22–61.
SOLUTION
As shown in Prob. 22–50, the velocity is inversely proportional to the period.
Since the the velocity is proportional of f, and
Hence, the amplitude of motion is
Ans.(xp)max
=
4.53 mm
(xp)max =
`
0.05
1–(4.17
1.20)2
`
=0.00453 m
(xp)max =
`
d0
1–
A
v0
vn
B
2
`
=
`
d0
1–
A
v
vR
B
2
`
v0
vn
1
t=f
d0=0.05 m
v=15 km>h=15(1000)
3600 m>s=4.17 m>s
v=15 km>h.
v
100 mm
2m 2m
s
Determine the amplitude of vibration of the trailer in
Prob. 22–60 if the speed
1251
22–62.
A
r
L
2
L
2
The motor of mass Mis supported by a simply supported
beam
of negligible mass. If block Aof mass mis clipped
onto the rotor
, which is turning at constant angular velocity
of
, determine the amplitude of the steady-state vibration.
Hint
:When the beam is subjected to a concentrated force of
P
at its mid-span, it deflects at this point.
Here
Eis Young’s modulus of elasticity,a property of the
material,
and Iis the moment of inertia of the beam’s cross-
s
ectional area.
d=PL3>48EI
v
SOLUTION
In
this case, .Then, .Thus, the natural
frequenc
y of the system is
Here
,.Thus,
Ans.Y=mrv2L3
48EI –Mv
2
L
3
Y=
m(v2r)
48EI>L3
1–v2
48EI>ML3
Y=FO>keq
1–av
vnb2
FO=man=m(v2r)
vn=
C
keq
m=R48EI
L3
M=B48EI
ML3
keq =P
d=P
PL3>48EI =48EI
L3
P=keqd
48EI –Mv
L
22–63.
SOLUTION
of the system is
Here, and , so that
Thus,
Ans.
or
Ans.
v2
100
=0.5
v=7.07 rad>s
v2
100 =1.5
v=12.2 rad>s
v2
100 =1;0.5
;0.4 =0.2
1–
¢
v
10
≤
2
(YP)max =
dO
1–
¢
v
vn
≤
2
(YP)max =;0.4 mdO=0.2 m
vn=Dkeq
m=A5000
50 =10 rad>s
The spring system is connected to a crosshead that oscillates
vertically when the wheel rotates with a constant angular
velocity of . If the amplitude of the steady-state vibration
is observed to be 400 mm, and the springs each have a
stiffness of , determine the two possible
values of at which the wheel must rotate.The block has a
mass of 50 kg.
V
k=2500 N>m
V
200 mm
kk
v
*22–64.
The spring system is connected to a crosshead that oscillates
vertically
when the wheel rotates with a constant angular
velocity of
. If the amplitude of the steady-state
vibration
is observed to be 400 mm, determine the two
possible
values of the stiffness kof the springs.The block
has a mass of 50 kg
.
v=5 rad>s
SOLUTION
In this case
,Thus, the natural circular frequency of the system is
Here
, and , so that
T
hus,
Ans.
or
Ans.
625
k
=0.5 k=1250 N>m
625
k=1.5 k=417 N>m
625
k=1;0.5
;0.4 =0.2
1–
¢
5
20.04k
≤
2
(YP)max =
dO
1–
¢
v
vn
≤
2
(YP)max =;0.4 mdO=0.2 m
vn=Dkeq
m=A2k
50 =20.04k
keq =2k
200 mm
kk
v
22–65.
A7–
lb
bl
oc
k
i
s suspen
d
e
d
from a spr
i
ng
h
av
i
ng a st
i
ffness of
.The support to which the spring is attached is
given simple harmonic motion which may be expressed as
, where tis in seconds. If the damping
factor is , determine the phase angle of forced
vibration.
fc>cc=0.8
d=(0.15 sin 2t)ft
k=75 lb>ft
22–66.
Ans.
MF =0.997
MF =1
C
B
1–av
vnb2
R
2
+c2ac
cnbav
vnbd2
=1
C
B
1–a2
18.57 b2
R
2
+c2(0.8)a2
18.57 b
R
2
d0=0.15, v=2
d=0.15 sin 2t
m=Q75
¢
7
32.2
≤
Determ
i
ne t
h
e magn
i
f
i
cat
i
on factor of t
h
e
bl
oc
k
, spr
i
ng,an
d
dashpot combination in Prob. 22–65.
1256
22–67.
SOLUTION
Since , the system is underdamped,
From Eq. 22-32
Applying the initial condition at , and .
Ans.y=[–0.0702e–3.57tsin (8.540)] m
D=-0.0702 m
–0.6 =De–0[8.542 cos 0° –0]
sin f=0f=0°
0=D[e–0sin (0 +f)]
since
DZ0
y=-0.6 m>sy=0t=0
y=De–
A
c
2m
B
t
C
vdcos (vdt+f)–c
2msin (vdt+f)
D
v=y
#=DCe–
A
c
2m
B
tvdcos (vdt+f)+
A
–c
2m
B
e–
A
c
2m
B
tsin (vdt+f)S
y=DCe–
A
c
2m
B
tsin (vdt+f)S
c
2m=50
2(7) =3.751
vd=vnB1–ac
ccb2
=9.258 B1–a50
129.6 b2
=8.542 rad>s
c6cz
cc=2mvn=2(7)(9.258) =129.6 N #s>m
vn=Ak
m=A600
7=9.258 rad>s
c=50 Ns>mk=600 N>mm=7kg
A block having a mass of 7 kg is suspended from a spring
that has a stiffness If the block is given an
upward velocity of from its equilibrium position at
determine its position as a function of time. Assume
that positive displacement of the block is downward and
that motion takes place in a medium which furnishes a
damping force where is in m>s.
vF =150 ƒvƒ2N,
t=0,
0.6 m>s
k=600 N>m.
*22–68.
The 200-lb electric motor is fastened to the midpoint of the
simply supported beam.It is found that the beam deflects
2 in. when the motor is not running. The motor turns an
eccentric flywheel which is equivalent to an unbalanced
weight of 1 lb located 5 in. from the axis of rotation. If the
motor is turning at 100 rpm, determine the amplitude of
steady-state vibration.The damping factor is ccc0.20.
Neglect the mass of the beam.
SOLUTION
=0.00224 ft
=
1.419
1200
C
1–10.47
13.90
2 2
+2(0.20) 10.47
13.90
2
C¿=
F
0
k
C
B
1–
¢
v
p
≤
2
R
2
+
B
2
¢
c
cc
≤¢
v
p
≤R
2
p=Ak
m=Q1200
200
32.2
=13.90 rad>s
F
O=mrv2=a1
32.2 ba5
12 b(10.47)2=1.419 lb
k=200
2
12
=1200 lb>ft
v=100a2p
60 b=10.47 rad>s
d=2
12 =0.167 ft
22–69.
Two identical dashpots are arranged parallel to each other,
as shown. Show that if the damping coefficient ,
then the block of mass mwill vibrate as an underdamped
system.
c6
2
mk
SOLUTION
The equivalent damping coefficient ceq of a single dashpot is
For the vibration to occur (underdamped system), . However,
.Thus,
Ans.c62mk
2c62mAk
m
ceq 6cc
=2mAk
m
cc=2mvn
ceq 6cc
ceq =F
y
#=2cy
#
y
#=2c
F=cy
+cy
=2cy
k
cc
1259
22–70.
T
he damping factor,may be determined experimentally
by
measuring the successive amplitudes of vibrating motion
of
a system. If two of these maximum displacements can be
approximated
by and as shown in Fig. 22–16, show that
is called the
logarithmic decrement.
ln x1x2
The quantity
ln
c
1x>x1>2=2p1c>cc2>2–1cc22.
x2,x1
c
>
cc,
SOLUTION
Using Eq.
22–32,
T
he maximum displacement is
At
, and
Hence
,
Since
then
so that ln
Using Eq.
22–33,
So that,
Q.E.D.ln ax1
x2b=
2pac
ccb
B1–ac
ccb2
vd=vnB1–ac
ccb2
=cc
2mA1–ac
crb2
cc=2mvn
ax1
x2b=cp
mvd
t2–t1=2p
vd
vdt2–vdt1=2p
x1
x2
=De–
A
c
2m
B
t1
De–
A
c
2m
B
t2=e–
A
c
2m
B
(t1–t2)
x2=De–
A
c
2m
B
t2
x1=De–
A
c
2m
B
t1
t=t2
t=t1
xmax =De–
A
c
2m
B
t
x=Dce–
A
c
2m
B
tsin (vdt+f)d
>
A
1–
a
c
cb
22–71.
c 25 lb s/ft
k 200 lb/ftk 200 lb/ft
9 in.
v
SOLUTION
Then
Solving for the positive root of this equation,
Ans.
v
=21.1 rad>s
v2=443.16
15.07(10–6)v4–3.858(10–3)v2–1.25 =0
0.5 =0.75
D
B
1–
¢
v
16.05
≤
2
R
2
+
¢
2(0.5016)v
16.05
≤
2
Y=dO
D
B
1–
¢
v
vn
≤
2
R
2
+
¢
2(c>cc)v
vn
≤
2
c
cc
=25
49.84 =0.5016
cc=2mvn=2
¢
50
32.2
≤
(16.05) =49.84 lb #s>ft
vn=
C
keq
m =B400
(50>32.2) =16.05 rad>s
If the amplitude of the 50-lb cylinder’s steady-state vibration
is 6 in., determine the wheel’s angular velocity
v.