PROBLEM 19.148
A 91-kg machine element supported by four springs, each of constant k = 175 N/m,
is subjected to a periodic force of frequency 0.8 Hz and amplitude 89 N. Determine
the amplitude of the fluctuating force transmitted to the foundation if (a) a dashpot
with a coefficient of damping c = 365 N ⋅ s/m is connected to the machine element
and to the ground, (b) the dashpot is removed.
SOLUTION
Forcing frequency: 2 (2 )(0.8) 1.6 rad/s
ff
f
ωπ π π
== =
Exciting force amplitude: 89 N
m
P=
Equivalent spring constant: (4)(175 N/m) 700 N/mk
=
=
Natural frequency: 700
n
k
ω
==
f
ω
π
()
()
()
22
2
2
2
12
1 1 (1.8123) 2.2844
ff
ncn
c
c
f
n
ωω
ωω
ω
ω
⎡⎤
⎡
⎤
−+
⎢⎥
⎢
⎥
⎣
⎦
⎣⎦
⎛⎞
−=− =−
⎜⎟
⎝⎠
Copyright © McGraw-Hill Education. Permission required for reproduction or display.
PROBLEM 19.148 (Continued)
(a) m
F
when 365 N s/m:c=⋅
365
504.78
0.72309
2 (2)(0.72309)(1.8123)
2.6209
c
f
cn
c
c
c
c
ω
ω
=
=
⎛⎞
⎛⎞ =
⎜⎟
⎜⎟
⎝⎠
⎝⎠
=
From Eq. (1):
2
22
89 1 (2.6209)
( 2.2844) (2.6209)
m
F+
=−+
89 7.8692
12.088
= 71.8 N
m
F=
(b) m
F
when 0:c=
()
2
89
2.2844
1f
n
m
m
P
F
ω
ω
==
−
39.0 N
m
F=
PROBLEM 19.149
A simplified model of a washing machine is shown.
A bundle of wet clothes forms a weight wb of 20 lb in
the machine and causes a rotating unbalance. The
rotating mass is 40 lb (including mb) and the radius of
the washer basket e is 9 in. Knowing the washer has
an equivalent spring constant k = 70 lb/ft and
damping ratio
ζ
= c/cc = 0.05 and during the spin
cycle the drum rotates at 250 rpm, determine the
amplitude of the motion and the magnitude of the
force transmitted to the sides of the washing machine.
SOLUTION
Forced circular frequency: (2 )(250) 26.18 rad/s
60
f
π
ω
==
System mass: 40 lb
32.2
W
mg
==
Spring constant: 70 lb/ftk
=
70 7.5067 rad/s
k
mf
fm f
Copyright © McGraw-Hill Education. Permission required for reproduction or display.
PROBLEM 19.149 (Continued)
where
()
()
2
22
22 2
40
32.2
22
()
319.28
[70 (26.18) ] [(0.9325)(26.18)]
319.28 319.28 0.40839 ft
781.796
( 781.42) (24.413)
m
m
ff
P
x
km c
ωω
=
−+
=−+
===
−+
(a) Amplitude of vibration. 4.90 in.
m
x=
0.40839sin( )
(26.18)(0.40839)cos( )
10.6917 cos( )
f
f
f
xt
xt
t
ωϕ
ω
ϕ
ωϕ
=−
=−
=−
Spring force: (70)(0.40839)sin( )
28.588sin( )
f
f
kx t
t
ω
ϕ
ωϕ
=−
=−
Damping force: (0.9325)(10.6917) cos( )
9.9701cos( )
f
f
cx t
t
ω
ϕ
ωϕ
=−
=−
(b) Total force: 28.588sin( ) 9.9701cos( )
ff
Ft t
ω
ϕωϕ
=−+−
Let cos sin( ) sin sin( )
sin( )
mfmf
mf
FF t F t
Ft
ψ
ωϕ ψωϕ
ωϕψ
=−+−
=−+
Maximum force. 222 22
cos sin
mm m
FF F
ψ
ψ
=+
22
(28.588) (9.9701)
916.65
=+
= 30.3 lb
m
F=
PROBLEM 19.150*
For a steady-state vibration with damping under a harmonic force, show that the mechanical energy dissipated
per cycle by the dashpot is 2,
mf
Ecx
π
ω
= where c is the coefficient of damping, m
x
is the amplitude of the
motion, and
f
ω
is the circular frequency of the harmonic force.
SOLUTION
Energy is dissipated by the dashpot.
From Equation (19.48), the deflection of the system is
sin( )
mf
xx t
ω
ϕ
=
−
The force on the dashpot.
cos( )
d
dmf f
Fcx
Fcx t
ω
ωϕ
=
=
−
The work done in a complete cycle with
2
f
f
π
τω
=
⎡
⎢
2
ff
ωω
⎢
⎥
⎣
⎦
mf
S
O
(a)
O
LUTION
Recalling
FmaΣ=
that
st
Wk
δ
=
PR
O
The
sim
p
equ
a
syst
e
of a
m
amp
l
st
:(Wk
δ
−
+
,
we write
O
BLEM 19
suspension
p
lified spring-
a
a
tion defining
e
m moves at
a
m
plitude
m
δ
a
l
itude of the
v
)d
x
xc
d
δ
⎛
+
−−
⎜
⎝
.151*
of an auto
m
a
nd-dashpot s
the vertical
d
a
speed
v
ove
r
a
nd wave len
g
v
ertical displa
c
x
d
d
m
d
tdt
d
δ
⎞
−=
⎟
⎠
m
obile can
b
ystem shown
.
d
isplacement
r
a road with
g
th L. (b) De
r
c
ement of the
m
2
2
d
x
d
t
b
e approxim
a
.
(a) Write th
e
of the mass
a sinusoidal
c
r
ive an expre
s
m
ass m.
a
ted by the
e
differential
m when the
c
ross section
s
sion for the
f
u
n
T
T
f
2
f
s
s
Copyright © McGraw-Hill Education. Permission required for reproduction or display.
PROBLEM 19.151* (Continued)
(b) From the identity 22
22
22
sin cos sin( )
sin
cos
AyB y AB y
B
AB
A
AB
ψ
ψ
ψ
+
=+ +
=+
=+
We can write the differential equation
2
22
2
1
()sin( )
tan
mff
f
dx dx
mckxkc t
dt
dt
c
k
δ
ωωψ
ω
ψ
−
++= + +
=
The solution to this equation is analogous to Equations 19.47 and 19.48, with
22
()
mm f
Pkc
δω
=+
sin( )
mf
xx t
ω
ϕψ
=−+
(where analogous to Equations (19.52))
()
22
22
2
()
()
mf
m
f
f
kc
x
c
km
δω
ω
ω
+
=
+
−
2
tan f
f
c
km
ω
ϕ
ω
=−
tan
f
c
k
ω
ψ
=
S
O
Si
n
co
m
Fo
r
O
LUTION
n
ce the origi
n
m
pressions a
n
r
load A,
n
s of coordin
a
n
d the effect o
f
PROBLE
M
Two blocks
A
of the same
c
is connected
coefficient o
P
sin
m
f
P
ω
=
and
B
x
of th
e
a
tes are chos
e
f
gravit
y
M
19.152*
A
and B, eac
h
c
onstant k. Bl
o
to the grou
n
f damping
c
.
f
t
Write the
d
e
two
b
locks f
r
e
n from the e
h
of mass m, a
o
cks A and B
a
n
d by two da
c
. Block A i
s
d
ifferential e
q
r
om their equ
i
quilibrium p
o
re supported
a
a
re connected
shpots, each
s
subjected
t
q
uations defin
i
i
librium positi
o
sition, we m
a
a
s shown by
t
by a dashpot,
dashpot havi
n
t
o a force o
f
i
ng the displ
a
ons.
a
y omit the i
t
hree springs
and block B
n
g the same
f
magnitude
a
cements
A
x
nitial spring
r
a
u
u
d
d
PR
O
Expr
for
w
S is
c
O
BLEM 19
.
ess in terms
o
w
hich oscillati
o
c
losed.
.
153
o
f L, C, and E
o
ns will take
p
the range of
v
p
lace in the c
i
v
alues of the
r
i
rcuit shown
w
r
esistance R
w
hen switch
m
o
q
o
q
l
c
l
e
d
p
P
R
Co
de
t
(1
−
S
O
Th
e
th
e
(a)
s
n
n
R
OBLEM 1
nsider the cir
c
t
ermine (a) th
e
1/ )
e
−
times i
t
O
LUTION
e
mechanical
e
mass.
Final val
u
9.154
c
uit of Probl
e
e
final value
o
t
s final value.
(
Electrical
analogue of c
u
e of the curr
e
e
m 19.153 wh
o
f the current
(
The desired
v
system
losing a swit
c
e
nt correspon
en the capaci
t
in the circuit,
v
alue of t is k
n
c
h S is the sud
d
ds to the fin
a
t
or C is remo
v
(b) the time
t
n
own as the ti
m
M
e
c
d
en applicati
o
a
l velocity of
v
ed. If switc
h
t
at which the
m
e constant o
c
hanical syste
m
o
n of a consta
n
the mass, an
d
h
S is closed
a
current will
h
f the circuit.)
m
n
t force of m
a
d
since the c
a
a
t time
t
=
0,
h
ave reached
a
gnitude P to
a
pacitance is