22–1.
A spring is stretched 175 mm by an 8-kg block. If the block is
displaced 100 mm downward from its equilibrium position
and given a downward velocity of 1.50 m>s, determine the
differential equation which describes the motion. Assume
that positive displacement is downward. Also, determine the
position of the block when t=0.22 s.
SOLUTION
$ where kyst =mg
22–2.
A spring has a stiffness of 800 N>m. If a 2-kg block is
attached to the spring, pushed 50 mm above its equilibrium
position, and released from rest, determine the equation
that describes the block’s motion. Assume that positive
displacement is downward.
SOLUTION
22–3.
A spring is stretched by a 1
5
-kg block. If the block
is displaced downward from its equilibrium
position and given a downward velocity of
determine the equation which describes the motion. What is
the phase angle? Assume that positive displacement is
downward.
0.75 m>s,
100 mm
200 mm
*22–4.
Ans
.
Ans
.
t=2p
vn
=0.452 s
vn=Ak
m=A60
10
32.2
=13.90 rad>s
4
12
When a 20-lb weight is suspended from a spring, the spring
is
stretched a distance of 4 in. Determine the natural
frequenc
y and the period of vibration for a 10-lb weight
attached to the same spring
.
22–5.
Ans.
Ans.
t=1
f=1
7.88 =0.127 s
f=vn
2p=49.52
2p=7.88 Hz
vn=Ak
m=A490.5
0.2 =49.52 = 49.5 rad>s
When a 3-kg block is suspended from a spring, the spring is
stretched a distance of 60 mm. Determine the natural
frequency and the period of vibration for a 0.2-kg block
attached to the same spring.
22–6.
An 8-
k
g
bl
oc
k
i
s suspen
d
e
d
from a spr
i
ng
h
av
i
ng a st
i
ffness
If the block is given an upward velocity of
when it is 90 mm above its equilibrium position,
determine the equation which describes the motion and the
maximum upward displacement of the block measured
from the equilibrium position. Assume that positive
displacement is measured downward.
0.4 m>s
k=80 N>m.
22–7.
SOLUTION
Ans.
From Eqs. 22–3 and 22–4,
Ans.
Position equation,
Ans.y=(0.0833 cos 19.7t)ft
C=2A2+B2=0.0833 ft =1 in.
A=0
0=Avn+0
B=-0.0833
–1
12 =0+B
y=-
1
12,y=0att=0
vn=Ak
m=A24
2
32.2
=19.66 = 19.7 rad>s
k=2(12) =24 lb>ft
A2-lb weight is suspended from a spring having a stiffness
If the weight is pushed 1 in. upward from its
equilibrium position and then released from rest, determine
the equation which describes the motion. What is the
amplitude and the natural frequency of the vibration?
k=2lb>in.
*22–8.
A6-lb weight is suspended from a spring having a stiffness
If the weight is given an upward velocity of
when it is 2 in. above its equilibrium position,
determine
the equation which describes the motion and the
maximum
upward displacement of the weight, measured
from
the equilibrium position. Assume positive displacement
is
downward.
20 ft>s
k=3lb>in.
SOLUTION
t=0, y=-20 ft>s,
y=-
1
6ft
vn=Ak
m=A36
6
32.2
=13.90 rad>s
k=3(12) =36 lb>ft
22–9.
A3-kg block is suspended from a spring having a stiffness
of If the block is pushed upward
from its equilibrium position and then released from rest,
determine the equation that describes the motion. What are
the amplitude and the frequency of the vibration? Assume
that positive displacement is downward.
50 mmk=200 N>m.
1199
22–10.
The uniform rod of mass m is supported by a pin atA and a
spring at B. If B is given a small sideward displacement and
released, determine the natural period of vibration.
SOLUTION
1
2.
A
L
1200
22–11.
SOLUTION
Using the result of Example 22–1,
We have
Ans.t=2p
vn
=2p
4.336 =1.45 s
vn=Ag
l=A28.2
18>12 =4.336 rad>s
While standing in an elevator,the man holds a pendulum
which consists of an 18-in. cord and a 0.5-lb bob.If the elevator
is descending with an acceleration determine the
natural period of vibration for small amplitudes of swing.
a=4ft>s2,
a4ft/s
2
Ans:
t=1.45 s
1201
Am
3k
1202
22–13.
The body of arbitrary shape has a mass m, mass center at G,
and a radius of gyration about Gof . If it is displaced a
slight amount from its equilibrium position and released,
determine the natural period of vibration.
u
kG
SOLUTION
a
However, for small rotation . Hence
u
$+gd
sin uLu
u
$+gd
k2
G+d2sin u=0
+©MO=IOa;–mgd sin u=
C
mk2
G+md2
D
u
$
O
u
G
d
1203
22–14.
The 20-lb rectangular plate has a natural period of vibration
t=0.3 s
, as it oscillates around the axis of rod AB.
Determine the torsional stiffness k, measured inlb
#
ft
>
rad,
of the rod. Neglect the mass of the rod.
Ans:
k=90.8 lb #ft>rad
k
B
A
SOLUTION
T=ku
t=
2
p
2
k(4.83)
=0.3
k=90.8 lb #ft>rad
Ans.
22–15.
A platform, having an unknown mass, is supported by four
springs, each having the same stiffness k. When nothing is
on the platform, the period of vertical vibration is measured
as 2.35 s; whereas if a 3-kg block is supported on the
platform, the period of vertical vibration is 5.23 s. Determine
the mass of a block placed on the (empty) platform which
causes the platform to vibrate vertically with a period of
5.62 s. What is the stiffness k of each of thesprings?
kk
SOLUTION
$
*22–16.
k2
k2
k1k1
A block of mass mis suspended from two springs having a
s
tiffness of and , arranged a) parallel to each other, and
b)
as a series. Determine the equivalent stiffness of a single
s
pring with the same oscillation characteristics and the
period of
oscillation for each case.
k2
k1
SOLUTION
(a)
When the springs are arranged in parallel, the equivalent spring stiffness is
Ans.
T
he natural frequency of the system is
Th
us, the period of oscillation of the system is
vn=
C
m=Bk1+k2
m
keq =k1+k2
1206
22–17.
k2
k2
k1k1
The 15-kg block is suspended from two springs having a
different stiffness and arranged a) parallel to each other,
and b) as a series. If the natural periods of oscillation of the
parallel system and series system are observed to be 0.5 s
and 1.5 s,respectively, determine the spring stiffnesses
and .k2
k1
SOLUTION
series can be determined by equating the stretch of the system to a single equivalent
spring when they are subjected to the same load.
Thus the natural frequencies of the parallel and series spring system are
Thus, the natural periods of oscillation are
(1)
(2)
Solving Eqs. (1) and (2),
or Ans.
or Ans.2067 N>mk
2=
302 N>m
302 N>mk1=2067 N>m
tS=2p
(vn)S
=2p D15
A
k1+k2
B
k1k2
=1.5
tP=2p
(vn)P
=2p B15
k1+k2
=0.5
(vn)S=D
A
keq
B
S
m=U
¢
k1k2
k1+k2
≤
15 =Dk1k2
15
A
k1+k2
B
(vn)P=D
A
keq
B
P
m=Bk1+k2
15
A
keq
B
S=k1k2
k1+k2
k2+k1
k1k2
=1
A
keq
B
S
F
k1
k2
(keq)S
Ans:
k1=2067 N>m
k2=302 N>m
or vice versa
22–18.
The uniform beam is supported at its ends by two springs
Aand B, each having the same stiffness k.When nothing is
supported on the beam, it has a period of vertical vibration
of 0.83 s. If a 50-kg mass is placed at its center, the period
of vertical vibration is 1.52 s. Compute the stiffness of each
spring and the mass of the beam.
SOLUTION
(1)
(2)
Eqs. (1) and (2) become
Ans.
Ans.k=609 N m
mB=21.2 kg
mB+50 =0.1170k
mB=0.03490k
(1.52)2
(2p)2=mB+50
2k
(0.83)2
(2p)2=mB
2k
t2
(2p)2=m
k
t=2pAm
k
A
kk
B
1208
22–19.
SOLUTION
Each spring force .
a
However, for small displacement , and . Hence
From the above differential equation, .
Ans.y=0.503 m =503 mm
19.74y2–4.905y–2.5211 =0
1=2p
B3.4686 +4.905y
0.024 +0.5y2
t=2p
p
p=B3.4686 +4.905y
0.024 +0.5y2
u
$
+3.4686 +4.905y
0.024 +0.5y2u=0
cos u=1sin uLux=0.6u
–4.8xcos u–(0.5886 +4.905y) sin u=(0.024 +0.5y2)u
$
–0.5(9.81)(ysin u)=(0.024 +0.5y2)u
$
+©MO=IOa;–2(4x)(0.6 cos u)–0.2(9.81)(0.3 sin u)
F
s=kx =4x
The slender rod has a mass of 0.2 kg and is supported at O
byapin and at its end Aby two springs, each having a
stiffness .The period of vibration of the rod can
be set by fixing the 0.5-kg collar Cto the rod at an
appropriate location along its length. If the springs are
originally unstretched when the rod is vertical, determine
the position yof the collar so that the natural period of
vibration becomes . Neglect the size of the collar.t=1s
k=4N>m
yO
600 mm
C
Ans:
y=503 mm
*22–20.
A uniform board is supported on two wheels which rotate
in opposite directions at a constant angular speed. If the
coefficient of kinetic friction between the wheels and board
is , determine the frequency of vibration of the board if it
is displaced slightly,a distance xfrom the midpoint between
the wheels, and released.
m
SOLUTION
Equation of Motion:
a
(1)
Kinematics:Since , then substitute this value into Eq.(1), we have
(2)
From Eq.(2),
=Amg
d
2=mg
d
x
## +mg
dx=0
a=d2x
dt2=x
##
a+mg
dx=0
:
+©F
x=m(aG)x;mcmg(d-x)
2dd–mcmg(d+x)
2dd=ma
N
A=mg(d–x)
2d
+c©F
y=m(aG)y;N
A+mg(d+x)
2d–mg =0
N
B=mg(d+x)
2d
+©MA=©(MA)k;N
B(2d)–mg(d+x)=0
dd
x
BA
vn. Applying Eq. 22–4, we have
vn, thus,