation (2.70),
1/ 2
é 1 + (2z r)2
ù
X =Yê
2 2
2ú
ë (1 - r ) + (2z r) û
From Figure 2.9, we can choose a value of r away from resonance and a low damping
ratio. Choose r = 2.5 and z =0.05.
So,
r = 2.5 =
wb
14.545
=
w
k / 1585
k = 53,650 N/m
c
z = 0.05 =
2 km
c = 922.2 kg/s
1/ 2
So,
é
ù
2
ê
ú
1 + [2(0.05)(2.5)]
X = (0.01) ê
ú
2
ê 1- 2.5 2 + [2(0.05)(2.5)]2 ú
êë
úû
(
( ))
At 100 km/h, b = 29.09 rad/s and r =
= 0.00196 m
wb
= 5.
k/m
Copyright © 2015 Pearson Education Ltd.
2- 36
2.58
Consider the base motion problem of Figure 2.13. a) Compute the damping ratio needed
to keep the displacement magnitude transmissibility less than 0.50 for a frequency ratio
of r = 1.5. b) What is the value of the force transmissibility ratio for this system?
Solution: Working with equation 2.71 make a plot of TR versus ζ and use equation 2.77
to compute the value of force transmissibility.
r=1 . 8 M (ζ )= 0.50
Solution:
1
2
 1+ 2ζr 
2
We know that TRζ  = 
 at resonance
2
2
 1  r 2 + 2ζr  
Using Octave plotting the TR and Mr
The force transmissibility of this problem is
F ζ t = r 2TRζ 


F 0.1596 = 1.82  0.4976 = 1.596
From the plot the value of ζ = 0.1596 keeps the displacement less than 0.5 as desired.
The value of force transmissibility is less than 1.596. Precise values can be found out by
equating the expression to 0.50.
Copyright © 2015 Pearson Education Ltd.
2- 37
2.59
Consider the effect of variable mass on an aircraft landing suspension system by
modeling the landing gear as a moving base problem similar to that shown in Figure
P2.56 for a car suspension. The mass of a regional jet is 13, 236 kg empty and its
maximum takeoff mass is 21,523 kg. Compare the maximum deflection for a wheel
motion of magnitude 0.50 m and frequency of 35 rad/s, for these two different masses.
Take the damping ratio to be  = 0.1 and the stiffness to be 4.22 x 106 N/m.
Solution: Using a Mathcad worksheet the following calculations result:
Note that if the suspension stiffness were defined around the full case, when empty the
plane would bounce with a larger amplitude then when full. Note Mathcad does not have
a symbol for a Newton so the units on stiffness above are kg/sec2 in order to allow
Mathcad to compute the units.
2.60
Consider the simple model of a building subject to ground motion suggested in Figure
P2.60.The building is modeled as a single degree of freedom spring-mass system where
the building mass is lumped atop of two beams used to model the walls of the building in
bending. Assume the ground motion is modeled as having amplitude of 0.1 m at a
frequency of 7.5 rad/s. Approximate the building mass by 105 kg and the stiffness of
each wall by 3.519 x 106 N/m. Compute the magnitude of the deflection of the top of the
building.
Figure P2.60 A simple model of a building subject to ground motion, such as an
earthquake.
Copyright © 2015 Pearson Education Ltd.
2- 38
Solution: The equation of motion is
m
x(t) + 2kx(t) = 0.1cos7.5t
The natural frequency and frequency ratio are
2k
w
7.5
= 8.389 rad/s and r =
=
= 0.894
m
w n 8.389
The amplitude of the steady state response is given by equation (2.70) with  = 0 in this
case:
1
X =Y
= 0.498 m
1- r2
Thus the earthquake will cause serious motion in the building and likely break.
wn =
Copyright © 2015 Pearson Education Ltd.
2- 39
Problems and Solutions Section 2.5(2.61 through 2.68)
2.61
A lathe can be modeled as an electric motor mounted on a steel table. The table plus the
motor have a mass of 60 kg. The rotating parts of the lathe have a mass of 4 kg at a
distance 0.12 m from the center. The damping ratio of the system is measured to be ζ =
0.06 (viscous damping) and its natural frequency is 7.5 Hz. Calculate the amplitude of
the steady-state displacement of the motor, assuming  r = 30 Hz.
Soltuion:
Solution: Given m = 60 kg, mo=4 kg, e = 0.12 m ζ = 0.06, ω n = 7.5 Hz,  r = 30 Hz
ωr 30
so, r= ω = 7.5 = 4
n
From equation 2.84
m0 e

r2
4  0.12 
42
X=
=
m
60
1  r 2 2 + 2ξr 2
1  4 2 2 + 2  0.06  42
= 0.0066183m = 0.66183cm
2.62
The system of Figure 2.19 produces a forced oscillation of varying frequency. As the
frequency is changed, it is noted that at resonance, the amplitude of the displacement is
10 mm. As the frequency is increased several decades past resonance the amplitude of
the displacement remains fixed at 1 mm. Estimate the damping ratio for the system.
Solution: Equation (2.84) is
me
r2
X = o
m (1 - r 2 ) 2 + (2zr ) 2
At resonance, X = 10 mm =
mo e 1
m 2z
10m
1
=
mo e 2z
Xm
When r is very large,
= 1 and X = 1 mm, so
mo e
m
=1
mo e
1
Therefore, 10(1) =
2z
z = 0.05
Copyright © 2015 Pearson Education Ltd.
2- 40
2.63
An electric motor (Figure P2.63) has an eccentric mass of 12 kg (12% of the total mass of
100 kg) and is set on two identical springs (k = 3000 N/m). The motor runs at 1800 rpm,
and the mass eccentricity is 100 mm from the center. The springs are mounted 250 mm
apart with the motor shaft in the center. Neglect damping and determine the amplitude of
the vertical vibration.
Solution:
N
, e = 0.1
m
rev  min
rad 
rad
ωr = 1800
2π

 = 188.50
min  60 sec
rev 
s
Given mo = 12kg,m = 100kg , k = 3000
Vertical vibration
ωn =
3000 / 100  = 5.4772 rad
s
ωr 188.50
r=
=
= 34.415 m
ωn 5.4772
From equation 2.84
 m e  r 2 
 = 0.012010
X =  o 
2 
 m  1  r 

2.64

Consider a system with rotating unbalance as illustrated in Figure P2.63. Suppose the
deflection at 1800 rpm is measured to be 0.05 m and the damping ratio is measured to be
ζ = 0.1. The out-of-balance mass is estimated to be 10%. Locate the unbalanced mass
by computing e.
Solution: Solution: Given x = 0.05 m, ζ = 0.1 and from the solution to the problem 2.53
the frequency ratio is calculated to be 34.414
2π  1800
60
r=
= 2π 30 = 34.414
3000 / 100
Solving the rotating unbalance from equation 2.84 for e yields

 mX
e = 
 m0




1  r  + 2ζr 
2 2
r2
2
=

10  0.05  1  34.414 
 + 2  0.1 34.414
2 2
34.4142
2
= 0.49958m
This sort of calculation can be introduced to discuss the application of machinery
diagnostics if time permits. Machinery diagnostics deals with determining the location
and extent of damage from measurements of the response and input.
Copyright © 2015 Pearson Education Ltd.
2- 41
2.65
A fan of 45 kg has an unbalance that creates a harmonic force. A spring-damper system
is designed to minimize the force transmitted to the base of the fan. A damper is used
having a damping ratio of z = 0.2. Calculate the required spring stiffness so that only
10% of the force is transmitted to the ground when the fan is running at 10,000 rpm.
Solution: The equation of motion of the fan is
m˙x˙ + c x˙ + kx = m0 ew 2 sin(wt + q)
The steady state solution as given by equation (2.84) is
m0 e
r2
x(t) =
sin w t
m (1 - r 2 )2 + (2z r)2
where r is the standard frequency ratio. The force transmitted to the ground is
m0 e
kr 2
m0 e
cw r 2
F(t) = kx + cx =
sin w t +
cos w t
m (1 - r 2 )2 + (2z r)2
m (1 - r 2 )2 + (2z r)2
Taking the magnitude of this quantity, the magnitude of the force transmitted becomes
1 + (2z r)2
m0 e r 2 k 2 + c 2w 2
F0 =
= m0 ew
m (1 - r 2 )2 + (2z r)2
(1 - r 2 )2 + (2z r)2
From equation (2.81) the magnitude of the force generated by the rotating mass Fr is
Fr = m0 ew 2
The limitation stated in the problem is that F0 = 0.1Fr, or
m0 ew
1 + (2z r)2
2
(1 - r ) + (2z r)
Setting  = 0.2 and solving for r yields:
2 2
4
2
= 0.1m0 ew 2
2
r -17.84r - 99 = 0
which yields only one positive solution for r2, which is
2
w2
k æ 10000 ´ 2p ö
1
r = 22.28 =
Þ =ç
÷
k
m è
60
ø 22.28
m
2
2
æ 10000 ´ 2p ö
1
Þ k = 45 ç
= 2.21 ´ 106 N/m
÷
60
è
ø 22.28
Copyright © 2015 Pearson Education Ltd.
2- 42
2.66
Plot the normalized displacement magnitude versus the frequency ratio for the out of
balance problem (i.e., repeat Figure 2.21) for the case of z = 0.05.
Solution: Working in Mathcad using equation (2.84) yields:
Copyright © 2015 Pearson Education Ltd.
2- 43
2.67
Consider a typical unbalanced machine problem as given in Figure P2.67 with a machine
mass of 150 kg, a mount stiffness of 1000 kN/m and a damping value of 600 kg/s. The
out of balance force is measured to be 374 N at a running speed of 3000 rev/min. a)
Determine the amplitude of motion due to the out of balance. b) If the out of balance
mass is estimated to be 1% of the total mass, estimate the value of the e.
Fo
ωr2
m = 150 kg, stiffness k = 1000 K N / m and c = 600 kg/s
100π  = 3.8476
ω
r=
=
k
10 6
m
150
 F0 
 374 
2
 ω2 
2


3.8476 

r
10000π 2 



X=
=
2
2
m
150
2 2
 


1  r  +  2ζr 
600
2 2
1   3.8476  +  2 
3.8476  
 2 150 106  


 

= 0.00002709m
Solution: Using equation 2.84 with mo e =


And
e=
2.68
F
374
=
= 0.00002526m
2
mω
150  10000  π 2
Plot the response of the mass in Problem 2.67 assuming zero initial conditions.
Copyright © 2015 Pearson Education Ltd.
2- 44
Solution: The steady state response is the particular solution given by equation (2.84)
and is plotted here in Mathcad:
Copyright © 2015 Pearson Education Ltd.
2-42
Problems and Solutions Section 2.6 (2.69 through 2.72)
2.69
Calculate damping and stiffness coefficients for the accelerometer of Figure 2.24 with
moving mass of 0.04 kg such that the accelerometer is able to measure vibration between
0 and 50 Hz within 5%. (Hint: For an accelerometer it is desirable for
=
constant.)
Solution: Use equation (2.90):
Given: m = 0.04 kg with error < 5%
0.2f = 50 Hz
f = 250 Hz
ω=2
Thus,
k=
= 98,696 N/m
= 1570.8 rad/s
<1.05
When r = .2,
0.95 <
This becomes
0.8317+0.1444
(± 5% error)
<1<1.016+0.1764
Therefore,
c = 87.956 Ns/m
2.70
The damping constant for a particular accelerometer of the type illustrated in Figure 2.26
is 50 N s/m. It is desired to design the accelerometer (i.e., choose m and k) for a
maximum error of 3% over the frequency range 0 to 75 Hz.
Solution: Given 0.2f = 75 Hz
equation (2.93) when r = 0.2:
f = 375 Hz
0.97 <
ω n= 2
<1.03
This becomes
0.8671 + 0.1505
Therefore,
0.3622 <
= 2356.2 rad/s. Using
(± 3% error)
<1<0.9777+0.1697
<0.9395
Choose
m = 0.015 kg
= 8.326 × 104 N/m
Copyright © 2015 Pearson Education Ltd.
2-43
2.71
The accelerometer of Figure 2.24 has a natural frequency of 120 kHz and a damping ratio
of 0.2. Calculate the error in measurement of a sinusoidal vibration at 60 kHz.
Solution:
Given: ω = 120 kHz,
= 60 kHz
So,
The error is
2.72
× 100% = 28.8%
Design an accelerometer (i.e., choose m, c and k) configured as in Figure 2.24 with very
small mass that will be accurate to 1% over the frequency range 0 to 50 Hz.
Solution:
Given: error < 1% , 0.2f = 50 Hz
f = 250 Hz
When r =0.2,
0.99 <
This becomes
0.9032 + 0.1568
Therefore,
0.6057 <
Choose
m = 0.01 kg , then
Thus
ω=2
<1.01
= 1570.8 rad/s
(± 1% error)
<1<0.9401 + 0.1632
<0.7854
= 24,674 N/m
implies that: c = 21.99 Ns/m
Copyright © 2015 Pearson Education Ltd.
2-44
Problems and Solutions Section 2.7 (2.73 through 2.89)
2.73
Consider a spring-mass sliding along a surface providing Coulomb friction, with stiffness
1.25 × 104 N/m and mass 10 kg, driven harmonically by a force of 50 N at 10 Hz.
Calculate the approximate amplitude of steady-state motion assuming that both the mass
and the surface that it slides on, are made of lubricated steel.
Solution: Given m = 10 kg,
rad
rad
N
= 125.66
k = 1.25 × 10 4 , F = 50 N, ω = 10 Hz = 20 × 2π
s
s
m
 1.25 ×10 4 
rad

 = 35.355
10
s


For lubricated steel
20π
= 0.1777m
r=
1.25 × 10 4 × 10
From equation 2.109
k
ωn =   =
m
2
X=
2.74
 4 μmg 

1 − 
πF
50
0 

=
2
1− r
1.25 × 10 4
F0
k
(
)
 4 × 0.07 × 10 × 9.81 
1− 

π × 50


= 0.0040688m
2
1 − 0.1777
2
(
)
A spring-mass system with Coulomb damping of 10 kg, stiffness of 2400 N/m, and
coefficient of friction of 0.15 is driven harmonically at 10 Hz. The amplitude at steady
state is 4 cm. Calculate the magnitude of the driving force.
.
Solution:
Given m = 10 kg, k = 2400 N/m and
rad
μ = 0.15, ω = 10 Hz = 20π
s
ωn =
r=
F0 =
2.75
rad
 2400 

 = 15.492
s
 10 
(k / m ) =
20π
= 4.0557
240
kX 1 − r 2
|
|
 4 μmg 

1 − 
 πF0 
2
(2400 × 0.04 × ((4.05577) − 1)) = 1483.248N
2
=
 4 × 0.15 ×10 × 9.81 

1 − 
F0 × π


2
A system of mass 10 kg and stiffness 1.8 x 104 N/m is subject to Coulomb damping. If
the mass is driven harmonically by a 90-N force at 25 Hz, determine the equivalent
viscous damping coefficient if the coefficient of friction is 0.12.
Copyright © 2015 Pearson Education Ltd.
2-45
Solution:
N
rad
rad
, F = 90 N, ω = 25 Hz = 25 × 2π
= 157.08
m
s
s
Given m = 10 kg, k = 1.8 × 10 4

rad
 = 42.426
s

50π
= 0.18512m
For lubricated steel μ = 0.12 , r =
1.8 × 10 4 × 10
From equation 2.109
k
ωn =   =
m
 1.8 ×10 4

 10
2
2
 4 μmg 
 4 × 0.12 × 10 × 9.81 

1 − 
1− 

π × 50
F0
90
 πF0 


X=
=
= 0.0049393m
2
4
2
k
1− r
1.8 × 10
1 − 0.18512
From equation 2.105 the equivalent viscous damping coefficient becomes
4 μmg 4 × 0.12 × 10 × 9.81
Ns
c eq =
=
= 19.319
πωX
π 50π 0.0049393
m
(
2.76
)
(
)
a. Plot the free response of the system of Problem 2.75 to initial conditions of x(0) = 0
and (0) = |F /m| = 9 m/s using thesolution in Section 1.10.
b. Use the equivalent viscous damping coefficient calculated in Problem 2.75 and plot
the free response of the “equivalent” viscously damped system to the same initial
conditions.
Solution: See Problem 2.75
(a)
x(0) = 0 and
(0) =
= 9 m/s
=38.73 rad/s
From section 1.10:
for
for
Let
= (0.1)(10)(9.81) = 9.81 N
To start,
Therefore,
and
So, x(t) =
This will continue until
= 0, which occurs at time :
Copyright © 2015 Pearson Education Ltd.
2-46
x(t) =
(t) =
Therefore,
and
So,
Again, when
= 0 at time , the motion will reverse:
x(t) =
(t) =
Therefore,
and
So,
This continues until
(b)
From Problem 2.75,
= 0 and
= 9.81 N
= 206.7 kg/s
Copyright © 2015 Pearson Education Ltd.
2-47
The equivalently damped system would be:
Also,
= 38.73 rad/s
0.2668
= 37.33 rad/s
The solution would be found from Equation 1.36:
=9
Therefore,
= 0.2411m and
= 0 rad
So,
Copyright © 2015 Pearson Education Ltd.
2-48
2.77
Referring to the system of Example 2.7.1; aspring–mass system with sliding friction
described by equation (2.97) with stiffness k = 1.5 ×104 N/m, driving harmonically a 10kg mass by a force of 90 N at 25 Hz, calculate how large the magnitude of the driving
force must be to sustain motion if the steel is lubricated. How large must this magnitude
be if the lubrication is removed?
Solution:
From Example 2.7.1 m = 10 kg, k = 1.5 ×
N/m,
rad/s
Lubricated Steel
Unlubricated Steel
= 90 N,
Lubricated:
= 8.74 N
Unlubricated:
= 37.5 N
2.78
Calculate the phase shift between the driving force and the response for the system of
Problem 2.77 using the equivalent viscous damping approximation.
Solution:
N/m,
From Problem 2.67: m = 10 kg, k = 1.5 ×
rad/s
= 90 N,
rad/s
From Equation (2.111), and since r>1
Since in Problem 2.67,
, this reduces to
rad = -90˚
Copyright © 2015 Pearson Education Ltd.
2-49
2.79
Derive the equation of vibration for the system of Figure P2.79 assuming that a viscous
dashpot of damping constant c is connected in parallel to the spring. Calculate the energy
loss and determine the magnitude and phase relationships for the forced response of the
equivalent viscous system.
Figure P2.79
Solution: Sum of the forces in Figure P2.79
sgn
sgn +kx= 0
Assume the mass is moving to the left
The solution of the form:
Substituting:
So,
Initial conditions
Copyright © 2015 Pearson Education Ltd.
2-50
(1)
This will occur until
So Equation (1) is valid from
For motion to the right
Initial conditions (From Equation (1)):
Solution:
Copyright © 2015 Pearson Education Ltd.
2-51
Forced Case:
sgn
Approximate Steady-state Response:
Energy Dissipated per Cycle:
This results in an equivalent viscously damped system:
where
The magnitude is:
Solving for X:
The phase is:
Copyright © 2015 Pearson Education Ltd.
2-52
2.80
A system of unknown damping mechanism is driven harmonically at 10 Hz with an
adjustable magnitude. The magnitude is changed, and the energy lost per cycle and
amplitudes are measured for five different magnitudes. The measured quantities are:
E(J)
0.25
0.45
X (M)
0.01
0.02
Is the damping viscous or Coulomb?
Solution:
0.8
0.04
1.16
0.08
3.0
0.15
For viscous damping,
For Coulomb damping,
For the data given, a plot of
vs
yields a curve, while
line. Therefore, the damping is likely Coulomb in nature
Copyright © 2015 Pearson Education Ltd.
vs X yields a straight
2-53
2.81
Calculate the equivalent loss factor for a system with Coulomb damping.
Solution:
Loss Factor:
For Coulomb damping:
Substituting for X (from Equation 2.109):
2.82
A spring-mass system (m = 10 kg, k = 4 × 103 N/m) vibrates horizontally on a surface
with coefficient of friction µ = 0.18. When excited harmonically at 5 Hz, the steady-state
displacement of the mass is 5 cm. Calculate the amplitude of the harmonic force applied.
rad
rad
Solution: Given m = 10 kg, k = 4 x 103 N/m μ = 0.18, ω = 5 Hz = 10π
= 31.414
s
s
 4 ×103 
rad

 = 20
s
 10 
From equation 2.109
F0
k
X=
ωn =
(k / m ) =
(1 − r )
2 2
r=
20π
 4 μmg 
+

 πkX 
2
= 4.0557
240
F0 =
|
kX 1 − r 2
|
 4 μmg 

1 − 
πF
0 

2
=
2



 4000 × 0.05 ×   10π  − 1 
  20 





 4 × 0.18 ×10 × 9.81 

1 − 
×
F
π
0


2
⇒ F0 = 294.34 N
Copyright © 2015 Pearson Education Ltd.
2-54
2.83
Calculate the displacement for a system with air damping using the equivalent viscous
damping method.
Solution:
The equivalent viscous damping for air is given by Equation (2.131):
From Equation 2.31:
Solving for X and taking the real solution:
Copyright © 2015 Pearson Education Ltd.
2-55
2.84
Calculate the semimajor and semiminor axis of the ellipse of equation (2.119). Then
calculate the area of the ellipse. Use c = 10 kg/s, ω = 2 rad/s and X = 0.01 m.
Solution: The equation of an ellipse usually appears when the plot of the ellipse is
oriented along with the x axis along the principle axis of the ellipse. Equation (2.1109) is
the equation of an ellipse rotated about the origin. If k is known, the angle of rotation can
be computed from formulas given in analytical geometry. However, we know from the
energy calculation that the stiffness does not effect the amount of energy dissipated. Thus
only the orientation of the ellipse is effected by the stiffness, not its area or axis. Thus we
can use this fact to answer the question. First re-write equation (2.119) with k = 0 to get:
This is the equation of an ellipse with major axis a and minor axis b given by
The area, and hence energy lost per cycle through the damper then becomes
= (3.14159)(10)(2)(.0001) = 0.006283 Joules.
Alternately, realized that Equation 2.119 is that of ellipse rotated by an angle defined
). Then match the ellipse to standard form, read off the
by tan2 = -2k/(
major and minor axis (say a andb) and calculate the area from
. See the following
web site for an elipse http://mathworld.wolfram.com/Ellipse.html
2.85
The area of a force deflection curve of Figure 2.29 is measured to be 2.5 N- m, and the
maximum deflection is measured to be 8 mm. From the “slope” of the ellipse the
stiffness is estimated to be 5 × 10 N/m. Calculate the hysteretic damping coefficient.
What is the equivalent viscous damping if the system is driven at 10 Hz?
Solution:
Given: Area = 2.5
, k = 5x
N/m, X = 8 mm,
Hysteric Damping Coefficient:
= Area =
2.5 =
Equivalent Viscous Damping:
c eq = 198 kg/s
Copyright © 2015 Pearson Education Ltd.
rad/s
2-56
2.86
The area of the hysteresis loop of a hysterically damped system is measured to be 5 N•m
and the maximum deflection is measured to be 1 cm. Calculate the equivalent viscous
damping coefficient for a 24-Hz driving force. Plot c versus ω for 2
ω 100
rad/s.
Solution:
Given Area = 5 N.m, X = 1cm ω = 2π 24 = 150.8
rad
s
Area of the hysteresis loop: A = πcωX 2
Hysteric Damping Coefficient
DE = Area = πkbX 2
N
m
Equivalent viscous damping is
kb 15915
c eq = =
= 105.54
ω (48π )
5 = πkb0.012 ⇒ kb = 15915
To plot rearrange so that
50000
ΔE
c eq =
=
2
πω
πωX
Plot:
Copyright © 2015 Pearson Education Ltd.
2-57
2.87
Calculate the nonconservative energy of a system subject to both viscous and hysteretic
damping.
Solution:
2.88
Derive a formula for equivalent viscous damping for the damping force of the form, F =
c( ) where n is an integer.
Solution:
Given:
Assume the steady-state response
The energy lost per cycle is given by Equation (2.99) as:
Substituting for
Let
:
Equating this to Equation 2.91 yields:
Copyright © 2015 Pearson Education Ltd.
2-58
2.89
Using the equivalent viscous damping formulation, determine an expression for the
steady-state amplitude under harmonic excitation for a system with both Coulomb and
viscous damping present.
Solution:
Equate to Equivalent Viscously Damped System
Amplitude:
Solving for X:
Copyright © 2015 Pearson Education Ltd.
2- 60
Problems and Solutions Section 2.8 (2.90 through 2.96)
2.90*.Numerically integrate and plot the response of an underdamped system determined
by m = 100 kg, k = 20,000 N/m, and c = 200 kg/s, subject to the initial conditions of x0 =
0.01 m and v0 = 0.1 m/s, and the applied force F(t) = 160cos5t. Then plot the exact
response as computed by equation (2.33). Compare the plot of the exact solution to the
numerical simulation.
Solution: Given: 100 xt  + 200 x t  + 20000 x t  = 160cos5t  with initial conditions
x0 = 0.01m; v0 = 0.1m / s
Total Analytical Solution in Octave:
First the m file containing the state equation to integrate is set up and saved as
Func_P290.m
functionxdot=f(t, x)
xdot=[-(200/100)*x(1)-(20000/100)*x(2)+(160/100)*cos(5*t); x(1)];
% xdot=[x(1)'; x(2)']=[-2*zeta*wn*x(1)-wn^2*x(2)+fo*cos(w*t) ; x(1)]
% which is a state space form of
% x" + 2*zeta*wn*x' + (wn^2)*x = fo*cos(w*t) (fo=Fo/m)
clear all;
Then the following m file is created and run:
%---- numerical simulation --x0=[0.1; 0.01];
%[xdot(0); x(0)]
tspan=[0 10];
[t,x]=ode45(@Func_P290,tspan,x0);
plot(t, x(:,2), '.');
hold on;
%--- exact solution ---t=0: .002: 10;
m=100; k=20000; c=200; Fo=160 ; w=5
wn=sqrt(k/m); zeta=c/(2*wn*m); fo=Fo/m; wd=wn*sqrt(1-zeta^2)
x0=0.01; v0= 0.1;
xe= exp(-zeta*wn*t) .* ( (x0-fo*(wn^2-w^2)/((wn^2-w^2)^2 ...
+(2*zeta*wn*w)^2))*cos(wd*t) ...
+ (zeta*wn/wd*( x0-fo*(wn^2-w^2)/((wn^2-w^2)^2+(2*zeta*wn*w)^2)) ...
- 2*zeta*wn*w^2*fo/(wd*((wd^2-w^2)^2 ...
+ (2*zeta*wn*w)^2))+v0/wd)*sin(wd*t) ) ...
+ fo/((wn^2-w^2)^2+(2*zeta*wn*w)^2)*((wn^2-w^2)*cos(w*t) ...
+ 2*zeta*wn*w*sin(w*t))
plot(t, xe, 'r');
hold off;
This produces the following plot:
Copyright © 2015 Pearson Education Ltd.
2- 61
Copyright © 2015 Pearson Education Ltd.
2- 62
2.91*.Numerically integrate and plot the response of an underdamped system determined
by m = 120 kg, and k = 3200 N/m subject to the initial conditions of x0 = 0.01 m and v0 =
0.1 m/s, and the applied force F(t) = 15cos10t , for various values of the damping
coefficient. Use this “program” to determine a value of damping that causes the transient
term to die out within 3 seconds. Try to find the smallest such value of damping
remembering that added damping is usually expensive.
Solution:
The solution is given by the following Octave code for C = 350 kg/s and corresponding
ζ = 0.28241; ωn = 5.1640 gives the desired result
clear all
clc
xo =[0.01;0.1];
ts=[0,10];
[t,x] = ode45(@Func1_p291,ts,xo);
figure(1)
plot(t,x(:,1),'r') ;
functionxd = func1_p291(t,x)
m= 120;
k=3200;
c=350;
omega_n=sqrt(k/m);
zeta = c/(2*sqrt(k*m));
Fo = 15;
omega = 10;
xd=[x(2);-x(1)*omega_n**2 - 2*x(2)*zeta*omega_n + Fo/m*sin(omega*t)];
end
Copyright © 2015 Pearson Education Ltd.
2- 63
2.92*.Compute the total response of a spring-mass system with the following values: k =
1000 N/m, m = 12 kg, subject to a harmonic force of magnitude F0= 100 N and frequency
of 8.162 rad/s, and initial conditions given by x0 = 0.01 m and v0 = 0.01 m/s, by
numerically integrating rather than using analytical expressions as was done in Problem
2.8. Plot the response.
Solution: The following Octave code illustrates the solution with given initial conditions
Function to be used in the code
function v = func1_p292(t,x)
m = 12;
k=1000;alpha = 0.005;
Fo= 100;
omega_n = sqrt(k/m);
omega = 8.162;
ceq = sqrt(8*m*alpha*Fo/m/3/pi);
v=[x(2);x(1).*-k/m + x(2).*-ceq/m + Fo/m*sin(omega*t)];
end
clear all
clc
xo =[0.01;0.01];
ts=[0,20];
[t,x] = ode45(@func1_p292,ts,xo);
figure(1)
plot(t,x(:,1),'r') ;
Copyright © 2015 Pearson Education Ltd.
2- 64
2.93*.A
A foot pedal for a musical instrument is model
modeled
ed by the sketch in Figure P2.93.
P2.93
With k = 2000 N/m, c = 25 kg/s, m = 25 kg and , F(t) = 50 cos 2p t N, numerically
simulate the response of the system assuming the system starts from rest. Also use the
small angle approximation.
F(t)
0.05 m
0.05 m
0.05 m
m
k
c
Figure 2.9
Solution: From problem 2.30, the equation of motion is
9 a 2 m q + 4 a 2 c cosqq + a 2 k sin q = -3a F(t)
wherek = 2000 kg, c = 25 kg/
kg/s , m = 25 kg , F(t) = 50 cos 2p t , a = 0.05 m
Placing the equation of motion in first order form and numerically integrating using
Mathcad yields
Copyright © 2015 Pearson Education Ltd.
2- 65
2.94*.Numerically
Numerically integrate and plot the response of an underdamped system determined
by m = 100 kg, k = 2000 N/m, and c = 200 kg/s, subject to the applied force F(t) = 150
cos 10t, for the following sets of initial conditions:
a) x0 = 0.0 m and v0 = 0.1 m/s
b) x0 = 0.01 m and v0 = 0.0 m/s
c) x0 = 0.05 m and v0 = 0.0 m/s
d) x0 = 0.0 m and v0 = 0.5 m/s
Plot these responses on the same graph and note the effects of the initial conditions on the
transient part of the response.
Solution: The following are the solutions in Mathcad. Of course the other codes and
Toolbox will yield the same results.
a)
b)
Copyright © 2015 Pearson Education Ltd.
2- 66
c)
d)
Copyright © 2015 Pearson Education Ltd.
2- 67
Note the profound effect on the transient, but of course no effect on the steady state.
Copyright © 2015 Pearson Education Ltd.
2- 68
2.95*. A DVD drive is mounted on a chassis and is modeled as a single degree-degreedegree
of-freedom
freedom spring, mass and damper. During normal operation, the drive (having a mass
of 0.4 kg) is subject to a harmonic force of 1 N at 10 rad/s. Because of material
considerations
rations and static deflection, the stiffness is fixed at 500 N/m and the natural
damping in the system is 10 kg/s. The DVD player starts and stops during its normal
operation providing initial conditions to the module of x0 = 0.001 m and v0 = 0.5 m/s.
Thee DVD drive must not have an amplitude of vibration larger then 0.008 m even during
the transient stage. First compute the response by numerical simulation to see if the
constraint is satisfied. If the constraint is not satisfied, find the smallest value of damping
that will keep the deflection less then 0.008 m.
Solution: The solution is given by the following Mathcad session:
This yields c =17 kg/s as a solution.
Copyright © 2015 Pearson Education Ltd.
2- 69
2.96
Use a plotting routine to examine the base motion problem (see Figure 2.13) by
plotting the particular solution (for an undamped system) for the three cases k =
1500 N/m, and k = 700 N/m. Also note the values of the three frequency ratios
and the corresponding amplitude of vibration of each case compared to the input.
Use the following values: b = 4.4 rad/s, m = 100 kg, and Y = 0.05 m.
Solution: The following Mathcad worksheet shows the plotting:
Note that k2, the softest system (smallest k) has the smallest amplitude, smaller
than the amplitude of the input as predicted by the magnitude plots in section 2.3.
Thus when r > 2 , the amplitude is the smallest.
Copyright © 2015 Pearson Education Ltd.
2- 69
Problems and Solutions Section 2.9 (2.97 through 2.103)
2.97*.Compute the response of the system in Figure P2.97 for the case that the damping
is linear viscous and the spring is a nonlinear soft spring of the form
k(x) = kx - k1 x 3
and the system is subject to a harmonic excitation of 300 N at a frequency of
approximately one third the natural frequency (= n/3) and initial conditions of x0 =
0.01 m and v0 = 0.1 m/s. The system has a mass of 100 kg, a damping coefficient of 170
kg/s and a linear stiffness coefficient of 2000 N/m. The value of k1 is taken to be 10000
N/m3. Compute the solution and compare it to the linear solution (k1 = 0). Which system
has the largest magnitude?
Figure P2.97
Solution: The following is a Mathcad simulation. The green is the steady state magnitude
of the linear system, which bounds the linear solution, but is exceeded by the nonlinear
solution. The nonlinear solution has the largest response.
Copyright © 2015 Pearson Education Ltd.
2- 70
Copyright © 2015 Pearson Education Ltd.
2- 71
2.98*.Compute the response of the system in Figure P2.97 for the case that the damping
is linear viscous and the spring is a nonlinear hard spring of the form
k(x) = kx + k1 x 3
and the system is subject to a harmonic excitation of 300 N at a frequency equal to the
natural frequency ( = n) and initial conditions of x0 = 0.01 m and v0 = 0.1 m/s. The
system has a mass of 100 kg, a damping coefficient of 170 kg/s and a linear stiffness
coefficient of 2000 N/m. The value of k1 is taken to be 10000 N/m3. Compute the
solution and compare it to the linear solution (k1 = 0). Which system has the largest
magnitude?
Solution: The Mathcad solution appears below. Note that in this case the linear
amplitude is the largest!
Copyright © 2015 Pearson Education Ltd.
2- 72
Copyright © 2015 Pearson Education Ltd.
2- 73
2.99*.Compute
Compute the response of the system in Figure P2.97 for the case that the damping
is linear viscous and the spring is a nonlinear soft spring of the form
k(x) = kx - k1 x 3
and the system is subject to a harmonic excitation of 300 N at a frequency equal to the
natural frequency ( = n) and initial conditions of x0 = 0.01 m and v0 = 0.1 m/s. The
system has a mass of 100 kg, a damping coefficient of 15 kg/s and a linear stiffness
coefficient of 2000 N/m. The value of k1 is taken to be 100 N/m3. Compute the solution
3
and compare it to the hard spring solution ( k(x) = kx + k1 x ).
Solution: The Mathcad solution is presented, first for a hard spring, then for a soft spring
Copyright © 2015 Pearson Education Ltd.
2- 74
Next consider the result for the soft spring and note that the nonlinear response is higher
in the transient then the linear case (opposite of the hardening spring), but nearly the
same in steady state as the hardening spring.
Copyright © 2015 Pearson Education Ltd.
2- 75
2.100*.Compute
Compute the response of the system in Figure P2.97 for the case that the damping
is linear viscous
scous and the spring is a nonlinear soft spring of the form
k(x) = kx - k1 x 3
and the system is subject to a harmonic excitation of 300 N at a frequency equal to the
natural frequency ( = n) and initial conditions of x0 = 0.01 m and v0 = 0.1 m/s. The
system has a mass of 100 kg, a damping coefficient of 15 kg/s and a linear stiffness
coefficient of 2000 N/m. The value of k1 is taken to be 1000 N/m3. Compute the solution
2
and compare it to the quadratic soft spring ( k(x) = kx + k1 x ).
Solution: The response to both the hardening and softening spring are given in the
following Mathcad sessions. In each case the linear response is also shown for
comparison. With the soft spring, the response is more variable, whereas the hardening
hardenin
spring seems to reach steady state.
Copyright © 2015 Pearson Education Ltd.
2- 76
2.101*. Compare the forced response of a system with velocity squared damping with
equation of motion given by:
()
mx + a sgn x x 2 + kx = F0 cos wt
using numerical simulation of the nonlinear equation to that of the response of the linear
system
m obtained using equivalent viscous damping as defined by equation (2.131):
8
ceq =
awX
3p
Use as initial conditions, x0 = 0.01 m and v0 = 0.1 m/s with a mass of 10 kg, stiffness of
25 N/m, applied force of 150 cos ((nt) and drag coefficient of  = 250.
Solution:
Copyright © 2015 Pearson Education Ltd.
2- 77
2.102*. Compare the forced response of a system with structural damping (see table 2.2)
using numerical simulation of the nonlinear equation to that of the response of the linear
system obtained using equivalent viscous damping as defined in Table 2.2. Use
Us as initial
conditions, x0 = 0.01 m and v0 = 0.1 m/s with a mass of 10 kg, stiffness of 25 N/m,
applied force of 150 cos (nt)) and solid damping coefficient of b = 25.
Solution: The solution is presented here in Mathcad
Copyright © 2015 Pearson Education Ltd.
3- 1
Chapter Three Solutions
Solutions for Section 3.1 (3.1 through 3.17)
3.1
Calculate the solution to
 + 200 x(t)
 + 2000 x(t) = 100d (t), x 0 = 0, v0 = 0
1000 x(t)
Solution
Thus the natural frequency, damping ratio and damped natural frequency are
wn =
2000
200
= 1.414 rad/s, z =
= 0.071,
1000
2 1000 × 2000
w d = 1.414 1- 0.0712 = 1.411 rad/s
Using equation (3.6), the response becomes
-zw t
F̂e n
x (t ) =
sin w d t = 0.071e-0.1t sin(1.411t)
mw d
3.2
Consider a spring-mass-damper system with m = 2 kg, c = 2 kg/s and k = 3000 N/m with
an impulsive force applied to it of 10,000 N for 0.01 s. Compute the resulting response.
Solution: A 10,000 N force acting over 0.01s provides a value
Fˆ = Ft  10000  0.01 = 100 N  .s
Using the values given the equation of motion is
2 xt  + 2 x t  + 3000 x t  = 100δ t 
Thus the natural frequency
0.5
k
ωn =   =
m
c
ζ=
=
2 km 


rad
 3000 

 = 38.73
s
 2 
2
= 0.013
2 3000  2 


ωd = ωn 1  ζ 2 = 38.33 1  0.0132 = 38.3267
So,


xt  = e  ξωnt  ACos ωd t + B sin ωd t 


 xt  =  1.291e 0.5t sin38.3267t 
Copyright © 2015 Pearson Education Ltd.
rad
s
3- 2
3.3
Calculate the solution to

x  2 x  4 x   (t   )
x(0)  0
x (0)  1
and plot the response.
.
Solution: Given:

x  2 x  4 x   (t   )
x(0)  0
x (0)  1
We have to find xt  and plot the response.
rad
k
c = 2 ,k = 4, ωn =   = 4 / 1 = 2
s
m
c
ζ=
= 2 / 2 4 = 0.5
2 km 






ωd = ωn 1  ζ 2 = 2 1  0.52 = 1.7321
xt  = xh t  + x p t 
rad
s
Homogenous solution from eq 1.36
xh t  = Ae
 ζω n
t
sin ωd t +

 v + ζωn x0 2 +  x0ωd 2 


x0ωd
 ; = tan 1 
A =  0
2
 v + ζω x  

ω
d
n 0 
 0


xh t  = 0.57735e  t sin 1.732t + 1.0472
Particular From equation 3.9
x p t  =
1 ζωn t  π 
1
e
sin ωd t  π  =
e t  τ sin 1.7321t  τ 
mωd
1.7321
But, sin(-t) = -sin(t).
So x p t  = 0.57733e  t π sin1.7321t  π 
The total solution becomes
x (t )= x h (t )+ x p (t )= 0.57735 e− t sin (1.732t +1.0472 )+ 0.57733 e− (t − π ) sin (1.7321 (t − π ))
Copyright © 2015 Pearson Education Ltd.
3- 3
And the plot is as shown below:
3.4
Calculate the solution to

x + 2 x + 3x = sint + d ( t - p )
 =1
x(0) = 0 x(0)
and plot the response.
(
)
()
()
Solution: Given: 
x + 2 x + 3x = sint + d t - p , x 0 = 0, x 0 = 0
k
c
= 1.732 rad/s, z =
= 0.5774, w d = w n 1 - z 2 = 1.414 rad/s
m
2 km
Total Solution:
x t = xh + x p1
0<t <p
wn =
()
()
x t = xh + x p1 + x p2
t >p
Homogeneous: Eq. (1.36)
-zw t
xh t = Ae n sin w d t + f = Ae-t sin 1.414t + f
()
(
)
(
Particular: #1 (Chapter 2)
Copyright © 2015 Pearson Education Ltd.
)
3- 4
(
)
x p1 (t) = X sin w t - q , where w = 1 rad/s . Note that f0 =
ÞX=
(w
f0
2
n
- w2
) + ( 2zw w )
2
n
2
F0
=1
m
é 2zw w ù
= 0.3536, and q = tan -1 ê 2 n 2 ú = 0.785 rad
êë w n - w úû
()
(
Þ x p1 t = 0.3536 sin t - 0.7854
)
Particular: #2 Equation 3.9
1 -zw n ( t - p )
1
- t -p
x p2 t =
e
sin w d t - t =
e ( ) sin1.414 t - p
mw d
1 1.414
- t -p
Þ x t = 0.7071e ( ) sin1.414 t - p
()
(
p2
)
( )(
()
)
(
(
)
)
The total solution for 0<t< becomes:
x t = Ae-t sin 1.414t + f + 0.3536 sin t - 0.7854
()
(
)
(
)
x ( t ) = - Ae sin(1.414t + f ) + 1.414 Ae cos (1.414t + f ) + 0.3536 cos ( t - 0.7854 )
0.25
x ( 0 ) = 0 = Asin f - 0.25 Þ A =
sin f
-t
-t
()
(
) tan1 f
)
0<t <p
x 0 = 1 = - Asin f + 1.414 Acos f + 0.25 Þ 0.75 = 0.25 - 1.414 0.25
Þ f = 0.34 and A = 0.75
Thus for the first time interval, the response is
()
(
)
(
x t = 0.75e-t sin 1.414t + 0.34 + 0.3536sin t - 0.7854
Next consider the application of the impulse at t = :
x t = xh + x p1 + x p2
()
x ( t ) = -0.433e
-t
(
)
(
)
(
- t -p
sin 1.414t + 0.6155 + 0.3536 sin t - 0.7854 - 0.7071e ( ) sin 1.414t - p
The response is plotted in the following (from Mathcad):
Copyright © 2015 Pearson Education Ltd.
)
t >p
3- 5
3.5
Calculate the response ofa critically damped system to a unit impulse.
Solution:
F̂
, while x0 = 0. So for a critically
m
damped system, we have from Eqs. 1.45 and 1.46 with x0 = 0:
The change in the velocity from an impulse is v0 =
x(t) = v0te
Þ x(t) =
3.6
-w n t
F̂ -w n t
te
m
Calculate the response of an overdamped system to a unit impulse..
Solution:
F̂
, while x0 = 0. So, for an overdamped
m
system, we have from Eqs. 1.41, 1.42 and 1.43:
The change in velocity for an impulse v0 =
é
-v0
v0
- w ( z 2 -1)t
-w (
ê
x t =e
e n
+
e n
2
2
êë 2w z - 1
2w n z - 1
n
F̂
-zw t
- w ( z 2 -1)t
- w ( z 2 -1)t ù
x t =
e n ée n
-e n
ëê
ûú
2mw n z 2 - 1
()
-zw n t
()
Copyright © 2015 Pearson Education Ltd.
z 2 -1)t
ù
ú
úû
3- 6
3.7
Derive equation (3.6) from equations (1.36) and (1.38).
Solution:
Equation 1.36: x(t) = Ae
Equation 1.38: A =
Since x0 = 0 and v0 =
(v
- zw n t
(
sin w d t + f
+ zw n x0
0
)
) + (x w )
2
0
w
2
d
2
d
é x0w d ù
, f = tan -1 ê
ú
ë v0 + zw n x0 û
F̂
, Equation 1.38 becomes
m
v
F̂
A= 0 =
w d mw d
f = tan -1 ( 0 ) = 0
So Equation 1.36 becomes
F̂ -zw n t
x t =
e
sin w d t which is Equation 3.6
mw d
()
3.8
( )
Consider a simple model of an airplane wing given in Figure P3.8. The wing is
approximated as vibrating back and forth in its plane, massless compared to the missile
carriage system (of mass m). The modulus and the moment of inertia of the wing are
approximated by E and I, respectively, and l is the length of the wing. The wing is
modeled as a simple cantilever for the purpose of estimating the vibration resulting from
the release of the missile, which is approximated by the impulse function F(t).
Calculate the response and plot your results for the case of an aluminum wing 3 m long
with m = 2000 kg,  = 0.02, and I = 1.0 m4. Model F as 2000 N lasting for 0.1s.
Solution: Given:
m= 2000 kg I = 1 m 4 L= 3 m F = 2000 N ∆ t = 0.1
x 0= 0
3EI 3  7.1  1010  0.5
F
= 7.7778  1010 ζ = 0.02
v0 =  t E = 70  109 k = 3 =
3
L
3
m
0.5
k
rad
rad A= x
2 0.5
ω n=
= 6236.1
0
ωn = 6234.8
m
s ωd = 1  ζ
s
()
B=



 
ζ  ωn  x0 + v0
 ζω t
= 1.6093  10 5 xt  = e n   ACos ωd  t + B sin ωd  t 
ωd
Solution: Function: xt  = 0.00005072
Plot:
39.4 t
sin1971.63t 
octave:115> m = 2000; I = 1;l=3;F =2000,dt=0.1;x0=0.0;v0=(F/m)*dt;
Copyright © 2015 Pearson Education Ltd.
3- 7
octave:116> E=70e10;k=(3*E*I)/l**3;
octave:117> omega_n =sqrt(k/m);zeta = 0.02; omega_d=omega_n*sqrt(1-zeta**2);
octave:118> A=x0;B = (zeta*omega_n*x0+v0)/omega_d;
octave:119> t=linspace(0,10,1000);
octave:120> xt = exp(-zeta*omega_n*t).*(A*cos(omega_d*t) + B*sin(omega_d*t));
octave:130> t=linspace(0,0.04,1000);
octave:131> xt = exp(-zeta*omega_n*t).*(A*cos(omega_d*t) + B*sin(omega_d*t));
octave:132> plot(t,xt)
octave:133> xlabel ("t")
octave:134> ylabel ("x(t)")
3.9
A cam in a large machine can be modeled as applying a 15,000 N-force over an interval
of 0.01 s. This can strike a valve that is modeled as having physical parameters: m = 10
kg, c = 20 N s/m, and stiffness k = 8000 N/m. The cam strikes the valve once every 1 s.
Calculate the vibration response, x(t), of the valve once it has been impacted by the cam.
The valve is considered to be closed if the distance between its rest position and its actual
position is less than 0.0001 m. Is the valve closed the very next time it is hit by the cam?
Copyright © 2015 Pearson Education Ltd.
3- 8
Solution: Given:
Copyright © 2015 Pearson Education Ltd.
3- 9
3.10
The vibration packages dropped from a height of h meters can be approximated by
considering Figure P3.10 and modeling the point of contact as an impulse applied to the
system at the time of contact. Calculate the vibration of the mass m after the system falls
and hits the ground. Assume that the system is underdamped.
FigureP3.10 Vibration model of a package being dropped onto the ground.
Solution: When the system hits the ground, it responds as if an impulse force acted on it.
-zw t
F̂e n
From Equation (3.6): x t =
sin w d t
mw d
()
where
F̂
= v0
m
Calculate v0:
x=
For falling mass:
1 2
at
2
So, v0 = gt * , where t* is the time of impact from height h
h=
1 *2
gt Þ t * =
2
2h
g
v0 = 2gh
Let t = 0 when the end of the spring hits the ground
The response is
()
x t =
2gh -zw n t
e
sin w d t
wd
Where n, d, and  are calculated from m, c, k. Of course the problem could be solved
as a free response problem with x0 = 0, v0 =
model as the unit velocity given.
2gh or an impulse response with impact
Copyright © 2015 Pearson Education Ltd.
3- 10
3.11
Calculate the response of
4 
x (t )  16 x (t )  16 x (t )  4 (t )
for zero initial conditions. The units are in Newtons. Plot the response.
Solution: Given:
4 
x (t )  16 x (t )  16 x (t )  4 (t )
x0= 0,
x 0 = 0
rad
ωn = 16 / 4 = 4 = 2
,
s
c
16
ζ=
=
=1
2 km  2 16  4
Ft
, x0 = 0
m
ω t
x = a1 + a2t e n
F = 4, v0 =
Ft
F
 xt  = te  2t = te  2t
m
m
 xt  = e 2t t
And the plot is as shown below:
a1 = 0, a2 =
Copyright © 2015 Pearson Education Ltd.
3- 11
3.12 Compute the response of
the system:
4 
x(t )  16 x (t )  16 x (t )  4 (t )
subject to the initial conditions x(0) = 0.01 m and v(0) = 0. The units are in Newtons.
Plot the response.
Solution
Copyright © 2015 Pearson Education Ltd.
3- 12
3.13
Calculate the response of the system
3
x(t )  6 x  t   12 x  t   3  t    (t  1)
subject to the initial conditions x(0) =0.02 m and v(0) = 2 m/s. The units are in Newtons.
Plot the response.
Solution:
Now compute the response to the initial conditions from Equation 1.36
Using the Heaviside function the total response is
 ζω t
 ζωn t  1

e n
e
 ζωnt











xt =
sin ωd t + Ae
sin ωd t + +
sin ωd t  1 Φ t  1
  3ωd

ωd


Octave Code
xo=0.02 ;
v0=2;
m = 3;
c = 6;
k = 12;
omega_n = sqrt(k/m);
zeta = c/(2*sqrt(k*m))
omega_d = omega_n*sqrt(1-zeta**2)
A = sqrt( ((v0 + zeta*omega_d*xo)**2 +(xo*omega_d)**2)/omega_d**2)
phi = atan((xo*omega_d)/(v0 + zeta*omega_d*xo))
t = linspace(0,10,10000);
Term2 = A*exp(-zeta*omega_n*t).*sin(omega_d*t + phi);
Term3 =[(exp(-zeta*omega_n*t)/(-3*omega_d)).*sin(omega_d*(t-1))];
x = (exp(-zeta*omega_n*t)/ omega_d)*sin(omega_d*t) + Term2 + Term3;
Copyright © 2015 Pearson Education Ltd.
3- 13
plot (t,x)
ylabel("x(t)")
xlabel("t")
3.14
A chassis dynamometer is used to study the unsprung mass of an automobile as
illustrated in Figure P3.14, and discussed in Example 1.4.1. Compute the maximum
magnitude of the center of the wheel due to an impulse of 10000 N applied over 0.01
seconds in the x direction. Assume the wheel mass is m = 15 kg, the spring stiffness is k
= 500,000 N/m, the shock absorber provides a damping ratio of = 0.3, and the rotational
inertia is J = 2.323 kg m2. Assume that the dynamometer is controlled such that x  r .
Compute and plot the response of the wheel system to an impulse of 5000 N over 0.01 s.
Compare the undamped maximum amplitude to that of the maximum amplitude of the
damped system (use r = 0.457 m).
Solution: Given:
Copyright © 2015 Pearson Education Ltd.
3- 14
F= 10000 N
m= 15 kg
ξ= 0.3
k = 500000 N / m
J = 2.323 Kg/ m
2
r= 0.457 m
ξ
0.5
2km 
J 

 m + 2  x + c x + k x = 0
r 

x 0  = 0
F
x0  =   t
m
Solution:
The solution to the above conditions is given below:
xt  = 0.029 e 41.50t sin131.97t 
Plot:
c=
Copyright © 2015 Pearson Education Ltd.
3- 15
3.15
Consider the effect of damping on the bird strike problem of Example 3.1.2. Recall from
the example that the bird strike causes the camera to vibrate out of limits. Adding
damping will cause the magnitude of the response to decrease but may not be able to
keep the camera from vibrating past the 0.01 m limit. If the damping in the aluminum is
modeled as = 0.02, approximately how long before the camera vibration reduces to the
required limit? (Hint: plot the time response and note the value for time after which the
oscillations remain below 0.01 m).
Solution: Given:
m camera= 3 kg
v0= 0
c= 2 ;
 0.020.023 

3  70  10 

12
3EI

 = 1.707  10 4 N / m
k= 3 =
3
l
l
9
ωn =
x0 =
k
= 75.432rad / s
m
mbird  vbird  =
mcamera  ωn
1  72  1000
= 0.088379m
3  75.432  3600
 mx t  + cx t  + kx t  = 0
x 0  = 0.04898 m
x 0  = 0
Therefore xt  =
Plot:
1.508t
0.0884cos75.417t + 0.001768sin75.417t 
Copyright © 2015 Pearson Education Ltd.
3- 16
The time after which the amplitude falls below 0.01m is after 1.4443 s.
3.16
Consider the jet engine and mount indicated in Figure P3.16 and model it as a mass on
the end of a beam as done in Figure 1.24. The mass of the engine is usually fixed. Find a
expression for the value of the transverse mount stiffness, k, as a function of the relative
speed of the bird, v, the bird mass, the mass of the engine and the maximum displacement
that the engine is allowed to vibrate.
Figure P3.16Model of a jet engine in transverse vibration due to a bird strike.
Solution: The equation of motion is
m
x(t) + kx(t) = F̂d (t)
From equations (3.7) and (3.8) the magnitude of the response is
F̂
X =
mw n
for an undamped system. If the bird is moving with momentum mbv then:
mv
mv
1 æ m vö
X = b Þ X = b Þk= ç b ÷
mw n
mè X ø
mk
2
This can be used to provide some guidance in designing the engine mount.
Copyright © 2015 Pearson Education Ltd.
3- 17
3.17
A machine part is regularly subject to a force of 350 N lasting 0.01 seconds, as part of a
manufacturing process. Design a damper, i.e. choose a value of the damping constant c,
such that the part does not deflect more that 0.01 m. The part has a mass of 100 kg and a
stiffness of 1250 N/m.
Solution The maximum amplitude occurs at t =0, which from equation (3.6) is
X = 0.01 =
F̂
=
mw d
F̂
m
k
æ c ö
1- ç
è 2 km ÷ø
m
Solving this for c yields:
F̂ 2
c = 2 km - 2
X
Here not that if the requirement on X makes the argument under the square root negative,
it means that the system is no longer underdamped and a different solution is in play. For
the values given c = 100 kg/s, which corresponds to a damping ratio of 0.141< 1 and the
underdamped equation does apply.
Copyright © 2015 Pearson Education Ltd.
3- 17
Solutions for Section 3.2 (3.18 through 3.29)
3.18
Calculate the analytical response of an overdamped single-degree-of-freedom
system to an arbitrary non-periodic excitation.
t
()
() (
)
Solution: From Equation (3.12): x t = ò F t h t - t dt
0
For an overdamped SDOF system (see Problem 3.6)
(
)
()
t
1
h t -t =
2mw n z - 1
2
()
x t = òF t
0
()
Þx t =
3.19
e
1
2mw n z - 1
2
e
-zw n
(
- zw n t - t
t
e
)æ
è
(
-zw n t - t
()
òF t e
2mw n z - 1 0
2
w n z 2 -1 ( t - t )
e
zw n t
)æ
è
-e
w n z 2 -1 ( t - t )
e
æ ew n
è
z 2 -1 ( t - t )
(
- w n z 2 -1 t - t
-e
-e
)ö
dt
(
)ö
ø
- w n z 2 -1 t - t
(
- w n z 2 -1 t - t
ø
)ö
ø
dt
dt
Calculate the response of an underdamped system to the excitation given in
Figure P3.19 where the pulse ends at  s.
f (t)
F0
t
0
2
Figure P3.19 Plot of a pulse input of the form f(t) = F0sint.
Solution:
t
()
()
(
)
1 -zw n t é
zw t
x t =
e
F t e n sin w d t - t ùû dt
ò
ë
mw d
0
()
()
t <p
F t = F0 sin t
For t £ p ,
()
x t =
( From Figure P3.16)
(
))
F0 -zw n t t
zw t
e
sin t e n sin w d t - t dt
ò
mw d
0
Copyright © 2015 Pearson Education Ltd.
(
3- 18
()
x t =
F0 -zw n t
e
´
mw d
é
1
zw t
ê
e n éë w d - 1 sint - zw n cost ùû - w d - 1 sin w d t - zw n cos w d t
2
ê 2 éë1+ 2w d + w n ùû
ë
ù
1
zw n t
é w d - 1 sin t - zw n cost ù + w d - 1 sin w d t - zw n cos w d t ú
+
e
ë
û
ú
2 éë1+ 2w d + w n 2 ùû
û
{
{
(
)
(
)
t
For t > p , : ò f (t )h(t - t )dt =
0
()
x t =
=
(
F0
mw d
F0
mw d
e
)
(
ò
-zw n t
p
}
}
)
t
f (t )h(t - t )dt + ò (0)h(t - t )dt
p
0
p
ò ( sin t e
zw nt
))
(
sin w d t - t dt
0
e
(
-zw n t
´
)
(
)
(
)
é
ìezw n t é w - 1 sin éw t - p ù - zw cos éw t - p ù ù ü
1
ï
n
ë d
û
ë d
ûûï
ë d
ê
í
ý
2
ê 2 é1 + 2w + w ù ï
w
1
sin
w
t
zw
cos
w
t
ïþ
d
n ûî
d
d
n
d
ë ë
ìezw n t é w + 1 sin éw t - t ù + zw cos éw t - p
1
ï
ë d
û
ë d
ë d
+
2 í
2 éë1 + 2w d + w n ùû ï
+ w d - 1 sin w d t - zw n cos w d t
î
(
(
)
)
(
(
)
)
(
)ùû ùû üïùú
ýú
ïþ û
Alternately, one could take a Laplace Transform approach and assume the under-damped
system is a mass-spring-damper system of the form
m
x ( t ) + cx ( t ) + kx ( t ) = F ( t )
The forcing function given can be written as
F ( t ) = F0 ( H ( t ) - H ( t - p )) sin ( t )
Normalizing the equation of motion yields
x ( t ) + 2zw n x ( t ) + w n2 x ( t ) = f0 ( H ( t ) - H ( t - p )) sin ( t )
where f0 =
F0
and m, c and k are such that 0 < z < 1.
m
Assuming initial conditions, transforming the equation of motion into the Laplace domain
yields
Copyright © 2015 Pearson Education Ltd.
3- 19
X ( s) =
(s
(
f0 1 + e - p s
2
)(
)
+ 1 s 2 + 2zw n s + w n2
)
The above expression can be converted to partial fractions
æ
ö
Cs + D
æ As + B ö
X ( s ) = f0 1 + e - p s ç 2
+ f0 1 + e - p s ç 2
÷
è s +1 ø
è s + 2zw n s + w n2 ÷ø
(
)
(
)
where A, B, C, and D are found to be
A=
C=
-2zw n
(1- w ) + ( 2zw )
2 2
n
n
2zw n
(1- w ) + ( 2zw
2 2
n
2 , B=
w n2 - 1
2
(1- w n2 ) + ( 2zw n )2
(1- w ) + ( 2zw )
, D=
)
(1- w ) + ( 2zw )
2
n
2
n
2 2
n
2
n
2
n
Notice that X ( s ) can be written more attractively as
æ As + B
ö
æ As + B
ö
Cs + D
Cs + D
X ( s ) = f0 ç 2
+ 2
+ f0 e - p s ç 2
+ 2
2÷
2÷
è s + 1 s + 2zw n s + w n ø
è s + 1 s + 2zw n s + w n ø
(
= f0 G ( s ) + e - p s G ( s )
)
Performing the inverse Laplace Transform yields
x ( t ) = f0 ( g ( t ) + H ( t - p ) g ( t - p ) )
where g(t) is given below
æ D - Czw n ö -zw n t
g ( t ) = A cos ( t ) + Bsin ( t ) + Ce-zw n t cos (w d t ) + ç
sin (w d t )
÷ø e
wd
è
w d is the damped natural frequency, w d = w n 1 - z 2 .
Let m=1 kg, c=2 kg/sec, k=3 N/m, and F0=2 N. The system is solved numerically. Both
exact and numerical solutions are plotted below
Copyright © 2015 Pearson Education Ltd.
3- 20
Figure 1 Analytical vs. Numerical Solutions
Below is the code used to solve this problem
% Establish a time vector
t=[0:0.001:10];
% Define the mass, spring stiffness and damping coefficient
m=1;
c=2;
k=3;
% Define the amplitude of the forcing function
F0=2;
% Calculate the natural frequency, damping ratio and normalized force amplitude
zeta=c/(2*sqrt(k*m));
wn=sqrt(k/m);
f0=F0/m;
% Calculate the damped natural frequency
wd=wn*sqrt(1-zeta^2);
% Below is the common denominator of A, B, C and D (partial fractions
% coefficients)
dummy=(1-wn^2)^2+(2*zeta*wn)^2;
% Hence, A, B, C, and D are given by
A=-2*zeta*wn/dummy;
B=(wn^2-1)/dummy;
C=2*zeta*wn/dummy;
Copyright © 2015 Pearson Education Ltd.
3- 21
D=((1-wn^2)+(2*zeta*wn)^2)/dummy;
% EXACT SOLUTION
%
************************************************************************
*
%
************************************************************************
*
for i=1:length(t)
% Start by defining the function g(t)
g(i)=A*cos(t(i))+B*sin(t(i))+C*exp(-zeta*wn*t(i))*cos(wd*t(i))+((DC*zeta*wn)/wd)*exp(-zeta*wn*t(i))*sin(wd*t(i));
% Before t=pi, the response will be only g(t)
if t(i)<pi
xe(i)=f0*g(i);
% d is the index of delay that will correspond to t=pi
d=i;
else
% After t=pi, the response is g(t) plus a delayed g(t). The amount
% of delay is pi seconds, and it is d increments
xe(i)=f0*(g(i)+g(i-d));
end;
end;
% NUMERICAL SOLUTION
%
************************************************************************
*
%
************************************************************************
*
% Start by defining the forcing function
for i=1:length(t)
if t(i)<pi
f(i)=f0*sin(t(i));
else
f(i)=0;
end;
end;
% Define the transfer functions of the system
% This is given below
%
1
% --------------------------Copyright © 2015 Pearson Education Ltd.
3- 22
% s^2+2*zeta*wn+wn^2
% Define the numerator and denominator
num=[1];
den=[1 2*zeta*wn wn^2];
% Establish the transfer function
sys=tf(num,den);
% Obtain the solution using lsim
xn=lsim(sys,f,t);
% Plot the results
figure;
set(gcf,'Color','White');
plot(t,xe,t,xn,'--');
xlabel('Time(sec)');
ylabel('Response');
legend('Forcing Function','Exact Solution','Numerical Solution');
text(6,0.05,'\uparrow','FontSize',18);
axes('Position',[0.55 0.3/0.8 0.25 0.25])
plot(t(6001:6030),xe(6001:6030),t(6001:6030),xn(6001:6030),'--');
3.20
Speed bumps are used to force drivers to slow down. Figure P3.20 is a model of a
car going over a speed bump. Using the data from Example 2.4.1 and an
undamped model of the suspension system (i.e., k = 4 x 105 N/m, m = 1007 kg),
find an expression for the maximum relative deflection of the car’s mass versus
the velocity of the car. Model the bump as a half sine of length 40 cm and height
20 cm. Note that this is a moving base problem.
Copyright © 2015 Pearson Education Ltd.
3- 23
Figure P3.20Model of a car driving over a speed bump.
Solution: This is a base motion problem, so the first step is to translate the
equation of motion into a useable form. Summing forces yields in the vertical
direction yields
m
x(t) + k x(t) - y(t) = 0
(
)
were the displacement y(t) is prescribed. Next defined the relative displacement
to be z(t) = x(t)-y(t), the relative motion between the car’s wheel and body. The
equation of motion becomes:
m
z(t) + my(t) + kz(t) = 0 Þ m
z(t) + kz(t) = -m
y(t)
Substitution of the form of y(t) into this last expression yields:
m
z(t) + kz(t) = mYw b2 sin w bt F(t) - F(t - t1 )
(
)
where  is the Heavyside step function and bis the frequency associated with
the bump. The relationship between the bump frequency and the car’s constant
velocity is
2p
p
wb =
v= v
2

where v is the speed of the car in m/s. For constant velocity, the time t1 = v ,
when the car finishes going over the bump.
Here, z(t) is From equation (3.13) with zero damping the solution is:
t
1
z(t) =
f (t - t )sin w nt dt
mw n ò0
Substitution of f(t) =y(t) yields:
Yw b2 t
z(t) =
sin(w bt - w bt ) sin w nt dt =
w n ò0
(
)
t < t1
(
sin w bt + (w n - w b )t
Yw b2 1 é sin w bt - (w n + w b )t
=
ê
w n 2 êë
-(w n + w b )
wn - wb
Yw b2
(
)
) ùú
t
úû0
1
w sin w bt - w b sin w nt
t < t1
w n w n2 - w b2 n
where the integral has been evaluated symbolically. Clearly a resonance situation
prevails. Consider two cases, high speed (w b >> w n ) and low speed ( (w b << w n ) )
as when the two frequencies are near each other and obvious maximum occurs.
For high speed, the amplitude can be approximated as
Yw b2 w b
Yw b2 w b
(
w
/
w
)sin
w
t
sin
w
t
»
sin w nt
n
b
b
n
w n w n2 - w b2
w n w n2 - w b2
For the values given, this has magnitude:
=
(
)
Copyright © 2015 Pearson Education Ltd.
3- 24
3
Z(v) »
æpö
Y ç ÷ v3
è ø
(
w n w n2 - w b2
)
This increases with the cube of the velocity. Thus the faster the car is going the
more sever the bump is (larger relative amplitude of vibration), hence serving to
slow motorist down. A plot of magnitude versus speed shows bump size is
amplified by the suspension system.
For slow speed, magnitude becomes
2
Z(v) »
æpö
Y ç ÷ v 2w n
è ø
(
w n w n2 - w b2
)
A plot of the approximate magnitude versus speed is given below
Copyright © 2015 Pearson Education Ltd.
3- 25
Clearly at speeds above the designed velocity there is strong amplification of the
bump’s magnitude, causing discomfort to the driver and passengers, encouraging
a slow speed when passing over the bump.
3.21
Calculate and plot the response of an undamped system to a step function with a
finite rise time of t1 for the case m = 1 kg, k = 1 N/m, t1 = 4 s and F0 = 20 N. This
function is described by
ì F0t
ï
F t = í t1
ïF
î 0
()
0 £ t £ t1
t > t1
Solution: Working in Mathcad to perform the integrals the solution is:
Copyright © 2015 Pearson Education Ltd.
3- 26
Copyright © 2015 Pearson Education Ltd.
3- 27
3.22
A wave consisting of the wake from a passing boat impacts a seawall. It is
desired to calculate the resulting vibration. Figure P3.22 illustrates the situation
and suggests a model. This force in Figure P3.22 can be expressed as
ì æ
tö
ï F0 ç 1 - ÷
F t = í è t0 ø
ï
0
î
0 £ t £ t0
()
t > t0
Calculate the response of the seal wall-dike system to such a load.
D ike
F(t)
Concrete
seawall
k
m
K EA
l
Water
F
m
F
Wake
Physical setting
0
Model
t
Input model
Figure P3.22 Model of a wave hitting a dike.
t
( ) ò F (t ) h (t - t ) dt
Solution: From Equation (3.12): x t =
0
(
)
(
)
1
sin w n t - t for an undamped system
mw n
From Problem 3.18, h t - t =
For t < t0 :
t
ù
æ
1 é
tö
ê ò F0 ç 1- ÷ sin w n t - t dt ú
mw n êë 0
è t0 ø
úû
t
ù
F0 é t
1
x t =
ê ò sin w n t - t dt - ò t sin w n t - t dt ú
mw n êë 0
t0 0
úû
After integrating and rearranging,
ù F
F é 1
x t = 0 ê sin w nt - t ú + 0 éë1- cos w nt ùû t < t0
kt0 ë w n
û k
()
(
x t =
()
(
)
)
(
)
()
t
For t > t0 : ò f (t )h(t - t )dt =
0
ò
t0
0
t
f (t )h(t - t )dt + ò (0)h(t - t )dt
t0
Copyright © 2015 Pearson Education Ltd.
3- 28
t
ù
æ
1 éê 0
tö
ú
x t =
F
1
sin
w
t
t
d
t
n
mw n ê ò0 0 çè
t0 ÷ø
ú
ë
û
t0
t0
ù
F é
1
x t = 0 ê ò sin w n t - t dt - ò t sin w n t - t dt ú
mw n ê 0
t0 0
úû
ë
After integrating and rearranging,
F
F
x t = 0 éësin w nt - sin w n (t - t0 ) ùû - 0 éë cos w nt ùû t > t0
kt0w n
k
()
(
()
(
)
)
(
)
()
3.23
Determine the response of an undamped system to a ramp input of the form F(t) =
F0t, where F0 is a constant. Plot the response for three periods for the case m = 2
kg, k = 200 N/m and F0 = 100 N.
Solution: Given:
F0 = 100 N
k = 200 N / m
m= 2 kg
Also mxt  + kxt  = F0 t
To find x (t )
Solution:
x (t )∫ F (τ )h (t − τ )d τ From problem 3.18 for an undamped system
1
Therefore xt  =
F0 τ sin ωn t  τ dτ = F0  τsin ωn t  τ dτ

mωn
mωn
after rearranging
 F
F  τ
1
F
xt  = 0   2 sin ωn τ  = 0 t  0 sin ωn t  Using the given values
mωn  ωn ωn
kωn
 k

xt  =


1
10t  sin10t 
20
Plot:
Copyright © 2015 Pearson Education Ltd.

3- 29
Octave code for plotting
octave:31>t = linspace(0,10,1000);
octave:32> x = 0.5 *t - 0.05*sin(10*t);
octave:33> plot(t,x)
3. 24
A machine resting on an elastic support can be modeled as a single-degree-offreedom, spring-mass system arranged in the vertical direction. The ground is
subject to a motion y(t) of the form illustrated in Figure P3.24. The machine has a
mass of 5000 kg and the support has stiffness 1.5x103 N/m. Calculate the
resulting vibration of the machine.
y(t) (mm)
0.5
0.2
0.6
t (s)
Figure P3.24
Solution: Given m = 5000 kg, k = 1.5x103 N/m, w n =
k
m
= 0.548 rad/s and that
the ground motion is given by:
ì
2.5t
0 £ t £ 0.2
ï
y(t) = í0.75 - 1.25t 0.2 £ t £ 0.6
ï
0
t ³ 0.6
î
The equation of motion is m
x + k(x - y) = 0 or m
x + kx = ky = F(t) The impulse
response function computed from equation (3.12) for an undamped system is
1
h(t - t ) =
sin w n (t - t )
mw n
This gives the solution by integrating a yh across each time step:
t
1 t
x(t) =
ky(t )sin w n (t - t )dt = w n ò y(t )sin w n (t - t )dt
ò
0
mw n 0
For the interval 0<t < 0.2:
t
x(t) = w n ò 2.5t sin w n (t - t )dt
0
Þ x(t) = 2.5t - 4.56 sin 0.548t mm 0 £ t £ 0.2
For the interval 0.2<t < 0.6:
0.2
t
0
0.2
x(t) = w n ò 2.5t sin w n (t - t )dt + w n ò (0.75 - 1.25t )sin w n (t - t )dt
= 0.75 - 0.5cos0.548(t - 0.2) - 1.25t + 2.28sin0.548(t - 0.2)
Combining this with the solution from the first interval yields:
Copyright © 2015 Pearson Education Ltd.
3- 30
x(t) = 0.75 + 1.25t - 0.5cos0.548(t - 0.2)
+6.48sin 0.548(t - 0.2) - 4.56sin 0.548(t - 0.2) mm 0.2 £ t £ 0.6
Finally for the interval t>0.6:
0.2
0.6
t
0
0.2
0
x(t) = w n ò 2.5t sin w n (t - t )dt + w n ò (0.75 - 1.25t)sin w n (t - t )dt + w n ò (0)sin w n (t - t )dt
= -0.5cos0.548(t - 0.2) - 2.28sin 0.548(t - 0.6) + 2.28sin0.548(t - 0.2)
Combining this with the total solution from the previous time interval yields:
x(t) = -0.5cos0.548(t - 0.2) + 6.84 sin0.548(t - 0.2) - 2.28sin 0.548(t - 0.6)
- 4.56sin 0.548t mm t ³ 0.6
3.25
Consider the step response described in Figure 3.7 and Example 3.2.1. Calculate
the analytical value oftp by noting that it occurs at the first peak, or critical point,
of the curve.
Solution: Assume t0 = 0. The response is given by Eq. (3.17):
()
x t =
F0
k
-
F0
k 1- z 2
e
-zw n t
(
cos w d t - f
)
()
cos (w t - f ) + e
( -w ) sin (w t - f ) ùû = 0
To find tp, compute the derivative and let x t = 0
()
x t =
- F0
k 1-z
2
é -zw n e-zw n t
ë
(
)
-zw n t
d
(
d
d
)
-zw
Þ tan (w t - f ) =
w
Þ -zw n cos w d t - f - w d sin w d t - f = 0
n
d
d
æ -zw n ö
w d t - f - p = tan -1 ç
÷ ( can be added or subtracted without changing the
è wd ø
tangent of an angle)
t=
æ
öù
1 é
-1 -zw n
êp + f + tan ç
÷ú
w d êë
è w d ø úû
æ z ö
But, f = tan ç
÷
çè 1 - z 2 ÷ø
So,
-1
Copyright © 2015 Pearson Education Ltd.
3- 31
æ z ö
æ z öù
1 é
-1
êp + tan -1 ç
÷ - tan ç
÷ú
çè 1 - z 2 ÷ø
çè 1- z 2 ÷ø ú
wd ê
ë
û
p
tp =
wd
t=
3.26
Calculate the value of the overshoot (o.s.), for the system of Example 3.2.1. Note
from the example that the overshoot is defined as occurring at the peak time
defined by t p = p / w d and is the difference between the value of the response at tp
and the steady state response at tp.
Solution:
The overshoot occurs at t p =
p
wd
Substitute into Eq. (3.17):
( )
x tp =
é æ p ö
ù
F0
F0
-zw p / w
e n d cos êw d ç ÷ - q ú
k k 1-z2
êë è w d ø
úû
The overshoot is
( )
()
o.s. = x t p - xss t
o.s. =
F0
k
-
F0
k 1- z 2
e
- zw n p / w d
( - cosq ) -
F0
k
æ z ö
2
Since q = tan -1 ç
÷ , then cosq = 1-z
2
çè 1 - z ÷ø
o.s. = o.s. =
3.27
F0
k 1-z
2
(e
-zw n p / w d
)(
1- z 2
)
F0 -zw np / w d
e
k
It is desired to design a system so that its step response has a settling time of 4 s
and a time to peak of 1 s. Calculate the appropriate natural frequency and
damping ratio to use in the design.
Solution:
Given: There are two unknowns in this problem and two equations to use. The
two equations are that for the peak time and for the settling time given are
t p = 1 s and t s= 4 s
Copyright © 2015 Pearson Education Ltd.
3- 32
Peak time is t p =
 ωd = πrad / s
π
3.5
and settling time is ts =
ωd
ξωn
rad
s
And ωn ζ = 0.875


But ωd = 1 ζ 2 ωn
 π 2 = ωn2  ζ 2ωn2
 ωn2 = π 2 + 0.875  = 10.635
2
3.28
 ωn = 3.2611rad / s
0.875
And ζ =
= 0.26831
3.2611
Plot the response of a spring-mass-damper system for this input of square input of
magnitude F0= 30 N, illustrated in Figure 3.8 of Example 3.2.1, for the case that
the pulse width is the natural period of the system (i.e., t1 = n). Recall that k =
1000 N/m,  = 0.1 and n = 3.16 rad/s.
Solution:
The values from Figure 3.7 will be used to plot the response.
F0 = 30 N, k = 1000 N/m, z = 0.1, w = 3.16 rad/s
p
From example 3.2.2 and Figure 3.7, with t1 =
we have for t = 0 to t1,
wn
()
x t =
F0
k
-
F0 e
-zw n t
k 1- z 2
(
cos w d t - f
)
x(t)= .03 - .03015e-.316t cos(3.144t - .1002)
æ z ö
where f = tan -1 ç
÷
çè 1 - z 2 ÷ø
0 <tt1
For t>t1,
()
x t =
ìï zw nt1
é æ
ù
p ö
e cos êw d ç t - ÷ - f ú - cos w d t - f
í
êë è w n ø
úû
k 1 - z 2 îï
F0 e
-zw n t
(
)
üï
ý
þï
x(t) = 0.0315e-.316t {1.3691cos(3.144t – 3.026) – cos(3.144t - .1002)}t > t1
The plot in Mathcad follows:
Copyright © 2015 Pearson Education Ltd.
3- 33
3.29
Consider the spring-mass
mass system described by
mx(t )  kx (t )  F0 sin t , x0  0.01 m and v0  0
Compute the response of this system for the values of m = 100 kg, k = 2000 N/m,
= 10 rad/s and F0 = 10 N, using the convolution integral approach outlined in
Example 3.2.4. Check your answer using the results of equation (2.25).
Solution: Given:
m = 100kg
k = 2000 N / m
ω = 10 rad / s
F0 = 10 N
ωn =
k
= 20 = 4.4721rad / s
m
ξ=0
The Total solution has the form
xt  = Asin ωn t  + Bcos ωn t  + x p t 
Where x p t  is the particular solution, which is given by
t
x p t  =  ht  τ F τ ddτ
0
t
 1 ξωn t  τ 

 x p t  =  
e
sin ωd t  τ  F τ dτ
mωd

0
After applying given values we have,
t
1


x p t  =  
sin 4.4721t  τ 10sinωτ dτ
100  4.4721

0
Copyright © 2015 Pearson Education Ltd.
3- 34
 x p t  = 0.00279 sin 4.4721t   0.001249 sin 10t 
Therefore
xt  = Asin 4.4721t  + Bcos4.4721t  + 0.00279 sin 4.4721t   0.001249 sin 10t 
And from initial conditions we have x0  = 0.01 and x 0  = 0
A= 0 B= 0.01
xt  = 0.01cos4.4721t  + 0.00279 sin 4.4721t   0.001249 sin 10t 
Actual Solution via theory from chapter 2 gives us
  10 
 


10 / 100
10 / 100

sin 4.4721t  + 
sin 10t 
xt  = x0 cos4.4741t  + 
2
2

 4.4721  4.4721  100 
  4.4721  100 
xt  = 0.01cos4.4741t + 0.002795sin4.4721t   0.001249sin 10t 
Thus, both methods give us the same result.
Copyright © 2015 Pearson Education Ltd.
3- 37
Solutions Section 3.3 (problems 3.30-3.38)
3.30
Derive equations (3.24). (3.25) and (3.26) and hence verify the equations for the Fourier
coefficient given by equations (3.21), (3.22) and (3.23).
Solution: For n ¹ m, integration yields:
T
ò
0
(
)
(
)
T
é sin n - m w T t sin n + m w T t ù
sin nw T t sin mw T tdt = ê
ú
w T 2 n + m úû
êë w T 2 n - m
0
(
)
(
)
é
é
æ 2p ö ù
æ 2p ö ù
sin ê n - m ç ÷ T ú sin ê n + m ç ÷ T ú
è T ø û
è T ø û
ë
ë
=
2 n - m wT
2 n + m wT
(
)
(
)
(
)
( )
sin éë( n - m) ( 2p ) ùû sin éë( n + m) ( 2p ) ùû
=
=0
2 ( n - m)w
2 ( n + m)w
T
T
Since m and n are integers, the sine terms are 0, so this is equal to 0.
Equation (3.24), for m = n:
T
ò
0
T
é1
ù
é æ 2p ö ù
1
T
T
sin nw T tdt = ê t sin 2nw T t ú = sin ê 2p ç ÷ T ú
4nw T
ë è T ø û
ë2
û 0 2 8np
(
2
=
)
T
T
T
sin éë 4np ùû =
2 8np
2
Since n is an integer, the sine term is 0, so this is equal to T/2.
T
So,
ò
0
ì 0
m¹ n
sin nw T t sin mw T tdt = í
îT / 2 m = n
Equation (3.25), for m ¹ n
Copyright © 2015 Pearson Education Ltd.
3- 38
T
ò
0
(
(
)
)
(
(
)
)
T
é sin n - m w T t sin n + m w T t ù
cos nw T t cos mw T tdt = ê
ú
2 h + m w T úû
êë 2 n - m w T
0
é
é
æ 2p ö ù
æ 2p ö ù
sin ê n - m ç ÷ T ú sin ê n + m ç ÷ T ú
è T ø û
è T ø û
ë
ë
=
2 n - m wT
2 n + m wT
(
)
(
)
( )
( )
sin éë( n - m) ( 2p ) ùû sin éë( n + m) ( 2p ) ùû
=
=0
2 ( n - m) w
2 ( n + m) w
T
T
Since m and n are integers, the sine terms are 0, so this is equal to 0.
Equation (3.25), for m = n becomes:
T
é1
ù
é æ 2p ö ù
1
T
T
cos
n
w
tdt
=
t
+
sin
2n
w
t
=
+
sin
ê
ú
ê 2n ç ÷ T ú
ò
T
T
ë è T ø û
0
ë 2 4nw T
û 0 2 8np
T
(
2
=
)
T
T
T
+
sin éë 4np ùû =
2 8np
2
Since n is an integer, the sine term is 0, so this is equal to T/2.
T
So,
ò
0
ì 0
m¹ n
cos nw T t cos mw T tdt = í
îT / 2 m = n
Equation (3.26), for m ¹ n :
T
ò
0
(
)
(
)
T
é cos n - m w T t cos n + m w T t ù
cos nw T t sin mw T tdt = ê
ú
2w T n + m úû
êë 2w T n - m
0
(
)
(
)
é
é
æ 2p ö ù
æ 2p ö ù
cos ê n - m ç ÷ T ú cos ê n + m ç ÷ T ú
è T ø û
è T ø û
1
1
ë
ë
=
+
2 n - m wT
2 n + m wT
2 m - n wT 2 m + n wT
(
)
(
=
(
(
(
)
)( )
)
(
(
)
(
)
)( )
)
(
)
(
T
ò
)
cos éë n - m 2p ùû cos éë n + m 2p ùû
1
1
+
=0
2 n - m wT
2 n + m wT
2 m - n wT 2 m + n wT
)
Since n is an integer, the cosine term is 1, so this is equal to 0.
So,
(
cos nw T t sin mw T tdt = 0
0
Copyright © 2015 Pearson Education Ltd.
(
)
3- 39
Equation (3.26) for n = m becomes:
T
é 1
ù
T
cos nw T t sin nw T tdt = ê
sin 2 nw T t ú =
sin 2 2p n = 0
ë 2nw T
û0 4np
T
ò
0
T
ò
Thus
cos nw T t sin nw T tdt = 0
0
3.31
Calculate bn from Example 3.3.1 for the triangular force given by
ì4
T
0£t £
ï t -1
2
ïT
F t =í
ï1 - 4 æ t - T ö T £ t £ T
ïî T çè
2 ÷ø 2
and show that bn = 0, n = 1,2,…,. Also verify the expression an by completing the
integration indicated. (Hint: Change the variable of integration from t to x = 2nt/T.)
()
2
Solution: From Equation (3.23), bn =
T
2é
bn = ê
T êë
æ4
ö
t
1
çè T
÷ø sin nw T tdt +
T /2
ò
0
2 é4
bn = ê
T êë T
T /2
ò
T /2
1 é 2
ê
p n êë p n
tdt . Computing the integral yields:
ù
é 4 æ T öù
ê1 - ç t - ÷ ú sin nw T tdt ú
2øû
úû
ë Tè
sin nw T tdt + 3 ò
T /2
T
ù
4
sin nw T tdt - ò t sin nw T tdt ú
T T /2
úû
2p n
t
T
pn
ò
T
0
0
Substitute x = nw T t =
bn =
ò
T /2
ò F ( t ) sin nw
T
ò
t sin nw T tdt -
0
T
T
pn
x sin xdx -
ò
0
2p n
sin xdx + 3 ò
pn
0
2
pn
2p n
ò
p
n
ù
x sin xdx ú
úû
ù
ú
0
pn
û
ù
1 é 2
2
=
-p ncos p n + cos p n - 1- 3 + 3cos p n -2p n + p ncos p n ú
ê
p n ëp n
pn
û
1
1
éë -2cos p n + 4cos p n - 4 + 4 - 2cos p n ùû =
é0 ù = 0
=
pn
pn ë û
=
1 é 2
sin x - x cos x
p n êë p n
sin xdx -
(
(
)
pn
pn
2p n
+ cos x 0 - 3cos x p n -
(
2
sin x - x cos x
pn
)
Copyright © 2015 Pearson Education Ltd.
(
)
2p n
)
3- 40
From equation (3.22), an =
2é
an = ê
T êë
ò
0
T /2
ò F ( t ) cos nw
ò
t cos nw T tdt -
0
ò
T
tdt
ò
T /2
ù
é 4 æ Töù
ê1- ç t - ÷ ú cos nw T tdt ú
2øû
úû
ë Tè
T
cos nw T tdt + 3 ò
T /2
0
1 é 2
ê
p n êë p n
T
0
T /2
Substitute x = nw T t =
an =
T
æ4
ö
t
1
çè T
÷ø cos nw T tdt +
T /2
2 é4
an = ê
T êë T
2
T
T
ù
4
cos nw T tdt - ò t cos nw T tdt ú
T T /2
úû
2p n
t
T
ù
x cos xdx ú
úû
0
0
pn
pn
2p n ù
pn
2p n
2
cos x + x sin x - sin x 0 + 3sin x p n cos x - sin x
ú
pn
pn
û
ù
2
cos p n - 1 1- cos p n ú
pn
û
pn
ò
pn
x cos xdx -
ò
2p n
cos xdx + 3 ò
=
1 é 2
p n êë p n
(
=
1 é 2
p n êë p n
(
=
2
écos p n - 1 - 1+ cos p n ùû
p n2 ë
cos xdx -
)
)
2p n
ò
(
(
)
)
2
ì 0
4
ï
= 2 2 éëcos p n - 1ùû = í -8
p n
ï 2 2
îp n
3.32
2
pn
n even
n odd
Determine the Fourier series for the rectangular wave illustrated in Figure P3.32.
Figure P3.32 Rectangular periodic signal
Solution: The square wave of period T is described by
Copyright © 2015 Pearson Education Ltd.
3- 41
ì1 0£t £p
F t =í
î-1 p £ t £ 2p
Determine the coefficients a0 ,an ,bn from direct integration:
()
2
a0 =
T
T
ò F ( t ) dt
0
p
2p
ù
2 é
=
ê ò 1 dt + ò -1 dt ú
2p êë 0
p
ûú
2p
1 p
= é t 0 - t p dt ù
ûú
p ëê
1
1
= éëp - 2p + p ùû = 0
Þ a0 = 0
p
p
2p 2p
F t cos nw T tdt, where w T =
=
=1
T
2p
()
( )
()
T
2
an = ò
T 0
()
p
2p
ù 1 é1
p
2p ù
2 é
1
=
ê ò cos ntdt - ò cos ntdt ú = ê sin nt 0 - sin nt p ú
2p êë 0
n
úû p ë n
û
p
1
ésin np - sin n2p + sin np ù = 0
=
û
pn ë
T
p
2p
ù
2
2 é
bn = ò F t sin w T tdt =
sin
ntdt
sin ntdt ú
êò
ò
T 0
2p êë 0
úû
p
p
2p ù
1 é -1
1
1
2
éë - cos np + 1 - 1- cos np ùû =
é1- cos np ùû
= ê cos nt 0 - cos nt p ú =
pën
n
pn ë
û pn
If n is even, cosn = 1. If n is odd, cosn = -1
ì 0 n even
ï
So, bn = í 4
n odd
ï
îp n
Thus the Fourier Series collapses to a sine series of the form
( )
(
)
( )
()
()
¥
F t = å bn sin nt =
n=1
¥
4
sin nt
n=1,3, np
å
The Vibration Toolbox can also be used:
t=0:pi/100:2*pi-pi/100;
f=-2*floor(t/pi)+1;
vtb3_3(f',t',100)
[a,b]=vtb3_3(f',t',100)
Copyright © 2015 Pearson Education Ltd.
3- 42
Note that vtb3_3 always gives some error on the order of delta t (.01 in this case). Using a
smaller delta t reduced the error.
3.33
Determine the Fourier series representation of the sawtooth curve illustrated in Figure
P3.33.
f (t)
1
t
2
4
6
8
Figure 3.33 Sawtooth periodic signal
Solution: The sawtooth curve of period T is
1
F t =
t
0 £ t £ 2p
2p
Determine coefficients a0 ,an ,bn :
()
2
a0 =
T
=
2
an =
T
T
ò ()
0
2
F t dt =
2p
2p
ò
0
æ 1 ö
æ 1 ö1 2
çè 2p t ÷ø dt = çè 2p 2 ÷ø 2 t
2p
0
1
é 4p 2 - 0 ù = 1
2 ë
û
4p
T
ò F ( t ) cos nw
T
tdt, where w T =
0
2p 2p
=
=1
T
2p
2p
2p
ù
ù
2 é æ 1 ö
1 é
=
t ÷ cos ntdt ú = 2 ê ò t cos ntdt ú
êò ç
2p êë 0 è 2p ø
úû 2p êë 0
úû
1
= 2
2p
bn =
2
T
2p
é1
ù
1
1
ê 2 cos nt + t sin nt ú = 2
n
ën
û 0 2p
T
ò ()
F t sin nw T tdt =
0
2
2p
é1
ù
1
ê 2 1-1 + 0 - 0 ú = 0
n
ën
û
(
)
(
)
2p
é 2p æ 1 ö
ù
ù
1 é
t
sin
ntdt
=
t sin ntdt ú
êò ç
ú
2 êò
÷
êë 0 è 2p ø
úû 2p êë 0
úû
2p
ù
ù
1 é1
1
1 é1
1
= 2 ê 2 sin nt - t cos nt ú = 2 ê 2 0 - 0 - 2p - 0 ú
n
n
2p ë n
2p ë n
û0
û
=
(
)
1 æ -2p ö -1
=
2p 2 çè n ÷ø p n
Copyright © 2015 Pearson Education Ltd.
(
)
3- 43
Fourier Series
1 ¥ æ -1 ö
F t = + å ç ÷ sin nt
2 n=1 è p n ø
()
()
F t =
3.34
1 1 ¥ 1
- å sin nt
2 p n=1 n
Calculate and plot the response of the base excitation problem with base motion specified
by the velocity
y (t )  4et /2 (t )m / s
Where  (t) is the unit step function and m = 10 kg,  = 0.01, and k = 1000 N/m. Assume
that the initial conditions are both zero.
Solution: Given:
y (t )  4et /2 (t )m / s
m= 10 kg
ξ= 0.01
k = 1000 N / m
Assume ωn = ωb
x 0  = 0
x 0  = 0
t


Y =  4e  t / 2 t dt = 8 1  e  t / 2 m
0
The force acting on the machine is of the form
F t  = cYωb cosωbt  + kYsinωbt 
And thus the displacement is of the form
1
2
 1 + 2r 
2
ξω t
xt  = Ae n sin ωd t +  + ωnY 
 cosωb t  θ1  θ 2 
2
2
 1  r 2 + 2r  
π
−1 2 ξ ω n ωb
= rad
Where θ1= tan
2
2
2
ωn −ω b

(

)
ωn
1
−1
−1
= 1.5508 rad
And θ2= tan 2 ξ ω = tan
2 (0.01 )
b
And the magnitude becomes
1
1
2
 1 + 2r 
2
 1
2
ωnY 
= 10  8
+ 1 = 4000.8m

2
2
2
 1  r 2 + 2r  
 0.02

( )

(
)

And
Copyright © 2015 Pearson Education Ltd.
3- 44

ωd = 1   2

0.5
ωn = 9.999
t
π


 xt  = Ae 10 sin 9.999t +   + 4000.8cos 9.999t   1.5508 
2


Where A = 4000.2 and  = 1.5879
t
π


 xt  = 4000.2e 10 sin 9.999t + 1.5879  + 4000.8cos 9.999  t   1.5508 
2


Plot:
Copyright © 2015 Pearson Education Ltd.
3- 45
3.35
Calculate and plot the total response of the spring-mass-damper system with m = 80 kg, 
= 0.1 and k = 1000 N/m to the signal of defined by
T
4
0t 
 T t  1
2
F t   
1  4  t  T  T  t  T
 T  2  2
with maximum force of 1 N. Assume that the initial conditions are zero and let T = 2 s.
Solution: Given: m = 80 kg,  = 0.1 and k = 1000 N/m to the signal of defined by
T
4
0t 
 T t  1
2
F t   
1  4  t  T  T  t  T
 T  2  2
with maximum force of 1 N.
Assume that the initial conditions are zero and let T = 2 s
Solution:
xt  =



0.353t
 0.353t
 0.00107
0.005072
0.353t


t + 0.00107cos3.5178t   0.000254sin3.517t  0 < t < π 

 0.001273 0.353t t + 0.00414cos3.517t   0.005726sin3.5178t  π < t < 2π 
0.353t
+ 0.00127
0.353t
Plot:
Copyright © 2015 Pearson Education Ltd.
3- 46
3.36
Calculate the total response of the system of Example 3.3.2 for the case of a base motion
driving frequency of b = 1.414 rad/s with amplitude Y = 0.05 m subject to initial
conditions x0 = 0.01 m and v0 = 3.0 m/s. The system is defined by m = 1 kg, c = 10 kg/s,
and k = 1000 N/m.
Solution: Given:
m= 1 kg
c= 1 0 kg / s
k = 1000 N / m
 =
c
10
=
= 0.15811
0.5
0.5
2km
21000
Assume ωn = 1000 rad / s
ω b= 1. 414 rad / s
x 0  = 3m / s
x0 = 0.01m
Y = 0.05m
The force acting on the machine is of the form
F t  = cYωb cosωbt  + kYsinωbt 
And thus the displacement is of the form
1
2
 1 + 2r 
2
ω t
xt  = Ae n sin ωd t +  + ωnY 
 cosωb t  θ1  θ 2 
2
2
 1  r 2 + 2r  
 20.15811 1000 1.414 
 2ω ω 
 = 0.01416rad
Where θ1 = tan 1  2 n b2  = tan 1 
2

ω

ω


1000

1.414
b 
 n




 ω 
1000
 = 1.5566rad
And θ2 = tan 1  n  = tan 1 



2

ω
2
0.15811
1.414
b 



And the magnitude becomes




1

2
2

 1.414  
1


1 +  2  0.15811
 


 1 + 2r 2
2
 1000  


 = 1.5654m
ωnY 
=
1000

0.05

2
2
2 2
   1.414  2   2  0.15811  1.414  2 
 1  r + 2r  
 1  
  +
 
1000
 
   1000   

And



ωd = 1  0.15811
 1000 = 31.22 rad / s
 x t  = Ae  t 0.15811 1000 sin 31.22t + + 1.5654cos 31.22t  0.01416  1.5654 
Where A = 1.46564 and  = 3.12541
2 0.5
 xt  = 1.5654cos1.579  31.22t +1.46564e 4.999t sin3.12541+ 31.22t 
Copyright © 2015 Pearson Education Ltd.
3- 47
3.37
Validate your solution to the square wave Problem 3.32 by calculating an andbn using
VTB3_3 in the Vibration Toolbox. Print the function and its Fourier series approximation
for 5, 20, then 100 terms. The Toolbox makes this easy. The purpose is to illustrate the
Gibbs effect in approximation by Fourier series.
3.38
Validate your solution to the saw tooth wave of Problem 3.33 by calculating an
andbnusing VTB3_3 in the Vibration Toolbox. Print the function and its Fourier series
approximation for 5, 20, and 100 terms. The Toolbox makes this easy. The purpose is to
illustrate the Gibbs effect in approximation by Fourier series.
Copyright © 2015 Pearson Education Ltd.
3- 50
Solutions for Section 3.4 (3.39 through 3.43)
3.39
Calculate the response of
where Φ(t) is the unit step function for the case with x 0 = v 0 = 0. Use the Laplace
transform method and assume that the system is underdamped.
Solution:
Given:
Take Laplace Transform:
Using inverse Laplace tables,
Copyright © 2015 Pearson Education Ltd.
3- 51
3.40
Using the Laplace transform method, calculate the response of the system
for the overdamped case (ζ > 1). Plot the response for m = 1 kg, k = 100 N/m,
and ζ = 1.5.
Solution:
From example 3.4.4,
Take Laplace Transform:
Using inverse Laplace tables,
Inserting the given values yields:
Copyright © 2015 Pearson Education Ltd.
3- 52
3.41
Calculate the response of the underdamped system given by
using the Laplace transform method. Assume a > 0 and that the initial conditions
are both zero.
Solution:
Given:
Rewrite:
Take Laplace Transform:
For an underdamped system, the inverse Laplace Transform is
Copyright © 2015 Pearson Education Ltd.
3- 53
3.42
Solve the following system for the response x(t) using Laplace transforms:
x(t ) + 2000 x(t ) =
100 
40δ (t )
where the units are in Newtons and the initial conditions are both zero.
Solution:
Given:
100 x(t ) + 2000 x(t ) = 40δ (t )
Taking the Laplace Transform on both sides we have
100 s 2 X (s ) − sx(0 ) − x' (0 ) + 2000( X (s )) = 40
But x(0 ) = 0 and x (0 ) = 0
(
)
(
)
⇒ X (s ) 100 s 2 + 2000 = 40
⇒ X (s ) =
⇒ x(t ) =
3.43
0.4 
20


2
20  s + 20 2
( (
))




0.4
sin 20t = 0.089442sin (4.4721t )
20
Use the Laplace transform approach to solve for the response of the spring-mass
system with equation of motion and initial conditions given by
Assume the units are consistent (Hint see the example in Appendix B).
Solution Taking the Laplace transform of the equation of motion yields
Substitution of the given initial conditions and solving algebraically for X(s)
yields:
where the right side is the partial fraction expansion. Following the partial
fraction expansion example in Appendix B yields
Copyright © 2015 Pearson Education Ltd.
3- 54
Using the 5th entry in Table 3.1 to invert this expression yields
This can be verified by referring back to the method of section 2.1.
Copyright © 2015 Pearson Education Ltd.
3- 54
Solutions Section 3.5 (3.44 through 3.48)
3.44
Calculate the mean-square response of a system to an input force of constant PSD, S 0 ,
10
and frequency response function H (ω ) =
( 5 + 3 jω )
Solution:
Copyright © 2015 Pearson Education Ltd.
3- 55
Copyright © 2015 Pearson Education Ltd.
3- 56
3.45
Consider the base excitation problem of Section 2.4 as applied to an automobile model of
Example 2.4.2 and illustrated in Figure 2.17.Recall that the model is a spring-massdamper system with values
. In this
problem let the road have a random stationary cross section producing a PSD of S 0 .
Calculate the PSD of the response and the mean-square value of the response.
Solution: Given:
From example 2.4.2:
So,
The PSD is found from equation (3.62):
The mean square value is found from equation (3.68):
Using equation (3.70) yields
Copyright © 2015 Pearson Education Ltd.
3- 57
3.46
To obtain a feel for the correlation functions, compute autocorrelation R xx (τ) for the
deterministic signal Asinω n t.
Solution: The autocorrelation is found from
Simplifying yields:
3.47
The autocorrelation of a signal is given by
Compute the mean square value of the signal.
Solution From equation equation (3.66) the mean square value is just R xx (0):
3.48
Verify that the average
compute the average.
is zero by using the definition given in equation (3.47) to
Solution:
The definition is
Let
Copyright © 2015 Pearson Education Ltd.
3- 58
Copyright © 2015 Pearson Education Ltd.
3- 57
Solutions Section 3.6 (3.49 through 3.50)
3.49
A power line pole with a transformer is modeled by
m
x + kx = - y
where x and y are as indicated in Figure P3.49. Assuming the initial conditions are zero
calculate the response of the relative displacement (x – y) if the pole is subject to an
earthquake base excitation of
ì æ
tö
ï Aç1 - ÷

y t = í è t0 ø
ï
0
î
()
0 £ t £ 2t0
t > 2t0
Figure P3.49 Vibration model of a power-line
pole with a transformer mounted on it.
Solution: Given: m
x + kx = - y
ì æ
tö
ï A ç 1- ÷ 0 £ t £ 2t0

y = í è t0 ø
ï
0
t > 2t0
î
x 0 = x 0 = 0
() ()
The response x(t) is given by Eq. (3.12) as
t
( ) ò F (t ) h (t - t ) dt
x t =
0
Copyright © 2015 Pearson Education Ltd.
3- 58
(
)
where h t - t =
(
)
1
sin w n t - t for an undamped system
mw n
For 0 £ t £ 2t0 ,
t
æ
t öæ 1 ö
x t = ò Aç1 - ÷ ç
÷ sin w n t - t dt
t
m
w
è
ø
è
0
0
nø
()
(
()
x t =
A
mw n2
)
é
ù
t
1
sin w nt - cos w nt ú
ê1- +
ë t0 t0w n
û
For t>2t0,
2t0
æ
öæ
( ) ò A çè 1 - tt ÷ø çè mw1
x t =
0
()
x t =
0
ö
÷ sin w n t - t dt
nø
(
)
ù
A é 1
sin w nt - sin w n t - 2t0 - cos w nt - cos w n t - 2t0 ú
2 ê
mw n ë t0w n
û
(
))
(
(
)
Find y(t) when 0 £ t £ 2t0 ,
æ
tö

y t = Aç1 - ÷
t0 ø
è
()
()
y t = At -
()
y t =
A 2
t + C1
2t0
A 2 A 3
t t + C1t + C2
2
6t0
Using IC's yields C1 = C2 = 0. Find y(t) when t > wt0:
()
y ( t ) = C
y (t ) = C t + C

y t =0
3
3
4
Using IC's yields C3 = C4 =0. The relative displacement x(t) – y(t) is therefore:
For 0 £ t £ 2t0
() ()
x t -y t =
ù A 2 A 3
A é
t
1
1+
sin
w
t
cos
w
t
- t +
t
ê
n
n ú
6t0
mw n2 ë t0 t0w n
û 2
Copyright © 2015 Pearson Education Ltd.
3- 59
For t > 2t0,
() ()
x t -y t =
3.50
ù
A é 1
sin w nt - sin w n t - 2t0 - cos w nt - cos w n t - 2t0 ú
2 ê
mw n ë t0w n
û
(
))
(
(
)
Calculate the response spectrum of an undamped system to the forcing function
ì
pt
0 £ t £ t1
ï F0 sin
t1
F t =í
ï0
t > t1
î
assuming the initial conditions are zero.
()
Solution: Let w = p / t1 .
The solution is the homogeneous solution xh(t) and the
()
()
()
()
or x t = xh t + x p t . Thus
particular solution x p t
æ F0 ö
x t = Acos w nt + B sin w nt + ç
sin w t
2÷
è k - mw ø
where A and B are constants and n is the natural frequency of the system:
Using the initial conditions x 0 = x 0 = 0 the constants A and B are
()
() ()
A = 0, B =
()
so that x t =
F0 / k
(
1- w / w n
)
2
(
- F0w
w n k - mw 2
)
ìï
üï
w
sin w nt ý , 0 £ t £ t1
ísin w t wn
îï
þï
Which can be written as (where d = F0 / k the static deflection)
( )=
ìï p t t
2p t üï
sin
sin
í
ý , 0 £ t £ t1
2
d
t þï
æ t ö îï t1 2t1
1- ç ÷
è 2t1 ø
and where t = 2p / w n . After t1 the solution is a free response
x t
1
()
x t = A'cos w nt + B'sin w nt, t > t1
where the constants A' and B' can be found by using the values of x(t = t1) and
x t = t1 , t > t0 . This gives
(
)
é t
2p t1 ù
x t = t1 = a ê - sin
ú = A'cos w nt1 + B'sin w nt1
2t
t
ë
û
1
ìï p p
2p t üï
x t = t1 = a í- - cos 1 ý = -w n A'sin w nt1 + w n B'cos w nt
t ïþ
ïî t1 t1
(
)
(
)
where
Copyright © 2015 Pearson Education Ltd.
3- 60
a=
d
æ t ö
1- ç ÷
è 2t1 ø
2
These are solved to yield
ap
ap
é1+ cos w nt1 ùû
A' =
sin w nt1 , B ' = w nt1
w nt1 ë
So that after t1 the solution is
x t
t / t1
é
æ t1 t ö
tù
=
sin
2
p
sin
2
p
ê
ú , t ³ t1
ç
÷
2
d
t
t
t
è
ø
ê
úû
2 1- t / 2t1 ë
()
{
(
(
)
)}
Copyright © 2015 Pearson Education Ltd.
3- 61
Solutions for Section 3.7 (3.51 through 3.58)
3.51
Using complex algebra, derive equation (3.89) from (3.86) with s = j.
Solution: From equation (3.86):
()
1
ms + cs + k
H s =
2
Substituting s = jw yields
( )
H jw =
1
(
m jw
)
2
( )
+ c jw + k
=
1
k - mw 2 - cjw
The magnitude is given by
(
H jw dr
)
éæ
= êç
êç m jw
êëè
( )
(
H jw =
1/ 2
ö æ
öù
1
ú
÷
=ç
÷ú
2
2
k
m
w
cj
w
÷
è
ø
+ cjw + k ø
úû
1
) ( )
1
( k - mw ) + ( cw )
2
3.52
2
which is Eq. (3.89)
2
Using the plot in Figure 3P.52, estimate the system’s parameters m, c, and k, as well as
the natural frequency.
Figure P3.52 The magnitude plot of a spring-mass-damper system
Solution: From Figure P3.52
Copyright © 2015 Pearson Education Ltd.
3- 62
1
= 2 Þ k = 0.5
k
w = w n = 0.25 =
k
Þm=8
m
1
» 4.6 Þ c = 0.087
cw
3.53
From a compliance transfer function of a spring- mass-damper system the stiffness is
determined to have a value of 0.5 N/m, a natural frequency of 0.25 rad/s and a damping
coefficient of 0.087 kg/s. Plot the inertance transfer function's magnitude and phase for
this system.
Solution: From the data give
k = 0.5,w = w n = 0.25 =
k
Þ m = 8, and c = 0.087
m
The inertance transfer function is given by Eq. (3.88):
s2
s H s = 2
ms + cs + k
2
()
Substitute s = jw to get the frequency response function. The magnitude is given by:
( ) ( )
jw
2
H jw =
w2
( k - mw ) + ( cw )
2
2
w2
=
(0.5 - 8w ) + (0.087w )
2
2
2
2
The phase is given by
æ Imaginary part of frequency response function ö
f = tan-1 ç
è Real part of frequency response function ÷ø
( ) H ( jw ) by ( k - mw ) - cjw to get
Multiply the numerator and denominator of jw
( jw ) H ( jw ) =
2
æ
cw 3
So, f = tan -1 ç
çè -w 2 k - mw 2
(
)
2
(
2
)
-w 2 k - mw + cjw 3
( k - mw ) + ( cw )
2
2
2
ö
æ 0.087w ö
÷ = tan -1 ç 2
÷ø
è 8w - 0.5 ÷ø
Copyright © 2015 Pearson Education Ltd.
3- 63
The magnitude and phase plots are shown on a semilog scale. The plots are given in the
following Mathcad session
3.54
From a compliance transfer function of a spring- mass-damper system the stiffness is
determined to have a value of 0.5 N/m, a natural frequency of 0.25 rad/s and a damping
coefficient of 0.087 kg/s. Plot the mobility transfer function's magnitude and phase for
the system.
Solution: From the data given
k = 0.5,w = w n = 0.25 =
k
Þ m = 8, and c = 0.087
m
The mobility transfer function is given by equation (3.87):
()
sH s =
s
ms + cs + k
2
Substitute s = jw to get the frequency response function. The magnitude is given by
( jw ) H ( jw ) =
w
( k - jw ) + ( cw )
2
2
=
2
w
(0.5 - 8w ) + (0.087w )
2
The phase is given by
Copyright © 2015 Pearson Education Ltd.
2
2
3- 64
æ Imaginary part of frequency response function ö
f = tan-1 ç
è Real part of frequency response function ÷ø
(
( )
)
Multiply the numerator and denominator of jw H jw by j and by - k - mw 2 j - cw to
get
( jw ) H ( jw ) =
(
æ w k - mw 2
So, f = tan ç
çè
cw 2
-1
) ö÷ = tan
÷ø
-1
(
)
jw k - mw 2 + cw 2
( k - mw ) + ( cw )
2
2
2
æ 0.5 - 8w 2 ö
ç
÷
è 0.087w ø
The magnitude and phase plots are shown on a semilog scale.
3.55
Calculate the compliance transfer function for a system described by
a
d 4 x(t)
d 3 x(t)
d 2 x(t) dx(t)
+
b
+
c
+
+ ex(t) = f (t)
dt
dt 4
dt 3
dt 2
where f(t) is the input force and x(t) is a displacement.
Copyright © 2015 Pearson Education Ltd.
3- 65
Solution:
The compliance transfer function is
( ).
F ( s)
X s
Taking the Laplace Transform yields
( as
So,
3.56
( )=
F ( s ) as
X s
4
4
) ()
()
+ bs3 + cs 2 + ds + e X s = F s
1
+ bs + cs 2 + ds + e
3
Calculate the frequency response function for the compliance for the system defined by
d 4 x(t)
d 3 x(t)
d 2 x(t) dx(t)
a
+
b
+
c
+
+ ex(t) = f (t) .
dt
dt 4
dt 3
dt 2
Solution: Taking the Laplace Transfer function for zero initial conditions forming
1
H s = 4
3
as + bs + cs2 + ds + e
Substitute s = jw to get the frequency response function:
1
H jw =
4
3
2
a jw + b jw + c jw + d jw + e
()
( )
( )
H jw =
( )
( )
( )
(
+ e ) + ( -bw
( )
)
+ dw )
aw 4 - cw 2 + e - j -bw 3 + dw
( aw
4
- cw 2
2
3
2
*3.57 Plot the magnitude of the frequency response function for the system of Problem 3.56 for
a  1, b  4, c  12, d  16, and e  10.
Solution The response function is for the problem 3.56 is given by
H (j ω )=
a ω 4 − c ω 2+ e − j (− b ω3 + d ω )
2
2
(a ω4 − c ω 2+ e ) + (− b ω 3+ d ω )
The magnitude of which is given by
H  jω =
aω
1
4
 
2
 cω 2 + e +  bω 3 + dω

2
Given a =1, b= 4, c= 12, d= 16, e = 10
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H  jω =
ω
1
4
 
2
 12ω 2 + 10 +  4ω3 + 16ω

2
Plot for H( ω ) vs ω :
3.58
An experimental (compliance) magnitude plot is illustrated in Fig. P3.58. Determine
w ,z ,c,m, and k. Assume that the units correspond to m/N along the vertical axis.
Mag H ( j)
0.16
0.14
0.12
0.10
0.08
0.06
0.04
0.02
0.00
0.1
1
log ()
10
Figure P3.58 An experimentally determined compliance magnitude plot
Solution: Referring to the plot, it starts at
H (w j) =
Thus:
0.05 =
1
k
1
Þ k = 20 N/m
k
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At the peak, n =  = 3 rad/s. Thus the mass can be determined by
k
m = 2 Þ m = 2.22 kg
wn
The damping is found from
1
c
3.03
= 0.11 Þ c = 3.03 kg/s Þ z =
=
= 0.227
cw
2 km 2 20 × 2.22
Copyright © 2015 Pearson Education Ltd.
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Problems and Solutions Section 3.8 (3.59 through 3.64)
3.59
Show that a critically damped system is bounded-input, bounded- output stable.
Solution:
For a critically damped system
Let f(t) be bounded by the finite constant M. Using the inequality for integrals and
Equation (3.96) yields:
The function h(t – τ) decays exponentially and hence is bounded by some constant times
1/t, say M 1 /t. This is just a statement the exponential decays faster then “one over t”
does. Thus the above expression becomes;
This is bounded, so a critically damped system is BIBO stable.
3.60
Show that an overdamped system is bounded-input, bounded- output stable.
Solution: For an overdamped system,
Let f(t) be bounded by M,
From equation (3.96),
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Since
then
Also, since -
is bounded.
then
is bounded.
Therefore, an overdamped system is BIBO stable.
3.61
Lagrange stable?
Is the solution of
Solution: Given
The total solution will be
From Eq. (1.3):
From Eq. (2.7):
and
Adding the solutions yields
Copyright © 2015 Pearson Education Ltd.
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Since
, and the homogeneous solution is marginally stable, this system is
Lagrange stable.
3.62
Calculate the response of the system described by
for
Solution: Given:
for the case that a = 4 and b = 0. Is the response bounded?
. From Eq. (3.99),
So,
Let
Substituting,
So,
The solution is
So,
Therefore,
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Since
3.63
, the response is bounded.
A crude model of an aircraft wing can be modeled as
Where the factor a is determined by the aerodynamics of the wing and is proportional to
the air speed. At what value of the parameter a will the system start to flutter?
Solution Write this as a homogenous differential equation by moving the forcing term
to the left side of the equation.
The system will start to flutter when the coefficient of the velocity term becomes negative
or for values of a > 25.
3.64
Consider the inverted pendulum of Figure P3.64 and compute the value of the stiffness k
that will keep the linear system stable. Assume that the pendulum rod is massless.
Figure P3.64 An inverted pendulum.
Solution Summing moments the equation of motion is
Using the small angle approximation this becomes
This the system is stable when the coefficient of θ is positive or kl2 > 2mgl.
Copyright © 2015 Pearson Education Ltd.
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Solutions from Section 3.9 (3.65-3.72)
*3.65 Numerically integrate and plot the response of an underdamped system determined
by m = 80 kg, k = 800 N/m, and c = 20 kg/s, subject to the initial conditions of x 0 = 0 and
v 0 = 0, and the applied force F(t) = 30 φ (t -1). Then plot the exact response as computed
by equation (3.17). Compare the plot of the exact solution to the numerical simulation.
Solution: Given m =80 kg, k= 800 kN/m, c = 20 kg/s and initial conditions x0= 0, v0 =
0, applied force is f (t ) = 10f (t − 1) , F(0) = 30 N
k
 800 
ωn =   = 
 = 3.162
m
 80 
20
c
ζ=
=
= 0.03953
2 (k × m ) 2 × (800 × 80 )
 ζ 

θ = artanh
2 
 1− ζ 
F0
30
=
= 0.0375
800
k
F0
= 0.03753
k 1− ζ 2
(
(
ωd =
)
)
ζωn = 0.125
(1 − ζ ) × ω
2
n
= 3.162
( (1 − 0.03953 ))= 3.157
2
F

F0
− ζω (t − t 0 )
cos(ωd (t − t o ) − θ ) (t − t 0 )
xa (t ) =  0 −
e n
 k k 1 − ζ 2

(
)
%Numerical Solutions in OCTAVE
%Problem #3.65
clc
clear
close all
%Numerical Solution
x0=[0;0];
tspan=[0 15];
[t,x]=ode45(@prob365,tspan,x0);
figure(1)
plot(t,x(:,1));
title('Problem #57');
xlabel('Time, sec.');
ylabel('Displacement, m');
hold on
%Analytical Solution
m=80;
Copyright © 2015 Pearson Education Ltd.
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73
c=20;
k=800;
F=30;
w=sqrt(k/m);
d=c/(2*w*m);
wd=w*sqrt(1-d^2);
to=1;
phi=atan(d/sqrt(1-d^2));
%for t<to
t=linspace(0,1,3);
x=0.*t;
plot(t,x,'*');
%for t>=to
t=linspace(1,15);
x=F/k-F/(k*sqrt(1-d^2)).*exp(-d.*w.*(t-to)).*cos(wd.*(t-to)-phi);
plot(t,x,'*');
legend('Numerical Solution', 'Analytical Solution')
%M-file for Prob #365
function dx=prob365(t,x);
[rows, cols]=size(x);dx=zeros(rows, cols);
m=80;
c=20;
k=800;
F=30;
if t<1
dx==0;
else
dx(1)=x(2);
dx(2)=-c/m*x(2) - k/m*x(1) + F/m;
end
Copyright © 2015 Pearson Education Ltd.
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74
Copyright © 2015 Pearson Education Ltd.
3*3.66.
75
Numerically integrate and plot the response of an underdamped system
determined by m = 150 kg, and k = 5000 N/m subject to the initial conditions of
x 0 = 0.01 m and v 0 = 0.1 m/s, and the applied force F(t) = φ (t) = 15(t -1), for
various values of the damping coefficient. Use this “program” to determine a
value of damping that causes the transient term to die out within 3 seconds. Try
to find the smallest such value of damping remembering that added damping is
usually expensive.
Solution:
%Numerical Solutions in Octave
clc
clear
close all
%Numerical Solution
x0=[0.01;0];
tspan=[0 15];
[t,x]=ode45(@prob,tspan,x0);
figure(1)
plot(t,x(:,1));
title('Problem 3.66');
xlabel('Time, sec.');
ylabel('Displacement, m');
hold on
%Analytical Solution
m=150;
c=0;
k=5000;
F=15;
w=sqrt(k/m);
d=c/(2*w*m);
wd=w*sqrt(1-d^2);
to=1;
phi=atan(d/sqrt(1-d^2));
%for t<to
t=linspace(0,1,10);
x0=0.01;
v0=0.0;
A=sqrt(v0^2+(x0*w)^2)/w;
theta=pi/2;
x=A.*sin(w.*t + theta);
plot(t,x,'*')
%for t>=to
t=linspace(1,15);
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x2=F/k-F/(k*sqrt(1-d^2)).*exp(-d.*w.*(t-to)).*cos(wd.*(t-to)-phi);
x1=A.*sin(w.*t + theta);
x=x1+x2;
plot(t,x,'*');
legend('Numerical Solution', 'Analytical Solution')
%Solutions
Octave Function for the problem
function dx=prob(t,x)
[rows, cols]=size(x);
dx=zeros(rows, cols);
m=150;c=0;k=5000;F=15;if t<1dx(1)=x(2);dx(2)=-c/m*x(2)- k/m*x(1);elsedx(1)=x(2);
dx(2)=-c/m*x(2) - k/m*x(1)+ F/m;
end
Copyright © 2015 Pearson Education Ltd.
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*3.67 Calculate the total response of the base isolation problem given in Example 3.3.2,
with the parameters ωb = 3 rad/s, m = 1 kg, c = 10 kg/s, k = 1000 N/m, and Y =
0.05 m, subject to initial conditions x 0 = 0.01 m and v 0 = 3.0 m/s, by numerically
integrating rather then using analytical expressions, and plot the response,
reproduce Figure 3.14 and compare the results to see that they are the same.
Solution: Both Mathcad and Matlab solutions follow:
%Numerical Solutions
%Problem #53
clc
Copyright © 2015 Pearson Education Ltd.
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78
clear
close all
%Numerical Solution
x0=[0;0];
tspan=[0 10];
[t,x]=ode45('prob53a',tspan,x0);
figure(1)
plot(t,x(:,1));
title('Problem #53');
xlabel('Time, sec.');
ylabel('Displacement, mm');
hold on
%Analytical Solution
t1=0.2;
t2=0.6;
%for t<to
t=linspace(0,t1);
x=2.5*t-4.56.*sin(0.548.*t);
plot(t,x,'*');
%for t1<t<t2
t=linspace(t1,t2);
x=0.75 - 1.25.*t + 6.84.*sin(0.548*(t-t1))- 4.56.*sin(0.548.*t);
plot(t,x,'*');
%for t2<t
t=linspace(t2,10);
x=6.84.*sin(0.548.*(t-t1))-2.28.*sin(0.548.*(t-t2))-4.56.*sin(0.548.*t);
plot(t,x,'*');
legend('Numerical', 'Analytical')
%Clay
%Vibrations
%Solutions
%Clay
%Vibrations
%Solutions
%M-file for Prob #52
function dx=prob(t,x);
[rows, cols]=size(x);dx=zeros(rows, cols);
m=1;
c=10;
k=1000;
Y=0.05;
wb=3;
a=c*Y*wb;
b=k*Y;
alpha=atan(b/a);
AB=sqrt(a^2+b^2)/m;
Copyright © 2015 Pearson Education Ltd.
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79
dx(1)=x(2);
dx(2)=-c/m*x(2)- k/m*x(1)+ a/m*cos(wb*t) + b/m*sin(wb*t);
*3.68 Numerically simulate the response of the system of a single-degree-of-freedom
spring-mass system subject to the motion y(t) given in Figure P3.68 and plot the
response. The mass is 5000 kg and the stiffness is 1.5 x 103 N/m.
Figure P3.68 The base motion for Problem 3.68.
Solution: The solution in Matlab is
%Clay
%Vibrations
%Numerical Solutions
%Problem #53
clc
clear
close all
%Numerical Solution
x0=[0;0];
tspan=[0 10];
[t,x]=ode45('prob53a',tspan,x0);
figure(1)
plot(t,x(:,1));
title('Problem #53');
xlabel('Time, sec.');
ylabel('Displacement, mm');
hold on
%Analytical Solution
t1=0.2;
t2=0.6;
%for t<to
t=linspace(0,t1);
Copyright © 2015 Pearson Education Ltd.
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80
x=2.5*t-4.56.*sin(0.548.*t);
plot(t,x,'*');
%for t1<t<t2
t=linspace(t1,t2);
x=0.75 - 1.25.*t + 6.84.*sin(0.548*(t-t1))- 4.56.*sin(0.548.*t);
plot(t,x,'*');
%for t2<t
t=linspace(t2,10);
x=6.84.*sin(0.548.*(t-t1))-2.28.*sin(0.548.*(t-t2))-4.56.*sin(0.548.*t);
plot(t,x,'*');
legend('Numerical', 'Analytical')
%Clay
%Vibrations
%Solutions
%M-file for Prob #53
function dx=prob(t,x);
[rows, cols]=size(x);dx=zeros(rows, cols);
m=5000;
k=1.5e3;
ymax=0.5;
F=k*ymax;
t1=0.2;
t2=0.6;
if t<t1
dx(1)=x(2);
dx(2)= - k/m*x(1)+ F/m*(t/t1);
elseif t<t2 & t>t1
dx(1)=x(2);
dx(2)= - k/m*x(1)+ F/(2*t1*m)*(t2-t);
else
dx(1)=x(2);
dx(2)= - k/m*x(1);
end
Copyright © 2015 Pearson Education Ltd.
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Copyright © 2015 Pearson Education Ltd.
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82
*3.69 Numerically simulate the response of and undamped system to a step function
with a finite rise time of t 1 for the case m = 1 kg, k = 1 N/m, t 1 = 4 s and F 0 = 20
N. This function is described by
plot the response.
Solution: The solution in Matlab is
%Clay
%Vibrations
%Numerical Solutions
%Problem #69
clc
clear
close all
%Numerical Solution
x0=[0;0];
tspan=[0 10];
[t,x]=ode45('prob54a',tspan,x0);
figure(1)
plot(t,x(:,1));
title('Problem #69');
xlabel('Time, sec.');
ylabel('Displacement, m');
hold on
%Analytical Solution
to=4;
%for t<to
t=linspace(0,to);
x=5*(t-sin(t));
plot(t,x,'*');
%for t>=to
t=linspace(to,10);
x=5*(sin(t-to)-sin(t))+20;
plot(t,x,'*');
legend('Numerical', 'Analytical')
%Clay
Copyright © 2015 Pearson Education Ltd.
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83
%Vibrations
%Solutions
%M-file for Prob #54
function dx=prob(t,x);
[rows, cols]=size(x);dx=zeros(rows, cols);
m=1;
k=1;
F=20;
to=4;
if t<to
dx(1)=x(2);
dx(2)= - k/m*x(1)+ F/m*(t/to);
else
dx(1)=x(2);
dx(2)= - k/m*x(1)+ F/m;
end
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84
3.70*. Numerically simulate the response of the system of Problem 3.22 for a 2 meter
2
concrete wall with cross section 0.03 m and mass modeled as lumped at the end of 1000
kg. Use F 0 = 100 N, and plot the response for the case t 0 =0.25 s.
Solution The solution in Matlab is:
%Numerical Solutions
%Problem #3.62
clc
clear
close all
%Numerical Solution
x0=[0;0];
tspan=[0 0.5];
[t,x]=ode45('prob55a',tspan,x0);
figure(1)
plot(t,x(:,1));
title('Problem #55');
xlabel('Time, sec.');
ylabel('Displacement, m');
hold on
%Analytical Solution
m=1000;
E=3.8e9;
A=0.03;
l=2;
k=E*A/l;
F=100;
w=sqrt(k/m);
to=0.25;
%for t<to
t=linspace(0,to);
x=F/k*(1-cos(w*t))+ F/(to*k)*(1/w*sin(w*t)-t);
plot(t,x,'*');
%for t>=to
t=linspace(to,0.5);
x=-F/k*cos(w*t)- F/(w*k*to)*(sin(w*(t-to))-sin(w*t));
plot(t,x,'*');
legend('Numerical', 'Analytical')
%Clay
%Vibrations
%Solutions
Copyright © 2015 Pearson Education Ltd.
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85
%M-file for Prob #3.62
function dx=prob(t,x);
[rows, cols]=size(x);dx=zeros(rows, cols);
m=1000;
E=3.8e9;
A=0.03;
l=2;
k=E*A/l;
F=100;
w=sqrt(k/m);
to=0.25;
if t<to
dx(1)=x(2);
dx(2)= - k/m*x(1) + F/m*(1-t/to);
else
dx(1)=x(2);
dx(2)= - k/m*x(1);
end
Copyright © 2015 Pearson Education Ltd.
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3.71
86
Numerically simulate the response of an undamped system to a ramp input of the
form F(t) = F 0 t, where F 0 is a constant. Plot the response for three periods for the
case m = 1 kg, k = 100 N/m, and F 0 = 50 N.
Solution The solution in Matlab is:
%Clay
%Vibrations
%Numerical Solutions
%Problem #56
clc
clear
close all
%Numerical Solution
x0=[0;0];
tspan=[0 2];
[t,x]=ode45('prob56a',tspan,x0);
figure(1)
plot(t,x(:,1));
title('Problem #56');
xlabel('Time, sec.');
ylabel('Displacement, m');
hold on
%Analytical Solution
t=linspace(0,2);
x=0.5*t-0.05*sin(10*t);
plot(t,x,'*');
legend('Numerical', 'Analytical')
%Clay
%Vibrations
%Solutions
%M-file for Prob #56
function dx=prob(t,x);
[rows, cols]=size(x);dx=zeros(rows, cols);
m=1;
k=100;
F=50;
dx(1)=x(2);
dx(2)= - k/m*x(1) + F/m*(t);
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Copyright © 2015 Pearson Education Ltd.
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3.72
88
Compute and plot the response of the system of following system using numerical
integration:
with initial conditions of x 0 = 0.01 m and v 0 = 1.0 m/s.
Solution The solution in Matlab is:
%Clay
%Vibrations
%Numerical Solutions
%Problem #57
clc
clear
close all
%Numerical Solution
x0=[0.01;1];
tspan=[0 5];
[t,x]=ode45('prob57a',tspan,x0);
figure(1)
plot(t,x(:,1));
title('Problem #57');
xlabel('Time, sec.');
ylabel('Displacement, m');
hold on
%Analytical Solution
m=10;
c=20;
k=1500;
w=sqrt(k/m);
d=c/(2*w*m);
wd=w*sqrt(1-d^2);
Y1=0.00419;
ph1=3.04;
wb1=25;
Y2=0.01238;
ph2=2.77;
wb2=15;
Y3=0.01369;
ph3=0.0268;
wb3=2;
A=0.1047;
phi=0.1465;
Copyright © 2015 Pearson Education Ltd.
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89
x=A.*exp(-d*w.*t).*sin(wd*t+phi)+ Y1.*sin(wb1*t-ph1) + Y2*sin(wb2*t-ph2)
+ Y3*sin(wb3*t-ph3);
plot(t,x,'*')
legend('Numerical', 'Analytical')
%Clay
%Vibrations
%Solutions
%M-file for Prob #57
function dx=prob(t,x);
[rows, cols]=size(x);dx=zeros(rows, cols);
m=10;
c=20;
k=1500;
dx(1)=x(2);
dx(2)= -c/m*x(2) - k/m*x(1) + 20/m*sin(25*t) + 10/m*sin(15*t) +
20/m*sin(2*t) ;
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Solutions Section 3.10 (3.73 through 3.79)
*3.73 Compute the response of the system in Figure 3.26 for the case that the damping
is linear viscous and the spring is a nonlinear soft spring of the form
and the system is subject to a excitation of the form (t 1 = 1.5 and t 2 = 1.6)
and initial conditions of x 0 = 0.01 m and v 0 = 1.0 m/s. The system has a mass of
100 kg, a damping coefficient of 30 kg/s and a linear stiffness coefficient of 2000
3
N/m. The value of k 1 is taken to be 300 N/m . Compute the solution and compare
it to the linear solution (k 1 = 0). Which system has the largest magnitude?
Compare your solution to that of Example 3.10.1.
Solution: The solution in Mathcad is
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Note that for this load the load, which is more like an impulse, the linear and nonlinear
responses are similar, whereas in Example 3.10.1 the applied load is a “wider” impulse
and the linear and nonlinear responses differ quite a bit.
*3.74 Compute the response of a spring mass system in for the case that the damping is
linear viscous and the spring is a nonlinear soft spring of the form
and the system is subject to a excitation of the form (t 1 = 1.5 and t 2 = 1.6)
and initial conditions of x 0 = 0.01 m and v 0 = 1.0 m/s. The system has a mass of
100 kg, a damping coefficient of 30 kg/s and a linear stiffness coefficient of 2000
N/m. The value of k 1 is taken to be 300 N/m3. Compute the solution and compare
it to the linear solution (k 1 = 0). How different are the linear and nonlinear
responses? Repeat this for t 2 = 2. What can you say regarding the effect of the
time length of the pulse?
Copyright © 2015 Pearson Education Ltd.
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Solution: The solution in Mathcad for the case t 2 = 1.6 is
Note in this case the linear response is almost the same as the nonlinear response.
Next changing the time of the pulse input to t 2 = 2 yields the following:
Copyright © 2015 Pearson Education Ltd.
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Note that as the step input last for a longer time, the response of the linear and the
nonlinear becomes much different.
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*3.75 Compute the response of a spring-mass-damper system for the case that the
damping is linear viscous and the spring stiffness is of the form
and the system is subject to a excitation of the form (t 1 = 1.5 and t 2 = 2.5)
initial conditions of x 0 = 0.01 m and v 0 = 1 m/s. The system has a mass of 100 kg,
a damping coefficient of 30 kg/s and a linear stiffness coefficient of 2000 N/m.
3
The value of k 1 is taken to be 450 N/m . Which system has the largest magnitude?
Solution: The solution is computed in Mathcad as follows:
Copyright © 2015 Pearson Education Ltd.
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Note that the linear response under predicts
the actual response
*3.76 Compute the response of a spring-mass-damper system for the case that the
damping is linear viscous and the spring stiffness is of the form
and the system is subject to a excitation of the form (t 1 = 1.5 and t 2 = 2.5)
initial conditions of x 0 = 0.01 m and v 0 = 1 m/s. The system has a mass of 100 kg,
a damping coefficient of 30 kg/s and a linear stiffness coefficient of 2000 N/m.
The value of k 1 is taken to be 450 N/m3. Which system has the largest magnitude?
Copyright © 2015 Pearson Education Ltd.
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Solution: The solution is calculated in Mathcad as follows:
In this case (compared to the
hardening spring of the previous
problem, the linear response over
predicts the time history.
*3.77 Compute the response of a spring-mass-damper system for the case that the
damping is linear viscous and the spring stiffness is of the form
and the system is subject to a excitation of the form (t 1 = 1.5 and t 2 = 2.5)
initial conditions of x 0 = 0.01 m and v 0 = 1 m/s. The system has a mass of 100 kg,
a damping coefficient of 30 kg/s and a linear stiffness coefficient of 2000 N/m.
The value of k 1 is taken to be 5500 N/m3. Which system has the largest magnitude
transient? Which has the largest magnitude in steady state?
Solution: The solution in Mathcad is given below. Note that the linear system response
is less than that of the nonlinear system, and hence underestimates the actual response.
Copyright © 2015 Pearson Education Ltd.
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*3.78 Compare the forced response of a system with velocity squared damping as
defined in equation (2.129) using numerical simulation of the nonlinear equation
to that of the response of the linear system obtained using equivalent viscous
damping as defined by equation (2.131). Use as initial conditions, x 0 = 0.01 m
and v 0 = 0.1 m/s with a mass of 10 kg, stiffness of 25 N/m, applied force of the
form (t 1 = 1.5 and t 2 = 2.5)
and drag coefficient of α = 25.
Solution: The solution calculated in Mathcad is given in the follow:
Copyright © 2015 Pearson Education Ltd.
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Note that the linear solution is very different from the nonlinear solution and dies out
more rapidly.
*3.79 Compare the forced response of a system with structural damping (see table 2.2)
using numerical simulation of the nonlinear equation to that of the response of the
linear system obtained using equivalent viscous damping as defined in Table 2.2.
Use the initial conditions, x 0 = 0.01 m and v 0 = 0.1 m/s with a mass of 10 kg,
stiffness of 25 N/m, applied force of the form (t 1 = 1.5 and t 2 = 2.5)
Copyright © 2015 Pearson Education Ltd.
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and solid damping coefficient of b = 8. Does the equivalent viscous damping
linearization, over estimate the response or under estimate it?
Solution: The solution is calculated in Mathcad as follows. Note that the linear solution
is an over estimate of the nonlinear response in this case.
Copyright © 2015 Pearson Education Ltd.
Problems and Solutions for Section 4.1 (4.1 through 4.19)
4.1
Consider the system of Figure P4.1. For c1 = c2 = c3 = 0, derive the equation of motion
and calculate the mass and stiffness matrices. Note that setting k3 = 0 in your solution
should result in the stiffness matrix given by Eq. (4.9).
k2
k1
k3
m2
m1
c2
c1
c3
x2
x1
Figure P4.1
Solution:
For mass 1:
(
)
Þ m 
x + (k + k ) x - k x
m1x1 = -k1x1 + k2 x2 - x1
1 1
1
2
1
2 2
=0
For mass 2:
(
m2 
x2 = -k3 x2 - k2 x2 - x1
(
)
)
Þ m2 
x2 - k2 x1 + k2 + k3 x2 = 0
So, M
x + Kx = 0
é m1
ê
êë 0
é k + k2
0ù
x+ê 1
ú 
m2 úû
êë -k2
-k2 ù
úx = 0
k2 + k3 úû
Thus:
ém
M=ê 1
êë 0
0ù
ú
m2 úû
é k + k2
K=ê 1
êë -k2
-k2 ù
ú
k2 + k3 úû
Copyright © 2015 Pearson Education Ltd.
4.2
Calculate the characteristic equation from problem 4.1 for the case
m1  8 kg m2  2 kg k1  24 N/m k2  3 N/m k3  3 N/m
and solve for the system's natural frequencies.
Solution: Given: m1 = 8 kg , m 2 = 2 kg , k 1 = 24 N / m , k 2 = 3 N / m , k 3 = 3 N / m
We know that
det  ω 2 M + K = 0


 ω m1 + k 1 + k 2  k 2
= 8ω 2 + 27  3
= 0


 
2
2

3
 2ω + 6
 ω m 2 + k 2 + k 3 
 k 2
  8ω 2 + 27  2ω 2 + 6  9 = 0
2


2

4
153 −102 ω +16 ω = 0
Solving for ω :
ω1 = 1.55384 rad / s
ω2 = 1.9901rad / s
4.3
Calculate the vectors u1 and u2 for problem 4.2.
Solution: To calculate the vectors u1 and u2
Solution:
Vector u1 is given by:
 8ω12 + 27  3
 u12  = 0

   
2
 3
 2ω1 + 6 u 22  0

2
 u12  = 0
  81.55384  + 27  3

   
0
2
 3

 21.55384  + 6 u 22 

 7.68465  3
 u11  = 0

   
1.17116 u 21  0
 3
 0.3906u 21 = u 11
 u1 = 0.3906 


1

Vector u 2 is given by:
Copyright © 2015 Pearson Education Ltd.
 8ω 22 + 27  3
 u12  = 0

   
2
 3
 2ω2 + 6 u 22  0

2
 u12  = 0
  81.9901 + 27  3

   
0 
2
 3

 21.9901 + 6 u 22 

  4.68398  3
 u12  = 0

   
 1.921 u 22  0
 3
 u12 = 0.6404u 22
 u 2 =  0.6404


1

4.4
For initial conditions x(0) = [1 0]T and x (0) = [0 0]T calculate the free response of the
system of Problem 4.2. Plot the response x1 and x2.
()
T
Solution: Given x(0) = [1 0]T, x 0 = éë 0 0 ùû , The solution is
x t = A1 sin w1t + f1 u1 + A2 sin w 2t + f2 u2
()
(
()
()
)
(
)
(
)
(
(
é x1 t ù é A1 sin w 1t + f1 - 0.101A2 sin w 2t + f2
ê
ú=ê
êë x2 t úû êë 0.909 A1 sin w1t + f1 + A2 sin w 2t + f 2
Using initial conditions,
(
)
1 = A1 sin f1 - 0.101A2 sin f2
0 = 0.909 A1 sin f1 + A2 sin f2
0 = 1.642 A1 cos f1 - 0.2536 A2 cos f2
0 = 6.033A1 cos f1 + 2.511A2 cos f2
) ùú
)úû
éë1ùû
éë 2 ùû
éë3ùû
éë 4 ùû
From [3] and [4],
f1 = f 2 = p / 2
From [1] and [2],
A1 = 0.916, and A2 = -0.833
So,
()
(
)
(
)
x ( t ) = 0.833sin (1.642t + p / 2 ) - 0.833sin ( 2.511t + p / 2 )
x1 t = 0.916 sin 1.642t + p / 2 + 0.0841sin 2.511t + p / 2
2
Copyright © 2015 Pearson Education Ltd.
()
x ( t ) = 0.833( cos1.642t - cos 2.511t )
x1 t = 0.916 cos1.642t + 0.0841cos 2.511t
2
4.5
Calculate the response of the system
x(t )  

 
 x(t )  0
8 0 
 24 2
0 1 
 2 2 




described in Example 4.1.7, to the initial condition x(0) = 0, x (0) = [1 0]T, plot the
response and compare the result to Figure 4.3.
Solution: x0 = 0 x 0 = 1 0
We know that
det  ω 2 M + K = 0
 ω 2 m1 + k 1 + k 2  k 2
= 8ω 2 + 24  2
= 0




2
2

2



ω
+
2

k
 ω m 2 + k 2 + k 3
 2
T





  8ω 2 + 24  ω 2 + 2  4 = 0
44  40 ω + 8ω = 0
2
4
Solving for ω :
ω1 = 1.2782rad / s , ω2 = 1.8346rad / s
Vector u1 is given by :
Copyright © 2015 Pearson Education Ltd.
 8ω12 + 24  2
 u12  = 0

   
2
 2
 ω1 + 2 u 22  0

2
 u12  = 0
  81.2782  + 24  2

   
0 
2
 2

 1.2782  + 2 u 22 

 10.9296  2
 u11  = 0

   
0.366205 u 21  0
 2
 0.1831u 21 = u 11
 u1 = 0.1831


1

u
Vector 2 is given by:
 8ω 22 + 24  2
 u12  = 0

   
2
 2
 ω 2 + 2 u 22  0

2
 u12  = 0
  81.8346  + 24  2

   
0 
2
 2

 1.8346  + 2 u 22 

  2.92606  2
 u12  = 0

   
 1.36576 u 22  0
 2
 u12 = 0.6835u 22  u 2 =  0.6835


1

But we know that
 x1 t   = u1 u 2  A1sinω1t + 1  




 x 2 t 
 A2 sinω2 t +  2 
0.1831  0.6835  A1sinω1t +  1   = 0.1831 A1sinω1t +  1   0.6835 A2 sinω2 t +  2 


 

1
1
  A2 sinω2 t +  2   A1sinω1t +  1 + A2 sinω2 t +  2 

At t = 0 we have
 x1 0  = 0.1831 A1sin1   0.6835A2 sin 2  = 0

 
  
 x 2 0  A1sin1 + A2 sin 2 
 0
And t = 0 for velocity we have x t 
 x1 0  = 0.1831A1 cos1 1.2782  0.6835A2 1.8346cos 2  = 1 

 
  
 x 2 0  A1 cos1 1.2782 + A2 1.8346cos 2 
 0
Copyright © 2015 Pearson Education Ltd.
Convert all of them to equations
0.1831A1sin1   0.6835A2 sin 2  = 0
A1sin1 + A2 sin 2  = 0
0.2340A1cos1   1.2539A2 cos2  = 1
1.2782A1cos1 + 1.8346A2 cos2  = 0
From the above equations we have
A 1= 0.9028
A 2= − 0.62902
1 = 0
2 = 0
Therefore
x1 t  = 0.18310.9028sin1.2782t  + 0.68350.62902sin1.8346t 
x2 t  = 0.9028sin1.2782t   0.62902sin1.8346t 
Plot: x1 t , x2 t 
Copyright © 2015 Pearson Education Ltd.
4.6
Write the equations of motion for the system of figure P4.1 for the case that k1 = k3 = 0
and identify the mass and stiffness matrix for this case.
Solution:
The equations of motion are
m1 
x1 + k2 x1 - k2 x2 = 0
m2 
x2 - k2 x1 + k2 x2 = 0
So, M
x + Kx = 0
é m 0 ù
é k
-k2
ê 1
ú x + ê 2
ê 0 m2 ú
ê -k2 k2
ë
û
ë
é m 0 ù
é
1
ú and K = ê
M=ê
ê 0 m2 ú
ê
ë
û
ë
4.7
ù
úx = 0
ú
û
k2
-k2
-k2 ù
ú
k2 ú
û
Calculate and solve the characteristic equation for the following system:
x(t )  10 

 
 x(t )  0
4
0


 1 1
0 1 
 1 1 




Solution:


We know that det  ω 2 M + K = 0
But
M = 4 0 ; K = 1
 1




 1 1 
0 1 
And  ω 2 M + K = ω 2 4 0 + 1
 1

 

0 1    1 1 
 5ω 2 + 4ω 4 = 0
 ω1 = 0rad / s ω2 = 1.11803 rad / s
Copyright © 2015 Pearson Education Ltd.
4.8
Compute the natural frequencies of the following system:
6 4
 3 1
x(t )  
 4 6  
 x(t )  0 .


 1 1 
Solution:
We know that
2
det (− ω M + K )= 0
But M = 6 4 ; K = 3
 1




 1 1 
4 6 
 1
And  ω 2 M + K = ω 2 6 4 + 3

 

4 6  1 1 
 2  32ω 2 + 20ω 4 = 0
 ω1 = 0.25525 rad / s
ω2 = 1.23889rad / s
4.9
Calculate the solution
é 1 ù
é 9 0 ù
é 27 -3 ù
ê
ú
x(t) + ê
ê
ú 
ú x(t) = 0, x(0) = ê 3 ú
ë 0 1 û
ë -3 3 û
êë 1 úû
x(0)
 =0
Compare the response that of Fig. 4.3.
()
()
T
Solution: Given: x 0 = éë1 / 3 1ùû , x 0 = 0
From Eq. (4.27) and example 4.1.7,
é1
é x1 t ù ê A1 sin
ê
ú = ê3
êë x2 t úû ê A sin
ë 1
Using initial conditions:
()
()
(
(
) 13 A sin ( 2t + f )ùúú
2t + f ) + A sin ( 2t + f ) ú
û
2t + f1 1
2
2
1 = A1 sin f1 - A2 sin f2
From [3] and [4]:
2
2
1 = A1 sin f1 + A2 sin f2
éë1ùû
éë 2 ùû
0 = 2 A1 cos f1 - 2 A2 cos f2
éë3ùû
0 = 2 A1 cos f1 + 2 A2 cos f2
éë 4 ùû
f1 = f2 =
p
2
Copyright © 2015 Pearson Education Ltd.
From [1] and [2]: A1 = 1, and A2 = 0
The solution is
()
x ( t ) = cos
1
x1 t = cos 2t
3
2t
In this problem, both masses oscillate at only one frequency. This happens because the
initial condition corresponds to the first mode shape and hence the second mode is not
excited.
2
Copyright © 2015 Pearson Education Ltd.
4.10
Calculate the solution to
é 1 ù
é 9 0 ù
é 27 -3 ù
ê - ú
x(t) + ê
 =0
ê
ú 
ú x(t) = 0, x(0) = ê 3 ú x(0)
0
1
-3
3
ë
û
ë
û
êë 1 úû
Compare the response that of Fig. 4.3 and if you worked problem 4.9, compare your
solution to that response too.
Solution:
()
Given: x(0) = [-1/3 1]T, x 0 = 0
From Eq. (4.27) and example 4.1.7,
(
)
é1
1
é x1 t ù ê A1 sin 2t + f1 - A2 2t + f2
3
ê
ú = ê3
êë x2 t úû ê A sin 2t + f + A sin 2t + f
1
2
2
ë 1
()
()
(
)
(
(
ù
)ú
ú
) úû
Using initial conditions:
-1 = A1 sin f1 - A2 sin f2
1 = A1 sin f1 + A2 sin f2
éë1ùû
éë 2 ùû
0 = 2 A1 cos f1 - 2 A2 cos f2
éë3ùû
0 = 2 A1 cos f1 + 2 A2 cos f2
éë 4 ùû
From [3] and [4]
f1 = f2 =
p
2
From [1] and [2]:
A1 = 0
A2 = 1
The solution is
()
()
1
x1 t = - cos 2t
3
x2 t = cos 2t
In this problem, both masses oscillate at only one frequency (not the same frequency as in
Problem 4.9, though.)
Copyright © 2015 Pearson Education Ltd.
4.11
Compute the natural frequencies and mode shapes of the following system
é 4 0 ù
é 4 -2 ù
x(t) + 10 ê
ê
ú 
ú x(t) = 0
ë 0 1 û
ë -2 1 û
4.12
Solution: The eigenvalues are 0 and 2 and the natural frequencies are 0 and 1.414 rad/s
The mode shapes are
é 1 ù é -1 ù
ê
ú, ê
ú
1
ë
û ë 1 û
Determine the equation of motion in matrix form, then calculate the natural frequencies
and mode shapes of the torsional system of Figure P4.12. Assume that the torsional
(
)
stiffness values provided by the shaft are equal k1 = k2 and that disk 1 has four times
the inertia as that of disk 2( J1 = 3J 2 ) .
Solution: Given:
k 1 = k 2 = k (Assumed constant k)
J 1 = 3J 2
 J 2 3 0θ + k 2  1θ = 0




 1 1 
0 1 
Now to calculate their natural frequencies
2
det (− ω J + K )= 0
But
J = J 2 3 0 ; K = k 2
 1 And  ω 2 M + K = ω 2 J 2 3 0 + k 2  1





 

 1 1 
0 1 
0 1   1 1 
k
k
rad / s ω2 = 1.19760
rad / s
 k 2  5 J 2 kω 2 + 3J 22 ω 4  ω1 = 0.4820
J2
J2
Calculate the mode Shapes
Mode shape 1:
 30.2324k + 2k  k
 u11  = 0

 
 0.2324k + k  u12 
 k
u11 = 0.7676u12  u1 = 0.7676 


1

Mode shape 2:
 31.434k + 2k  k
 u11  = 0

 
 1.434k + k  u12 
 k
Copyright © 2015 Pearson Education Ltd.
u11 = 0.434u12  u 2 =  0.434


1

4.13
Two subway cars of Fig. P4.13 have 2100 kg mass each and are connected by a coupler.
The coupler can be modeled as a spring of stiffness k = 270,000 N/m. Write the equation
of motion and calculate the natural frequencies and (normalized) mode shapes.
Solution: Given:
m1 = m 2 = m = 2100 kg , k = 270000 N / m  m 0  x + k

0
 270000 x = 0
2100 0  x + 270000




2100
0
 270000 270000 

m
 kx = 0


 k k 
Now to calculate their natural frequencies
det  ω 2 M + K = 0
But
det  ω 2 M + K = 2100ω 2 + 270000  270000
= 0


2
 270000
 2100ω + 270000
2
4
 1134000000ω + 4410000ω = 0




 ω1 = 0 rad / s
ω 2 = 16.035 rad / s
Calculate the mode Shapes
Mode shape 1:
 270000 u11  = 0 u11 = u12  u1 = 1
270000

 

1
 270000 270000  u12 
Mode shape 2:
 269618  270000 u11  = 0  0.9985 u 11 = u 12  u 2 =  0.9985

 


1

 270000  269618 u12 
Normalizing the mode shapes we have
1
u1 =
1
2
1
And
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

1
u1 =  0.9985

2

 1+ 0.9985




1


2 
 1+ 0.9985 
4.14

 = 0.70657 


 
 0.70763 




Suppose that the subway cars of Problem 4.13 are given the initial position of x10 = 0, x20
= 0.05 m and initial velocities of v10 = v20 = 0. Calculate the response of the cars.
Solution:
Given:
T
x (0 )= [0 0.05 ] x 0 = 0
From the previous problem we have
u1 = 1 u 2 =  0.9985



1
1

ω1 = 0 rad / s
ω2 = 16.035rad / s
The solution is
xt  = c1 + c 2 t u1 + Asin16.035t +  u 2
xt  = c 2 u1 + A 16.035cos u 2
x0 = c1 u1 + Asin u 2
Using initial conditions we have
c1 + Asin  = 0
c1  Asin  = 0.05
c 2 +16.035Acos  = 0
c 2  16.035Acos  = 0 From the last two equations we have
c 2 = 0 and  =
π
rad
2
And from the two before we have c1 = 0.025m and A = 0.025m
The solution is
x1 t  = 0.025  0.025cos16.035t  x 2 t  = 0.025+ 0.025cos16.035t 
Copyright © 2015 Pearson Education Ltd.
4.15
A slightly more sophisticated model of a vehicle suspension system is given in Figure
P4.15. Write the equations of motion in matrix form. Calculate the natural frequencies
for k1 =103 N/m, k2 = 104 N/m, m2 = 60 kg, and m1 = 2400 kg.
Solution: Given:
The equations of motion are
2400 x 1 + 1000 x1  1000 x 2 = 0
60 x 2  1000 x1 + 11000 x 2 = 0
 1000 x = 0
In matrix form this becomes: 2400 0  x + 1000




60
0
 1000 11000 
But Natural frequencies are given by
2
det (− ω M + K )= 0
 2400ω 2 +1000  1000
= 10000000  26460000ω 2 +144000ω 4 = 0


2
 1000
 60ω +11000
 ω1 = 0.61539 rad / s
ω 2 = 13.5414 rad / s
4.16
Examine the effect of the initial condition of the system of Figure 4.1(a) on the responses
x1 and x2 by repeating the solution of Example 4.1.7 given by
é1
ù
1
é x1 t ù ê A1 sin 2t + f1 - A2 sin 2t + f2 ú
3
ê
ú = ê3
ú
êë x2 t úû ê A sin 2t + f + A sin 2t + f ú
1
2
2
ë 1
û
(
(
()
()
)
)
(
(
)
)
first for x10 = 0,x20 = 1 with x10 = x20 = 0 and then for x10 = x20 = x10 = 0 and x20 = 1. Plot
the time response in each case and compare your results against Figure 4.3.
Solution: From Eq. (4.27) and example 4.1.7,
é1
1
é x1 t ù ê A1 sin 2t + f1 - A2 sin 2t + f2
3
ê
ú = ê3
x
t
êë 2 úû ê A sin 2t + f + A sin 2t + f
1
2
2
ë 1
(
(
()
()
()
T
)
)
(
(
()
)
ù
)ú
ú
ú
û
(a) x 0 = éë0 1ùû , x 0 = 0 . Using the initial conditions:
0 = A1 sin f1 - A2 sin f 2
1 = A1 sin f1 + A2 sin f2
éë1ùû
éë 2 ùû
0 = 2 A1 cos f1 - 2 A2 cos f2
éë3ùû
0 = 2 A1 cos f1 + 2 A2 cos f2
éë 4 ùû
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From [3] and [4]
f1 = f 2 =
p
2
From [1] and [2]
A1 = A2 =
1
2
The solution is
()
1
1
x1 t = cos 2t - cos 2t
6
6
1
1
x2 t = cos 2t + cos 2t
2
2
This is similar to the response of Fig. 4.3
()
()
()
T
(b) x 0 = 0, x 0 = éë0 1ùû . Using these initial conditions:
éë1ùû
0 = A1 sin f1 - A2 sin f 2
0 = A1 sin f1 + A2 sin f 2
éë 2 ùû
0 = 2 A1 cos f1 - 2 A2 cos f2
éë3ùû
1 = 2 A1 cos f1 + 2 A2 cos f2
éë 4 ùû
From [1] and [2] f1 = f2 = 0
From [3] and [4] A1 =
2
1
, and A2 =
4
4
The solution is
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x1 t =
()
2
1
sin 2t - sin 2t
12
12
()
2
1
sin 2t + sin 2t
4
4
x2 t =
This is also similar to the response of Fig. 4.3
4.17
Consider the system defined by
é 24 + k
é 9 0 ù
2
x+ê
ê
ú 
ê -k2
ë 0 1 û
ë
-k2 ù
úx = 0
k2 ú
û
Using the initial conditions x1(0) = 1 mm, x2(0) = 0, and x1(0) = x2 (0) = 0, resolve and
plot x1(t) for the cases that k2 takes on the values 0.3, 3, 30, and 300. In each case
compare the plots of x1 and x2. What can you conclude?
Solution: Let k2 = 0.3, 30, 300 for the example(s) in Section 4.1. Given
()
T
()
x 0 = éë1 0 ùû m
mm, x 0 = éë 0 0 ùû
m1 = 9, m2 = 1, k1 = 24
Equation of motion becomes:
T
é 24 + k
é 9 0 ù
2
ê
x

+
ê
ú
ê -k2
ë 0 1 û
ë
-k2 ù
úx = 0
k2 ú
û
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(a) k2 = 0.3
(
)
det -w M + K =
2
-9w 2 + 24.3
-0.3
-0.3
-w + 0.3
2
= 9w 4 - 27w 2 + 7.2 = 0
w 2 = 0.2598,2.7042
w 1 = 0.5439
w 2 = 1.6444
Mode shapes:
Mode 1, w 12 = 0.2958
é 21.6374
-0.3 ù é u11 ù é0 ù
ê
úê ú = ê ú
-0.3
0.004159
ë
û êë u12 úû ë0 û
21.6374u11 - 0.3u12 = 0
u11 = 0.01386u12
é0.01386 ù
u1 = ê
ú
ë 1 û
Mode 2, w 22 = 2.7042
é -0.03744 -0.3 ù é u21 ù é 0 ù
ê
úê ú = ê ú
2.4042 û êëu22 úû ë 0 û
ë -0.3
-0.3u21 = 2.4042u22
u22 = -0.1248u 21
é 1 ù
u2 = ê
ú
ë -0.1248 û
x t = A1 sin w 1t + f1 u1 + A2 sin w 2t + f2 u2
()
The solution is
(
)
(
)
Using initial conditions
(
)
1 = A1 0.01386 sin f1 + A2 sin f2
(
)
0 = A ( 0.01386 ) ( 0.5439 ) cos f + A (1.6444 ) cos f
0 = A ( 0.5439 ) cos f + A (1.6444 ) ( -0.1248 ) cos f
éë1ùû
éë 2 ùû
0 = A1 sin f1 + A2 -0.1248 sin f2
1
1
1
1
2
2
From [3] and [4],
f1 = f 2 = p / 2
From [1] and [2],
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2
2
éë3ùû
éë 4 ùû
A1 = 0.1246
A2 = 0.9983
So,
()
(
)
x ( t ) = 0.1246 éëcos ( 0.5439t ) - cos (1.6444t ) ùû mm
x1 t = 0.001727 cos(0.5439t) + 0.9983cos 1.6444t mm
2
(b) k2 = 30
(
)
det -w 2 M + K =
-9w 2 + 54
-30
-30
-w + 30
2
= 9w 4 - 32w 2 + 720 = 0
w 2 = 2.3795,33.6205
33.6205
w 1 = 1.5426
w 2 = 5.7983
Mode shapes:
Mode 1, w 12 = 2.3795
é32.5845
-30 ù é u11 ù é0 ù
ê
úê ú = ê ú
27.6205 û êëu12 úû ë0 û
ë -30
30u11 = 27.6205u12
u11 = 0.9207u12
é0.9207 ù
u1 = ê
ú
ë 1 û
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Mode 2, w 22 = 33.6205
é -248.5845
-30 ù é u21 ù é0 ù
ê
úê ú = ê ú
-3.6205 û êëu22 úû ë0 û
ë -30
30u21 = -3 / 6205u22
u21 = -0.1207u22
é -0.1207 ù
u2 = ê
ú
ë 1 û
The solution is
()
(
)
(
)
x t = A1 sin w 1t + f1 u1 + A2 sin w 2t + f2 u2
Using initial conditions,
(
)
1 = A1 0.9207 sin f1 + A2 (-0.1207 sin f2
0 = A1 sin f1 + A2 sin f2
(
)(
)
(
0 = A (1.5426 ) cos f + A ( 5.7983) cos f
)(
)
0 = A1 0.9207 1.5426 cos f1 + A2 -0.1207 5.7983 cos f2
1
1
2
2
From [3] and [4]
f1 = f 2 = p / 2
From [1] and [2]
A1 = 0.9602
A2 = -0.9602
So,
()
(
)
(
)
x ( t ) = 0.9602 éë cos (1.5426t ) - cos (5.7983t ) ùû mm
x1 t = 0.8841cos 1.5426t + 0.1159 cos 5.7983t mm
2
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éë1ùû
éë 2 ùû
éë3ùû
éë 4 ùû
(c) k2 = 300
(
)
det -w M + K =
2
-9w 2 + 324
-300
-300
-w + 300
2
= 9w 4 - 3024w 2 + 7200 = 0
w 2 = 2.3981,333.6019
333.6019
w 1 = 1.5486
w 2 = 18.2648
Mode shapes:
Mode 1, w 12 = 2.3981
é302.4174
-300 ù é u11 ù é 0 ù
ê
úê ú = ê ú
297.6019 û êëu12 úû ë 0 û
ë -300
302.4174 11 = 300u12
302.4174u
u11 = 0.9920u12
é0.9920 ù
u1 = ê
ú
ë 1 û
Mode 2, w 22 = 333.6019
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é -2678.4174
-300 ù é u21 ù é0 ù
ê
úê ú = ê ú
-33.6019
33.6019 û êëu22 úû ë0 û
ë -300
300u21 = 33.6019u22
u21 = -0.1120u22
é -0.1120 ù
u2 = ê
ú
ë 1 û
The solution is
()
(
)
(
)
x t = A1 sin w 1t + f1 u1 + A2 sin w 2t + f2 u2
Using initial conditions
1 = A1 0.9920 sin f1 + A2 -0.1120 sin f2
(
)
(
)
0 = A1 sin f1 + A2 sin f2
(
)(
)
(
)(
0 = A (1.5486 ) cos f + A (18.2648 ) cos f
0 = A1 0.9920 1.5486 ccos f1 + A2 -0.1120 18.2648
1
1
2
2
)
éë1ùû
éë 2 ùû
éë3ùû
éë 4 ùû
From [3] and [4]
f1 = f2 = p / 2
From [1] and [2],
A1 = 0.9058 and A2 = -0.9058.
So,
x1 t = 0.8986
0.8986cos 1.5486t + 0.1014 cos 18.2648t mm
()
(
)
(
)
x ( t ) = 0.9058 éëcos (1.5486t ) - cos (18.2648t ) ùû mm
2
The case k2 = 3 is Figure 4.3. Conclusion:As the value of k2 increases the effect on mass 1
is small, but mass 2 oscillates similar to mass 1 with a superimposed higher frequency
oscillation.
Copyright © 2015 Pearson Education Ltd.
4.18
Consider the system defined by
é m
0 ù é x1 ù é k1 + k2 -k2 ù é x1 ù é0 ù
ê 1
ú ê ú+ ê
úê ú=ê ú
ê 0 m2 ú êë x2 úû ê -k2
k2 ú êë x2 úû ë0 û
ë
û
ë
û
Determine the natural frequencies in terms of the parameters m1, m2, k1 and k2. How do
these compare to the two single-degree-of-freedom frequencies w 1 = k1 / m1 and
w 2 = k2 / m2 ?
Solution:
The equation of motion is
M
x + Kx = 0
é m1 0 ù
é k1 + k2
ê
ú x + ê
êë 0 m2 ûú
êë -k2
-k2 ù
úx = 0
k2 úû
The characteristic equation is found from Eq. (4.19):
(
)
det -w 2 M + K = 0
-m1w 2 + k1 + k2
-k2
-k2
-m2w 2 + k2
(
(
))
m1m2w 4 - k1m2 + k2 m1 + m2 w 2 + k1k2 = 0
2
w 1,2
(
)
(
)
(
)
(
)
2
k1m2 + k2 m1 + m2 ± éë k1m2 + k2 m1 + m2 ùû - 4m1m2 k1k2
=
2m1m2
So,
w 1,2 =
2
k1m2 + k2 m1 + m2 ± éë k1m2 + k2 m1 + m2 ùû - 4m1m2 k1k2
2m1m2
In two-degree-of-freedom systems, each natural frequency depends on all four
parameters (m1, m2, k1, k2), while a single-degree-of-freedom system's natural frequency
depends only on one mass and one stiffness.
Copyright © 2015 Pearson Education Ltd.
4.19
Consider the problem of Example 4.1.7, which the first degree-of-freedom response
given by x1 (t )  0.6(cos 2t  cos 2t ). Use a trig identity to show the x1(t) experiences a
beat. Plot the response to show the beat phenomena in the response.
Solution Given: Applying the Trigonometric identity
 a +b   a  b 
cosa + cosb  = 2cos
cos
 to the solution x1 t  = 0.6 cos 2t + cos2t 
 2   2 
We have
 2+2   2 2 
x1 t  = 1.2cos 
t cos
t  = 1.2cos 1.7071t cos0.2929t 
 2
  2

Plot:
  
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
Solutions for Section 4.2 (4.20 through 4.35)
4.20
Calculate the square root of the matrix
 25 −32 
M =

 −32 41 
Solution: Given:
1
Let M 2 = a

− b
2
Then M = a

− b
− b

c 
− ab − bc  = 25
− b  = a 2 + b 2
− 32
− b  a
 

 

c  − b c  − ab − bc b 2 + c 2  − 32 41 
a 2 + b 2 = 25
ab + bc = 32
b 2 + c 2 = 41
On solving these equations we arrive a number of solutions of which the solution
a = 3 , b = 4 , c = 5 ensures the solution is positive definite.
4.21
Normalize the vectors
first with respect to unity (i.e.,
(i.e.,
) and then again with respect to the matrix M
), where
Solution:
(a) Normalize the vectors
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Normalized:
Normalized:
Normalized:
(b) Mass normalize the vectors
Mass normalized:
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Mass normalized:
4.22
Consider the vibrating system described by
Compute the mass normalized stiffness matrix, the eigenvalues, the normalized
eigenvectors, the matrix P and show that PTMP =I and PTKP is the diagonal matrix of
eigenvalues Λ.
Solution The square root of M is just the roots of the diagonal elements so
Next compute the eigenvectors and normalize them:
Copyright © 2015 Pearson Education Ltd.
Next show that it works:
4.23
Calculate the matrix
for the system defined by
and see that it is symmetric.
Solution:
Since
is symmetric.
Using the numbers given in problem 4.2 yields
This is obviously symmetric.
Copyright © 2015 Pearson Education Ltd.
4.24
Consider the vibrating system described by
Compute the mass normalized stiffness matrix, the eigenvalues, the normalized
eigenvectors, the matrix P and show that PTMP =I and PTKP is the diagonal matrix of
eigenvalues Λ.
Solution:
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4.25
Discuss the relationship or difference between a mode shape of equation (4.54) and an
eigenvector of .
Solution:
The relationship between a mode shape, u, of
is given by
and an eigenvector, v, of
If v is normalized, then u is mass normalized.
This is shown by the relation
4.26
Calculate the units of the elements of matrix
.
Solution:
M-1/2 has units kg-1/2
K has units N/m = kg/s2
So,
has units
Copyright © 2015 Pearson Education Ltd.
4.27
Calculate the spectral matrix Λ and the modal matrix P for the vehicle model of Problem
4.15 described by
Solution: The equation of motion is:
Calculate eigenvalues:
The spectral matrix is
Calculate eigenvectors and normalize them:
Copyright © 2015 Pearson Education Ltd.
The modal matrix is
4.28
Calculate the spectral matrix Λ and the modal matrix P for the system given by
Solution: The equation of motion is
Calculate eigenvalues:
The spectral matrix is
Calculate eigenvectors and normalize them:
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The modal matrix is
Copyright © 2015 Pearson Education Ltd.
4.29
Calculate
for the torsional vibration problem given by
What are the units of
?
Solution: The equation of motion is,
The units of
4.30
are
Consider the system in the Figure P4.30 for the case where m 1 = 1 kg, m 2 = 9 kg, k 1 = 240
N/m and k 2 =300 N/m. Write the equations of motion in vector form and compute each of the
following
a) the natural frequencies
b) the mode shapes
c) the eigenvalues
d) the eigenvectors
e) show that the mode shapes are not orthogonal
f) show that the eigenvectors are orthogonal
g) show that the mode shapes and eigenvectors are related by
h) write the equations of motion in modal coordinates
Note the purpose of this problem is to help you see the difference between these various
quantities.
Copyright © 2015 Pearson Education Ltd.
Solution Given: m1 = 1kg , m 2 = 9kg , k1 = 240 N / m , k 2 = 300 N / m
From the FBD we have
− 300 x = 0
1 0 x + 540
 

 

0 
0 9 − 300 300 
a) We know that
(
)
det − ω 2 M + K = 0


2

⇒ − ω 1 0 + 540
− 300= 0


 

0 9 − 300 300 

⇒ 72000 − 5160ω 2 + 9ω 4 = 0 ⇒ ω1 = 3.7829rad / s and ω2 = 23.643rad / s
b) To calculate the vectors
and
Vector
is given by:
2
− ω1 + 540 − 300
 u11  = 0

   
− 300
− 9ω12 + 300 u 21  0

2
 u11  = 0
⇒ − (3.7829) + 540 − 300

   
0 
2
− 300

− 9(3.7829) + 300 u 21 

⇒ 525.69 − 300  u11  = 0 ⇒ 0.57068u 21 = u 11 ⇒ u1 = 0.57068



   
1

− 300 171.207 u 21  0
Vector u 2 is given by :
− ω 22 + 540 − 300
 u12  = 0 ⇒ − 18.9914 − 300
 u12  = 0

   
   

− 4730.92 u 22  0
− 300
2
− 300
 u 22  0
−
9
300
ω
+
2


⇒ u 22 = −0.063313u12 ⇒ u 2 = 1



− 0.063313
4.31
Consider the following system:
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where M is in kg and K is in N/m. (a) Calculate the eigenvalues of the system. (b)
Calculate the eigenvectors and normalize them.
Solution: Given:
Calculate eigenvalues:
The spectral matrix is
Calculate eigenvectors and normalize them:
Copyright © 2015 Pearson Education Ltd.
4.32
The torsional vibration of the wing of an airplane is modeled in Figure P4.32. Write the
equation of motion in matrix form and calculate analytical forms of the natural
frequencies in terms of the rotational inertia and stiffness of the wing.
Figure P4.32 A crude model of the torsional vibration of a wing consisting of a two-shaft, two
disk system similar to Problem 4.12 used to estimate the torsional natural frequencies of the wing
where the engine inertias are approximated by the disks.
Solution: Recall from Chapter 1 that
Equation of motion:
Natural frequencies:
Solving for λ yields
Copyright © 2015 Pearson Education Ltd.
The natural frequencies are
4.33
Calculate the value of the scalar a such that x 1 = [a -2 2]T and x 2 = [2 0 2]T are
orthogonal.
Solution: For the column vectors
and
to satisfy orthogonality
T
x1 x 2 = 0
⇒ [a − 2 2]2 = 0 ⇒ 2a + 4 = 0 ⇒ a = −2 ⇒ x1 = − 2
 
 
0 
− 2
2 
2
4.34
Normalize the vectors of Problem 4.33. Are they still orthogonal?
Solution: Upon normalizing the vectors x1 and x2 we have
1
1
Normalized x1 = X 1 =
− 2 and normalized x 2 = X 2 =
 2
12  
8 
− 2
0 
2 
2
Now to check if X 1T X 2 = 0 ⇒
1
12 × 8
[− 2 − 2 2]2 = 0
 
0 
2
Thus they are still orthogonal.
Copyright © 2015 Pearson Education Ltd.
4.35
Which of the following vectors are normal? Orthogonal?
 1 
 2
 
=
x1  =
0  x2
 1 
 
 2 
 0.1
0.2  x
=
 3
 0.3
0.4 
 
 0.3
0.4 
Solution:
Given:
2
2
 1 
 1 
 1 
x1 =   x 2 = 0.1 x3 = 0.4 | x1| = 
 +0+
 = 1 (Thus it is normal)
2
2
2






 
0.2
0.3
0 


 
0.3
0.4
 
 1 
 2 

| x 2 | = (0.1)2 + (0.2)2 + (0.3)2 = 0.3742 (Thus it is not normal)
| x3 | = (0.4)2 + (0.3)2 + (0.4)2 = 0.64031 (Thus it is not normal)
To check orthogonality:
T
T
T  1 
 1 
x1T x 2 =   0.1 ≠ 0 x 2T x 3 = 0.1 0.4 ≠ 0 x 3T x1 = 0.4   ≠ 0
   
   2
 2  0.2
0.2 0.3

0.3  


0 





0.4 0 

  0.3
0.3 0.4
 
 
 1 
 1 
 2 
 2 

Thus neither of them are orthogonal
Copyright © 2015 Pearson Education Ltd.
Problems and Solutions for Section 4.3 (4.36 through 4.46)
4.36
Decouple the following equation of motion into two decoupled equations of motion:
x(t )  

 
 x(t )  0
3
0


 4 2 
0 1 
 2 2 




Solution: The given equations are:
 2 xt  = 0
3 0 xt  + 4




 2 2 
0 1 
To decouple the solution first multiply the inverse of the matrix 3 0


0 1 
which results in
xt  + 1.333  0.6667 xt  = 0


2
 2

Diagonalizing the matirx B = 1.333  0.6667 using octave code we get


2
 2

[evec,eval] = eig(B)
evec =
-0.60889
-0.79325
0.39832
-0.91725
eval =
Diagonal Matrix
0.46482
0
0
2.86852
Here the decoupled equations are
ẍ 1( t)+0.46482 x 1 (t )= 0
ẍ 2( t)+2.86852 x 2 (t )= 0
4.37
Solve the system of Problem 4.12 given by
Copyright © 2015 Pearson Education Ltd.
é 3 0 ù 
é 2 -1ù
J2 ê
úq + k ê
úq = 0
ë0 1 û
ë -1 1 û
Using modal analysis for the case where the rods have equal stiffness (i.e.,
T
k1 = k2 ), J1 = 3J 2 , and the initial conditions are x(0) = éë 0 1 ùû and x (0) = 0.
Solution: From Problem 4.12 and Figure P4.12, with k = k1 = k2 and J1 = 3J 2 :
é 3 0 ù 
é 2 -1ù
J2 ê
úq + k ê
úq = 0
ë0 1 û
ë -1 1 û
Calculate eigenvalues and eigenvectors:
é 1
ù
0ú
ê
J
=J
ê 3
ú
êë 0 1 úû
é 2
ê
k ê 3
-1/ 2
-1/ 2

K = J KJ
=
J 2 ê -1
ê
ë 3
-1/ 2
l1 =
-1/ 2
2
(5 - 13) k Þ
6J 2
-1 ù
ú
5k
k2
3ú
2

Þ det K - l I = l l+ 2 =0
ú
3J 2
3J 2
1 ú
û
(
w 1 = l1 , and
(
)
(5 + 13) k
6J 2
Þ w 2 = l2
)
é 5 + 13 k
ù
-k
ê
ú
ê
6J
5 - 13 k
3J 2 ú é v11 ù
2
ê
úê ú = 0
l1 =
Þ
ê
ú
6J 2
5 + 13 k ú êë v12 úû
-k
ê
ê
ú
6J 2
3J 2
êë
úû
é0.7992 ù
Þ v11 = 1.3205v12 Þ v1 = ê
ú
ë 0.6011 û
(
)
(
)
Copyright © 2015 Pearson Education Ltd.
(
)
é -1 - 13 k
ù
-k
ê
ú
ê
6J 2
5 + 13 k
3J 2 ú é v211 ù
ê
úê
l2 =
Þ
=0
ê
ú ê v úú
6J 2
1 - 13 k ú ë 22 û
-k
ê
ê
ú
6J 2
3J 2
êë
úû
é -0.6011ù
Þ v21 = -0.7522v22 Þ v 2 = ê
ú
ë 0.7992 û
(
)
(
)
é 0.7992 -0.6011ù
v 2 ùû = ê
ú
ë 0.6011 0.7992 û
Calculate S and S-1:
1 é0.4614 -0.3470 ù
S = J -1/ 2 P =
ê
ú
J 2 ë 0.6011 0.7992 û
Now, P = éë v1
é 1.3842 0.6011 ù
S -1 = PT J 1/ 2 = J 21/ 2 ê
ú
ë -1.0411 0.7992 û
Modal initial conditions:
é0 ù
é 0.6011 ù
r 0 = S -1q 0 = S -1 ê ú = J 21/ 2 ê
ú
ë1û
ë 0.7992 û
()
()
()
()
()
w 12 r102 + r102
()
w 22 r202 + r202
r 0 = S -1q 0 = 0
Modal solution:
r1 t =
r2 t =
w1
w2
é
wr ù
sin êw 1t + tan -1 1 10 ú
r10 û
ë
é
wr ù
sin êw 2t + tan -1 2 10 ú
r20 û
ë
é
pù
r1 t = 0.6011J 21/ 2 sin êw1t + ú = 0.6011J 21/ 2 cos w 1t
2û
ë
()
é
pù
r2 t = 0.7992 J 21/ 2 sin êw 2t + ú = 0.6011J 21/ 2 cos w 2t
2û
ë
é 0.6011J 21/ 2 cos w 1t ù
r t =ê
ú
1/ 2
êë0.7992J 2 cos w 2t úû
()
()
Copyright © 2015 Pearson Education Ltd.
Convert to physical coordinates:
q ( t ) = Sr ( t ) = J
1/ 2
2
1/ 2
é0.4614 -0.3470 ù é 0.6011J 2 cos w1t ù
ú
ê
úê
1/ 2
ë 0.6011 0.7992 û êë 0.7992 J 2 cos w 2t úû
é 0.2774 cos w 1t - 0.2774 cos w 2t ù
q (t ) = ê
ú
êë 0.3613cos w 1t + 0.6387 cos w 2t úû
where w 1 = 0.4821
k
k
and w 2 = 1.1976
.
J2
J2
Copyright © 2015 Pearson Education Ltd.
4.38
Consider the system
é 9 0 ù
é 27 -3 ù
x(t) + ê
ê
ú 
ú x(t) = 0
0
1
-3
3
ë
û
ë
û
()
of Example 4.3.1. Calculate a value of x(0) and x 0 such that both masses of the
system oscillate with a single frequency of 4 rad/s.
Solution:
From example 4.3.1
1
1
1
S=
1 / 3 1 / 3 and S =
3 1 
2 
2 


1
1 
3  1
From equations 4.67 and 4.68
r1 t  =
r2 t  =
ω r
2 2
1 10
+ r102
ω1
ω r
2 2
2 20
+ r202
ω2
 sin ω t + tan


1
 sin ω t + tan


2
1
 ω1 r10

 r10

 


 ω2 r20

 r20

 


1
Choose x0 , x 0 so that r1 t  = 0 , this will cause the frequency 2 to drop out.
For r1 t  = 0 its coefficient must be zero
ω r
2 2
1 10
ω1
+ r102
=0
choose r1 t  = 0, r10 = 0
ω12 r102 + r102 = 0
3
r = 0 as calculated in the example 4.3.1
Let r20 =
2 20
T
 3 
So r 0 = 0
 , r0 = 0
 2 
Solve for X 0 and x 0 
 0 
1 1  
 =  0.5 
3
x0  = sr 0  =
12  3 3   2   1.5


1  1
x 0  = S {r0  = 0
1
Copyright © 2015 Pearson Education Ltd.
4.39
Consider the system of Figure P4.39 consisting of two pendulums coupled by a spring.
Determine the natural frequency and mode shapes. Plot the mode shapes as well as the
solution to an initial condition consisting of the first mode shape for k = 20 N/m, l = 0.5
m and m1 = m2 = 10 kg, a = 0.2 m along the pendulum
Solution: Given:
k = 20N / m , m1 = m2 = 10kg , a = 0.2m l = 0.5m . Also I = ml 2 and g = 9.81m / s .
The equations of motion are as follows:
I 1 θ́1= k a2 (θ 2 − θ1 )− m1 glθ 1
I 2 θ́2= − k a2 (θ2 −θ1 )−m2 gl θ2
 θ1  = 0
 ml 2 θ1  + mgl + ka 2  ka 2
   
  
2
2

mgl + ka  θ2  0
θ2   ka
Substitution of the given values yields:
2
2
θ
2.5 0 θ + 10  9.8  0.5 + 20  0.2   20  0.2 




2.5
0
2
2

10  9.8  0.5 + 20  0.2  
 20  0.2 
= 2.5 0 θ + 49.8  0.8θ = 0




2.5
0
 0.8 49.8 
1. To find the natural frequencies


We solve the equation det  ω 2 J + K = 0


2

  ω 2.5 0  + 49.8  0.8= 0


 

2.5  0.8 49.8 
0

 2479.4  249ω 2 + 6.25ω 4 = 0
 ω1 = 4.42719 rad / s
ω 2 = 4.49889 rad / s
2. Mode shapes:
From the above analysis we have λ1= 19.6 and λ2= 20.24
Eigenvectors:
λ1 = 19.6
1
 0.8 v11  = 0  v1 =
0.8
1
2

   
1
 0.8 0.8  v12  0
λ2= 20.24
Copyright © 2015 Pearson Education Ltd.
1
 0.8  0.8 v11  = 0  v 2 =
1 
2 

   
 1
 0.8  0.8 v12  0
Mode Shapes:
1
u1 = M 2 v1 = 0.4472


0.4472
1
u 2 = M 2 v 2 = 0.4472 


 0.4472
4.40
Compute and plot the response of
é 1 0 ù
é 12 -2 ù
x(t) + ê
ê
ú 
ú x(t) = 0
ë 0 10 û
ë -2 12 û
subject to x(0) = [1 1]T and x(0)

= 0 . Compare your result to Example 4.3.2 and
Figure 4.6.
Solution: From examples 4.3.2 and 4.2.5, with m2 = 10 kg,
é1 0 ù
é12 -2 ù
M
x + Kx = ê
x+ê
ú 
úx = 0
0
10
-2
12
ë
û
ë
û
Calculate eigenvalues and eigenvectors:
Copyright © 2015 Pearson Education Ltd.
M
-1/ 2
é1
ê
=ê
0
êë
0 ù
ú
1 ú
10 úû
é 12
-0.6325 ù
K = M -1/ 2 KM -1/ 2 = ê
ú
1.2 û
ë -0.6325
det( K - l I ) = l 2 - 13.2l + 14 = 0
l1 = 1.163 w1 = 1.078 rad/s
l2 = 12.04 w 2 = 3.469 rad/s
é0.0583 -0.9983ù
v 2 ùû = ê
ú
ë0.9983 0.0583 û
Calculate S and S-1:
é 0.0583 -0.9983ù
S = M -1/ 2 P = ê
ú
ë 0.9983 0.0583 û
é 0.0583 3.1569 ù
S -1 = PT M 1/ 2 = ê
ú
ë -0.9983 0.1842 û
Modal initial conditions:
é1ù é 3.2152 ù
r 0 = S -1x 0 = S -1 ê ú = ê
ú
ë1û ë -0.8141û
r 0 = S -1x 0 = 0
P = éë v1
()
()
()
()
Modal solution (from Eqs. (4.67) and (4.68):
é
pù
r1 t = 3.2152 sin ê1.078t + ú = 3.2152 cos1.078t
2û
ë
()
()
r2 t = -0.8141cos3.469t
Covert to physical coordinates:
é 0.0583 -0.9983ù é 3.2152 cos1.078t ù
x t = Sr t = ê
úê
ú
ë 0.3157 0.0184 û ë -0.8141cos3.469t û
é0.1873cos1.078t + 0.8127 cos3.469t ù
x t =ê
ú
ë 1.015cos1.078t - 0.0150 cos3.469t û
()
()
()
Copyright © 2015 Pearson Education Ltd.
These figures are similar to those of Figure 4.6, except the responses are reversed (2
looks like x2 in Figure 4.6, and 1 looks like x1 in Figure 4.6)
4.41
Use modal analysis to calculate the solution of
x(t )  

 
 x(t )  0
2 0
 3 1
0 8 
 1 1 




()
T
( )
()
for the initial conditions x 0 = éë0 1ùû mm and x 0 = éë0 0 ùû
T
( mm/s) .
SolutionGiven:
From the equation
 1 x = 0
2 0 x + 3

 

0 8    1 1 
And the initial conditions
x0 = 0 , x 0 = 0
 
 
1 
0 
We have λ1= 0.0809 and λ2= 1.54404 ω1 = 0.284528 rad / s ω2 = 1.2426 rad / s
The eigenvectors are :
v1 = 0.173503  and v 2 = 0.984833




 0.984833
0.173505
Therefore P =  0.17350  0.98483


 0.98483 0.17350 
Copyright © 2015 Pearson Education Ltd.
And S 1 =   0.12268  0.69638  r t  = 2.78552cos1.24259t 




  0.34819 0.06134 
0.49074cos0.28452t 
Converting to physical space:
xt  =  0.34174cos0.28452t   0.34174cos1.24259t 


0.030103cos0.28452t   0.96989cos1.2425t  
4.42
For the matrices
M
(
-1/ 2
é 1
ê
=ê 2
êë 0
)
ù
0ú
1 é 1 1ù
ê
ú
ú and P =
2 ë -1 1û
ú
4û
T
calculate M -1/ 2 P, M -1/ 2 P , and PT M -1/ 2 and hence verify that the computations in
Eq. (4.70) make sense.
Solution
Given
M
-1/ 2
é 1
ê
=ê 2
êë 0
ù
0ú
1 é 1 1ù
and
P
=
ê
ú
ú
-1 1û
2
ë
4 úû
Now
é 0.5
0.5 ù
M -1/2 P = ê
ú
êë -2 2 2 2 úû
So
é 0.5 -2 2 ù
T
ú
M -1/2 P = ê
ê 0.5 2 2 ú
ë
û
é 0.5 -2 2 ù
T
-1/2
ú
P M
=ê
ê 0.5 2 2 ú
ë
û
(
)
(
Thus, M -1/ 2 P
4.43
)
T
= PT M -1/ 2 [Equation (4.71)]
Consider the 2-degree-of-freedom system defined by:
é 9 0 ù
é 27 -3 ù
x(t) + ê
ê
ú 
ú x(t) = 0 .
0
1
-3
3
ë
û
ë
û
Calculate the response of the system to the initial conditions
Copyright © 2015 Pearson Education Ltd.
é1ù
1 ê ú
x0 =
3 x 0 = 0
2 êê 1 úú
ë û
What is unique about your solution compared to the solution of Example 4.3.1?
Solution: Following the calculations made for this system in Example 4.3.1,
w 1 = l1 = 1.414 rad/s, w 2 = l2 = 2 rad/s
é1 1 ù
1 é1 1 ù
1 ê
-1/ 2
ú and S -1 = PT M 1/ 2 = 1 é3 1 ù
P=
3
3
ê
úÞS = M P=
ê
ú
2 ë1 -1û
2 ê 1 -1ú
2 ë3 -1û
êë
úû
Next compute the modal initial conditions
é1ù
r 0 = S -1x 0 = ê ú , and r 0 = S -1x 0 = 0
ë0 û
Modal solution (from Eqs. (4.67) and (4.68)):
é cos1.414t ù
r t =ê
ú
0
ë
û
Note that the second coordinate modal coordinate has zero initial conditions and is hence
not vibrating. Convert this solution back into physical coordinates:
()
()
()
()
()
é1 1 ù
1 ê
ú é cos1.414t ù
x t = Sr t =
3
3
ê
ú
0
2 ê 1 -1ú ë
û
êë
úû
é 0.236 cos1.414t ù
Þx t =ê
ú
ë0.707 cos1.414t û
()
()
()
The unique feature about the solution is that both masses are vibrating at only one
frequency. That is the frequency of the first mode shape. This is because the system is
excited with a position vector equal to the first mode of vibration.
Copyright © 2015 Pearson Education Ltd.
4.44Consider the 2-degree-of-freedom system defined by:
é 9 0 ù
é 27 -3 ù
x(t) + ê
ê
ú 
ú x(t) = 0 .
ë 0 1 û
ë -3 3 û
Calculate the response of the system to the initial conditions
é1ù
1 ê ú
x 0 = 0, and x 0 =
3
2 êê -1úú
ë û
What is unique about your solution compared to the solution of Example 4.3.1
and to Problem 4.40, if you also worked that?
Solution: From example 4.3.1,
w 1 = l1 = 1.414 rad/s, w 2 = l2 = 2 rad/s, P =
1 é1 1 ù
ê
ú
2 ë1 -1û
é1 1 ù
1 ê
ú , and S -1 = PT M 1/ 2 = 1 é3 1 ù
ÞS = M P=
3
3
ê
ú
2 ê 1 -1ú
2 ë3 -1û
êë
úû
Modal initial conditions:
é0 ù
r 0 = S -1x 0 = 0, and r 0 = S -1x 0 = ê ú
ë1 û
Modal solution (from Eqs. (4.67) and (4.68)):
é
ù
0
ù
ê
ú é 0
r t =ê 1
=ê
ú
ú
sin 2t
0.5sin 2t û
êë w 2
úû ë
Convert to physical coordinates:
é1 1 ù
1 ê
ú é 0 ù = é 0.118sin 2t ù
x t = Sr t =
3
3
ê
ú ê
ú
2 êê 1 -1úú ë0.5sin 2t û ë -0.354 sin 2t û
ë
û
Compared to Example 4.3.1, only the second mode is excited, because the initial velocity
is proportional to the second mode shape, and the displacement is zero. Compared to the
previous problem, here it is the second mode rather then the first mode that is excited.
-1/ 2
()
()
()
()
()
()
()
Copyright © 2015 Pearson Education Ltd.
4.45
Consider the system of defined by
é 25,000 -15,000 ù
é 100 0 ù
x(t) + ê
ú x(t) = 0
ê
ú 
êë -15,000 25,000 úû
ë 0 100 û
Solve for the free response of this system using modal analysis and the initial conditions
T
x(0) = éë 1 0 ùû and x(0)
 = 0.
Solution: Given:
k1 = 10,000 N/m m1 = m2 = 100 kg
()
()
k2 = 15,000 N/m x 0 = éë1 0 ùû
k3 = 10,000 N/m x 0 = 0
Equation of motion:
T
M
x + Kx = 0
é100 0 ù
é 25,000 -15,000 ù
x+ê
ê
ú 
úx = 0
ë 0 100 û
ë -15,000 25,000 û
Calculate eigenvalues and eigenvectors:
é 0.1 0 ù
M -1/ 2 = ê
ú
ë 0 0.1û
é 250 -150 ù
K = M -1/ 2 KM -1/ 2 = ê
ú
ë -150 250 û
det K - l I = l 2 - 500l + 40,000 = 0
(
l1 = 100
)
w 1 = 10 rad/s
l2 = 400 w 2 = 20 rad/s
l1 = 100
é 150 -150 ù é v11 ù é0 ù
ê
úê ú = ê ú
ë -150 150 û êë v12 úû ë0 û
1 é1ù
v1 =
ê ú
2 ë1û
Copyright © 2015 Pearson Education Ltd.
l2 = 400
é -150 -150 ù é v21 ù é0 ù
ê
úê ú = ê ú
ë -150 -150 û êë v22 úû ë0 û
1 é1ù
v2 =
ê ú
2 ë -1û
1 é1 1 ù
Now, P = éë v1 v 2 ùû =
ê
ú
2 ë1 -1û
Calculate S and S-1:
S = M -1/ 2 P =
S -1 = PT M 1/ 2
1 é0.1 0.1 ù
ê
ú
2 ë0.1 -0.1û
1 é10 10 ù
=
ê
ú
2 ë10 -10 û
Modal initial conditions:
()
r 0 = S -1x(0) =
1 é10 ù
ê ú
2 ë10 û
()
()
()
w 12 r102 + r102
()
w 22 r202 + r202
r 0 = S -1x 0 = 0
Modal solutions:
r1 t =
r2 t =
w1
w2
é
wr ù
sin êw1t + tan -1 1 10 ú
r10 û
ë
é
wr ù
sin êw 2t + tan -1 2 20 ú
r20 û
ë
So
r1 t = 7.071sin 10t + p / 2 = 7.071cos10t
()
(
)
r ( t ) = 7.071sin ( 20t + p / 2 ) = 7.071cos 20t
2
é 7.071cos10t ù
r t =ê
ú
ë 7.071cos 20t û
Convert to physical coordinates:
()
Copyright © 2015 Pearson Education Ltd.
1 é0.1 0.1ù é 7.071cos10t ù
ê
úê
ú
2 ë0.1 0.1û ë7.7071cos 20t û
é0.5 cos10t + cos 20t ù
ú
x t =ê
êë0.5 cos10t - cos 20t úû
()
()
()
(
(
x t = Sr t =
4.46
)
)
Consider the model of a vehicle given in Problem 4.15 illustrated in Figure P4.15 defined
by
é 1000 -1000 ù
é 2000 0 ù
x+ê
úx = 0
ê
ú 
-1000
11,000
0
50
ê
úû
ë
û
ë
Suppose that the tire rolls over a bump modeled as the initial conditions of
x(0) = [0
0.01]T and x (0) = 0 . Use modal analysis to calculate the response of the car
x1(t). Plot the response for three cycles.
Solution: From Problem 4.15,
é 2000 0 ù é 1000 -1000
M
x + Kx = ê
x+ê
ú 
50 û êë -1000 11,000
ë 0
Calculate the eigenvalues and eigenvectors:
é 0.0224
é 0.5
0 ù 
-1/ 2
-1/ 2
M -1/ 2 = ê
=ê
ú , K = M KM
0.1414 û
ë 0
ë -3.1623
(
)
Þ det K - l I = l 2 - 220.05l + 100 = 0 Þ
é0.9999
v 2 ùû = ê
ë0.0144
-1
Calculate S and S :
é0.0224
S = M -1/ 2 P = ê
ë0.0020
Modal initial conditions:
P = éë v1
ù
úx = 0
úû
-3.1623ù
ú
0.1414 û
l1 = 0.4545 w 1 = 0.6741 rad/s
l2 = 220.05 w 2 = 14.834 rad/s
-0.0144 ù
ú
0.9999 û
é 44.7167 0.1018 ù
-0.003ù
-1
T
1/ 2
ú, S = P M = ê
ú
0.1414 û
ë -0.6441 7.0703 û
é 0 ù é0.001018 ù
-1
r 0 = S -1x 0 = S -1 ê
ú=ê
ú , r 0 = S x 0 = 0
ë0.01û ë 0.07070 û
é0.001018cos0.6741t ù
Modal solution (from equations (4.67) and (4.68)): r t = ê
ú
ë 0.07070 cos14.834t û
Convert to physical coordinates:
()
()
()
()
Copyright © 2015 Pearson Education Ltd.
()
é 0.0224 -0.0003ù é 0.001018 cos0.6741t ù é 2.277 ´ 10-5 cos0.6741t - 2.277 ´ 10-5 cos14.834t ù
x t = Sr t = ê
ú
úê
ú=ê
-6
-3
0.0020
0.1414
0.07070
cos14.834t
ë
ûë
û ë 2.074 ´ 10 cos0.6741t + 9.998 ´ 10 cos14.834t û
()
()
Copyright © 2015 Pearson Education Ltd.
Problems and Solutions for Section 4.4 (4.47 through 4.59)
4.47
A vibration model of the drive train of a vehicle is illustrated as the three-degreeof-freedom system of Figure P4.47. Calculate the undamped free response [i.e.
M(t) = F(t) = 0, c1 = c2 = 0] for the initial condition x(0) = 0, x (0) = [0 0 1]T.
Assume that the hub stiffness is 12,000 N/m and that the axle/suspension is
20,000 N/m. Assume the rotational element J is modeled as a translational mass
of 75 kg.
Solution: Given:
From the equation of motion
0  x +120001
1 0  x = 0
75 0




 2
0 100 0 
 1 3
0 0
0  2 2 
3000
And the initial conditions
x0 = 0 , x 0 = 0
 
 
0 
0
0
1 
We have
λ1= 0 And λ2= 93.5418 λ3 = 434.458 ω1 = 0 rad / s ω2 = 9.6717 rad /s
ω3 = 20.8437 rad / s The eigenvectors are :
v1 = 0.153695 v2 = 0.880361  v3 =  0.4488 






0.177471
0.422218 
0888952 
0.97205 
 0.216276
 0.0913382
1
1
q 0  = M 2 x 0  = 0 q 0  = M 2 x 0  = 0



0

54.7723
q t  = c1 + c 4 t v1 + c 2 sin ω 2 t + 2 v 2 + c 3 sin ω3 t + 3 v 3
 ωi viT q0 
vT q 0 
 and ci = i
Where i = tan  T
i = 2,3 2= 3= 0 c 2= −1.2248
v
q

0

ω
cos

i
i
 i

c 3= 0.240015 c 1= 0 and c 4 = 53.2414
 qt  = 53.2414tv1 +1.2248sin 9.6717t v2 + 0.240015sin 20.8437t v3
1
Copyright © 2015 Pearson Education Ltd.
1
xt  = M 2 qt  = 0.944882t 1 +   0.1245sin 9.6717t + 0.012438 sin 20.8437t 
 



1  0.05171 
 0.021336 
1  0.0048 
0.0004002 
4.48
Calculate the natural frequencies and normalized mode shapes of
é4 0 0 ù
é 4 -1 0 ù
ê
ú
ê
ú
x + ê -1 2 -1ú x = 0
ê 0 2 0 ú 
ê0 0 1 ú
ê 0 -1 1 ú
ë
û
ë
û
Solution: Given the indicated mass and stiffness matrix, calculate eigenvalues:
é 0.5
é 1
0
0ù
-0.3536
0 ù
ê
ú
ê
ú
M -1/ 2 = ê 0 0.7071 0 ú Þ K = M -1/ 2 KM -1/ 2 = ê -0.3536
1
-0.7071ú
ê0
ê 0
0
1 úû
-0.7071
1 úû
ë
ë
(
)
det K - l I = l 3 - 3l 2 + 2.375l - 0.375 = 0
l1 = 0.2094, l2 = 1, l3 = 1.7906
The natural frequencies are:
w 1 = 0.4576 rad/s
w 2 = 1 rad/s
w 3 = 1.3381 rad/s
The corresponding eigenvectors are:
é -0.3162 ù
é 0.8944 ù
é 0.3162 ù
ê
ú
ê
ú
ê
ú
v1 = ê -0.7071 ú v 2 = ê 0 ú v 3 = ê -0.7071ú
ê -0.6325 ú
ê -0.4472 ú
ê 0.6325 ú
ë
û
ë
û
ë
û
The relationship between eigenvectors and mode shapes is
u = M -1/ 2 v
The mode shapes are:
é -0.1581ù
é 0.4472 ù
é 0.1581ù
ê
ú
ê
ú
ê
ú
u1 = ê -0.5 ú , u2 = ê 0 ú , u3 = ê -0.5 ú
ê -0.6325 ú
ê -0.4472 ú
ê 0.6325 ú
ë
û
ë
û
ë
û
The normalized mode shapes are
Copyright © 2015 Pearson Education Ltd.
û1 =
4.49
é0.192 ù
é 0.707 ù
é 0.192 ù
ê
ú
ê
ú
ê
ú
= ê0.609 ú , û 2 = ê 0 ú , û3 = ê -0.609 ú .
T
u1 u1 ê 0.77 ú
ê -0.707 ú
ê 0.77 ú
ë
û
ë
û
ë
û
u1
The vibration is the vertical direction of an airplane and its wings can be modeled
as a three-degree-of-freedom system with one mass corresponding to the right
wing, one mass for the left wing, and one mass for the fuselage. The stiffness
connecting the three masses corresponds to that of the wing and is a function of
the modulus E of the wing. The equation of motion is
é 1 0 0 ù é x1 ù
é 3 -3 0 ù é x1 ù é 0 ù
ê
ú ê ú EI ê
úê ú ê ú
m ê 0 4 0 ú ê x2 ú + 3 ê -3 6 -3ú ê x2 ú = ê 0 ú
l
ê 0 0 1 ú ê x ú
ê 0 -3 3 ú ê x ú ê 0 ú
ë
ûë 3û
ë
ûë 3û ë û
The model is given in Figure P4.49. Calculate the natural frequencies and mode
shapes. Plot the mode shapes and interpret them according to the airplane's
deflection.
x 1(t)
x 2(t)
l
l
m
EI
(a)
x 3(t)
4m
EI
m
(b)
x 2(t)
x 1(t)
m
x 3(t)
m
4m
k
3E I
k
l3
3E I
l3
(c)
Figure P4.49 Model of the wing vibration of an airplane: (a) vertical wing vibration; (b) lumped
mass/beam deflection model; (c) spring–mass model.
Copyright © 2015 Pearson Education Ltd.
Solution: Given the equation of motion indicated above, the mass-normalized
stiffness matrix is calculated to be
é1 0 0 ù
é 3
-1.5 0 ù
1 ê
EI ê
ú 
ú
- 12
- 12
- 12
M =
ê 0 0.5 0 ú , K = M KM = m3 ê -1.5 1.5 -1.5ú
mê
ú
ê 0
-1.5
3 úû
ë0 0 1 û
ë
EI
Computing the matrix eigenvalue by factoring out the constant
yields
m3
EI
EI
det( K - l I ) = 0 Þ l1 = 0, l2 = 3 3 , l3 = 4.5 3
m
m
and eigenvectors:
é0.4082 ù
é -0.7071ù
é 0.5774 ù
ê
ú
ê
ú
ê
ú
v1 = ê 0.8165 ú v 2 = ê 0 ú v 3 = ê -0.5774 ú
ê0.4082 ú
ê 0.7071 ú
ê 0.5774 ú
ë
û
ë
û
ë
û
The natural frequencies are 1 = 0, 2 = 1.7321
EI
rad/s, and 3 = 2.1213
m3
EI
rad/s.
m3
The relationship between the mode shapes and eigenvectors u is just u = M-1/2v.
The fist mode shape is the rigid body mode. The second mode shape corresponds
to one wing up and one down the third mode shape corresponds to the wings
moving up and down together with the body moving opposite. Normalizing the
mode shapes yields (calculations in Mathcad):
Copyright © 2015 Pearson Education Ltd.
1 0 0
Mh
3
0 2 0
K
0 0 1
la
3
3
0
6
0
3
3
eigenvals Kh
Kh
Mh
1
K . Mh
3
1.5 0
Kh =
4.5
la =
1.
3
1.5 1.5
0
3
1.5
1.5 3
0
v1
eigenvec Kh , la2
0.408
v1 =
eigenvec Kh , la1
v2
0.707
0.816
v2 =
0.408
v3
0
0.707
0.577
eigenvec Kh , la0
v3 =
0.577
0.577
0.408
la2
w
2.967 10
la1
w =
la0
u1n
8
u1
Mh
1.
v1
u1 =
0.408
1.732
2.121
0.577
u1
u1n =
u1
0.577
0.707
u2
Mh
1.
v2
u2 =
0.577
Mh
1.
v3
u3 =
0.289
u2n
0.707
u2
u2
u2n =
0.577
u3
0
0.707
0.667
u3
u3n
0
0.707
0.577
u3
0.408
u3n =
0.333
0.667
These are plotted:
4.50
Solve for the free response of the system of Problem 4.49. Where E = 6.9  109
N/m2, l = 2 m, m = 4000 kg, and I = 5.2  10-6m4. Let the initial displacement
()
correspond to a gust of wind that causes an initial condition of x 0 = 0, x(0) =
T
[0.2 0 0] m. Discuss your solution.
Solution: From problem 4.49 and the given data
Copyright © 2015 Pearson Education Ltd.
0
4000 0
 x +


16,000 0
0



0
4000
0
4
 1.346 0
1.346
 10 x = 0


 1.346
 1.346 2.691
0
 1.346 1.346 
0.2
0 x 0 = 0 Iq + 3.365
 1.6825

T
0 m
 1.6825 1.68187
0
x0 =
 1.6825
Calculate eigenvalues and eigenvectors:
~
det K  λI = 0 λ2 = 0 ω1 = 0 rad / s λ2 = 3.365


q = 0

 1.6825
3.365 
0
ω2 = 1.83439 rad / s λ3 = 5.04729
ω3 = 2.24662rad / s
v1 = 0.408215 v 2 =  0.7071 v 3 = 0.57737 






0.81653 
0

 0.577303
0.408215
0.7071 
0.57737 
The solution is given by
q  t  =  c1 +c4 t  v1 +c2sin  ω2t + 2  v2 +c3sin  ω3t + 3  v3
 ω vT q  0  
viT q  0 
i=
2,3
Where i = tan 1  i T i
c
=
i = 2,3

i
v
q
0
sin



i
i


π
Thus, 2 = 3 = ,c2 = 8.94427,c3 = 7.30327
2
3
3
So, q  0  = c1v1 +  ci sini vi q  0  = c4 vi +  ωi ci cosi vi
i= 2
i= 2
3
T
T
Premultiplyby v i : ∑ v i q (0 )= 5.16355= c 1
i= 2
3
∑ v Ti q́ (0 )= 0= c 4
i= 2
So, q  t  = 5.16355v1  8.94427cos 1.834t  v2 + 7.30327cos  2.246 t  v3
Convert to physical coordinates:
1
xt  = M 2 q t 
xt  = 0.0333 + 0.1  cos1.834t + 0.06667  cos2.246tm

 



0.0333 0 
 0.0333
0.0333  0.1
0.06667 
4.51
Consider the two-mass system of Figure P4.51. This system is free to move in the
x1 - x2 plane. Hence each mass has two degrees of freedom. Derive the linear
Copyright © 2015 Pearson Education Ltd.
equations of motion, write them in matrix form, and calculate the eigenvalues and
eigenvectors for m = 20 kg and k = 200 N/m.
Solution: Given :
m= 20 kg ,k = 200 N / m
Mass 1
x 1 − direction: m x́ 1= − 4 k x 1 +k (x 3 − x1 )= −5 k x 1+ k x3
x 2 − direction: m x́ 2= − 3 k x 2 −k x 2= − 4 k x 2
Mass 2
x 3 − direction: m x́ 3= − 4 k x 3 +k (x 3 − x 1 )= − k x 1 −5 k x 3
x 4 − direction: m x́ 4 = − 4 k x 4 −2 k x 4= − 6 k x 4
In matrix form the values given:
 200 0  x = 0
20 0 0 0  x + 1000 0

 

800 0
0 
0 20 0 0  0
0 0 20 0   200 0
1000 0 

 
0 0 0 20 0
0
0
1200
~
K = M 1 / 2 KM 1 / 2 = 50
0  10 0 


40 0
0 
0
 10 0 50
0 


0
0 0
60
~
det K  λI = λ 4  200λ 3 +14,800λ 2  480,000λ + 5,760,000 = 0
λ1 = 40, λ2 = 40, λ3 = 60, λ4 = 60
~
The corresponding eigenvectors are found from solving K  λi vi = 0 for each
value of the index and normalizing:
v1 = 0 v2 = 0.7071 v3 = 0.7071  v4 = 0
 




 
1 
0

0

0 
0 
0.7071
 0.7071
0 
 




 
0
1 
0

0

These vectors are not unique.



4.52

Consider again the system discussed in Problem 4.51. Use modal analysis to
calculate the solution if the mass on the left is raised along the x2 direction exactly
0.02 m and let go.
Solution: Given: ∆ x 2= 0.02 m
Copyright © 2015 Pearson Education Ltd.
0  x + 1000 0
 200
 
20 0 0  0
800 0
0 20 0   200 0
1000
0 0 20 0
0
0
1 / 2
M
= 0.223607 1 0 0 0


0 1 0 0 
0 0 1 0 


0 0 0 1 
~
K = M 1 / 2 KM 1 / 2 = 50
0  10 0 


40 0
0 
0
 10 0 50
0 


0 0
60
0
20

0
0

0
0
λ1= 40
λ2= 40
λ3 = 60
λ 4= 60
0
x = 0

0 
0 
1200
0
ω1 = 6.3246
ω2 = 6.3246
ω3 = 7.7460
ω 4= 7.7460
rad/s
rad/s
rad/s
rad/s
v1 = 0 v 2 = 0.7071 v 3 = 0.7071  v 4 = 0
 




 
1 
0

0

0 
0
0.7071
 0.7071
0 
 




 
0
1 
0

0

Also, 0 0.02 0 0 m and x0 = 0
x0  =
Use the mode summation methods to find the solution
Transform the initial conditions:
0 0.0894427 0 0
q0 = M 1 / 2 x0  =
q 0  = M 1 / 2 x 0 = 0
The solution is given by Eq.(4.103),
4
x  t  =  d i sin  ωi t + i  ui
i=1
Where
 ω vT q  0  
i = tan 1  i T i
 i=1,2,3,4 (Eq.(4.97))
v
q
0


i


Copyright © 2015 Pearson Education Ltd.
di =
viT q  0 
sini
ui= M
− 1/ 2
i=1,2,3,4 (eq.(4.98))
vi
Substituting known values yields
π
1 = 2 = 3 = 4 = rad
2
d 1= 0.0894427
d 2= d 3 = d 4= 0
u 1 = 0
 u2 = 0.158114  u3 = 0.158114 






0.223607
0

0

0

0.158114 
 0.158114






0

0

0

u4 = 0



0

0



0.223607
The solution is
xt  = 0



0.02cos6.3246t 
0



0

4.53
The vibration of a floor in a building containing heavy machine parts is modeled
in Figure P4.53. Each mass is assumed to be evenly spaced and significantly
larger than the mass of the floor. The equation of motion then becomes
(m = m
1
2
)
= m3 = m .
Copyright © 2015 Pearson Education Ltd.
é 9
1 13 ù
ê
ú
ê 64 6 192 ú é x1 ù
EI ê 1
1
1 úê ú
mI
x+ 3 ê
êx ú = 0
6
3 6 úê 2ú
l
ê
ú
9 ú ë x3 û
ê 13 1
êë 192 6 64 úû
Calculate the natural frequencies and mode shapes. Assume that in placing box
m2 on the floor (slowly) the resulting vibration is calculated by assuming that the
initial displacement at m2 is 0.04 m. If l = 2 m, m = 200 kg, E = 0.6  109 N/m2, I
= 4.17  10-5 m4. Calculate the response and plot your results.
Solution:
The equations of motion can be written as
EI  9
1 13 
m 1 0 0  x1  + 3 
  x1  = 0

   l  64 6 192   
0 1 0  x2 

  x2 
0 0 1   
1
1
1

 
x 
 x3 
6
3 6  3 




 13 1 9 
192 6 64 
Or mI x́+ Kx= 0 where I is the 3x3 identity matrix
The natural frequencies of the system are obtained using the characteristic
equation K  ω 2 M = 0
Using the given mass and stiffness matrices yields the following
characteristic equation
2
3
59 EI m3 4 41 (EI ) m 2 7 (EI )
3 6

m ω −

96l 3
ω +
786 l 6
ω −
6912 l 9
=0
Substituting for E , I , m , and l yields the following answers for the
natural frequency
13 



137 EI
7 EI
13 + 137 EI
, ω2 = ±
, ω3 = ±
3
3
ml
96ml
48ml 3
The plus minus sign shown above will cause the exponential terms to change to
trigonometric terms using Euler's formula. Hence, the natural frequencies of the
system are 0.65 rad/s, 1.068rad/s and 2.837 rad/s.
ω1 = ±
Let the mode shapes of the system be u1 , u2 , and u 3 . The mode shapes should
satisfy the following equation
Copyright © 2015 Pearson Education Ltd.
K  ω M u
i1  = 0, i = 1,2,3
 
ui 2 
 
ui 3 
Notice that the system above does not have a unique solution for u1 since
2
1
[K −ω 21 M ] had to be singular in order to solve for the natural frequency
ω.
Solving the above equation yields the following relations
2 3
u i2 1 96 m ωi l −7 EI
=
, i= 1,2,3
u i3 3 13 mωi2 l 3 + EI
And ui 1= ui 3 ,i= 1,3 but for the second mode shape this is different u21= u23
Substituting the values given yields
2 3
u 12 1 96 m ωi l −7 EI
=
= −1.088
u 13 3 13 m ω2i l3+ EI
2 3
u 22 1 96 m ω2 l −7 EI
=
=0
u 23 3 13 mω 22 l 3 + EI
2 3
u 32 1 96 m ω3 l −7 EI
=
= 1.838
u 33 3 13 m ω23 l 3 + EI
If we let ui 3= 1, i= 1,2,3 , then
u1 = 1
, u2 =  1, u3 = 1





 
 1.088
1.838
0 


 


1

1

0 
These mode shapes can be normalized to yield
u1 = 0.5604 , u2 =  0.7071, u3 = 0.4312 






 0.6098
0

0.7926 






0.5604 
0.7071 
0.4312 
The second box, m2 , is placed slowly on the floor; hence, the initial velocity can
be safely assumed zero. The initial displacement at m2 is given to be
0.04 m.
Hence, the initial conditions in vector form are given as
x0  = 0
x 0  = 0


 and
 
 0.04
0


 
0

0
The equations of motion given by Mx t + Kxt  = 0 can be transformed into the
modal coordinates by applying the following transformation
xt  = Sr t  = M 1 / 2 Pr t  where P is the basis formed by the mode shapes of the
system, given by
P = u1 u 2 u 3 
Copyright © 2015 Pearson Education Ltd.
Hence, the transformation S is given by
S =  0.04  0.05 0.03 


0.056
0.043 0
 0.04 0.05
0.03 
The initial conditions will be also transformed
r 0 = S 1 x0  =  0.339943


0



 0.45258 
Hence, the modal equations are
With the above initial conditions.
The solution will then be
r t  = 0.339943cos0.65t 


0



0.45258cos2.837t 
The solution can then be determined by
xt  = 0.01359cos 0.65t   0.013577co s2.837 t  


 0.0146cos 0.65t   0.02534cos 2.837t  


0.01359cos 0.65t   0.0135774c os2.837 t 
Plot:
Copyright © 2015 Pearson Education Ltd.
4.54
Recalculate the solution to Problem 4.53 for the case that m2 is increased in mass
to 1000 kg. Compare your results to those of Problem 4.53. Do you think it
makes a difference where the heavy mass is placed?
Solution: Given the data indicated the equation of motion becomes :
1 13 
9
0  x + 3.197  10  4 
x=0
200 0
64 6 192 




1000 0 
0


0
1 1 
1
0
200
6
3 6 




 13 1 9 
192 6 64 
T
x (0 )= [0 0.04 0 ] , x́ (0 )= 0
Calculate eigenvalues and eigenvectors
M 1 / 2 = 0.07071 0
0



0.0316228 0
0

0
0
0.07071
Copyright © 2015 Pearson Education Ltd.
~
K = M 1 / 2 KM 1 / 2 = 2.24789 1.19145 1.08232   107


1.19145 1.06567 1.19145 
1.08232 1.19145 2.24789
~
det K  λI  0


−8
λ1= 1.67876× 10 ω 1= 1.29567 ×10
−5
rad/s
λ2= 1.16557× 10 ω 2= 3.414 ×10 rad/s
−7
−4
λ3 = 4.228 ×10 ω3= 6.570231 ×10 rad/s
v1 = 0.332509  v2 = 0.7071  v 3 = 0.62405 






 0.882539
0

0.470238
0.332509 
 0.7071
0.62405 
Use the mode summation method to find the solution. Transform the initial
conditions :
−7
−4
1
0 1.265491 0 q (0 )= M 2 x (0 )
1
q́ (0 )= M 2 x́ (0 )= 0
The solution is given by Eq.(4.103),
4
x  t  =  d i sin  ωi t + i  ui
i=1
Where
 ωi viT q  0  
i = tan  T
 i=1,2,3 (Eq.(4.97))
 vi q  0  
viT q  0 
i=1,2,3, (eq.(4.98))
di =
sini
1
ui= M
− 1/ 2
vi
Substituting known values yields
π
1 = 2 = 3 = rad
2
d 1= −1.11633
d 2= 0
d 3= 0.59481
u1 = 0.0235119  u2 = 0.05  u3 = 0.044127 






 0.0279083
0

0.0148702
0.0235119 
 0.05
0.044127 
The solution is
Copyright © 2015 Pearson Education Ltd.



xt  = 1.116330.0235119  cos 1.29567  10 5 t + 0.594810.044127  cos 6.570231  10 4 t




 0.0279083
0.0148702 
0.0235119 
0.044127 
4.55
Repeat Problem 4.49 for the case that the airplane body is 10 m instead of 4 m as
indicated in the figure. What effect does this have on the response, and which
design (4m or 10 m) do you think is better as to vibration?
Solution: Given:
é1 0 0 ù
é 3 -3 - ù
EI ê
ê
ú
ú
m ê 0 10 0 ú 
x + 3 ê -3 6 -3ú x = 0
l
ê0 0 1 ú
ê 0 -3 3 ú
ë
û
ë
û
Calculate eigenvalues and eigenvectors:
é1
0
0ù
ê
ú
M
= m ê0 0.3612 0 ú
ê0
0
1 úû
ë
é 3
-0.9487
0 ù
EI
ê
ú
K = M -1/ 2 KM -1/ 2 = 3 ê -0.9487
0.6
-0.9487 ú
ml
ê 0
-0.9487
3 úû
ë
Again choose the parameters so that the coefficient is 1 and compute the
eigenvalues:
det K - l I = l 3 - 6.6l 2 + 10.8l = 0
-1/ 2
-1/ 2
(
)
l1 = 0
l2 = 3
l3 = 3.6
é -0.2887 ù
é 0.7071 ù
é 0.6455 ù
ê
ú
ê
ú
ê
ú
v1 = ê -0.9129 ú v 2 = ê 0 ú v 3 = ê -0.4082 ú
ê -0.2887 ú
ê -0.7071ú
ê 0.6455 ú
ë
û
ë
û
ë
û
The natural frequencies are
Copyright © 2015 Pearson Education Ltd.

w 1 = 0 rad/s
w 2 = 1.7321 rad/s
w 3 = 1.8974 rad/s
The relationship between eigenvectors and mode shapes is
u = M -1/ 2 v
u1 = m
-1/ 2
é -0.2887 ù
é 0.7071 ù
é 0.6455 ù
ê
ú
ú
ê
ú
-1/ 2 ê
ê -0.2887 ú u 2 = m ê 0 ú u3 = ê -0.1291ú
ê -0.2887 ú
ê -0.7071ú
ê 0.6455 ú
ë
û
ë
û
ë
û
It appears that the mode shapes contain less "amplitude" for the wing masses.
This seems to be a better design from a vibration standpoint.
Copyright © 2015 Pearson Education Ltd.
4.56
Often in the design of a car, certain parts cannot be reduced in mass. For
example, consider the drive train model illustrated in Figure P4.47. The mass of
the torque converter and transmission are relatively the same from car to car.
However, the mass of the car could change as much as 1000 kg (e.g., a two-seater
sports car versus a family sedan). With this in mind, resolve Problem 4.47 for the
case that the vehicle inertia is reduced to 2000 kg. Which case has the smallest
amplitude of vibration?
Solution: Let k1 = hub stiffness and k2 = axle and suspension stiffness. From
Problem 4.44, the equation of motion becomes
é75 0
é 1 -1 0 ù
0 ù
ê
ú
ê
ú
0 ú 
x + 10,000 ê -1 3 -2 ú x = 0
ê 0 100
ê0
ê 0 -2 2 ú
0 2000 úû
ë
ë
û
()
()
T
x 0 = 0 and x 0 = éë0 0 1ùû m/s.
Calculate eigenvalues and eigenvectors.
M
-1/ 2
é0.1155 0
0 ù
ê
ú
=ê 0
0.1
0 ú
ê 0
0 0.0224 úû
ë
K = M
(
-1/ 2
KM
-1/ 2
é 133.33 -115.47
0 ù
ê
ú
= ê -115.47
300
-44.721ú
ê 0
-44.721
10 úû
ë
)
det K - l I = l 3 - 443.33l 2 + 29,000l = 0
l1 = 0
w1 = 0 rad/s
l2 = 70.765 w 2 = 8.9311 rad/s
l3 = 363.57 w 3 = 19.067 rad/s
é -0.1857 ù
é 0.8758 ù
é 0.4455 ù
ê
ú
ê
ú
ê
ú
v1 = ê -0.2144 ú v 2 = ê 0.4063 ú v 3 = ê -0.8882 ú
ê -0.9589 ú
ê -0.2065 ú
ê 0.1123 ú
ë
û
ë
û
ë
û
Use the mode summation method to find the solution. Transform the initial
conditions:
Copyright © 2015 Pearson Education Ltd.
()
()
q ( 0 ) = M x ( 0 ) = éë0
q 0 = M 1/ 2 x 0 = 0
1/ 2
0 44.7214 ùû
T
The solution is given by
() (
)
(
)
(
)
q t = c1 + c4t v1 + c2 sin w 2t + f2 v 2 + c3 sin w 3t + f3 v 3
where
æ w v T q(0) ö
f i = tan -1 ç i T i
÷ , i = 2,3

è v i q(0)
ø
v Ti q(0)
ci =
, i = 2,3
w i cos fi
Thus 2 = 3 =0, c2 = -1.3042 and c3 = 0.2635. Next apply the initial conditions:
3
3
i= 2
i= 2
 = c4 v1 + å ci sin fi v i
q(0) = c1v1 + å ci sin fi v i and q(0)
Pre multiply each of these by v1T to get:

c1 = 0 = v1T q(0) and c4 = -42.8845 = v1T q(0)
So
q(t) = -42.8845tv1 - 1.3042sin(8.9311t)v 2 + 0.2635sin(19.067t)v 3
Next convert back to the physical coordinates by
x(t) = M
-1
2
q(t)
é1ù é -0.1319 ù
é 0.01355 ù
ê ú ê
ú
ê
ú
= 0.9195t ê1ú + ê -0.05299 ú sin8.9311t + ê -0.02340 ú sin19.067t m
ê1ú ê 0.007596 ú
ê 0.0006620 ú
ë û ë
û
ë
û
Comparing this solution to problem 4.44, the car will vibrate at a slightly higher
amplitude when the mass is reduced to 2000 kg.
4.57
Use mode summation methodto compute the analytical solution for the response
of the 2-degree-of-freedom system:
é 1 0 ù
é 540 -300 ù
x(t) + ê
ê
ú 
ú x(t) = 0
ë 0 4 û
ë -300 300 û
to the initial conditions of
é 0 ù
é0 ù

x0 = ê
,
x
=
0
ú
ê0 ú .
ë 0.01û
ë û
Copyright © 2015 Pearson Education Ltd.
Solution: Following the development of equations (4.97) through (4.103) for the
mode summation for the free response and using the values of computed in
problem 1, compute the initial conditions for the “q” coordinate system:
é1 0 ù
é1 0 ù é 0 ù é 0 ù
é1 0 ù é0 ù é0 ù
M 1/ 2 = ê
Þ q( 0) = ê
=ê
, q(0) = ê
ú
ú
ê
ú
ú
úê ú = ê ú
ë0 2 û
ë 0 2 û ë 0.01û ë 0.02 û
ë0 2 û ë0 û ë0 û
From equation (4.97):
æ xö
æ xö p
f1 = tan -1 ç ÷ = f2 = tan -1 ç ÷ =
è 0ø
è 0ø 2
From equation (4.98):
v1T q ( 0 )
v T2 q ( 0 )
T
d1 =
= v1 q ( 0 ) ,d2 =
= v T2 q ( 0 )
p
p
sin 2
sin 2
( )
( )
Next compute q ( t ) from (4.92) and multiply by M 1/ 2 to get x ( t ) or use (4.103)
directly to get
q ( t ) = d1 cos (w 1t ) v1 + d2 cos (w 2 t ) v 2 = cos (w 1t ) v1T q ( 0 ) v1 + cos (w 2 t ) v1T q ( 0 ) v1
é 0.0054 ù
é -0.0054 ù
= cos ( 5.551t ) ê
+ cos ( 24.170t ) ê
ú
ú
ë 0.0184 û
ë 0.0016 û
Note that as a check, substitute t = 0 in this last line to recover the correct initial
condition q ( 0 ) . Next transform the solution back to the physical coordinates
é 0.0054 ù
é -0.0054 ù
x ( t ) = M -1/ 2 q ( t ) = cos ( 5.551t ) ê
+ cos ( 24.170t ) ê
ú
úm
ë 0.0092 û
ë 0.0008 û
4.58
For a zero value of an eigenvalue and hence frequency, what is the corresponding
time response? Or asked another way, the form of the modal solution for a nonzero frequency is Asin(w n t + f ) , what is the form of the modal solution that
corresponds to a zero frequency? Evaluate the constants of integration if the
modal initial conditions are: r1 (0) = 0.1, and r1 (0) = 0.01.
Solution:A zero eigenvalue corresponds to the modal equation:
r1 (t ) = 0 Þ r1 (t ) = a + bt
Applying the given initial conditions:
r1 (0) = a + b(0) = 0.1 Þ a = 0.1
r1 (0) = b = 0.01
Þ r1 (t ) = 0.1 + 0.01t
4.59
Consider the system described by
é 1 0 ù
é 400 -400 ù
x(t) + ê
ê
ú 
ú x(t) = 0
ë 0 4 û
ë -400 400 û
Copyright © 2015 Pearson Education Ltd.
T
subject to the initial conditions x(0) = éë 1 0 ùû , x (0) = 0. Plot the
displacements versus time.
Solution:
a  400
1 0

0 4
 1 1 
 a
 1 1 
M  
 400 200 

 200 100 
 0

 1
P  augment ( v1 v2)
 0.447

 0.894
 0.1 

 0 
x0  
1
s1  S  
0
2  5 5
 0.447 0.894

 0.447 0.224 
P  
1
r0  S
s2  r0  0.02
1
0
0
 0.447 1.789

 0.894 0.894
 0.894

 0.224 

0
0
x2( t )  s1  r0  s2  r0  cos  10 5 t 
1
0
1
1
0.1
x1 ( t )
0
0.2
0.4
0.6
0.8
 0.1
 0.2
t
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T
 0.02 0.04 

 0.02 0.04 
s1 r0  
1
0.2
x2 ( t )
 0.447 1.789

 0.894 0.894
T
SI  P  MR  
0
1
 0.447 0.894

 0.894 0.447 
P

x1( t )  s1  r0  s2  r0  cos  10 5 t 
1
1
1
S
0

1
s1  r0  0.02
0
v2  
2  11.18
s2  S 
s2  r0  0.08
0
 1 0

 0 1
T
P P  
 0.045 

 0.089
s1  r0  0.02
 0.894

 0.447 
v2  eigenvecs ( Kt)  
 x0  
 0.447
s1  

 0.447
1
 0.894 0.447

 0.447 0.894
P 
0 0 

 0 500 
 K MR
eigenvecs ( Kt)  
 0.447 0.894

 0.894 0.447
T
1
 1 

0
v1  
P  Kt P  
1
Kt  MR
eigenvals ( Kt)  
v1  eigenvecs ( Kt)  
S  MR
MR  
 500 

 0 
Kt  
T
1 0

0 2
K  
Problems and Solutions for Section 4.5 (4.60 through 4.72)
4.60
Consider the following two-degree-of-freedom system and compute the response
assuming modal damping rations of z 1 = 0.01 and z 2 = 0.1:
x(t )  

 
 x(t )  0, x 0  
 , x 0  0
10 0 
 20 3
 0.1 
 0 1
 3 3 
0.05 






Plot the response
Solution: Given:
A two degree of freedom system with damping ratios ξ 1= 0.01 and ξ 2= 0.1
Solution:
Equation of motion:
10 0 x t  + 20  3 xt  = 0 x0 = 0.1  , x 0 = 0






0 1 
 3 3 
0.05
1
M 2 = 0.316228 0


1
0
~
K = 2
 0.948683


 0.948683 3

P = v1 v 2  =  0.856228  0.516598


 0.516598 0.856228 
1
S = M 2 P =  0.270763  0.163363


 0.516598 0.856228 
S 1 =  2.70763  0.516598


 1.63363 0.856228 
From the above results we have
r 0  = S 1 x0 =  0.296593


 0.120551
And r0 = S 1 x 0 = 0
Also ω1 = 1.89007rad / s ; ω2 = 1.19483rad / s
 ωd 1 = 1.88988rad / s ; ωd 2 = 1.18288rad / s
 A1 = 0.296608 1 = 1.5608rad
And A2 = 0.121164  2 = 1.47013rad
Copyright © 2015 Pearson Education Ltd.
Thus as
xt  = Sr t  =  0.01979 0.119483t sin1.47013 + 1.18288t   0.08031 0.0189007 t sin1.5608 + 1.88988t 


0.10374 0.119483t sin1.47013 + 1.18288t   0.15322 0.0189007 t sin1.5608 + 1.88988t  
4.61
Consider the example of the automobile drive train system discussed in Problem 4.47,
modeled by
é75 0
é 1 -1 0 ù
0 ù
ê
ú
ê
ú
0 ú 
x + 10,000 ê -1 3 -2 ú x = 0
ê 0 100
ê0
ê 0 -2 2 ú
0 3000 úû
ë
ë
û
()
()
T
x 0 = 0 and x 0 = éë0 0 1ùû m/s
Add 10% modal damping to each coordinate, calculate and plot the system response.
Solution:
75 0

0 100
0 0
Given:
0
 2 0 x = 0
 x + 100002



0
 2

 2 6
0
3000
 2 4 
x0 = 0 and x 0 = 0 0 1 m / s
T
Copyright © 2015 Pearson Education Ltd.
Calculate eigenvalues and eigenvectors
M 1 / 2 = 0.115470 0. 0.



0.1 0.
0.

0.
0. 0.018257 
~
7
K = M 1 / 2 KM 1 / 2 = 266.666
 230.940 0.
  10


 36.514
 230.940 600.
0.
 36.514 13.333 
~
det K  λI  0
λ1= 9.84735 ω1= 3.13805 rad / s ω d 1= 3.10667 rad / s


λ2= 150.522ω 2= 12.2687 rad / s ωd 2= 12.146 rad /s
λ3 = 719.631ω 3= 26.8259 rad / s ωd 3= 26.8259 rad / s
v1 = 0.08514 v 2 =  0.88706 v3 = 0.45373 






0.09469
 0.44612
 0.88994
0.99185
0.11874 
0.04600 
Use the mode summation method to find the solution. Transform the initial conditions :
Copyright © 2015 Pearson Education Ltd.
1
2
q (0 )= M x (0 )= 0
1
T
q 0  = M 2 x0  = 0 0 54.7723
The solution is given by Eq.(4.103),
3
 ξ ωt
x  t  =  c1 + c4t  u1 +  ci e i i sin  ωdi t + i  ui
i= 2
Where


ωdi viT q  0 
i = tan  T
 i=2,3 (Eq.(4.97))
T
 vi q  0  + ξi ωi vi q  0  
viT q  0 
ci =
i=2,3, (eq.(4.98))
ωdi cos i   ξi ωi sini
1
ui= M − 1/ 2 v i
Substituting known values yields
T
T
2 = 3 = 0 rad c 1= v1 q (0 )= 0 c 2= 0.535457 c 3= 0.0939214 c 4 = v 1 q́ (0 )= 54.3259
[
0.052392
u
=
−0.088994
u1 = 0.009832 u 2 =  0.102428 3
0.000840




0.009469
 0.044612
0.018108
0.002167 
The solution is
3
ξ ω t
xt  = c1 + c4 t u1 +  ci e i i sin ωdi t + i u i m
i= 2
 xt  = c1 + c4 t u1 + c 2 e
 ξ2
ω2t
sin ωd 2 t +
 xt  = 54.3259t 0.009832 + 0.535457e


0.009469
0.018108
u
2
2
1.22687t
+ c3 e
ω3t
 ξ3
]
sin ωd 3t +
3
u 3
sin 12.14t  0.102428 + 0.0939214e 2.682t sin26.82t 0.052392 




 0.044612
 0.088994
0.002167 
0.000840 
Copyright © 2015 Pearson Education Ltd.
Plot:
Response without rigid body mode
4.62
Consider the following two-degree-of-freedom system and compute the response
assuming modal damping rations of z 1 = 0.05 and z 2 = 0.01 :
é 50 0 ù
é 20 -9 ù
é 0.05 ù
x(t) + ê
ê
ú 
ú x(t) = 0, x 0 = ê
ú , x 0 = 0
ë 0 1 û
ë -9 9 û
ë 0.09 û
Plot the response.
Solution: (computed in Mathcad)
é 0.14 ù
é 0.02 ù
-0.03t
x(t) = 0.368e -0.023 sin(0.464t +1.521) ê
sin(3.03t +1.561) ê
ú - 0.038e
ú
ë 0.143 û
ë -0.99 û
Copyright © 2015 Pearson Education Ltd.
0.1
0.05
x( t ) 0
x( t ) 1
0
10
20
30
40
50
 0.05
 0.1
t
4.63
Consider the model of an airplane discussed in problem 4.49, Figure P4.49 modeled by
é 13,455 -13,455
ù
é 3000
0
0
0 ù
ê
ú
ê
ú
x

+
0
12,000
0
-13,455
26,910
-13,455
ê
úx = 0
ê
ú
ê
ê 0
0
3000 úû
0
-13,455 13,455 úû
ë
ë
T
subject to the initial conditions x(0) = éë 0.02 0 0 ùû m and x (0) = 0 . (a) Calculate the
response assuming that the damping provided by the wing rotation is i = 0.01 in each
mode. (b) If the aircraft is in flight, the damping forces may increase dramatically to i =
0.1. Recalculate the response and compare it to the more lightly damped case of part (a).
Solution:
From Problem 4.47, with damping
é3000
é 13455 -13455
0
0 ù
0 ù
ê
ú
ê
ú
12,000
0 ú 
x + Cx + ê -13,455 26910 -13,455ú x = 0
ê 0
ê 0
ê 0
0
3,000 úû
-13,455 13,455 úû
ë
ë
x ( 0 ) = éë 0.02 0 0 ùû m
x ( 0 ) = 0
T
l1 = 0
w 1 = 0 rad/s
l2 = 4.485 w 2 = 2.118 rad/s
l3 = 6.727 w 3 = 2.594 rad/s
Copyright © 2015 Pearson Education Ltd.
é -0.4082 ù
é 0.7071 ù
é 0.5774 ù
ê
ú
ê
ú
ê
ú
v1 = ê -0.8165 ú v 2 = ê 0 ú v 3 = ê -0.5774 ú
ê -0.4082 ú
ê -0.7071ú
ê 0.5774 ú
ë
û
ë
û
ë
û
The solution is given by
() (
)
3
q t = c1 + c2t v1 + å di e
-z i w i t
i=2
where
()
(
æ
ö
w di v Ti q 0
fi = tan ç T
÷
T
è v i q 0 + z iw i v i q 0 ø
-1
di =
()
()
v Ti q 0
)
sin w dit + fi v i
()
i = 2,3
(Eq. (4.114))
i = 2,3
sin fi
Now,
()
3
q 0 = c1v1 + å di sin fi v i
i=2
()
3
q 0 = c2 v1 + å éë -z iw i di sin fi + w di di cos fi ùû v i
i=2
T
1
Premultiply by v :
()
v q ( 0 ) = 0 = c
v1T q 0 = 4.4721 = c1
T
1
2
(a) z1 = z 2 = z 3 = 0.01
w d 2 = 2.1177 rad/s, w d 3 = 2.593 rad/s
f 2 = -1.5808 rad,
f3 = 1.5608 rad
d2 = 7.7464,
d3 = 6.3249
Mode shapes:
u i = M -1/ 2 v i
é -0.007454 ù
é 0.01291 ù
é 0.01054 ù
ê
ú
ê
ú
ê
ú
u1 = ê -0.007454 ú u 2 = ê
0
ú u3 = ê -0.005270 ú
ê -0.007454 ú
ê -0.01291ú
ê 0.01054 ú
ë
û
ë
û
ë
û
The solution is given by
() (
)
3
x t = c1 + c2t u1 + å di e
i=2
-z i w i t
(
)
sin w dit + fi ui
Copyright © 2015 Pearson Education Ltd.
é1ù é0.100ù
x(t ) = 0.0333êê1úú + êê 0 úú e - 0.0212t sin (2.1178t - 1.5808)
ëê1ûú ëê0.100ûú
é 0.0667 ù
+ êê- 0.0333úú e - 0.0259t sin (2.5937t + 1.5608)
êë 0.0677 úû
b) z 1 = z 2 = z 3 = 0.1
Same thing as part (a), but now the following values are obtained
w d 2 = 2.1072 rad/sec
w d 3 = 2.5807 rad/sec
f2 = -1.6710 rad
f3 = 1.4706 rad
d2 = 7.7850
d3 = 6.3564
Notice that the rigid mode is not effected by changing the damping ratio, and hence
c = 4.4721
Consequently, the solution becomes
é1ù é -0.1005ù
ê ú ê
ú
x t = 0.0333 ê1ú + ê 0 ú e-0.2118t sin 2.1072t - 1.6710
ê1ú ê 0.1005 ú
ë û ë
û
é 0.0670 ù
ê
ú
+ ê -0.0335 ú e-0.2594t sin 2.5807t + 1.4706
ê 0.0670 ú
ë
û
()
(
)
(
Below is the plot of the displacement of the left wing
Copyright © 2015 Pearson Education Ltd.
)
4.64
Repeat the floor vibration problem of Problem 4.53 given by
é 9 / 64 1/ 6 13 / 192 ù
ú
-4 ê
200
x + 3.197 ´ 10 ê 1/ 6
1/ 3
1/ 6 ú x = 0
êë 13 / 192 1/ 6 9 / 64 úû
x ( 0) = éë 0 0.05 0 ùû m and x ( 0 ) = 0.
T
by assigning modal damping ratios of
z1 = 0.01 z 2 = 0.1 z 3 = 0.2
Solution: The equation of motion will be of the form:
é 9 / 64 1/ 6 13 / 192
ê
200
x + Cx + 3.197 ´ 10-4 ê 1/ 6
1/ 3
1/ 6
êë 13 / 192 1/ 6 9 / 64
x ( 0 ) = éë 0 0.05 0 ùû m and x ( 0 ) = 0.
T
Copyright © 2015 Pearson Education Ltd.
ù
ú
úx = 0
úû
M -1/ 2 = 0.7071
K = M
(
-1/ 2
KM
-1/ 2
é 2.2482 2.6645 1.0825 ù
ê
ú
= ê 2.6645 5.3291 2.6645 ú ´ 10-7
ê1.0825 2.6645 2.2482 ú
ë
û
)
det K - l I = l 3 - 9.8255 ´ 10-7 l 2 + 1.3645 ´ 10-13 l - 4.1382 ´ 10-21 = 0
l1 = 4.3142 ´ 10-8 w 1 = 2.0771 ´ 10-4 rad/s
l2 = 1.1657 ´ 10-7 w 2 = 3.34143 ´ 10-4 rad/s
l3 = 8.2283 ´ 10-7
w 3 = 9.0710 ´ 10-4 rad/s
é 0.5604 ù
é -0.7071ù
é0.4312 ù
ê
ú
ê
ú
ê
ú
v1 = ê -0.6098 ú v 2 = ê 0 ú v 3 = ê0.7926 ú
ê 0.5604 ú
ê 0.7071 ú
ê0.4312 ú
ë
û
ë
û
ë
û
Use the mode summation method to find the solution. First transform the initial
conditions:
()
()
()
()
q 0 = M 1/ 2 x 0 = éë0 0.7071 0 ùû
q 0 = M 1/ 2 x 0 = 0
The solution is given by Eq. (4.115):
()
3
x t = å di e
where
i =1
(
v q ( 0)
-z i w i t
)
sin w dit + fi u i
æ
ö
w di Ti
fi = tan ç T
÷
T
è v i q 0 + z iw i v i q 0 ø
-1
di =
()
v Ti q' 0
sin fi
()
T
()
i = 1,2,3
i = 1,2,3, u i = M -1/ 2 v i
z1 = 0.01, z 2 = 0.1,
z 3 = 0.2
Substituting
w d1 = 2.0770 ´ 10-4 rad/s, w d2 = 3.3972 ´ 10-4 rad/s w d3 = 8.8877 ´ 10-4 rad/s
f1 = 1.5808 rad, f 2 = 1.6710 rad, f3 = 1.3694 rad
d1 = 0.4312, d2 = 0,
The mode shapes are
d3 = 0.5720
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é 0.03963 ù
é -0.05ù
é 0.03049 ù
ê
ú
ê
ú
ê
ú
u1 = ê -0.04312 ú u 2 = ê 0 ú u3 = ê 0.05604 ú
ê 0.03963 ú
ê 0.05 ú
ê 0.03049 ú
ë
û
ë
û
ë
û
The solution is
é 0.01709 ù
-4
ê
ú
x t = ê -0.01859 ú e-2.0771´10 t sin 2.0770 ´ 10-4 t - 1.5808
ê 0.01709 ú
ë
û
é0.01744 ù
-4
ê
ú
+ ê0.03206 ú e-2.0771´10 t sin 8.8877 ´ 10-4 t + 1.3694 m
ê0.01744 ú
ë
û
(
()
)
(
4.65
)
Repeat Problem 4.64 with constant modal damping of z1 , z 2 , z 3 = 0.1. If you worked
the previous problem compare this solution with the solution of Problem 4.64.
Solution: Use the equations of motion and initial conditions from Problem 4.64. The
mode shapes, natural frequencies and transformed initial conditions remain the same.
However the constants of integration are effected by the damping ratio so the solution
()
3
x t = å di e
-z iw i t
i =1
(
)
sin w dit + fi ui
()
æ
ö
w di v Ti q 0
has new constants determined by fi = tan ç T
÷
T
è v i q 0 + z iw i v i q 0 ø
-1
di =
()
v Ti q' 0
sin fi
()
()
i = 1,2,3
i = 1,2,3
u i = M -1/ 2 v i
z1 = z 2 = z 3 = 0.1
Substituting yields
w d1 = 2.0667 ´ 10-4 rad/s, w d2 = 3.3972 ´ 10-4 rad/s, w d3 = 9.0255 ´ 10-4 rad/s
f1 = -1.6710 rad, f2 = -1.6710 rad, f3 = 1.4706 rad
d1 = 0.4334, d2 = 0.0,
d3 = 0.5633
Mode shapes:
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é 0.03963 ù
é -0.05ù
é 0.03049 ù
ê
ú
ê
ú
ê
ú
u1 = ê -0.04312 ú u 2 = ê 0 ú u3 = ê 0.05604 ú
ê 0.03963 ú
ê 0.05 ú
ê 0.03049 ú
ë
û
ë
û
ë
û
The solution is
é 0.01717 ù
-4
ê
ú
x t = ê -0.01869 ú e-2.0771´10 t sin 2.0667 ´ 10-4 t - 1.6710
ê 0.01717 ú
ë
û
é0.01717 ù
-4
ê
ú
+ ê0.03157 ú e-9.0710´10 t sin 9.0255 ´ 10-4 t + 1.4706 m
ê0.01717 ú
ë
û
()
(
)
(
)
The primary difference between problems 4.58 and 4.59 is the settling time; the
responses in Problem 4.59 decay faster than those of Problem 4.58.
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4.66
Consider the damped system of Figure P4.66. Determine the damping matrix and use the
formula of Eq. (4.119) to determine values of the damping coefficient cI for which this
system would be proportionally damped.
Figure P4.66 A damped 2 degree-of-freedom
system fixed at each end.
Solution:
From Fig. P4.66,
é m1 0 ù
éc + c
x+ê 1 2
ê
ú 
êë 0 m2 úû
êë -c2
é k1 + k2
-c2 ù
ú x + ê
c2 + c3 úû
êë -k2
-k2 ù
úx = 0
k2 + k3 úû
From Eq. (4.119)
C = a M + bK
é c1 + c2
ê
êë -c2
(
-c2 ù éa m1 + b k1 + k2
ú=ê
c2 + c3 úû êë
- b k2
)
ù
ú
a m1 + b k2 + k3 úû
- b k2
(
)
To be proportionally damped,
c2 = b k2
c1 = a m1 + b k1
c3 = a m2 + b k3
Alternately, compute KM C symbolically and show that the condition for symmetry:
-1
Requiring the off diagonal elements to be equal enforces symmetry. This requires
m1k2 c3 = m2 k2 c1 + (m2 k1 - m1k3 )c2
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4.67
Let k3 = 0 in Figure P4.66. Also let m1 = 1,m2 = 4,k1 = 2, k2 = 1 and calculate c1, c2 and c3
such that 1 = 0.02 and 2 = 0.2.
Solution:
Given:
From the question we have
 1 x = 0
1 0  x + c1 + c 2  c 2  x + 3






 1 1 
0 4 
c 2 + c 3 
 c 2
Also ξ1 = 0.02; ξ 2 = 0.2
Solution:
To calculate the Natural frequencies
~
K = 3.
 0.5


 0.5 0.25 
~
 det K  λI = 0
 λ1 = 0.161913 ω1 = 0.402384rad / s ωd1 = 0.386288rad / s
λ2 = 3.08809
ω 2 = 1.7573rad / s
ωd2 = 1.75659rad / s
α β ωi
α
β 0.4023
But ξ i =
+
2ω i 2  0.02 = 20.4023 +
2


α
β 1.75659
+
 α = 0.0219117 and β = 0.234815
21.75659 
2
and c1 + c 2  c 2  = αM + βK = 0.682533
 0.234815  c1 = 0.4477




 0.234815 0.147168 
c 2 + c 3 
 c 2
c 2 = 0.234815 c3 = 0.087647
Negative damping implies this design is impossible.
and 0.2 =
4.68
Calculate the constants  and  for the two-degree-of-freedom system given by
é1 0 ù
é 3 -1ù
x + a M + b K x + ê
ê
ú 
úx = 0
ë0 4 û
ë -1 1 û
such that the system has modal damping of z1 = z 2 = 0.3.
(
)
Solution:
From Problem 4.31 with proportional damping added,
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é1 0 ù
é 3 -1ù
ê
ú x + a M + b K x + ê
úx = 0
ë0 4 û
ë -1 1 û
(
)
Calculate natural frequencies:
é 3
-0.5ù
K = M -1/ 2 KM -1/ 2 = ê
ú
ë -0.5 0.25 û
det K - l I = l 2 - 3.25l + 0.5 = 0
(
)
l1 = 0.1619 w1 = 0.4024 rad/s
l2 = 3.0881 w 2 = 1.7573 rad/s
From Eq. (4.124)
zi =
So, 0.3 =
b ( 0.4024 )
a
+
2
2 ( 0.4024 )
and 0.3 =
b (1.7573)
a
+
2
2 (1,7573)
bw i
a
+
2w i
2
Solving for  and  yields
a = 0.1966
b = 0.2778
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4.69
Equation (4.124) represents n equations in only two unknowns and hence cannot be used
to specify all the modal damping ratios for a system with n > 2. If the floor vibration
system of Problem 4.53 has measured damping of 1 = 0.01 and 2 = 0.05, determine 3.
Solution:
From Problem 4.53
(
)
det K - l I = l 3 - 9.8255 ´ 10-7 l 2 + 1.3645 ´ 10 -14 l - 4.1382 ´ 10 -22 = 0
l1 = 4.3142 ´ 10-9 w 1 = 2.0771 ´ 10-5 rad/s
l2 = 1.1657 ´ 10-7 w 2 = 3.4143 ´ 10-4 rad/s
l3 = 8.2283 ´ 10-7
w 3 = 9.0710 ´ 10-4 rad/s
Eq. (4.124)
zi =
bw i
a
+
2w i
2
Since the problem contains three modes only, and since the first and second modal
damping ratios are give as z 1 = 0.01 and z 2 = 0.05 then the following linear system can
be set up
(
)
(
)
b 2.0771 ´ 10-5
a
+
= 0.01
2
2 2.0771 ´ 10-5
(
)
b 3.4143 ´ 10-4
a
+
= 0.05
2
2 3.4143 ´ 10-4
(
)
which can be solve to yield a = 2.9 ´ 10-7 and b = 290.397 . Hence, the modal damping
of the third mode can be obtained using 4.124
z3 =
4.70
bw 3
a
+
= 0.132
2w 3
2
If you worked the previous problem, calculate the damping matrix for the system of
Problem 4.69. What are the units of the elements of the damping matrix?
Solution:
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From Problem 4.69,
a = -8.8925 ´ 10-7
b = 3.0052 ´ 102
From Problem 4.69
é 200
0
0 ù
ê
ú
M = ê 0 2000 0 ú
ê 0
0
200 úû
ë
é 9 / 64 1 / 6 13 / 192 ù
ê
ú
K = 3.197 ´ 10 ê 1 / 6
1/ 3
1/ 6 ú
êë13 / 192 1 / 6 9 / 64 úû
-4
So,
C = a M + bK
é 0.01334 0.01602 0.006506 ù
ê
ú
C = ê 0.01602 0.03025 0.01602 ú
ê0.006506 0.01602 0.01334 ú
ë
û
The units are kg/s
4.71
Does the following system decouple? If so, calculate the mode shapes and write the
equation in decoupled form.
é1 0 ù
é 5 -3ù
é 5 -1ù
ê
ú x + ê
ú x + ê
úx = 0
ë0 1 û
ë -3 3 û
ë -1 1 û
Solution:
The system will decouple if
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C = a M + bK
é 5 -3ù éa + 5b
-b ù
ê
ú=ê
ú
a + bû
ë -3 3 û ë - b
Clearly the off-diagonal terms require
b=3
Therefore, the diagonal terms require
5 = a + 15
3=a +3
These yield different values of , so the system does not decouple. An easier approach is
to compute CM-1K to see if it is symmetric:
é 5 -3ù é 1 0 ù é 5 -1ù é 9 -2 ù
CM -1K = ê
úê
úê
ú=ê
ú
ë -3 3 û ë 0 1 û ë -1 1 û ë -12 6 û
Since this is not symmetric, the system cannot be decoupled.
4.72
Show that if the damping matrix satisfies C = a M + b K , then the matrix CM -1 K is
symmetric and hence that CM -1 K = KM -1C .
Solution: Compute the product CM -1 K where C has the form: C = a M + b K .
CM -1 = (a M + b K ) M -1 = a I + b KM -1 Þ CM -1 K = a K + b KM -1 K
KM -1C = KM -1 (a M + b K ) = a K + b KM -1 K
Þ KM -1C = CM -1 K
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Problems and Solutions for Section 4.6 (4.73 through 4.83)
4.73
Calculate the response of the system of Figure P4.73 discussed in Example 4.6.1
if F 2 (t) = δ(t) and the initial conditions are set to zero. This might correspond to a
two-degree-of-freedom model of a car hitting a bump.
Figure P4.73 A damped two-degree-offreedom system
Solution: From example 4.6.1, with F 1 (t) = δ(t), the modal equations are
Also from the example,
The solution to an impulse is given by equations (3.7) and (3.8):
This yields
The solution in physical coordinates is
4.74
Calculate the response of the system of Figure P4.73 discussed in Example 4.6.1
if F 1 (t) = δ(t) and the initial conditions are set to zero. This might correspond to a
two-degree-of-freedom model of a car hitting a bump.
Solution: From example 4.6.1, with F 1 (t) = δ(t), the modal equations are
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Also from the example,
The solution to an impulse is given by equations (3.7) and (3.8):
This yields
4.75
For an undamped two-degree-of-freedom system, show that resonance occurs at
one or both of the system’s natural frequencies.
Solution:
Undamped two-degree-of-freedom system:
Let
Note: placing F 1 on mass 1 is one way to do this. A second force could be placed
on mass 2 with or without F 1 .
Proceeding through modal analysis,
Or,
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where b 1 and b 2 are constants from the matrix PTM-1/2.
If F 1 (t) = a cos ωt and ω = ω1 then the solution for r 1 is (from Section 2.1),
The solution for r 2 is
If the initial conditions are zero,
Converting to physical coordinates X(t) = M-1/2Pr(t) yields
where c i is a constant from M-1/2P.
So, if the driving force contains just one natural frequency, both masses will be
excited at resonance. The driving force could contain the other natural frequency
(ω = ωn2 ), which would cause r 1 and r 2 to be
and
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so both masses still oscillate at resonance.
Also, if F 1 (t) = a 1 cos ω1 t + a 2 cos ω2 t where ω1 = ωn1 and ω2 = ωn2 , then both
r 1 and r 2 would be at resonance, so x 1 (t) and x 2 (t) would also be at resonance.
4.76
Use modal analysis to calculate the response of the drive train system of Problem
4.44 given by
to a unit impulse on the car body (i.e., and location x 3 ). Use the modal damping
of 10% in each mode. Calculate the solution in terms of physical coordinates, and
after subtracting the rigid-body modes, compare the responses of each part.
Solution:
Recall k 1 = hub stiffness and k 2 = axle and suspension stiffness. Then
The initial conditions are 0.
Also
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From equation (4.129):
Modal force vector:
The modal equations are
The solution for r 1 is
The solutions for r 2 and r 3 are given by equations 3.7 and (3.8)
This yields
The solution in physical coordinates is
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The magnitude of the components is much smaller than that in problem 54, but
they do oscillate at the same frequencies.
4.77
Consider the machine tool of Figure 4.28 with a floor mass of m = 1000 kg,
subject to a force of 10 sint (in Newtons) so that the equation of motions is


(103 ) 0.8 0 0 x(t ) + (104 )  30 −30 0  x(t ) =




 0 
 0 4 0
 −30 38 −8
 0 






−8 88 
 0 0 2 
 0
10sin t 
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Calculate the response. How much does this floor vibration affect the machine’s
toolhead compared to the solution given in Example 4.8.3?
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4.78
Consider the airplane of Figure P4.49 with modal damping of with 0.1 in each
mode. Suppose that the airplane hits a gust of wind, which applies an impulse of
3δ(t) at the end of the left wing and δ(t) at the end of the right wing. Calculate the
resulting vibration of the cabin [x 2 (t)].
Solution: From Problems 4.46 and 4.57
Also:
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From equation (4.129):
Modal force vector:
The modal equations are
The solution for r 1 is
The solutions for r 2 and r 3 are given by equations (3.7) and (3.8)
This yields
The solution in physical coordinates is
For x 2 :
4.79
Consider again the airplane of Figure P4.49 with 10% modal damping in each
mode. Suppose that this is a propeller-driven airplane with an internal
combustion engine mounted in the nose. At a cruising speed the engine mounts
transmit an applied force to the cabin mass (4m at x 2 ) which is harmonic of the
form 50 sin 10t. Calculate the effect of this harmonic disturbance at the nose and
on the wind tips after subtracting out the translational or rigid motion.
Copyright © 2015 Pearson Education Ltd.
Solution: From Problem 4.50
Also,
The initial conditions are 0. From equation (4.129):
Modal force vector:
The modal equations are
The solutions are
The solutions in physical coordinates is
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The wing tips are x 1 and x 3 , so
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4.80
Consider the automobile model of Problem 4.15 illustrated in Figure P4.15 with
equations of motion:
Add modal damping to this model of ζ 1 = 0.01 and ζ 2 = 0.2 and calculate the
response of the body [x 2 (t)] to a harmonic input at the second mass of 10 sin3t N.
Solution: From problem 4.15
Also,
The initial conditions are all 0. From equation (4.129):
Modal force vector:
The modal equations are
The solutions are
The solutions in physical coordinates is
The response of the body is
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4.81
Determine the modal equations for the following system and comment on
whether or not the system will experience resonance.
Solution: Here M = I so that the eigenvectors and mode shapes are the same.
yields:
Computing the natural frequencies from
ω 1 = 0.618 rad/s and ω2 =1.681 rad/s
Next solve for the mode shapes and normalize them to get
The modal equations then become:
The driving frequency is equal to the natural frequency of mode one so the system
exhibits resonance.
4.82
Consider the following system and compute the solution using the mode
summation method.
Solution: From Example 4.2.4
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4.83 Consider the following two systems and in each case determine if a resonance
response occurs.
(a)
(b)
where m 1 = 4 kg, k 1 = 25 N/m, m 2 = 9 kg, and k 2 = 5 N/m.
Solution
(a) First, substitute the numerical values to determine M and K:
Next, form the mass normalized stiffness matrix and compute the
Copyright © 2015 Pearson Education Ltd.
eigenvalues:
Solving the eigenvalue problem for this matrix yields λ 1 = 0.456956 and λ 2
= 7.5986 so that ω1 = 0.676 rad/s and ω2 = 2.756556 rad/s, neither of which
is equal to the driving frequency in this case. Thus there is no resonance in
case (a).
(b) Next, consider case (b). Clearly the second natural frequency is exactly equal
to the driving frequency, so we are tempted to think that resonance will
occur. However, a check of the modal equations will show that this second
mode is not excited. Finishing steps 3 through 5 in Window 4.5 yields that
Using S to transform the physical coordinate system into modal coordinates
by substituting x(t) = Sr(t) into the equations of motion yields
Multiplying by ST yields the following modal equations:
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Note that because of the nature of the vector force distribution, the second
mode has no force applied to it. With no force applied, the mode cannot
cause resonance. This is because the vector of the force distribution is
orthogonal to the second mode shape, causing f 2 (t) in equation (4.130) to be
zero. An independent check of this result is given in Example 4.10.1 by
solving for the time response directly using numerical integration.
It is important to state that if the vector of force distribution in case (b) were
changed to any other set of numbers, case (b) would be resonant. For instance, the
following force will cause resonance in the system of the previous example:
In order for resonance to be avoided when the driving frequency coincides
with one of the natural frequencies, the vector b must be orthogonal to the
mode shape corresponding to that natural frequency. This very seldom
happens unless by design. The most common case of resonance is when the
driving frequency is near a natural frequency. The rest of the modal analysis
procedure remains exactly the same. Here the vector holding the location of
the applied force is named b. This could be a vector or a matrix depending
on how many different forces are applied to a system.
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Problems and Solutions for Section 4.7 (4.84 through 4.87)
4.84
Use Lagrange's equation to derive the equations of motion of the lathe of Fig. 4.21 for the
undamped case.
Solution: Let the generalized coordinates be q1 ,q 2 and q 3 .
The kinetic energy is
1
1
1
T = J1q12 + J 2q22 + J 3q32
2
2
2
The potential energy is
2
2
1
1
U = k1 q 2 - q 2 + k2 q 3 - q 2
2
2
There is a nonconservative moment M(t) on inertia 3. The Lagrangian is
2
1
1
1
1
1
L = T - U = J1q12 + J 2q22 + J 3q32 - k1 q 2 - q1 - k2 q 3 - q 2
2
2
2
2
2
Calculate the derivatives from Eq. (4.136):
¶L
d æ ¶L ö
= J1q1
= J1q1
¶q1
dt çè ¶q1 ÷ø
(
)
(
)
(
¶L
= J 2q2
¶q2
d æ ¶L ö
= J 2q2
dt çè ¶q2 ÷ø
¶L
= J 3q3
¶q3
d æ ¶L ö
= J 3q3
dt çè ¶q3 ÷ø
)
¶L
= -k1q1 + k1q 2
¶q1
¶L
= -k1q1 - k1 + k2 q 2 + k2q 3
¶q 2
(
)
¶L
= -k2q 2 - k2q3
¶q 3
Using Eq. (4.136) yields
J1q1 + k1q1 - k2q 2 = 0
(
)
J 2q2 - k1q1 + k1 + k2 q 2 - k2q 3 = 0
()
J 3q3 - k2q 2 + k2q 3 = M t
In matrix form this yields
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(
)
2
é J1
ê
ê0
ê0
ë
4.85
0
J2
0
é k1
0ù
ú  ê
0 ú q + ê -k1
ê 0
J 3 úû
ë
0 ù
é 0
ú
ê
-k2 ú q = ê 0
êM t
k2 úû
ë
-k1
k1 + k2
()
-k2
ù
ú
ú
ú
û
Use Lagrange's equations to rederive the equations of motion for the automobile of
Example 4.8.2 illustrated in Figure 4.25 for the case c1 = c2 = 0 .
Solution: Let the generalized coordinates be x and .
The kinetic energy is
1
1
T = mx 2 + Jq 2
2
2
The potential energy is (ignoring gravity)
2
2
1
1
U = k1 x - l1q + k2 x + l2q
2
2
The Lagrangian is
2
1
1
1
1
L = T - U = mx 2 + Jq 2 - k1 x - l1q - k2 x + l2q
2
2
2
2
Calculate the derivatives from Eq. (4.136):
¶L
d æ ¶L ö
= mx
= m
x
¶x
dt çè ¶x ÷ø
(
)
(
)
(
¶L
= Jq
¶q
)
(
d æ ¶L ö
= Jq
dt çè ¶q ÷ø
¶L
= - k1 + k2 x + k1l1 - k1l2 q
¶x
¶L
= k1l1 - k2 l2 x - k1l22 q
¶q
(
) (
)
) ( )
(
Using Eq. (4.136) yields
m
x + k1 + k2 x + k1l1 - k2 l2 q = 0
(
Jq + ( k l
1 2
) (
- k l ) x - (k l
11
2
11
)
)
+ k2 l22 q = 0
In matrix form this yields
k2 l2 - k1l1 ù é x ù
é m 0 ù é xù é k1 + k2
=0
ê
ú ê ú + ê
2
2úê ú
ë 0 J û ëq û êë k2 l2 - k1l1 k1l1 + k2 l2 úû ëq û
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)
2
4.86
Use Lagrange's equations to rederive the equations of motion for the building model
presented in Fig. 4.9 of Ex. 4.4.3 for the undamped case.
Solution:
Let the generalized coordinates be x1, x2, x3 and x4.
The kinetic energy is
T=
1
1
1
1
m1 x12 + m2 x22 + m3 x32 + m4 x42
2
2
2
2
The potential energy is (ignoring gravity)
U=
(
)
(
2
1 2 1
1
k1 x1 + k2 x2 - x1 + k3 x3 - x2
2
2
2
)
2
(
1
+ k4 x4 - x3
2
)
2
The Lagrangian is
1 2 1 2 1 2 1 2
mx + mx + mx + mx
2 1 2 2 2 3 2 4
2 1
1
1
- k1 x12 - k2 x2 - x1
k x -x
2
2
2 3 3 2
L = T -U =
(
)
(
)
2
Calculate the derivatives from Eq. (4.136):
¶L
= m1 x1
¶x1
d æ ¶L ö
= m1 
x1
dt çè ¶x1 ÷ø
¶L
= m2 x2
¶x1
d æ ¶L ö
= m2 
x2
dt çè ¶x1 ÷ø
¶L
= m3 x3
¶x1
d æ ¶L ö
= m3 
x3
dt çè ¶x1 ÷ø
¶L
= m4 x4
¶x1
d æ ¶L ö
= m4 
x4
dt çè ¶x1 ÷ø
(
1
- k4 x4 - x3
2
Copyright © 2015 Pearson Education Ltd.
)
2
¶L
= - k1 + k2 x1 + k2 x2
¶x1
(
)
¶L
= k2 x1 - k2 + k3 x2 + k3 x3
¶x2
(
)
¶L
= k2 x2 - k2 + k4 x3 - k4 x4
¶x3
(
)
¶L
= k4 x3 - k4 x4
¶x4
Using Eq. (4.136) yields
(
)
m x - k x + ( k + k ) x - k x
m x - k x + ( k + k ) x - k x
m1 x1 + k1 + k2 x1 - k2 x2 = 0
2 2
2 1
2
3
2
3 3
=0
3 3
3 2
3
4
3
4 4
=0
m4 x4 - k4 x3 + k4 x4 = 0
In matrix form this yields
é m1
ê
ê0
ê0
ê
êë 0
0
0
m2
0
0
m3
0
0
é k1 + k2
0ù
ú
ê
0ú
ê -k2
x+ê

ú
0
0
ú
ê
êë 0
m4 úû
-k2
0
k2 + k3
-k3
-k3
k3 + k4
0
-k4
Copyright © 2015 Pearson Education Ltd.
0 ù
ú
0 ú
x=0
-k4 ú
ú
k4 úû
4.87
Consider again the model of the vibration of an automobile of Fig. 4.25. In this case
include the tire dynamics as indicated in Fig. P4.87. Derive the equations of motion
using Lagrange formulation for the undamped case. Let m3 denote the mass of the car
acting at c.g.
Figure P4.87
Solution:
Let the generalized coordinates be x1 , x2 , x3 and q . The kinetic energy is
T=
1
1
1
1
m1 x12 + m2 x22 + m3 x32 + Jq 2
2
2
2
2
The potential energy is (ignoring gravity)
2
1
1
1
1
U = k1 x3 -  1q - x1 + k2 (x3 +  2q - x2 )2 + k3 x12 + k4 x22
2
2
2
2
The Lagrangian is thus:
1
1
1
1
1
L = T - U = m1 x12 + m2 x22 + m3 x32 + Jq 2 - k1 x3 - l1q - x1
2
2
2
2
2
2
1
1
1
- k2 x3 + l2q - x2 - k3 x12 - k4 x22
2
2
2
(
)
(
(
)
Calculate the derivatives indicated in Eq. (4.146):
¶L
= m1 x1
¶x1
d æ ¶L ö
= m1 
x1
dt çè ¶x1 ÷ø
¶L
= m2 x2
¶x2
d æ ¶L ö
= m1 x2
dt çè ¶x2 ÷ø
¶L
= m3 x3
¶x3
d æ ¶L ö
= m3 
x3
dt çè ¶x3 ÷ø
¶L
= Jq
¶q
d æ ¶L ö
= Jq
dt çè ¶q÷ø
Copyright © 2015 Pearson Education Ltd.
)
2
¶L
= - ( k1 + k3 ) x1 + k1 x3 - k1l1q
¶x1
¶L
= - ( k2 + k4 ) x2 + k2 x3 + k2 l2q
¶x2
¶L
= k1 x1 + k2 x2 - ( k1 + k2 ) x3 + ( k1l1 - k2 l2 )q
¶x3
¶L
= -k1l1 x1 + k2 l2 x2 + ( k1l1 - k2 l2 ) x3 - k1l12 + k2 l22 q
¶q
(
)
Using Eq. (4.146)
d æ ¶L ö ¶L
= 0, i = 1, 2, 3, 4
dt çè ¶qi ÷ø ¶qi
yields
m1 
x1 + ( k3 + k1 ) x1 - k1 x3 + k1l1q = 0
m2 
x2 + ( k4 + k2 ) x2 - k2 x3 - k2 l2q = 0
m3 
x3 - k1 x1 - k2 x2 + ( k1 + k2 ) x3 - ( k1l1 - k2 l2 )q = 0
(
)
Jq + k1l1 x1 - k2 l2 x2 - ( k1l1 - k2 l2 ) x3 + k1l12 + k2 l22 = 0
in matrix form
é m1
ê0
ê
ê0
ê
ë0
0
m2
0
0
0
0
m3
0
0 ù ì x1 ü é( k1 + k3 )
ê
0 ú ïï 
x2 ïï ê 0
úí ý+
0 ú ï 
x3 ï ê -k1
ú ï  ï ê
J û î q þ êë k1l1
0
( k2 + k 4 )
-k2
-k1
-k2
( k1 + k2 )
-k2 l2
(k2 l2 - k1l1 )
Copyright © 2015 Pearson Education Ltd.
ù ì x1 ü
k1l1
ú
-k2 l2 ú ïï x2 ïï
=0
(k2 l2 - k1l1 ) ú í x3 ý
úï ï
k1l12 + k2 l22 úû ïî q ïþ
(
)
Problems and Solutions for Section 4.9 (4.88 through 4.98)
4.88
Consider the mass matrix
and calculate M-1, M-1/2, and the Cholesky factor of M. Show that
Solution: Given
The matrix, P, of eigenvectors is
The eigenvalues of M are
From Equation
From Equation
The following Mathcad session computes the Cholesky decomposition.
Copyright © 2015 Pearson Education Ltd.
4.89
Consider the matrix and vector
Use a code to solve Ax = b for ε = 0.1, 0.01, 0.001, 10-4, and 1.
Solution:
octave:183> epsilon = 0.1
epsilon = 0.10000
octave:184> A = [1 -epsilon;-epsilon epsilon]
A=
1.00000 -0.10000
-0.10000 0.10000
octave:185> b = [10;10]
b=
10
10
octave:186> x = inv(A)*b
x=
22.222
122.222
octave:187> epsilon = 0.01
epsilon = 0.010000
octave:188> A = [1 -epsilon;-epsilon epsilon]
A=
1.000000 -0.010000
-0.010000 0.010000
octave:189> x = inv(A)*b
x=
20.202
1020.202
octave:190> epsilon = 0.001
epsilon = 0.0010000
octave:191> A = [1 -epsilon;-epsilon epsilon]
A=
1.0000000 -0.0010000
-0.0010000 0.0010000
octave:192> x = inv(A)*b
x=
2.0020e+01
1.0020e+04
octave:193> epsilon = 1e-4
epsilon = 1.0000e-04
octave:194> A = [1 -epsilon;-epsilon epsilon]
A=
1.0000e+00 -1.0000e-04
Copyright © 2015 Pearson Education Ltd.
-1.0000e-04 1.0000e-04
octave:195> x = inv(A)*b
x=
2.0002e+01
1.0002e+05
octave:196> epsilon = 1
epsilon = 1
octave:197> A = [1 -epsilon;-epsilon epsilon]
A=
1 -1
-1 1
octave:198> x = inv(A)*b
warning: inverse: matrix singular to machine precision, rcond = 0
x=
Inf
Inf
It can be observed that as the epsilon value is increased the matrix is becoming ill
conditioned.
4.90
Calculate the natural frequencies and mode shapes of the system of Example
4.8.3. Use the undamped equation and the form given by equation (4.161).
Solution:
The following MATLAB program will calculate the natural frequencies and
mode shapes for Example 4.8.3 using Equation (4.161).
m=[0.4 0 0;0 2 0;0 0 8]*1e3;
k=[30 –30 0;-30 38 –8;0 –8 88]1e4;
[u, d]=eig(k,m);
w=sqrt (d);
The matrix d contains the square of the natural frequencies, and the matrix u
contains the corresponding mode shapes.
4.91
Compute the natural frequencies and mode shapes of the undamped version of the
system of Example 4.8.3 using the formulation of equation (4.164) and (4.168).
Compare your answers.
Copyright © 2015 Pearson Education Ltd.
Solution:
The following MATLAB program will calculate the natural frequencies and
mode shapes for Example 4.8.3 using Equation (4.161).
m=[0.4 0 0;0 2 0;0 0 8]*1e3;
k=[30 –30 0;-30 38 –8;0 –8 88]1e4;
mi=inv(m);
kt=mi*k;
[u, d]=eig(k,m);
w=sqrt (d);
The number of floating point operations needed is 439.
The matrix d contains the square of the natural frequencies, and the matrix u
contains the corresponding mode shapes.
The following MATLAB program will calculate the natural frequencies and
mode shapes for Example 4.8.3 using Equation (4.168).
m=[0.4 0 0;0 2 0;0 0 8]*1e3;
k=[30 –30 0;-30 38 –8;0 –8 88]1e4;
msi=inv(sqrt(m));
kt=msi*k*msi;
[p, d]=eig(kt);
w=sqrt (d);
u=msi*p;
The number of floating point operations needed is 461.
The matrix d contains the square of the natural frequencies, and the matrix u
contains the corresponding mode shapes.
The method of Equation (4.161) is faster.
4.92
Use a code to solve for the modal information of the following system
Solution: See Toolbox or use the following Mathcad code:
Copyright © 2015 Pearson Education Ltd.
4.93
Write a program to normalization the vector x = [0.4450 0.8019 1]T.
Solution:
The following MATLAB program will perform the normalization of
Example 4.4.2.
x=[.4450 .8019 1];
mag=sqrt(sum(x.^2));
xnorm=x/mag;
The variable mag is the same as α, and xnorm is the normalized vector.
The original vector x can be any length.
4.94
Use a code to calculate the natural frequencies and mode shapes obtained for the
system
x(t ) + 
0

 
 x(t ) =
1 0 
 24 −4 
0 4 
 −4 24 




Solution: The Octave Code to evaluate Natural Frequencies and mode shapes
obtained for the given system:
clear
clc
close all
M = [1 0 ; 0 4]
K = [24 -4 ; -4 24]
Mr = sqrt(M)
Kd = inv(Mr)*K*inv(Mr)
Copyright © 2015 Pearson Education Ltd.
%Finding Eigen Values and Eigen Vectors of Kd
[Eig_Vec, Eig_Val] = eig(Kd)
omega1 = sqrt(Eig_Val(1))
omega2 = sqrt(Eig_Val(4))
P = [Eig_Vec(:,1), Eig_Vec(:,2)]
%Orthogonal Testing of P Matrix P'P = I
P'*P
D = P'*Kd*P
Output from the code
M= 1 0
0 4
K=
24 -4
-4 24
Mr =
1 0
0 2
Kd =
24 -2
-2 6
Eig_Vec =
-0.10912 -0.99403
-0.99403 0.10912
Eig_Val = Diagonal Matrix
5.7805
0
0 24.2195
omega1 = 2.4043
omega2 = 4.9213
P=
-0.10912 -0.99403
-0.99403 0.10912
ans =
1.00000 0.00000
0.00000 1.00000
D=
5.78046 -0.00000
0.00000 24.21954
4.95
Following the modal analysis solution of Window 4.5, write a program to
compute the time response of the system of Problem 4.94.
Solution: The following MATLAB program will compute and plot the time
response of the system of Example 4.3.2.
Copyright © 2015 Pearson Education Ltd.
t=(0:.1:10)’;
m=[1 0;0 4];
k=[12 –2;-2 12];
n=max(size(m));
x0=[1 1]’;
xd0=[0 0]’;
msi=inv(sqrtm(m));
kt=msi*k*msi;
[p, w]=eig(kt);
for i=1: n-1
for j=1: n-I
if w(j,j)>w(j+1,j+1)
dummy=w(j,j);
w(j,j)=w(j+1,j+1);
w(j+1,j+1)=dummy;
dummy=p(:,j);
p(:,j)=p(:,j+1);
p(:,j+1)=dummy;
end
end
end
pt=p’;
s=msi*p;
si=pt*sqrtm(m);
r0=si*x0;
rd0=si*xd0;
r=[];
for i=1: n,
wi=sqrt(w(i,i));
rcol=(swrt((wi*r0(i))^2+rd0(i)^2/wi)*…
sin(wi*t+atan2(wi*r0(i),rd0(i)));
r(:,i)=rcol;
end
x=s*r;
plot(t,x);
end
4.96
Use a code to solve the damped vibration problem
by calculating the natural frequencies, damping ratios, and mode shapes.
Copyright © 2015 Pearson Education Ltd.
Solution: See Toolbox or use the following Mathcad code (all will do this)
4.97
Consider the vibration of the airplane of Problems 4.46 and 4.47 as given in
Figure P4.46. The mass and stiffness matrices are given as
Copyright © 2015 Pearson Education Ltd.
where m = 4000 kg, l = 2 m, I = 5.2 × 10-6 m4, E = 6.9 × 109 N/m2, and the
damping matrix C is taken to be C = (0.002)K. Calculate the natural frequencies,
normalized mode shapes, and damping ratios.
Solution: octave:228> clc
octave:229> clear
octave:230> m =4000
m = 4000
octave:231> l=2
l= 2
octave:232> I = 5.2E-6
I = 5.2000e-06
octave:233> E = 6.9E9
E = 6.9000e+09
octave:234> M=m*[1 0 0; 0 4 0; 0 0 1]
M=
4000
0
0
0 16000
0
0
0 4000
octave:235> K = (E*I/l^3)*[3 -3 0;-3 6 -3;0 -3 3]
K=
13455 -13455
0
-13455 26910 -13455
0 -13455 13455
octave:236> C = 0.002*K
C=
26.91000 -26.91000 0.00000
-26.91000 53.82000 -26.91000
0.00000 -26.91000 26.91000
octave:237> Mr = sqrt(M)
Mr =
63.24555 0.00000 0.00000
0.00000 126.49111 0.00000
0.00000 0.00000 63.24555
octave:238> kh = inv(Mr)*K*inv(Mr)
kh =
3.36375 -1.68187 0.00000
-1.68187 1.68187 -1.68187
0.00000 -1.68187 3.36375
octave:239> ch = inv(Mr)*C*inv(Mr)
ch =
0.0067275 -0.0033637 0.0000000
-0.0033637 0.0033637 -0.0033637
0.0000000 -0.0033637 0.0067275
octave:240> [evec,eval] = eig(kh)
Copyright © 2015 Pearson Education Ltd.
evec =
4.0825e-01 -7.0711e-01 -5.7735e-01
8.1650e-01 3.1572e-16 5.7735e-01
4.0825e-01 7.0711e-01 -5.7735e-01
eval =
Diagonal Matrix
-8.2020e-16
0
0
0 3.3638e+00
0
0
0 5.0456e+00
octave:241> lamda1 = eval(1)
lamda1 = -8.2020e-16
octave:242> lamda2 = eval(5)
lamda2 = 3.3638
octave:243> lamda3 = eval(9)
lamda3 = 5.0456
octave:244> omega1 = sqrt(lamda1)
omega1 = 0.0000e+00 + 2.8639e-08i
octave:245> omega2 = sqrt(lamda2)
omega2 = 1.8341
octave:246> omega3 = sqrt(lamda3)
omega3 = 2.2462
octave:247> V1 = evec(1,:)
V1 =
0.40825 -0.70711 -0.57735
octave:248> V1 = evec(:,1)
V1 =
0.40825
0.81650
0.40825
octave:249> V2 = evec(:,2)
V2 =
-7.0711e-01
3.1572e-16
7.0711e-01
octave:250> V3 = evec(:,3)
V3 =
-0.57735
0.57735
-0.57735
octave:251> P=[V1,V2,V3]
P=
4.0825e-01 -7.0711e-01 -5.7735e-01
8.1650e-01 3.1572e-16 5.7735e-01
4.0825e-01 7.0711e-01 -5.7735e-01
octave:252> lc = P'*ch*P
Copyright © 2015 Pearson Education Ltd.
lc =
0.000000 -0.000000 0.000000
-0.000000 0.006728 0.000000
0.000000 0.000000 0.010091
octave:253> zeta2 = lc(5)/(2*omega2)
zeta2 = 0.0018341
octave:254> zeta3 = lc(9)/(2*omega)
error: 'omega' undefined near line 1 column 18
octave:254> zeta3 = lc(9)/(2*omega3)
zeta3 = 0.0022462
octave:255> %Mode shapes
octave:255> u1 = inv(Mr)*V1
u1 =
0.0064550
0.0064550
0.0064550
octave:256> u2 = inv(Mr)*V2
u2 =
-1.1180e-02
2.4960e-18
1.1180e-02
octave:257> u3 = inv(Mr)*V3
u3 =
-0.0091287
0.0045644
-0.0091287
octave:258> %Normalized mode shapes
octave:260> norm(u1)
ans = 0.011180
octave:261> u1n = u1/norm(u1)
u1n =
0.57735
0.57735
0.57735
octave:262> u2n = u2/norm(u2)
u2n =
-7.0711e-01
1.5786e-16
7.0711e-01
octave:263> u3n = u3/norm(u3)
u3n =
-0.66667
0.33333
-0.66667
4.98
Consider the proportionally damped, dynamically coupled system given by
Copyright © 2015 Pearson Education Ltd.
 9 −1
 6 −2 
 49 −2 
=
M =
C =
K 



 −1 1 
 −2 2 
 −2 2 
and calculate the mode shapes, natural frequencies, and damping ratios.
Solution: octave:199> clear
octave:200> clc
octave:201> M = [9 -1; -1 1]
M=
9 -1
-1 1
octave:202> C=[6 -2 ; -2 2]
C=
6 -2
-2 2
octave:203> K=[49 -2 ; -2 2]
K=
49 -2
-2 2
octave:204> L = chol(M)
L=
3.00000 -0.33333
0.00000 0.94281
octave:205> kh = inv(L)*K*inv(L')
kh =
5.31509 -0.45711
-0.45711 2.25000
octave:206> ch = inv(L)*C*inv(L')
ch =
0.53731 -0.45711
-0.45711 2.25000
octave:207> eigenvals = eig(kh)
eigenvals =
2.1833
5.3818
octave:208> lambda1 = eigenvals(1)
lambda1 = 2.1833
octave:211> omega1 = sqrt(lambda1)
omega1 = 2.3199
octave:212> lambda2 = eigenvals(2)
lambda2 = 5.3818
octave:213> omega2 = sqrt(lambda2)
omega2 = 2.3199
octave:214> [v,l] = eig(kh)
v=
-0.14443 -0.98952
Copyright © 2015 Pearson Education Ltd.
-0.98952 0.14443
l=
Diagonal Matrix
2.1833
0
0 5.3818
octave:216> v1 = v(:,1)
v1 =
-0.14443
-0.98952
octave:217> v1 = v(:,2)
v1 =
-0.98952
0.14443
octave:218> v1 = v(:,1)
v1 =
-0.14443
-0.98952
octave:219> v2 = v(:,2)
v2 =
-0.98952
0.14443
octave:220> P = [v1,v2]
P=
-0.14443 -0.98952
-0.98952 0.14443
octave:221> P'*ch*P
ans =
2.08362 -0.68280
-0.68280 0.70369
octave:222> P'*P
ans =
1.00000 0.00000
0.00000 1.00000
octave:223> P'*ch*P
ans =
2.08362 -0.68280
-0.68280 0.70369
octave:224> lc=P'*ch*P
lc =
2.08362 -0.68280
-0.68280 0.70369
octave:225> zeta1 =lc(1)/(2*omega1)
zeta1 = 0.44908
octave:226> zeta2 =lc(4)/(2*omega1)
zeta2 = 0.15166
Copyright © 2015 Pearson Education Ltd.
Problems and Solutions Section 4.10 (4.99 through 4.106)
4.99* Recall the system of Example 1.7.3 for the vertical suspension system of a car
, with m = 1361 kg, k = 2.668 x 105 N/m, and
modeled by
c = 3.81 x 104 kg/s subject to the initial conditions of x(0) = 0 and v(0) = 0.01
m/s2. Solve this and plot the solution using numerical integration.
Solution: Use a Runge Kutta routine such as the one given in Mathcad here or
use the toolbox:
Copyright © 2015 Pearson Education Ltd.
4.100* Solve for the time response of Example 4.4.3 (i.e., the four-story building of
Figure 4.9) modeled by
subject to an initial displacement of x(0) = [0.001 0.010 0.020 0.025]T and zero
initial velocity. Compare the solutions obtained with using a modal analysis
approach to a solution obtained by numerical integration.
Solution: The following code provides the numerical solution.
Copyright © 2015 Pearson Education Ltd.
which compares very well with the plots given in Figure 4.11 obtained by plotting
the modal equations. One could also plot the modal response and numerical
response on the same graph to see a more rigorous comparison.
4.101* Reproduce the plots of Figure 4.13 for the two-degree of freedom system of
Example 4.5.1 given by
x(t ) + 
0, x(0) =
0

 
 x(t ) =
  , x (t ) =
9 0 
 27 −3
1 
0 1 
 −3 3 
0 




 
using an numerical integration code.
Solution: Matlab code for
clear
clc
x0=[0;0.1;1;0];
Copyright © 2015 Pearson Education Ltd.
ts=[0 100];
% options = odeset('RelTol',1e-4)
[t,x] = ode45('f',ts,x0);
plot(t,x(:,1),t,x(:,2),'-')
grid on
function v = f(t,x)
M = [9 0;0 1];
C=[0 0.0; 0.0 0.0];
K = [27 -3;-3 3];
B = [0;0];
A1 = [zeros(2) eye(2); -inv(M)*K -inv(M)*C];
v= A1*x
end
Copyright © 2015 Pearson Education Ltd.
4.102* Consider example 4.8.3 and a) using the damping ratios given, compute a
damping matrix in physical coordinates, b) use numerical integration to compute
the response and plot it, and c) use the numerical code to design the system so that
all 3 physical coordinates die out within 5 seconds (i.e., change the damping
matrix until the desired response results).
Solution: A Mathcad solution is presented. The damping matrix is found, as in
the previous problem, by keeping track of the various transformations. Using the
notation of the text, the damping matrix is constructed from:
as computed using the code that follows. With this form of the matrix the
damping ratios are adjusted until the desired criteria are met:
In changing the damping ratios it is best to start with the rubber component which
is the first mode-damping ratio. Doubling it nails the first two coordinates but
does not affect the third coordinate enough. Hence the second mode-damping
ratio must be changed (doubled here) to attack this mode. This could be
accomplished by adding a viscoelastic strip as described in Chapter 5 to the metal.
Thus the ratios given in the code above do the trick as the following plots show.
Note also how much the damping matrix changes.
Copyright © 2015 Pearson Education Ltd.
4.103* Compute and plot the time response of the system (Newtons):
x1   3
−0.5  x1   3 −1  x1  1
9 0   
+

0 1   
 +
  =
  sin(4t )

  x2   −0.5 0.5   x2   −1 1   x2  1
subject to the initial conditions:
using numerical integration.
.Solution:
Matlab code for
clear
clc
Copyright © 2015 Pearson Education Ltd.
x0=[0;0.1;1;0];
ts=[0 30];
% options = odeset('RelTol',1e-4)
[t,x] = ode45('f',ts,x0);
plot(t,x(:,1),t,x(:,2),'-')
grid on
function v = f(t,x)
M = [9 0;0 1];
C=[3 -0.5; -0.5 0.5];
K = [3 -1;-1 1];
B = [1;1];
t1=1;t2=3;
w=4;
A1 = [zeros(2) eye(2); -inv(M)*K -inv(M)*C];
f = inv(M)*B;
%v= A1*x + [0;0;f]*(stepfun(t,t1)-stepfun(t,t2));
v= A1*x + [0;0;f]*sin(4*t);
end
4.104* Consider the following system excited by a pulse of duration 0.1 s (in Newtons):
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x1   0.3 −0.05  x1   3 −1  x1  0 
 4 0   
+
 0 1   
 +
  =   [Φ (t − 1) − Φ (t − 3)]

  x2   −0.05 0.05   x2   −1 1   x2  1 
and subject to the initial conditions:
Compute and plot the response of the system using numerical integration. Here Φ
indicates the Heaviside Step Function introduced in Section 3.2.
Solution
Matlab code for
clear
clc
x0=[0;-0.1;0;0];
ts=[0 30];
% options = odeset('RelTol',1e-4)
[t,x] = lsode('f',ts,x0);
plot(t,x(:,1),t,x(:,2),'- -')
function v = f(t,x)
M = [2 0;0 1];
C=[0.3 -0.05; -0.05 0.05];
K = [3 -1;-1 1];
B = [0;1];
t1=1;t2=1.1;
A1 = [zeros(2) eye(2); -inv(M)*K -inv(M)*C];
f = inv(M)*B;
v= A1*x + [0;0;f]*(stepfun(t,t1)-stepfun(t,t2));
end
Copyright © 2015 Pearson Education Ltd.
4.105*Use numerical integration to compute and plot the time response of the system
(Newtons):
subject to the initial conditions:
Solution: Following the codes of Example 4.10.2 yields the solution directly.
Copyright © 2015 Pearson Education Ltd.
Copyright © 2015 Pearson Education Ltd.
4.106* Use numerical integration to compute and plot the time response of the system
(Newtons):
subject to the initial conditions:
Solution: Again follow Example 4.10.2 for the various codes. Mathcad is given.
Copyright © 2015 Pearson Education Ltd.
Copyright © 2015 Pearson Education Ltd.