1230
22–41.
If the block-and-spring model is subjected to the periodic
force , show that the differential equation of
motion is , where xis
measured from the equilibrium position of the block. What
is the general solution of this equation?
x
$+(k>m)x=(F
0>m) cos vt
F=F
0cos vt
F F0cos vt
k
Equilibrium
position
x
m
SOLUTION
$+k
:
+©Fx=max;F0cos vt–kx =mx
$
1231
22–42.
A block which has a mass m is suspended from a spring
having a stiffness k. If an impressed downward vertical force
F=FO
acts on the weight, determine the equation which
describes the position of the block as a function of time.
SOLUTION
+
c
ΣF
y=
ma
y ;
k(yst +y)–mg –F0=–my
$
my
$+ky +kyst –mg =F0
However, from equilbrium
kyst –mg =0
, therefore
my
$+ky =F0
y
$
+
k
m
y=
F
0
m
where vn=
Ak
m
y
$+vn
2y=
F
0
m
[1]
The general solution of the above differential equation is of the form of
y=y
c
+y
p.
yc=A sin
v
nt+B cos
v
nt
yP=C
[2]
y
$
P=0
[3]
Substitute Eqs. [2] and [3] into [1] yields :
0+vn
2C=
F
0
m
C=
F
0
mp2=
F
0
k
The general solution is therefore
y=A sin vnt+B cos vnt+
F
0
k
Ans.
The constants A and B can be found from the initial conditions.
Ans:
y=A
sin
vnt+B
cos
vnt+
F
O
k
22–43.
SOLUTION
The initial condition when ,, and
v0=Avn–0+d0v0
–d0v0
y0=0+B+0B=y0
v=v0
y=y0
t=0
v=y
#=Avncos vnt–Bvnsin vnt+d0v0
1–av0
vnb2cos v0t
y=Asin vnt+Bcos vnt+d0
1–av0
vnb2sin v0t
A4-lb weight is attached to a spring having a stiffness
The weight is drawn downward a distance of
4 in. and released from rest. If the support moves with a
vertical displacement in., where tis in
seconds, determine the equation which describes the
position of the weight as a function of time.
d=10.5 sin 4t2
k=10 lb>ft.
is
*22–44.
A 4-kg block is suspended from a spring that has a stiffness
of The block is drawn downward 50 mm from
the equilibrium position and released from rest when
If the support moves with an impressed displacement of
where tis in seconds, determine the
equation that describes the vertical motion of the block.
Assume positive displacement is downward.
d=110 sin 4t2mm,
t=0.
k=600 N
>
m.
SOLUTION
vn=Ak
m=A600
4=12.25
1234
22–45.
Use a
bl
oc
k
-an
d
-spr
i
ng mo
d
e
l
lik
e t
h
at s
h
own
i
n F
i
g. 22–14a,
but suspended from a vertical position and subjected to a
periodic support displacement determine the
equation of motion for the system, and obtain its general
solution. Define the displacement ymeasured from the
static equilibrium position of the block when t=0.
d=d0sin v0t,
1
–
(v
>
v
n
)2
22–46.
SOLUTION
A 5-kg block is suspended from a spring having a stiffness
of If the block is acted upon by a vertical force
where tis in seconds, determine the
equation which describes the motion of the block when it is
pulled down 100 mm from the equilibrium position and
released from rest at Assume that positive displacement
is downward.
t=0.
F=17 sin 8t2N,
300 N>m.
k300 N/m
1236
22–47.
SOLUTION
compress and stretch .Thus, the force in each spring is
.The mass moment of inertia of the rod about point Ais
. Referring to the free-body diagram of the rod shown in Fig. a,
Since is small, and .Thus, this equation becomes
(1)
The particular solution of this differential equation is assumed to be in the form of
(2)
Taking the time derivative of Eq. (2) twice,
(3)
Substituting Eqs. (2) and (3) into Eq. (1),
Ans.C=3FO
3
2
(mg +Lk)–mLv2
C=3FO>mL
3
2ag
L+k
mb–v2
C
B
3
2ag
L+k
m
≤
–v2
R
sin vt=3FO
mL sin vt
–Cv2sin vt+3
2ag
L+k
m
≤
(Csin vt)=3FO
mL sin vt
u
$
p=-Cv2sin vt
up=Csin vt
u
$
+3
2
¢
g
L+k
m
≤
u=3FO
mL sin vt
1
3mLu
$
+1
2(mg +kL)u=FOsin vt
cos u1sin u0u
=1
3mL2u
#
+©MA=IAa;FOsin vtcos u(L)–mg sin uaL
2b–2akL
2ubcos uaL
2b
IA=1
3mL2
Fsp =ks =kL
2u
s=rAGu=L
2u
The uniform rod has a mass of m. If it is acted upon by a
periodic force of , determine the amplitude of
the steady-state vibration.
F=F0sin vt
kk
L
2
L
2
A
*22–48.
SOLUTION
Fr
ee-body Diagram: When the block is being displaced by amount xto the right, the
restoring force
that develops in both springs is .
Equation of Motion:
[1]
Kinematics:
Since , then substituting this value into Eq. [1], we have
[2]
Since
the friction will eventually dampen out the free vibration, we are only
interested
in the particular solution of the above differential equation which is in the
form of
T
aking second time derivative and substituting into Eq. [2], we have
T
hus,
[3]
T
aking the time derivative of Eq. [3], we have
T
hus,
Ans.
A
yp
B
max =2.07 ft
>
s
yp=x
#
p=-2.0663 sin 3t
xp=0.6888 cos 3t
C=0.6888 ft
–9Ccos 3t+21.47Ccos 3t=8.587 cos 3t
xp=Ccos 3t
x
$+21.47x=8.587 cos 3t
a=d2x
dt2=x
$
a+21.47x=8.587 cos 3t
:
+©Fx=0; –2(10x)+8 cos 3t=30
32.2 a
Fsp =kx =10x
The 30-lb block is attached to two springs having a stiffness
of
A periodic force , where tis in
seconds
, is applied to the block. Determine the maximum
speed
of the block after frictional forces cause the free
vibrations to dampen out.
F=(8 cos 3t)lb10 lb>ft.
F8 cos 3t
k10 lb/ft
k10 lb/ft
22–49.
SOLUTION
Using Eq. 22–22, the amplitude is
Ans.(xp)max =0.0295 m =29.5 mm
(xp)max =3d0
1–av0
vnb23=30.015
1–
Q
12.57
17.93
R
23
vn=Ak
m=A1285.71
4=17.93
d0=0.015 m
v0=2Hz=2(2p)=12.57 rad>s
k=F
¢y=18
0.014 =1285.71 N>m
The light elastic rod supports a 4-kg sphere.When an 18-N
vertical force is applied to the sphere, the rod deflects
14 mm. If the wall oscillates with harmonic frequency of
2 Hz and has an amplitude of 15 mm, determine the
amplitude of vibration for the sphere.
0.75 m
1239
22–50.
SOLUTION
Equation
of Motion: When the rod is in equilibrium, , and
. writing the moment equation of motion about point Bby
referring to the free-body diagram of the rod,
Fig. a,
T
hus, the initial stretch of the spring is .When the rod rotates about
point
Bthrough a small angle , the spring stretches further by .Thus, the
force
in the spring is . Also, the velocity of end C
of
the rod is .Thus,.The mass moment of inertia of
the
rod about Bis . Again, referring to Fig. aand
writing the moment equation of motion about
B,
Since
is small, .Thus, this equation becomes
Ans.
Comparing this equation to that of the standard form,
T
hus,
F
or the system to be underdamped,
4c622mk
ceq 6cc
cc=2mvn=2mAk
m=22mk
vn=Ak
mceq =4c
u
$
+4c
mu
#
+k
mu=0
cos u1u
u
$
+4c
mcos uu
#
+k
m(cos u)u=0
=-ma2u
$
©MB=IBa;kamg
2k+aubcos u(a)+
A
2au
#
B
cos u(2a)–mg cos uaa
2b
IB=1
12 m(3a)2+maa
2b2
=ma2
Fc=cy
#
c=c(2au
#
)vc=y
#
c=2au
#
FA=k(s0+s1)=k
¢
mg
2k+au
≤
s1=auu
sO=FA
k=mg
2k
+©MB=0; –FA(a)–mgaa
2b=0FA=mg
2
u
$
=0
Fc=cy
#
c=0u=0°
Find the differential equation for small oscillations in terms
of
for the uniform rod of mass m. Also show that if
, then the system remains underdamped. The
rod is in a horizontal position when it is in equilibrium.
c62mk>2
u
A
B
a
C
c
k
2
u
a
m
m
1240
22–51.
The 40-kg block is attached to a spring having a stiffness of
800 N>m
. A force
F=(100 cos 2t) N
, where t is in seconds
is applied to the block. Determine the maximum speed of
the block for the steady-state vibration.
k 800 N/m
Ans:
(v
p
)
max
=0.3125 m>s
1241
*22–52.
SOLUTION
Since ,
(1)
Substitute ypinto Eq. (1)
C=Q–v02R
C(–v02+k
m) cos v0t=kd0
mcos v0t
yP=Ccos v0t(Particular sol.)
yC=Asin vny+Bcos vny(General sol.)
y
$+k
my=kd0
mcos v0t
W=kdst
+T©Fy=may;kd0cos v0t+W–kdst –ky =my
$
Use a block-and-spring model like that shown in
Fig. 22–14abut suspended from a vertical position and
subjected to a periodic support displacement of
determine the equation of motion for the
system, and obtain its general solution. Define the
displacement ymeasured from the static equilibrium
position of the block when t=0.
d=d0cos v0t,
kd0
m
a
m
–v02
b
1242
22–53.
The fan has a mass of 25 kg and is fixed to the end of a
horizontal beam that has a negligible mass.The fan blade is
mounted eccentrically on the shaft such that it is equivalent
to an unbalanced 3.5-kg mass located 100 mm from the axis
of rotation. If the static deflection of the beam is 50 mm as a
result of the weight of the fan, determine the angular
velocity of the fan blade at which resonance will occur. Hint:
See the first part of Example 22.8.
SOLUTION
Resonance occurs when
Ans.v=vn=14.0 rad>s
C
k
m=
C
4905
25 =14.01 rad>s
k=F
¢y=25(9.81)
0.05 =4905 N>m
V
Ans:
v
=14.0 rad>s
1243
22–54.
SOLUTION
The force caused by the unbalanced rotor is
Using Eq. 22–22, the amplitude is
Ans.(xp)max =14.6 mm
(xp)max =435
4905
1–a10
14.01b24=0.0146 m
(xp)max =4F0
k
1–av
pb24
F0=mr v2=3.5(0.1)(10)2=35 N
C
k
m=
C
4905
25 =14.01 rad>s
k=F
¢y=25(9.81)
0.05 =4905 N>m
10 rad>s
V
In Prob. 22–53, determine the amplitude of steady-state
vibration of the fan if its angular velocity is .
Ans:
(x
p
)
max
=14.6 mm
1244
22–55.
? Hint: See the first part of Example 22.8.18 rad>s
SOLUTION
The force caused by the unbalanced rotor is
Using Eq. 22–22, the amplitude is
Ans.(xp)max =35.5 mm
(xp)max =4113.4
4905
1–a18
14.01 b24=0.0355 m
(xp)max =4F0
k
1–av
pb24
F0=mrv2=3.5(0.1)(18)2=113.4 N
C
k
m=
C
4905
25 =14.01 rad>s
k=F
¢y=25(9.81)
0.05 =4905 N>m
V
What will be the amplitude of steady-state vibration of the
fan in Prob. 22–53 if the angular velocity of the fan blade is
Ans:
(x
p
)
max
=35.5 mm
*22–56.
SOLUTION
Fs=kx =15(x+xst –0.1 cos 15t)
+©MO=IOa; 4(9.81)(0.6) –Fs(1.2) =4(0.6)2u
$
The small block at Ahas a mass of 4 kg and is mounted on
the bent rod having negligible mass. If the rotor at Bcauses
a harmonic movement , where tis in
seconds, determine the steady-state amplitude of vibration
of the block.
dB=(0.1 cos 15t)m
0.6 m
1.2 m
AO
V
1246
22–57.
The electric motor turns an eccentric flywheel which is
equivalent to an unbalanced 0.25-lb weight located 10 in.
from the axis of rotation. If the static deflection of the
beam is 1 in. due to the weight of the motor,determine
the angular velocity of the flywheel at which resonance
will occur.The motor weights 150 lb.Neglect the mass of
the beam.
SOLUTION
Resonance occurs when Ans.v=vn=19.7 rad>s
k=F
d
=150
1>12 =1800 lb>ft
vn=Ak
m=A1800
150>32.2 =19.66
V
Ans:
v
=19.7 rad>s
1247
22–58.
SOLUTION
Hence
From Eq. 22–21, the amplitude of the steady state motion is
Ans.C=4F
0>k
1–av0
vnb24=42.588>1800
1–a20
19.657 b24=0.04085 ft =0.490 in.
k=F
d=150
1>12 =1800 lb>ft
vn=Ak
m=A1800
150>32.2 =19.657
F=2.588 sin 20t
W
h
at w
ill
b
e t
h
e amp
li
tu
d
e of stea
d
y-state v
ib
rat
i
on of t
h
e
motor in Prob. 22–57 if the angular velocity of the flywheel
is ?20 rad>s
V
Ans:
C=0.490 in.
1248
22–59.
Determine the angular velocity of the flywheel in Prob.22–
5
7
which will produce an amplitude of vibration of 0.25 in.
SOLUTION
From Eq. 22.21, the amplitude of the steady-state motion is
Ans.v=19.0 rad>s
0.25
12 =40.006470av2
1800 b
1–av
19.657 b24
C=4F
0>k
1–av
vnb24
k=F
d
=150
1>12 =1800 lb>ft
vn=Ak
m=A1800
150>32.2 =19.657
F=0.006470v2sin vt
V
Ans:
v
=19.0 rad>s
1249
*22–60.
The 450-kg trailer is pulled with a constant speed over the
s
urface of a bumpy road, which may be approximated by a
cosine
curve having an amplitude of 50 mm and wave length of
4m
.If the two springs swhich support the trailer each have a
stiffness
of determine the speed which will cause
t
he greatest vibration (resonance) of the trailer.Neglect the
w
eight of the wheels.
v800 N>m,
SOLUTION
T
he amplitude is
T
he wave length is
F
or maximum vibration of the trailer, resonance must occur, i.e.,
T
hus, the trailer must travel , in , so that
Ans.vR=l
t
=4
3.33
=1.20 m.s
t=3.33 sl=4m
v0=vn
t=2p
vn
=2p
1.89 =3.33 s
vn=Ak
m=A1600
450
=1.89 rad>s
k=2(800) =1600 N>m
l=4m
d0=50 mm =0.05 m
v
100mm
2m 2m
s
Ans:
vR=1.20 m.s