Problem 9.32
Suppose that the equation of motion of a forced harmonic oscillator is
given by
RxC!2
nxD.F0=m/ cos !0t
. Obtain the expression for the
response of the oscillator, and compare it to the response presented in
Eq. (9.36) (which is for a forced harmonic oscillator with the equation
of motion given in Eq. (9.31)).
Therefore, the response is given by
Notice that we can find the response due to
cos !0t
forcing by replacing the “
sin
” with a “
cos
” in the response
due to sin !0tforcing.
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Dynamics 2e 1971
Problem 9.33
A uniform bar of mass
m
and length
L
is pinned to a slider at
O
. The slider is
forced to oscillate horizontally according to
y.t/ DYsin !st
. The system lies
in the vertical plane.
(a) Derive the equation of motion of the bar for small angles .
(b) Determine the amplitude of steady-state vibration of the bar.
Solution
The FBD at the right shows the bar in an arbitrary postition.
Balance Principles.
To obtain the equation of motion of the bar, we can sum
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permission of McGraw-Hill, is prohibited.
where we have used
!2
nD3g=.2L/
from the equation of motion. Therefore, the amplitude of steady-state
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Dynamics 2e 1973
Problem 9.34
Determine the amplitude of vibration of the unbalanced motor we
studied in Example 9.6 if the forcing frequency of the motor is
0:95!n.
„ ƒ‚ …
A
„ ƒ‚ …
B
DpA2CB2sin.!ntC/;
where the angle
can be determined, but is not needed to find the amplitude of vibration. Thus we see that
the amplitude is
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1974 Solutions Manual
10
g,
100
g, and
1000
g. Using these, we find that the amplitude of vibration for each of the three unbalanced
masses is
A Closer Look.
A plot showing each of these three amplitudes as
a function of time is shown at the right. The green curve corresponds
to
muD10
g, the blue curve corresponds to
muD100
g, and the red
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Dynamics 2e 1975
Problem 9.35
Derive the equations of motion for the unbalanced motor introduced
in this section by applying Newton’s second law to the center of
mass of the system shown in Fig. 1 of Example 9.5.
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where we have used the fact that
h
is constant in taking the two time derivatives. Substituting
mRyG
from this
expression into Eq. (2), we obtain the equation of motion as
r
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Dynamics 2e 1977
Problem 9.36
Consider a sign mounted on a circular hollow steel pole of length
LD5
m, outer diameter
doD5cm
, and inner diameter
diD4cm
.
Aerodynamic forces due to wind provide a harmonic torsional excitation
with frequency
f0D3Hz
and amplitude
M0D10 Nm
about the
´
axis. The mass center of the sign lies on the central axis
´
of the pole.
The mass moment of inertia of the sign is
I´D0:1 kgm2
. The torsional
stiffness of the pole can be estimated as
ktDGstd4
od4
i=.32L/
,
where
Gst
is the shear modulus of steel, which is
79 GPa
. Neglecting the
inertia of the pole, calculate the amplitude of steady-state vibration of
the sign.
TeDM0sin .2f0t/and TpDktD“Gst d4
32L #;
where
kt
is the torsional stiffness of the pole,
is the angle of rotation of the sign about the
´
axis from its
static equilibrium position, and we have substituted in the given relationship for kt.
Kinematic Equations. Since we need the equation of motion, we let ˛sDR
.
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where
!0
is the forcing frequency,
keff
is the effective spring constant, and
.F0/eff
is the effective forcing
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Dynamics 2e 1979
Problem 9.37
An unbalanced motor is mounted at the tip of a rigid beam of mass
mb
and length
L
. The beam is restrained by a torsional spring
of stiffness
kt
and an additional support of stiffness
k
located at
the half length of the beam. In the static equilibrium position,
the beam is horizontal, and the torsional spring does not exert
any moment on the beam. The mass of the motor is
mm
, and
the unbalance results in a harmonic excitation
F .t/ DF0sin !0t
in the vertical direction. Derive the equation of motion for the
system, assuming that is small.
Solution
The FBD at the right shows all the forces and moments acting
on the system as it oscillates.
Balance Principles.
Referring to the FBD, we can sum mo-
ments about point Oto obtain
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