PROBLEM 19.166 (Continued)
Amplitude of vibration. Use Eq. (19.53).
()
()
()
22
2
12
1
m
f
f
cn
n
P
k
m
c
c
x
ω
ω
ω
ω
=
⎡⎤
⎡
⎤
+
−
⎢⎥
⎢
⎥
⎣
⎦
⎣⎦
2
ω
⎛⎞
m
PROBLEM 19.167
The compressor shown has a mass of 250 kg and operates at 2000 rpm. At this operating condition the force
transmitted to the ground is excessively high and is found to be 2
f
mr
ω
where mr is the unbalance and
ω
f is the
forcing frequency. To fix this problem, it is proposed to isolate the compressor by mounting it on a square
concrete block separated from the rest of the floor as shown. The density of concrete is 2400 kg/m3 and the
spring constant for the soil is found to be 80 × 106 N/m. The geometry of the compressor leads to choosing a
block that is 1.5 m by 1.5 m. Determine the depth h that will reduce the force transmitted to the ground by 75%.
SOLUTION
Forced circular frequency corresponding to 2000 rpm.
(2 )(2000) 209.44 rad/s
60
f
π
ω
==
In the first case the natural frequency is very large so that the transmitted force is 2.
f
mr
ω
After the problem is fixed, the transmitted force is
ω
m
P
f
f
Copyright © McGraw-Hill Education. Permission required for reproduction or display.
PROBLEM 19.167 (Continued)
Equating expressions (1) and (2) dividing by 2,
f
mr
ω
()
()
2
2
6
22
10.25
1
14
5
11
(209.44) 93.664 rad/s
55
80 10 N/m 9119 kg
(93.664 rad/s)
f
n
f
n
f
n
nf
n
n
k
m
k
m
ω
ω
ω
ω
ω
ω
ωω
ω
ω
=
−
−=
=
== =
=
×
== =
Required properties of the attached concrete block.
Mass: 250 kg 8869 kgm
−
=
3
3
mass 8869 kg
volume 3.6954 m
density 2400 kg/m
== =
2
area 1.5 m 1.5 m 2.25 m=×=
3
2
volume 3.6954 m
depth area 2.25 m
== 1.642 mh
=
Copyri
g
g
h
t
© McGra
w
w
-Hill Educ
a
a
tion. Permi
s
n
n
ω
s
sion require
d
PROBLE
A small bal
l
a tightly str
e
on a horiz
o
displaceme
n
cord and re
cord to re
m
equation of
period of vi
b
d
for reprod
u
M 19.168
l
of mass m a
t
e
tched elastic
o
ntal plane.
T
n
t in a direc
t
leased. Assu
m
m
ain constant,
motion of t
h
b
ration.
u
ction or dis
p
t
tached at the
cord of lengt
h
T
he ball is gi
v
t
ion perpendi
c
m
ing the tens
i
(a) write th
e
h
e ball, (b) d
e
n
p
lay.
midpoint of
h
l can slide
v
en a small
c
ular to the
i
on T in the
e
differential
e
termine the
T
)
)
f
n
f
n
τ
PROBLEM 19.169
A certain vibrometer used to measure vibration amplitudes consists
essentially of a box containing a slender rod to which a mass m is attached;
the natural frequency of the mass-rod system is known to be 5 Hz. When
the box is rigidly attached to the casing of a motor rotating at 600 rpm, the
mass is observed to vibrate with an amplitude of 0.06 m. relative to the box.
Determine the amplitude of the vertical motion of the motor.
SOLUTION
Natural frequency: 5Hz
2 31.416 rad/s
n
nn
f
f
ωπ
=
==
Forcing frequency: 600 rpm 10 Hz
2 62.832 rad/s
f
ff
f
f
ωπ
==
==
2.000
f
ω
=
44
m
S
O
Fr
o
an
d
B
u
O
LUTION
o
m Problem 1
d
u
t
PROBLE
M
If either a
acceleration
the string is
location of t
h
difficulty c
a
and B are pl
and the dist
then adjuste
Show that t
h
that g
=
4
π
2
l
9.52, the leng
t
M
19.170
simple or a
of gravity g,
not truly we
h
e mass cent
e
a
n be eliminat
aced so that t
h
a
nce l is me
a
d so that the
p
h
e period
τ
o
b
l
/
τ
2
.
t
h of an equi
v
A
l
B
l
π
A
A
τ
compound
p
difficulties a
r
ightless, whil
e
r is difficult t
o
ed by using
a
h
ey are obvio
u
a
sured with g
r
p
eriod of osc
i
b
tained is eq
u
v
alent simple
p
2
k
rr
=+
2
k
RR
=+
B
τ
=
p
endulum is
u
r
e encountere
e in the case
o
establish. I
n
a
reversible, o
u
sly not at th
e
r
eat precision
.
i
llation
τ
is th
u
al to that of
a
p
endulum is:
u
sed to dete
r
d. In the cas
e
of the comp
o
n
the case of a
r Kater, pend
u
e
same distanc
.
The positio
n
e same when
a
true simple
r
mine experi
m
e
of the simpl
e
o
und pendulu
m
compound p
e
u
lum. Two k
n
e from the m
a
n
of a counte
r
either knife
e
pendulum of
m
entally the
e
pendulum,
m
, the exact
e
ndulum, the
n
ife edges A
a
ss center G,
r
weight D is
e
dge is used.
length l and
u
u
Copyright © McGraw-Hill Education. Permission required for reproduction or display.
PROBLEM 19.170 (Continued)
Thus, and
A
GGA BGGB
′
′
=
=
That is, and
A
ABB
′
′
=
=
Noting that
2
AB
lll
l
g
τπ
=
=
=
or
2
2
4l
g
π
τ
=