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PROBLEM 19.93
The motion of the uniform rod AB is guided by the cord BC and by the
small roller at A. Determine the frequency of oscillation when the end B
of the rod is given a small horizontal displacement and released.
SOLUTION
Position 1. (Maximum deflection):
Let m
θ
be the small angle between the cord CB and the vertical. As the rod is moved from the equilibrium
position the center of gravity G moves up an amount .
m
y
22
2
2
1
1
11
(1 cos ) 1 1 22
11
1
4
0,
Bm mm
mG B m
mm
yl l l
yy y l
Vmgy mgl
T
θ
θθ
θ
θ
⎛⎞
=− ≈ −+ =
⎜⎟
⎝⎠
== =
==
=
Position 2. (Maximum velocity): At the equilibrium position the motion of the rod is a translation.
2
11
mm
vl l
ωθ
==
Conservation of energy: 222
11 2 2
11
:42
mm
TV T V mgl ml
θ
θ
+=+ =
so that
222
ππ
S
O
P
o
s
Lo
o
H
o
Lo
o
O
LUTION
s
ition c
(Ma
x
o
king from a
b
o
rizontal displ
a
o
king from ri
g
x
imum defle
c
b
ove:
a
cement of C:
g
ht:
p
tion)
Cm
xb
θ
=
C
m
x
h
φ
=
=
PROBLE
M
A uniform ro
d
at A and by
p
eriod of osc
i
displacement
m
b
h
θ
=
M
19.94
d
of length L
a vertical w
i
i
llation of the
and then rele
a
is supported
b
i
re CD. Deri
v
rod if end B i
a
sed.
b
y a ball-and
-
v
e an expres
s
s given a sm
a
-
socket joint
s
ion for the
a
ll horizontal
Copyright © McGraw-Hill Education. Permission required for reproduction or display.
PROBLEM 19.94 (Continued)
But for simple harmonic motion,
222
2
11
()
46
3
2
mnm
mnm
n
mgbL mL
h
bg
hL
θωθ
θωθ
ω
=
=
=
Period of vibration. 2
n
n
π
τ
ω
= 2
23
n
hL
bg
τπ
=
Copyright © McGraw-Hill Education. Permission required for reproduction or display.
PROBLEM 19.95
A section of uniform pipe is suspended from two vertical cables attached
at A and B. Determine the frequency of oscillation when the pipe is given
a small rotation about the centroidal axis OO′ and released.
SOLUTION
22
mm m m
aa
AA BB l l
θ
αα θ
′′
== = =
Position 1
11
0 (1 cos )
c
TVmgymgl
α
===−
For small angles
22
2
2
1cos 2sin
22
8
mm
mm
a
l
αα
α
θ
−= ≈=
22
12
8m
a
Vmgl
l
θ
⎛⎞
=⎜⎟
⎜⎟
⎝⎠
Position 2 222
22
111 0
2212
mm
TI ma V
θθ
⎛⎞
== =
⎜⎟
⎝⎠
mnm
θ
ωθ
=
11 2 2
TV T V+=+
2222
2
1
024
8nm
a
mgl ma
l
ω
θ
⎛⎞
++
⎜⎟
⎜⎟
⎝⎠
23
n
g
l
ω
=
13
2
n
g
fl
π
=
Copyright © McGraw-Hill Education. Permission required for reproduction or display.
PROBLEM 19.96
Three collars, each of mass m, are connected by pins to bars
AC and BC, each of length l and negligible mass. Collars A and
B can slide without friction on a horizontal rod and are
connected by a spring of constant k. Collar C can slide without
friction on a vertical rod and the system is in equilibrium in the
position shown. Knowing that collar C is given a small
displacement and released, determine the frequency of the
resulting motion of the system.
SOLUTION
2
22
11
27
22 6
x
Tmxmym
⎡⎤
=+=
⎢⎥
⎣⎦
Equilibrium of BC:
3
022 2
D
mg l l
Mku
⎛⎞
Σ== ⋅−
⎜⎟
⎜⎟
⎝⎠
Where 0
, natural length
23
mg
ul lu
k
=
==+
() ()
22
0
122
22
k
k
Vklxl xu
⎡⎤
=−−=+
⎣⎦
,
g
Vmgy=− where
2
2
23
22
ll
lx y
⎛⎞
⎛⎞
=− + +
⎜⎟
⎜⎟
⎜⎟
⎝⎠
⎝⎠
And
()
(
)
22
22 334
30, 2
ll xlx
ylyxlx y
−+ − −
++−==
So 2
118 H.O.T.
2
333
yx x
l
⎛⎞ ⎛ ⎞
=+− +
⎜⎟ ⎜ ⎟
⎝⎠ ⎝ ⎠
Then
2
24
2 2 constant 333
x
mgx
Vkx kux mg l
=++ − +
4
233
7
6
2mg
l
nm
k
ω
+
=
112 8 Hz
27 73
n
kg
fml
π
=+
S
O
Po
O
LUTION
o
sition
c (M
a
a
ximum defle
c
PROBLE
M
A thin plat
e
expression f
o
(sin )
(1
c
m
r
r
θ
−
c
tion)
M
19.97*
e
of length
l
o
r the period
o
2
2
sin
c
os ) 2
mm
m
m
r
r
θθ
θ
θ
≈
≈
1
1
0
m
T
VWy
mgr
θ
=
=
=
l
rests on a
o
f small oscill
2
2
m
θ
half cylinde
r
ations of the
p
r
of radius r
.
p
late.
.
Derive an
Copyright © McGraw-Hill Education. Permission required for reproduction or display.
PROBLEM 19.98*
As a submerged body moves through a fluid, the particles of the fluid flow around
the body and thus acquire kinetic energy. In the case of a sphere moving in an
ideal fluid, the total kinetic energy acquired by the fluid is 2
1
4Vv
ρ
, where
ρ
is the
mass density of the fluid, V is the volume of the sphere, and v is the velocity of the
sphere. Consider a 500-g hollow spherical shell of radius 80 mm, which is held
submerged in a tank of water by a spring of constant 500 N/m. (a) Neglecting
fluid friction, determine the period of vibration of the shell when it is displaced
vertically and then released. (b) Solve Part a, assuming that the tank is accelerated
upward at the constant rate of 8 m/s2.
SOLUTION
This is not a damped vibration. However, the kinetic energy of the fluid must be included.
(a)
P
osition d 2
2
2
0
1
2m
T
Vkx
=
=
P
osition c 22
1 spere fluid
1
11
24
0
s
mm
TT T mv Vv
V
ρ
=+= +
=
Conservation of energy and simple harmonic motion.
22 2
11 2 2
11 1
: 0 0
24 2
s
mm m
TV T V mv Vv kx
ρ
+=+ + +=+
()
22 2
2
1
2
2
1
2
33
22
11 1
22 2
500 N/m
(0.5 kg)
11 4
(1000 kg/m ) (0.08 m)
22 3
11.0723 kg
2
500 N/m 318 s
(0.5 kg) (1.0723 kg)
mmmn
smnm
n
s
n
n
vxx
mVx kx
k
mV
V
V
V
ω
ρω
ωρ
ωρ
ρπ
ρ
ω
−
==
⎛⎞
+=
⎜⎟
⎝⎠
=+
=+
⎛⎞
=⎜⎟
⎝⎠
=
==
+
Period of vibration. 22
318
n
n
π
π
τω
== 0.352 s
n
τ
=
(b) Acceleration does not change mass. 0.352 s
n
τ
=
Copyright © McGraw-Hill Education. Permission required for reproduction or display.
PROBLEM 19.99
A 4-kg collar can slide on a frictionless horizontal rod and is attached to a
spring of constant 450 N/m. It is acted upon by a periodic force of
magnitude sin ,
mf
P
Pt
ω
= where 13
m
P
=
N. Determine the amplitude of
the motion of the collar if (a) 5
f
ω
=
rad/s, (b) 10
f
ω
= rad/s.
SOLUTION
Eq. (19.33)
2
1
m
m
f
n
P
k
x
ω
ω
=⎛⎞
−⎜⎟
⎝⎠
1
450 N/m 10.607 s
4 kg
n
k
m
ω
−
== =
13 N 0.28889 m
450 N/m
m
P
k==
2
0.28889 m
110.607
m
f
x
ω
=⎛⎞
−⎜⎟
⎝⎠
(a) 2
0.28889 m
5: 0.03714 m
5
110.607
fm
x
ω
== =
⎛⎞
−⎜⎟
⎝⎠
(
)
In Phase 37.1 mm
m
x=
(b) 2
0.28889 m
10: 0.25984 m
10
110.607
fm
x
ω
== =
⎛⎞
−⎜⎟
⎝⎠
(
)
In Phase 260 mm
m
x=
Copyright © McGraw-Hill Education. Permission required for reproduction or display.
PROBLEM 19.100
A 4-kg collar can slide on a frictionless horizontal rod and is attached to a
spring of constant k. It is acted upon by a periodic force of magnitude
sin ,
mf
P
Pt
ω
= where 9
m
P
=
N and 5
f
ω
= rad/s. Determine the value
of the spring constant k knowing that the motion of the collar has an
amplitude of 150 mm and is (a) in phase with the applied force, (b) out of
phase with the applied force.
SOLUTION
Eq. (19.33)
2
2,
1
m
mn
f
n
P
k
k
xm
ω
ω
ω
==
⎛⎞
−⎜⎟
⎝⎠
()
2
2, or
mm
m
mf
f
PP
xk
xm
km
ω
ω
==
+
−
(a) In phase
()
()
1
9 N
0.15 m 4 kg 5 s
k−
=+160.0 N/m
=
(b) Out of phase 0.15 m
m
x=−
()
()
1
9 N 40.0 N/m
0.15 m 4 kg 5 s
k−
==
−+
Copyright © McGraw-Hill Education. Permission required for reproduction or display.
PROBLEM 19.101
A collar of mass m which slides on a frictionless horizontal rod is attached
to a spring of constant k and is acted upon by a periodic force of magnitude
sin .
mf
P
Pt
ω
= Determine the range of values of
f
ω
for which the
amplitude of the vibration exceeds three times the static deflection caused
by a constant force of magnitude .
m
P
SOLUTION
Eq. (19.33) 2
2
1
m
m
f
n
P
k
x
ω
ω
=
−
2
mst n
P
k
km
δω
==
2
2
3
1
st
s
t
f
n
δ
δ
ω
ω
≥
−
2
2
1
13
f
n
ω
ω
−
≤
2
2
2
3
f
n
ω
ω
>
Also 2
2
3
1
st
s
t
f
n
δ
δ
ω
ω
<−
−
22
22
14
1
33
ff
nn
ωω
ωω
−
<− <
2
2
242 4
333 3
f
f
n
kk
mm
ωω
ω
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