PROBLEM 19.23
Two springs of constants 1
k and 2
kare connected in series to a block A that
vibrates in simple harmonic motion with a period of 5 s. When the same two
springs are connected in parallel to the same block, the block vibrates with a
period of 2 s. Determine the ratio 12
/kk
of the two spring constants.
Solution
Equivalent Springs
Series: 12
12
s
kk
kkk
=+
Parallel: 12p
kkk
=
+
2
2.125 3.516
k=∓
2
k
k
S
u
c
O
LUTION
PR
O
The
p
b
loc
k
(a) t
h
and
B
1
1
m
ω
O
BLEM 19
p
eriod of vib
r
k
A is remo
v
h
e mass of bl
o
B
have been r
e
1
6 kg
22
0.8 s
C
m
π
π
τ
=+
==
.24
r
ation of the s
y
v
ed, the perio
o
ck C, (b) the
e
moved.
1
0.8 s
2rad/s
0.8
τ
π
=
=
y
stem shown
d is observe
d
period of vib
r
is observed t
o
d
to be 0.7 s
r
ation when b
o
o
be 0.8 s. If
. Determine
o
th blocks A
ω
ω
ω
π
Copyright © McGraw-Hill Education. Permission required for reproduction or display.
PROBLEM 19.25
The 100-lb platform A is attached to springs B and D, each of which has a
constant 120 lb/ft.k
=
Knowing that the frequency of vibration of the
platform is to remain unchanged when an 80-lb block is placed on it and a
third spring C is added between springs B and D, determine the required
constant of spring C.
SOLUTION
Frequency of the original system.
Springs B and D are in parallel.
()
2
2
100 lb
32.2 ft/s
22
2(120 lb/ft) 240 lb/ft
240 lb/ft
77.28(rad/s)
eBD
e
n
A
n
kkk
k
m
ω
ω
=+= =
==
=
Frequency of new system.
Springs A, B, and C are in parallel. (2)(120)
eBDC e
kkk k k
′
=
++= +
2
2
2
22
(240 )(32.2 ft/s )
() (100 lb 80 lb)
( ) (0.1789)(240 )
()
77.28 (0.1789)(240 )
191.97 lb/ft
eC
n
AB
nC
nn
C
C
kk
mm
k
k
k
ω
ω
ωω
′+
′==
++
′=+
′
=
=+
= 192.0 lb/ft
C
k=
Copyright © McGraw-Hill Education. Permission required for reproduction or display.
PROBLEM 19.26
The period of vibration for a barrel floating in salt water is found to be
0.58 s when the barrel is empty and 1.8 s when it is filled with 55 gallons of
crude oil. Knowing that the density of the oil is 900 kg/m3, determine (a) the
mass of the empty barrel, (b) the density of the salt water,
ρ
sw. [Hint: the
force of the water on the bottom of the barrel can be modeled as a spring
with constant k =
ρ
swgA.]
SOLUTION
Area of bottom of barrel:
22
2
(0.572 m) 0.2570 m
44
D
A
ππ
== =
Mass of oil:
3
3
oil
1m
(55 gal) (900 kg/m ) 187.378 kg
264.172 gal
m⎛⎞
==
⎜⎟
⎜⎟
⎝⎠
Barrel empty: 1
1
1
0.58 s
22
10.833 rad/s
0.58 s
n
τ
ππ
ωτ
=
== =
1n
b
k
m
ω
= (1)
Barrel full: 2
2
2
1.8 s
22
3.4907 rad/s
1.8 s
n
τ
ππ
ωτ
=
== =
2
oil
n
b
kk
mmm
ω
== + (2)
(a) Mass mb of empty barrel.
Divide Eq. (1) by Eq. (2) and square both sides.
22
1oil
22
2
oil
oil
(10.833) 9.6310
(3.4907)
9.6310
187.378 kg 21.710 kg
9.6310 1 8.6310
nb
b
n
bb
b
mm
m
mm m
m
m
ω
ω
+
===
=+
== =
−
21.7 kg
b
m=
Spring constant: 223
1(21.710)(10.833) 2.5477 10 N/m
bn
km
ω
== = ×
Copyright © McGraw-Hill Education. Permission required for reproduction or display.
PROBLEM 19.26 (Continued)
(b) Density of the salt water.
sw
3
sw 22
2.5477 10 N/m
(9.81 m/s )(0.2570 m )
kgA
k
gA
ρ
ρ
=
×
==
3
sw 1011 kg/m
ρ
=
PROBLEM 19.27
From mechanics of materials it is known that for a simply supported
beam of uniform cross section a static load P applied at the center will
cause a deflection 348 ,
A
P
LEI
δ
= where L is the length of the beam,
E is the modulus of elasticity, and I is the moment of inertia of the
cross-sectional area of the beam. Knowing that L = 15 ft,
E = 6
30 10 psi,× and 34
210 ft,I−
=× determine (a) the equivalent
spring constant of the beam, (b) the frequency of vibration of a 1500-lb
block attached to the center of the beam. Neglect the mass of the beam
and assume that the load remains in contact with the beam.
SOLUTION
()
(
)
(
)
62 22 34
48 30 10 lb/in. 144 in. /ft 2 10 ft
48
PEI
−
××
222
n
πππ
Copyright © McGraw-Hill Education. Permission required for reproduction or display.
PROBLEM 19.28
From mechanics of materials it is known that when a static load P is
applied at the end B of a uniform metal rod fixed at end A, the length of
the rod will increase by an amount /,
P
LAE
δ
=
where L is the length of
the undeformed rod. A is its cross-sectional area, and E is the modulus of
elasticity of the metal. Knowing that L = 450 mm and E = 200 GPa and
that the diameter of the rod is 8 mm, and neglecting the mass of the rod,
determine (a) the equivalent spring constant of the rod, (b) the frequency
of the vertical vibrations of a block of mass m = 8 kg attached to end B
of the same rod.
SOLUTION
(a) e
e
Pk
PL
AE
AE
PL
AE
kL
δ
δ
δ
=
=
⎛⎞
=⎜⎟
⎝⎠
=
232
52
92
52 9 2
6
(8 10 m)
44
5.027 10 m
0.450 m
200 10 N/m
(5.027 10 m )(200 10 N/m )
(0.450 m)
22.34 10 N/m
e
e
d
A
A
L
E
k
k
ππ
−
−
−
×
==
=×
=
=×
××
=
=× 22.3 MN/m
e
k=
(b)
6
22.3 10
8
2
2
e
k
m
n
f
π
π
×
=
=
265.96 Hz= 266 Hz
n
f=
P
De
lo
Ne
S
π
R
OBLEM 1
noting by
st
δ
t
a
d is
glect the mas
s
O
LUTION
9.29
t
he static defl
e
s
of the beam,
e
ction of a be
and assume t
h
k
m
am under a g
i
1
2
f
π
=
h
at the load r
e
st
W
W
k
W
m
g
δ
=
=
i
ven load, sho
st
g
δ
e
mains in con
t
w that the fre
q
t
act with the
b
q
uency of vib
b
eam.
ω
ration of the
Copyright © McGraw-Hill Education. Permission required for reproduction or display.
PROBLEM 19.30
A 40-mm deflection of the second floor of a building is measured directly under a newly installed 3500-kg
piece of rotating machinery, which has a slightly unbalanced rotor. Assuming that the deflection of the floor is
proportional to the load it supports, determine (a) the equivalent spring constant of the floor system, (b) the
speed in rpm of the rotating machinery that should be avoided if it is not to coincide with the natural
frequency of the floor-machinery system.
SOLUTION
(a) Equivalent spring constant.
3500(9.81) N
40 mm
es
e
Wk
mg
k
δ
δ
=
=
= 858 N/mm
e
k=
(b) Natural frequency.
(858.38 1000 N/m)
(3500 kg)
2
2
2.4924 Hz
1 Hz 1 cycle/s
60 rpm
e
k
m
n
n
f
f
π
π
×
=
=
=
=
=
(60 rpm)
Speed (2.424 Hz) Hz
= Speed 149.5 rpm
=
Da
Copyri
g
ta:
g
h
t
© McGra
w
P
I
d
i
n
t
e
w
-Hill Educ
a
P
ROBLEM
f 700 m
m
h
=
d
etermine the
n
finite. Negle
e
nsion or co
m
n
d
h
k
=
=
=
a
tion. Permi
s
19.31
m
and
50
0
d
=
mass m for w
ct the mass o
f
m
pression.
mh h
⎜⎟
⎝⎠
0.5 m
0.7 m
600 N/m
s
sion require
d
0
mm
and ea
c
hich the peri
o
f
the rod and
a
d
for reprod
u
c
h spring has
o
d of small os
c
a
ssume that e
a
u
ction or dis
p
a constant k
c
illations is (
a
a
ch spring can
p
lay.
600 N/m,=
a
) 0.50 s, (b)
act in either
x
2