8
V IBRATION AND
T IME R ESPONSE
CHAPTER OUTLINE
8/1
8/2
8/3
8/4
8/5
8/6
Introduction
Free Vibration of Particles
Forced Vibration of Particles
Vibration of Rigid Bodies
Energy Methods
Chapter Review
8/1 I N T R O D U C T I O N
An important and special class of problems in dynamics concerns
the linear and angular motions of bodies which oscillate or otherwise
respond to applied disturbances in the presence of restoring forces. A
few examples of this class of dynamics problems are the response of an
engineering structure to earthquakes, the vibration of an unbalanced
rotating machine, the time response of the plucked string of a musical
instrument, the wind-induced vibration of power lines, and the flutter
of aircraft wings. In many cases, excessive vibration levels must be reduced to accommodate material limitations or human factors.
In the analysis of every engineering problem, we must represent
the system under scrutiny by a physical model. We may often represent
a continuous or distributed-parameter system (one in which the mass
and spring elements are continuously spread over space) by a discrete
or lumped-parameter model (one in which the mass and spring elements are separate and concentrated). The resulting simplified model
is especially accurate when some portions of a continuous system are
relatively massive in comparison with other portions. For example, the
physical model of a ship propeller shaft is often assumed to be a massless but twistable rod with a disk rigidly attached to each end—one
disk representing the turbine and the other representing the propeller.
As a second example, we observe that the mass of springs may often be
neglected in comparison with that of attached bodies.
Not every system is reducible to a discrete model. For example, the
transverse vibration of a diving board after the departure of the diver
569
570 Chapter 8 Vibr at i on and Ti m e Res p ons e
is a somewhat difficult problem of distributed-parameter vibration. In
this chapter, we will begin the study of discrete systems, limiting our
discussion to those whose configurations may be described with one
displacement variable. Such systems are said to possess one degree of
freedom. For a more detailed study which includes the treatment of two
or more degrees of freedom and continuous systems, you should consult
one of the many textbooks devoted solely to the subject of vibrations.
The remainder of Chapter 8 is divided into four sections: Article
8 /2 treats the free vibration of particles and Art. 8 /3 introduces the
forced vibration of particles. Each of these two articles is subdivided
into undamped- and damped-motion categories. In Art. 8 /4 we discuss
the vibration of rigid bodies. Finally, an energy approach to the solution
of vibration problems is presented in Art. 8 /5.
The topic of vibrations is a direct application of the principles of
kinetics as developed in Chapters 3 and 6. In particular, construction of
a complete free-body diagram drawn for an arbitrary positive value of
the displacement variable, followed by application of the appropriate
governing equations of dynamics, will yield the equation of motion.
From this equation of motion, which is a second-order ordinary differential equation, you can obtain all information of interest, such as the
motion frequency, period, or the motion itself as a function of time.
8 /2 F R E E V I B R AT I O N O F P A R T I C L E S
When a spring-mounted body is disturbed from its equilibrium
position, its ensuing motion in the absence of any imposed external
forces is termed free vibration. In every actual case of free vibration,
there exists some retarding or damping force which tends to diminish the motion. Common damping forces are those due to mechanical
and fluid friction. In this article we first consider the ideal case
where the damping forces are small enough to be neglected. Then
we treat the case where the damping is appreciable and must be
accounted for.
k
m
Equilibrium
position
x
mg
Equation of Motion for Undamped Free Vibration
kx
N
(a)
Fs
Hard
Linear Fs = kx
Soft
x
(b)
Figure 8/1
We begin by considering the horizontal vibration of the simple
frictionless spring-mass system of Fig. 8 /1a. Note that the variable x
denotes the displacement of the mass from the equilibrium position,
which, for this system, is also the position of zero spring deflection.
Figure 8 /1b shows a plot of the force Fs necessary to deflect the
spring versus the corresponding spring deflection for three types of
springs. Although nonlinear hard and soft springs are useful in some
applications, we will restrict our attention to the linear spring. Such
a spring exerts a restoring force −kx on the mass—that is, when the
mass is displaced to the right, the spring force is to the left, and vice
versa. We must be careful to distinguish between the forces of magnitude Fs which must be applied to both ends of the massless spring
to cause tension or compression and the force F = −kx of equal magnitude which the spring exerts on the mass. The constant of proportionality k is called the spring constant, modulus, or stiffness and has
the units N /m or lb /ft.
A rt i c l e 8 / 2 F re e V i b ra t i o n o f P a r t i cl e s
The equation of motion for the body of Fig. 8 /1a is obtained by first
drawing its free-body diagram. Applying Newton’s second law in the
form ©Fx = mẍ gives
−kx = mẍ
mẍ + kx = 0
or
(8/1)
The oscillation of a mass subjected to a linear restoring force as described by this equation is called simple harmonic motion and is characterized by acceleration which is proportional to the displacement but
of opposite sign. Equation 8 /1 is normally written as
ẍ + ␻n 2x = 0
(8/2)
␻n = √k/m
(8/3)
where
is a convenient substitution whose physical significance will be clarified
shortly.
Solution for Undamped Free Vibration
Because we anticipate an oscillatory motion, we look for a solution
which gives x as a periodic function of time. Thus, a logical choice is
x = A cos ␻nt + B sin ␻nt
(8/4)
x = C sin (␻nt + ␺)
(8/5)
or, alternatively,
Direct substitution of these expressions into Eq. 8/2 verifies that each expression is a valid solution to the equation of motion. We determine the
constants A and B, or C and ␺, from knowledge of the initial displacement
x0 and initial velocity ẋ0 of the mass. For example, if we work with the solution form of Eq. 8/4 and evaluate x and ẋ at time t = 0, we obtain
x0 = A
and
ẋ0 = B␻n
Substitution of these values of A and B into Eq. 8 /4 yields
x = x0 cos ␻nt +
ẋ0
sin ␻nt
␻n
(8/6)
The constants C and ␺ of Eq. 8 /5 can be determined in terms of
given initial conditions in a similar manner. Evaluation of Eq. 8 /5 and
its first time derivative at t = 0 gives
x0 = C sin ␺
and
ẋ0 = C␻n cos ␺
571
572 Chapter 8 Vibr at i on and Ti m e Res p ons e
Solving for C and ␺ yields
␺ = tan−1 (x0␻n /ẋ0 )
C = √x02 + (ẋ0 /␻n ) 2
Substitution of these values into Eq. 8 /5 gives
x = √x02 + (ẋ0 /␻n ) 2 sin [␻nt + tan −1 (x0␻n /ẋ0 )]
(8/7)
Equations 8/6 and 8/7 represent two different mathematical expressions
for the same time-dependent motion. We observe that C = √A2 + B2
and ␺ = tan−1(A /B).
Graphical Representation of Motion
The motion may be represented graphically, Fig. 8 /2, where x is
seen to be the projection onto a vertical axis of the rotating vector of
length C. The vector rotates at the constant angular velocity ␻n = √k/m,
which is called the natural circular frequency and has the units radians
per second. The number of complete cycles per unit time is the natural
frequency ƒn = ␻n /2␲ and is expressed in hertz (1 hertz (Hz) = 1 cycle
per second). The time required for one complete motion cycle (one rotation of the reference vector) is the period of the motion and is given by
␶ = 1 /ƒn = 2␲ /␻n.
x
+x
C
ω nt
ψ
A
B
ω nt
2π
τ = —–
ωn
C
x
x0
0
t
0
−C
−x
Figure 8/2
We also see from the figure that x is the sum of the projections onto
the vertical axis of two perpendicular vectors whose magnitudes are A
and B and whose vector sum C is the amplitude. Vectors A, B, and C
rotate together with the constant angular velocity ␻n. Thus, as we have
already seen, C = √A2 + B2 and ␺ = tan−1(A /B).
Equilibrium Position as Reference
As a further note on the free undamped vibration of particles, we
see that, if the system of Fig. 8 /1a is rotated 90° clockwise to obtain the
system of Fig. 8 /3 where the motion is vertical rather than horizontal,
573
A rt i c l e 8 / 2 F re e V i b ra t i o n o f P a r t i cl e s
the equation of motion (and therefore all system properties) is unchanged if we continue to define x as the displacement from the equilibrium position. The equilibrium position now involves a nonzero
spring deflection ␦st. From the free-body diagram of Fig. 8 /3, Newton’s
second law gives
−k(␦st + x) + mg = mẍ
k
δ st
k(δ st + x)
x
Equilibrium
position
m
At the equilibrium position x = 0, the force sum must be zero, so that
m
−k␦st + mg = 0
mg
Thus, we see that the pair of forces −k␦st and mg on the left side of the
motion equation cancel, giving
Figure 8/3
mẍ + kx = 0
which is identical to Eq. 8 /1.
The lesson here is that by defining the displacement variable to
be zero at equilibrium rather than at the position of zero spring deflection, we may ignore the equal and opposite forces associated with
equilibrium.*
k
Equation of Motion for Damped Free Vibration
Every mechanical system possesses some inherent degree of friction, which dissipates mechanical energy. Precise mathematical models
of the dissipative friction forces are usually complex. The dashpot or
viscous damper is a device intentionally added to systems for the purpose of limiting or retarding vibration. It consists of a cylinder filled
with a viscous fluid and a piston with holes or other passages by which
the fluid can flow from one side of the piston to the other. Simple dashpots arranged as shown schematically in Fig. 8 /4a exert a force Fd
whose magnitude is proportional to the velocity of the mass, as depicted in Fig. 8 /4b. The constant of proportionality c is called the viscous damping coefficient and has units of N ∙ s/m or lb-sec /ft. The
direction of the damping force as applied to the mass is opposite that of
the velocity ẋ. Thus, the force on the mass is −cẋ.
Complex dashpots with internal flow-rate-dependent one-way
valves can produce different damping coefficients in extension and in
compression; nonlinear characteristics are also possible. We will restrict our attention to the simple linear dashpot.
The equation of motion for the body with damping is determined
from the free-body diagram as shown in Fig. 8 /4a. Newton’s second law
gives
−kx − cẋ = mẍ
or
mẍ + cẋ + kx = 0
m
c
Equilibrium
position
x
kx
cx·
N
(a)
Fd
x·
(8/8)
(b)
*For nonlinear systems, all forces, including the static forces associated with equilibrium,
should be included in the analysis.
mg
Figure 8/4
574 Chapter 8 Vibr at i on and Ti m e Res p ons e
In addition to the substitution ␻n = √k/m, it is convenient, for reasons
which will shortly become evident, to introduce the combination of
constants
␨ = c/(2m␻n )
The quantity ␨ (zeta) is called the viscous damping factor or damping
ratio and is a measure of the severity of the damping. You should verify
that ␨ is nondimensional. Equation 8 /8 may now be written as
ẍ + 2␨␻nẋ + ␻n2x = 0
(8/9)
Solution for Damped Free Vibration
In order to solve the equation of motion, Eq. 8 /9, we assume solutions of the form
x = Ae␭t
Substitution into Eq. 8 /9 yields
␭2 + 2␨␻n␭ + ␻n2 = 0
which is called the characteristic equation. Its roots are
␭1 = ␻n (−␨ + √␨2 − 1)
␭2 = ␻n (−␨ − √␨2 − 1)
Linear systems have the property of superposition, which means that
the general solution is the sum of the individual solutions each of
which corresponds to one root of the characteristic equation. Thus, the
general solution is
x = A1e␭1 t + A2 e␭2 t
= A1e( −␨+√␨ −1)␻n t + A2 e(−␨−√␨ −1)␻nt
2
2
(8/10)
Categories of Damped Motion
Because 0 ≤ ␨ ≤ ∞, the radicand (␨2 − 1) may be positive, negative, or even zero, giving rise to the following three categories of
damped motion:
I. ␨ > 1 (overdamped). The roots ␭1 and ␭2 are distinct, real, and negative numbers. The motion as given by Eq. 8/10 decays so that x approaches zero for large values of time t. There is no oscillation and
therefore no period associated with the motion.
II. ␨ = 1 (critically damped). The roots ␭1 and ␭2 are equal, real, and negative numbers (␭1 = ␭2 = −␻n). The solution to the differential equation for the special case of equal roots is given by
x = (A1 + A2t)e −␻nt
A rt i c l e 8 / 2 F re e V i b ra t i o n o f P a r t i cl e s
x, mm
30
Conditions: m = 1 kg, k = 9 N/m
x0 = 30 mm, x· 0 = 0
20
c = 15 N·s/m (ζ = 2.5), overdamped
c = 6 N·s/m (ζ = 1), critically damped
10
0
0
1
2
3
4
t, s
Figure 8/5
Again, the motion decays with x approaching zero for large time, and
the motion is nonperiodic. A critically damped system, when excited
with an initial velocity or displacement (or both), will approach equilibrium faster than will an overdamped system. Figure 8/5 depicts
actual responses for both an overdamped and a critically damped
system to an initial displacement x0 and no initial velocity (ẋ0 = 0).
III. ␨ < 1 (underdamped). Noting that the radicand (␨2 − 1) is negative
and recalling that e(a+b) = eaeb, we may rewrite Eq. 8/10 as
x = 5A1 ei√1−␨ ␻nt + A2 e −i√1−␨ ␻nt 6 e−␨␻nt
2
2
where i = √−1. It is convenient to let a new variable ␻d represent
the combination ␻n√1 − ␨2. Thus,
x = 5A1ei␻d t + A2 e−i␻d t 6 e−␨␻nt
Use of the Euler formula e±ix = cos x ± i sin x allows the previous
equation to be written as
x = 5A1 (cos ␻dt + i sin ␻dt) + A2 (cos ␻dt − i sin ␻dt)6e−␨␻nt
= 5 (A1 + A2 ) cos ␻d t + i(A1 − A2 ) sin ␻dt6e −␨␻nt
= 5A3 cos ␻dt + A4 sin ␻d t6 e−␨␻nt
(8/11)
where A3 = (A1 + A2) and A4 = i(A1 − A2). We have shown with
Eqs. 8 /4 and 8 /5 that the sum of two equal-frequency harmonics,
such as those in the braces of Eq. 8 /11, can be replaced by a single trigonometric function which involves a phase angle. Thus,
Eq. 8 /11 can be written as
x = 5C sin (␻d t + ␺)6 e−␨␻nt
or
x = Ce−␨␻n t sin (␻d t + ␺)
(8/12)
575
576 Chapter 8 Vibr at i on and Ti m e Res p ons e
x, mm
Conditions: m = 1 kg, k = 36 N/m
c = 1 N·s /m (ζ = 0.0833)
x0 = 30 mm, x· 0 = 0
–
ζ
ωnt
Ce
30
20
Equation 8 /12 represents an exponentially decreasing harmonic
function, as shown in Fig. 8 /6 for specific numerical values. The
frequency
␻d = ␻n√1 − ␨2
τd
10
0
0
–10
1
3
2
4 t, s
– Ce – ζ ωnt
–20
is called the damped natural frequency. The damped period is given
by ␶d = 2␲ /␻d = 2␲/(␻n √1 − ␨2 ).
It is important to note that the expressions developed for the constants C and ␺ in terms of initial conditions for the case of no damping
are not valid for the case of damping. To find C and ␺ if damping is present, you must begin anew, setting the general displacement expression
of Eq. 8 /12 and its first time derivative, both evaluated at time t = 0,
equal to the initial displacement x0 and initial velocity ẋ0, respectively.
–30
Figure 8/6
Determination of Damping by Experiment
x
We often need to experimentally determine the value of the damping ratio ␨ for an underdamped system. The usual reason is that the
value of the viscous damping coefficient c is not otherwise well known.
To determine the damping, we may excite the system by initial conditions and obtain a plot of the displacement x versus time t, such as that
shown schematically in Fig. 8 /7. We then measure two successive amplitudes x1 and x2 a full cycle apart and compute their ratio
2π
τd = —
ωd
x1
t1
x1
Ce−␨␻nt1
=
= e␨␻n␶d
x2
Ce−␨␻n(t1 +␶d)
x2
t2
The logarithmic decrement ␦ is defined as
t
x = Ce– ζω nt sin (ωdt +ψ )
␦ = ln a
x1
2␲␨
2␲
b = ␨␻n␶d = ␨␻n
=
x2
√1 − ␨2
␻n√1 − ␨2
From this equation, we may solve for ␨ and obtain
Figure 8/7
␨=
␦
√(2␲) 2 + ␦2
For a small damping ratio, x1 ≅ x2 and ␦ << 1, so that ␨ ≅ ␦ /2␲. If x1
and x2 are so close in value that experimental distinction between them
is impractical, the above analysis may be modified by using two observed amplitudes which are n cycles apart.
A rt i c l e 8 / 2 F re e V i b ra t i o n o f P a r t i cl e s
577
Sample Problem 8/1
A body weighing 25 lb is suspended from a spring of constant k = 160 lb /ft.
At time t = 0, it has a downward velocity of 2 ft /sec as it passes through the
position of static equilibrium. Determine
k = 160 lb /ft
(a) the static spring deflection ␦st
(b) the natural frequency of the system in both rad /sec (␻n) and
cycles /sec (ƒn)
W = 25 lb
(c) the system period ␶
(d) the displacement x as a function of time, where x is measured from the
position of static equilibrium
(e) the maximum velocity vmax attained by the mass
Equilibrium
position
(ƒ) the maximum acceleration amax attained by the mass.
Solution.
mg
25
␦st =
=
= 0.1562 ft or 1.875 in.
k
160
␻n =
(b)
k
160
=
= 14.36 rad/sec
B m B 25/32.2
ƒn = (14.36)a
␶=
(c)
2
k(δ st + x)
δst
1
b = 2.28 cycles/sec
2␲
1
1
=
= 0.438 sec
ƒn
2.28
≡
mg
Ans.
Ans.
Ans.
= 0.1393 sin 14.36t ft
Ans.
As an exercise, let us determine x from the alternative Eq. 8 /7:
=
B
02 + a
x0␻n
bd
ẋ0
2
(0)(14.36)
2
b sin c 14.36t + tan−1 a
bd
14.36
2
= 0.1393 sin 14.36t ft
(e) The velocity is ẋ = 14.36(0.1393) cos 14.36t = 2 cos 14.36t ft /sec. Because the cosine function cannot be greater than 1 or less than −1, the maximum velocity vmax is 2 ft /sec, which, in this case, is the initial velocity.
Ans.
(ƒ) The acceleration is
ẍ = −14.36(2) sin 14.36t = −28.7 sin 14.36t ft/sec2
The maximum acceleration amax is 28.7 ft /sec2.
caution in the matter of units. In
the subject of vibrations, it is quite
easy to commit errors due to mixing of feet and inches, cycles and
radians, and other pairs which frequently enter the calculations.
2 Recall that when we refer the mo-
2
sin 14.36t
14.36
x = √x02 + (ẋ0 /␻n ) 2 sin c ␻nt + tan−1 a
Helpful Hints
1 You should always exercise extreme
ẋ0
sin ␻nt
␻n
= (0) cos 14.36t +
mg
Ans.
(d) From Eq. 8 /6:
x = x0 cos ␻nt +
kx
x
(a) From the spring relationship Fs = kx, we see that at equilibrium
mg = k␦st
1
Fs = kδst
Ans.
tion to the position of static equilibrium, the equation of motion, and
therefore its solution, for the present system is identical to that for
the horizontally vibrating system.
578 Chapter 8 Vibr at i on and Ti m e Res p ons e
Sample Problem 8/2
x
c
The 8-kg body is moved 0.2 m to the right of the equilibrium position and
released from rest at time t = 0. Determine its displacement at time t = 2 s. The
viscous damping coefficient c is 20 N ∙ s/m, and the spring stiffness k is 32 N/m.
8 kg
Solution. We must first determine whether the system is underdamped, critically damped, or overdamped. For that purpose, we compute the damping ratio ␨.
␻n = √k/m = √32/8 = 2 rad/s
␨=
k
Equilibrium
position
c
20
=
= 0.625
2m␻n
2(8)(2)
x
mg
Since ␨ < 1, the system is underdamped. The damped natural frequency is
␻d = ␻n√1 − ␨2 = 2√1 − (0.625) 2 = 1.561 rad /s. The motion is given by Eq. 8/12
and is
cx· = 20x·
x = Ce −␨␻nt sin (␻dt + ␺) = Ce −1.25t sin (1.561t + ␺)
kx = 32x
The velocity is then
N
ẋ = −1.25Ce−1.25t sin (1.561t + ␺) + 1.561Ce−1.25t cos (1.561t + ␺)
Evaluating the displacement and velocity at time t = 0 gives
x0 = C sin ␺ = 0.2
ẋ0 = −1.25C sin ␺ + 1.561C cos ␺ = 0
Helpful Hint
Solving the two equations for C and ␺ yields C = 0.256 m and ␺ = 0.896 rad.
Therefore, the displacement in meters is
x = 0.256e −1.25t sin (1.561t + 0.896) m
1 Evaluation for time t = 2 s gives x2 = −0.01616 m.
1 We note that the exponential factor e−1.25t is 0.0821 at t = 2 s. Thus,
␨ = 0.625 represents severe damping,
although the motion is still oscillatory.
Ans.
Sample Problem 8/3
The two fixed counterrotating pulleys are driven at the same angular
speed ␻0. A round bar is placed off center on the pulleys as shown. Determine
the natural frequency of the resulting bar motion. The coefficient of kinetic friction between the bar and pulleys is ␮k.
Solution. The free-body diagram of the bar is constructed for an arbitrary
ω0
Central
position
a
––
2
y
A
G
displacement x from the central position as shown. The governing equations are
[©Fx = mẍ ]
␮k NA − ␮k NB = mẍ
[©Fy = 0]
NA + NB − mg = 0
1 [©MA = 0]
aNB − a
μk NA
NA
a
+ xb mg = 0
2
ω0
a
x
a
––
2
mg
B
μk NB
NB
Helpful Hints
Eliminating NA and NB from the first equation yields
2
1 Because the bar is slender and does
2␮k g
ẍ +
x=0
a
not rotate, the use of a moment
equilibrium equation is justified.
We recognize the form of this equation as that of Eq. 8 /2, so that the natural
frequency in radians per second is ␻n = √2␮k g/a and the natural frequency in
cycles per second is
2 We note that the angular speed ␻0
ƒn =
1
√2␮k g/a
2␲
Ans.
does not enter the equation of motion. The reason for this is our assumption that the kinetic friction
force does not depend on the relative
velocity at the contacting surface.
A r ti c l e 8 / 2 P r o b l e ms
PROBLEMS
(Unless otherwise indicated, all motion variables are referred to the equilibrium position.)
579
8/5 For the spring-mass system shown, determine the
static deflection ␦st, the system period ␶, and the
maximum velocity vmax which result if the cylinder
is displaced 100 mm downward from its equilibrium
position and released from rest.
Undamped, Free Vibrations
8/1 When a 3-kg collar is placed upon the pan which is
attached to the spring of unknown constant, the
additional static deflection of the pan is observed to
be 42 mm. Determine the spring constant k in N /m,
lb /in., and lb /ft.
Problem 8/5
8/6 The cylinder of the system of Prob. 8 /5 is displaced
100 mm downward from its equilibrium position and
released at time t = 0. Determine the position y, velocity v, and acceleration a when t = 3 s. What is the
maximum acceleration?
Problem 8/1
8/2 Determine the natural frequency of the spring-mass
system in both radians per second and cycles per second (Hz).
8/7 Determine the natural frequency in cycles per second
for the system shown. Neglect the mass and friction
of the pulleys. Assume that the block of mass m remains horizontal.
x
k
k = 54 lb/in.
64.4 lb
Problem 8/2
8/3 For the system of Prob. 8 /2, determine the position x of the mass as a function of time if the mass
is released from rest at time t = 0 from a position 2
inches to the left of the equilibrium position. Determine the maximum velocity and maximum acceleration of the mass over one cycle of motion.
8/4 For the system of Prob. 8 /2, determine the position x
as a function of time if the mass is released at time
t = 0 from a position 2 inches to the right of the equilibrium position with an initial velocity of 9 in. /sec to
the left. Determine the amplitude C and period ␶ of
the motion.
m
Problem 8/7
580 Chapter 8 Vibr at i on and Ti m e Res p ons e
8/8 The vertical plunger has a mass of 2.5 kg and is
supported by the two springs, which are always in
compression. Calculate the natural frequency ƒn of
vibration of the plunger if it is deflected from the
equilibrium position and released from rest. Friction
in the guide is negligible.
k1 = 3.6 kN/m
Problem 8/10
Fixed
k2 = 1.8 kN/m
8/11 For the cylinder of Prob. 8 /10, determine the vertical displacement x, measured positive down in millimeters from the equilibrium position, in terms of
the time t in seconds measured from the instant of
release from the position of 25 mm added deflection.
2.5 kg
Problem 8/8
8/12 Determine the natural frequency in radians per
second for the system shown. Neglect the mass and
friction of the pulleys.
8/9 Determine the period ␶ for the system shown. The
cable is always taut, and the mass and friction of the
pulley are to be neglected.
k
k
m
m
k
Problem 8/9
8/10 In the equilibrium position, the 30-kg cylinder
causes a static deflection of 50 mm in the coiled
spring. If the cylinder is depressed an additional
25 mm and released from rest, calculate the resulting natural frequency ƒn of vertical vibration of the
cylinder in cycles per second (Hz).
Problem 8/12
8/13 An old car being moved by a magnetic crane pickup is
dropped from a short distance above the ground. Neglect any damping effects of its worn-out shock absorbers and calculate the natural frequency ƒn in cycles
per second (Hz) of the vertical vibration which occurs
after impact with the ground. Each of the four springs
on the 1000-kg car has a constant of 17.5 kN/m. Because the center of mass is located midway between
the axles and the car is level when dropped, there is
no rotational motion. State any assumptions.