Engineering
Mechanics Dynamics 14th Edition Hibbeler Solutions Manual
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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
$
mg - k(y + yst) = my
+ T ΣFy = may;
where kyst = mg
k
$
y +
y = 0
m
Hence
p =
=
k
Bm
B
Where k =
8(9.81)
0.175
448.46
= 7.487
8
$
y + (7.487)2y = 0
6
= 448.46 N>m
$
y + 56.1y = 0
Ans.
The solution of the above differential equation is of the form:
y = A sin pt + B cos pt
(1)
#
v = y = Ap cos pt - Bp sin pt
(2)
At t = 0, y = 0.1 m and v = v0 = 1.50 m>s
From Eq. (1)
0.1 = A sin 0 + B cos 0
B = 0.1 m
v0
1.50
=
= 0.2003 m
p
7.487
From Eq. (2)
v0 = Ap cos 0 - 0
Hence
y = 0.2003 sin 7.487t + 0.1 cos 7.487t
At t = 0.22 s,
y = 0.2003 sin [7.487(0.22)] + 0.1 cos [7.487(0.22)]
A =
Ans.
= 0.192 m
1190
Ans:
$
y + 56.1 y = 0
y 0 t = 0.22 s = 0.192 m
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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
p =
k
800
=
= 20
Am
A 2
x = A sin pt + B cos pt
x = - 0.05 m
when t = 0,
-0.05 = 0 + B;
B = -0.05
v = Ap cos pt - Bp sin pt
v = 0 when t = 0,
0 = A(20) - 0;
A = 0
Thus,
Ans.
x = - 0.05 cos (20t)
Ans:
x = -0.05 cos (20t)
1191
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exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
22–3.
A spring is stretched 200 mm by a 15-kg block. If the block
is displaced 100 mm downward from its equilibrium
position and given a downward velocity of 0.75 m>s,
determine the equation which describes the motion. What is
the phase angle? Assume that positive displacement is
downward.
SOLUTION
k =
15(9.81)
F
= 735.75 N>m
=
y
0.2
vn =
k
735.75
=
= 7.00
Am
A 15
y = A sin vn t + B cos vn t
y = 0.1 m when t = 0,
0.1 = 0 + B;
B = 0.1
v = A vn cos vn t - Bvn sin vn t
v = 0.75 m>s when t = 0,
0.75 = A(7.00)
A = 0.107
Ans.
y = 0.107 sin (7.00t) + 0.100 cos (7.00t)
f = tan - 1 a
B
0.100
b = tan - 1 a
b = 43.0°
A
0.107
Ans.
Ans:
y = 0.107 sin (7.00t) + 0.100 cos (7.00t)
f = 43.0°
1192
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exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
*22–4.
When a 20-lb weight is suspended from a spring, the spring
is stretched a distance of 4 in. Determine the natural
frequency and the period of vibration for a 10-lb weight
attached to the same spring.
SOLUTION
k =
vn =
t =
20
4
12
= 60 lb>ft
k
60
=
= 13.90 rad>s
10
Am
A 32.2
Ans.
2p
= 0.452 s
vn
Ans.
Ans:
vn = 13.90 rad>s
t = 0.452 s
1193
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exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
22–5.
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.
SOLUTION
k =
vn =
3(9.81)
F
=
= 490.5 N>m
¢x
0.060
k
490.5
=
= 49.52 = 49.5 rad>s
Am
A 0.2
f =
vn
49.52
=
= 7.88 Hz
2p
2p
t =
1
1
=
= 0.127 s
f
7.88
Ans.
Ans.
Ans:
vn = 49.5 rad>s
t = 0.127 s
1194
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exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
22–6.
An 8-kg block is suspended from a spring having a stiffness
k = 80 N>m. If the block is given an upward velocity of
0.4 m>s 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.
SOLUTION
vn =
k
80
=
= 3.162 rad>s
Am
A8
y = - 0.4 m>s,
x = - 0.09 m at t = 0
x = A sin vn t + B cos vn t
- 0.09 = 0 + B
B = -0.09
y = Avn cos vn t - Bvn sin vn t
-0.4 = A(3.162) - 0
A = -0.126
Thus,
x = - 0.126 sin (3.16t) - 0.09 cos (3.16t) m
Ans.
C = 2A2 + B2 = 2( -0.126)2 + ( - 0.09) = 0.155 m
Ans.
Ans:
x = 5 - 0.126 sin (3.16t) - 0.09 cos (3.16t) 6 m
C = 0.155 m
1195
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exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
22–7.
A 2-lb weight is suspended from a spring having a stiffness
k = 2 lb>in. 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?
SOLUTION
k = 2(12) = 24 lb>ft
vn =
k
24
=
= 19.66 = 19.7 rad>s
2
Am
A 32.2
y = -
1
,
12
Ans.
y = 0 at t = 0
From Eqs. 22–3 and 22–4,
-
1
= 0 + B
12
B = -0.0833
0 = Avn + 0
A = 0
C = 2A2 + B2 = 0.0833 ft = 1 in.
Ans.
y = (0.0833 cos 19.7t) ft
Ans.
Position equation,
Ans:
vn = 19.7 rad>s
C = 1 in.
y = (0.0833 cos 19.7t) ft
1196
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exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
*22–8.
A 6-lb weight is suspended from a spring having a stiffness
k = 3 lb>in. If the weight is given an upward velocity of
20 ft>s 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.
SOLUTION
k = 3(12) = 36 lb>ft
vn =
k
36
=
= 13.90 rad>s
Am
A 6
32.2
t = 0,
y = - 20 ft>s,
1
y = - ft
6
From Eq. 22–3,
-
1
= 0 + B
6
B = -0.167
From Eq. 22–4,
-20 = A(13.90) + 0
A = - 1.44
Thus,
Ans.
y = [-1.44 sin (13.9t) - 0.167 cos (13.9t)] ft
From Eq. 22–10,
C = 2A2 + B2 = 2(1.44)2 + (- 0.167)2 = 1.45 ft
Ans.
Ans:
y = [-1.44 sin (13.9t) - 0.167 cos (13.9t)] ft
C = 1.45 ft
1197
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exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
22–9.
A 3-kg block is suspended from a spring having a stiffness
of k = 200 N>m. If the block is pushed 50 mm 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.
SOLUTION
vn =
k
200
=
= 8.16 rad>s
Am
A 3
Ans.
x = A sin vn t + B cos vn t
x = -0.05 m when t = 0,
- 0.05 = 0 + B;
B = -0.05
v = Ap cos vn t - Bvn sin vn t
v = 0 when t = 0,
0 = A(8.165) - 0;
A = 0
Hence,
x = - 0.05 cos (8.16t)
Ans.
C = 2A2 + B2 = 2(0)2 + ( - 0.05) = 0.05 m = 50 mm
Ans.
Ans:
vn = 8.16 rad>s
x = - 0.05 cos (8.16t)
C = 50 mm
1198
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exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
22–10.
The uniform rod of mass m is supported by a pin at A and a
spring at B. If B is given a small sideward displacement and
released, determine the natural period of vibration.
A
L
Solution
1
Equation of Motion. The mass moment of inertia of the rod about A is IA = mL2.
3
Referring to the FBD. of the rod, Fig. a,
L
1
a + ΣMA = IAa ;
-mg a sin u b - (kx cos u)(L) = a mL2 ba
2
3
B
k
However; x = L sin u. Then
- mgL
1
sin u - kL2 sin u cos u = mL2a
2
3
Using the trigonometry identity sin 2u = 2 sin u cos u,
- mgL
KL2
1
sin u sin 2u = mL2a
2
2
3
$
Here since u is small sin u ≃ u and sin 2u ≃ 2u. Also a = u . Then the above
equation becomes
$
mgL
1
mL2 u + a
+ kL2 bu = 0
3
2
$
3mg + 6kL
u +
u = 0
2mL
Comparing to that of the Standard form, vn =
t =
A
3mg + 6kL
. Then
2mL
2p
2mL
= 2p
vn
A 3mg + 6kL
Ans.
Ans:
t = 2p
1199
2mL
A 3mg + 6kL
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22–11.
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 a = 4 ft>s2, determine the
natural period of vibration for small amplitudes of swing.
a
4 ft/s2
SOLUTION
Since the acceleration of the pendulum is (32.2 - 4) = 28.2 ft>s2
Using the result of Example 22–1,
We have
vn =
t =
g
28.2
=
= 4.336 rad>s
Al
A 18>12
2p
2p
= 1.45 s
=
vn
4.336
Ans.
Ans:
t = 1.45 s
1200
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*22–12.
Determine the natural period of vibration of the uniform
bar of mass m when it is displaced downward slightly and
released.
O
k
L
—
2
L
—
2
Solution
Equation of Motion. The mass moment of inertia of the bar about O is I0 =
1
mL2.
12
Referring to the FBD of the rod, Fig. a,
a + ΣM0 = I0a ;
However, y =
L
1
- ky cos ua b = a mL2 ba
2
12
L
sin u. Then
2
-ka
1
L
L
sin u b cos u a b =
mL2a
2
2
12
Using the trigonometry identity sin 2u = 2 sin u cos u, we obtain
1
kL2
mL2a +
sin 2u = 0
12
8
$
Here since u is small, sin 2u ≃ 2u. Also, a = u . Then the above equation becomes
$
1
kL2
mL2 u +
u = 0
12
4
$
3k
u = 0
u +
m
Comparing to that of the Standard form, vn =
t =
3k
. Then
Am
m
2p
= 2p
vn
A 3k
Ans.
Ans:
t = 2p
1201
m
A 3k
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22–13.
The body of arbitrary shape has a mass m, mass center at G,
and a radius of gyration about G of kG. If it is displaced a
slight amount u from its equilibrium position and released,
determine the natural period of vibration.
O
d
u
G
SOLUTION
a + ©MO = IO a;
$
-mgd sin u = C mk2G + md2 D u
$
u +
gd
k2G
+ d2
sin u = 0
However, for small rotation sin u Lu. Hence
$
u +
gd
2
kG
+ d2
u = 0
From the above differential equation, vn =
t =
2p
=
vn
2p
gd
A k2G + d2
gd
B k2G
= 2p
+ d2
.
k2G + d2
C gd
Ans.
Ans:
t = 2p
1202
k 2G + d 2
C gd
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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 in lb # ft>rad,
of the rod. Neglect the mass of the rod.
A
k
B
Solution
T = ku
ΣMz = Iza ;
- ku =
$
u + k(4.83)u = 0
t =
2p
2k(4.83)
4 ft
$
1 20
a
b(2)2 u
12 32.2
2 ft
= 0.3
k = 90.8 lb # ft>rad
Ans.
Ans:
k = 90.8 lb # ft>rad
1203
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exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
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 the springs?
k
k
Solution
$
+ T ΣFy = may; mtg - 4k(y + yts) = mty Where 4k yts = mtg
4k
$
y +
y = 0
mt
Hence P =
t =
4k
A mt
2p
mt
= 2p
P
A 4k
For empty platform mt = mP, where mP is the mass of the platform.
2.35 = 2p
mP
A 4k
(1)
When 3-kg block is on the platform mt = mP + 3.
5.23 = 2p
mP + 3
A 4k
(2)
When an unknown mass is on the platform mt = mP + mB.
5.62 = 2p
mP + mB
A
4k
(3)
Solving Eqs. (1) to (3) yields :
k = 1.36 N>m mB = 3.58 kg
Ans.
mP = 0.7589 kg
Ans:
k = 1.36 N>m
mB = 3.58 kg
1204
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exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
*22–16.
A block of mass m is suspended from two springs having a
stiffness of k1 and k2, arranged a) parallel to each other, and
b) as a series. Determine the equivalent stiffness of a single
spring with the same oscillation characteristics and the
period of oscillation for each case.
k1
k2
k1
k2
SOLUTION
(a) When the springs are arranged in parallel, the equivalent spring stiffness is
Ans.
keq = k1 + k2
The natural frequency of the system is
vn =
keq
Cm
=
(a)
k1 + k2
B m
(b)
Thus, the period of oscillation of the system is
t =
2p
=
vn
2p
k1 + k2
B m
= 2p
m
Ans.
C k1 + k2
(b) When the springs are arranged in a series, the equivalent stiffness of the system
can be determined by equating the stretch of both spring systems subjected to
the same load F.
F
F
F
+
=
k1
k2
keq
1
1
1
+
=
k1
k2
keq
k2 + k1
1
=
k1k2
keq
keq =
k1k2
k1 + k2
Ans.
The natural frequency of the system is
k1k2
b
k2 + k1
vn =
=
m
Cm
S
keq
a
Thus, the period of oscillation of the system is
t =
2p
=
vn
2p
k1k2
b
a
k2 + k1
m
S
= 2p
C
m(k1 + k2)
k1k2
Ans.
Ans:
keq = k1 + k2
m
t = 2p
A k1 + k2
k1k2
keq =
k1 + k2
m(k1 + k2)
t = 2p
A
k1k2
1205
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exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
22–17.
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 k1
and k2.
k1
k2
k1
k2
SOLUTION
The equivalent spring stiffness of the spring system arranged in parallel is
A keq B P = k1 + k2 and the equivalent stiffness of the spring system arranged in a
series can be determined by equating the stretch of the system to a single equivalent
spring when they are subjected to the same load.
F
F
F
+
=
k1
k2
(keq)S
(a)
(b)
k2 + k1
1
=
k1k2
A keq B S
A keq B S =
k1k2
k1 + k2
Thus the natural frequencies of the parallel and series spring system are
(vn)P =
(vn)S =
A keq B P
D m
A keq B S
D m
=
=
B
k1 + k2
15
¢
k1k2
≤
k1 + k2
U
15
=
k1k2
D 15 A k1 + k2 B
Thus, the natural periods of oscillation are
tP =
tS =
2p
15
= 2p
= 0.5
(vn)P
B k1 + k2
(1)
15 A k1 + k2 B
2p
= 2p
= 1.5
(vn)S
k1k2
D
(2)
Solving Eqs. (1) and (2),
k1 = 2067 N>m or 302 N>m
Ans.
k2 = 302 N>m or 2067 N>m
Ans.
Ans:
k1 = 2067 N>m
k2 = 302 N>m
or vice versa
1206
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exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
22–18.
The uniform beam is supported at its ends by two springs
A and 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.
A
B
k
k
SOLUTION
t = 2p
m
Ak
m
t2
=
k
(2p)2
(0.83)2
(2p)2
(1.52)2
(2p)2
=
mB
2k
(1)
=
mB + 50
2k
(2)
Eqs. (1) and (2) become
mB = 0.03490k
mB + 50 = 0.1170k
Ans.
mB = 21.2 kg
Ans.
k = 609 N m
Ans:
mB = 21.2 kg
k = 609 N>m
1207
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exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
22–19.
The slender rod has a mass of 0.2 kg and is supported at O
by a pin and at its end A by two springs, each having a
stiffness k = 4 N>m. The period of vibration of the rod can
be set by fixing the 0.5-kg collar C to the rod at an
appropriate location along its length. If the springs are
originally unstretched when the rod is vertical, determine
the position y of the collar so that the natural period of
vibration becomes t = 1 s. Neglect the size of the collar.
O
y
C
600 mm
SOLUTION
k A k
Moment of inertia about O:
IO =
1
(0.2)(0.6)2 + 0.5y2 = 0.024 + 0.5y2
3
Each spring force Fs = kx = 4x.
a + ©MO = IO a;
- 2(4x)(0.6 cos u) - 0.2(9.81)(0.3 sin u)
$
- 0.5(9.81)(y sin u) = (0.024 + 0.5y2) u
$
- 4.8x cos u - (0.5886 + 4.905y) sin u = (0.024 + 0.5y2) u
However, for small displacement x = 0.6u, sin u L u and cos u = 1. Hence
$ 3.4686 + 4.905y
u+
u = 0
0.024 + 0.5y2
From the above differential equation, p =
t =
1 =
3.4686 + 4.905y
B 0.024 + 0.5y2
.
2p
p
2p
3.4686 + 4.905y
B 0.024 + 0.5y2
19.74y 2 - 4.905y - 2.5211 = 0
Ans.
y = 0.503 m = 503 mm
Ans:
y = 503 mm
1208
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exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
*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 m, determine the frequency of vibration of the board if it
is displaced slightly, a distance x from the midpoint between
the wheels, and released.
x
A
B
d
d
SOLUTION
Freebody Diagram: When the board is being displaced x to the right, the restoring
force is due to the unbalance friction force at A and B C (Ff)B 7 (Ff)A D .
Equation of Motion:
a + ©MA = ©(MA)k ;
NB (2d) - mg(d + x) = 0
NB =
+ c ©Fy = m(aG)y ;
NA +
mg(d + x)
- mg = 0
2d
NA =
+ ©F = m(a ) ;
:
x
G x
mc
Kinematics: Since a =
mg(d - x)
2d
mg(d + x)
mg(d-x)
d - mc
d = ma
2d
2d
mg
x = 0
a +
d
(1)
d2x
##
= x, then substitute this value into Eq.(1), we have
dt2
##
x+
From Eq.(2), vn 2 =
mg(d + x)
2d
mg
x = 0
d
(2)
mg
mg
, thus, vn =
. Applying Eq. 22–4, we have
d
A d
f =
vn
1
=
2p
2p
mg
d
Ans.
Ans:
f =
1209
mg
1
2p A d
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exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
22–21.
If the wire AB is subjected to a tension of 20 lb, determine
the equation which describes the motion when the 5-lb
weight is displaced 2 in. horizontally and released from rest.
A
6 ft
Solution
L′ K L
6 ft
+ ΣF = m a ; - 2T x = mx$
d
x
x
L
B
2T
$
x +
x = 0
Lm
P =
2(20)
2T
=
= 6.55 rad>s
5
A Lm
A 6 ( 32.2
)
x = A sin pt + B cos pt
x =
1
1
ft at t = 0, Thus B = = 0.167
6
6
v = A p cos pt - B p sin pt
v = 0 at t = 0, Thus A = 0
So that
Ans.
x = 0.167 cos 6.55t
Ans:
x = 0.167 cos 6.55t
1210
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exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
22–22.
The bar has a length l and mass m. It is supported at its ends
by rollers of negligible mass. If it is given a small displacement
and released, determine the natural frequency of vibration.
R
A
B
l
SOLUTION
Moment of inertia about point O:
IO =
1 2
l2 2
1
ml + ma R2 b = ma R2 - l2 b
12
4
6
B
c + ©MO = IOa;
mg a
B
R2 -
l2
1 $
b u = - ma R2 - l2 bu
4
6
1
$
3g(4R 2 - l2)2
u +
u = 0
6R 2 - l2
1
From the above differential equation, vn =
3g(4R2 - l2)2
D
6R2 - l2
.
Ans.
Ans:
vn =
1211
3g(4R2 - l 2)1>2
C
6R2 - l 2
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exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
22–23.
The 20-kg disk, is pinned at its mass center O and supports
the 4-kg block A. If the belt which passes over the disk is
not allowed to slip at its contacting surface, determine the
natural period of vibration of the system.
O
Solution
k =300
50 N/m
mm
k 200 N/m
Equation of Motion. The mass moment of inertia of the disk about its mass
1
1
center O is I0 = mr 2 = (20) ( 0.32 ) = 0.9 kg # m2. When the disk undergoes a
2
2
small angular displacement u, the spring stretches further by s = ru = 0.3u. Thus,
A
the total stretch is y = yst + 0.3u. Then Fsp = ky = 200(yst + 0.3u). Referring to
the FBD and kinetic diagram of the system, Fig. a,
a + ΣM0 = Σ(mk)0; 4(9.81)(0.3) - 200(yst + 0.3u)(0.3) = 0.90a + 4[a(0.3)](0.3)
(1)
11.772 - 60yst - 18u = 1.26a
When the system is in equilibrium, u = 0°. Then
a + ΣM0 = 0;
4(9.81)(0.3) - 200(yst)(0.3) = 0
60yst = 11.772
Substitute this result into Eq. (1), we obtain
- 18u = 1.26a
a + 14.2857u = 0
$
Since a = u , the above equation becomes
$
u + 14.2857u = 0
Comparing to that of standard form, vn = 214.2857 = 3.7796 rad>s.
Thus,
t =
2p
2p
=
= 1.6623 s = 1.66 s
vn
3.7796
Ans.
Ans:
t = 1.66 s
1212
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exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
*22–24.
The 10-kg disk is pin connected at its mass center. Determine
the natural period of vibration of the disk if the springs have
sufficient tension in them to prevent the cord from slipping
on the disk as it oscillates. Hint: Assume that the initial
stretch in each spring is dO.
150 mm
k 80 N/m
O
k 80 N/m
Solution
Equation of Motion. The mass moment of inertia of the disk about its mass center O
1
1
is I0 = Mr 2 = (10) ( 0.152 ) = 0.1125 kg # m2. When the disk undergoes a small
2
2
angular displacement u, the top spring stretches further but the stretch of the spring
is being reduced both by s = ru = 0.15u. Thus, (Fsp)t = Kxt = 80(d0 - 0.15u) and
(Fsp)b = 80(d0 - 0.15u). Referring to the FBD of the disk, Fig. a,
a + ΣM0 = I0a; - 80(d0 + 0.15u)(0.15) + 80(d0 - 0.15u)(0.15) = 0.1125a
- 3.60u = 0.1125a
a + 32u = 0
$
Since a = u , this equation becomes
$
u + 32u = 0
Comparing to that of standard form, vn = 232 rad>s. Then
t =
2p
2p
=
= 1.1107 s = 1.11 s
vn
232
Ans.
Ans:
t = 1.11 s
1213
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exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
22–25.
If the disk in Prob. 22–24 has a mass of 10 kg, determine the
natural frequency of vibration. Hint: Assume that the initial
stretch in each spring is dO.
150 mm
k 80 N/m
O
k 80 N/m
Solution
Equation of Motion. The mass moment of inertia of the disk about its mass center O
1
1
is I0 = mr 2 = (10) ( 0.152 ) = 0.1125 kg # m2 when the disk undergoes a small
2
2
angular displacement u, the top spring stretches but the bottom spring compresses,
both by s = ru = 0.15u. Thus, (Fsp)t = (Fsp)b = ks = 80(0.15u) = 12u. Referring to
the FBD of the disk, Fig. a,
a + ΣM0 = I0a; - 12u(0.3) = 0.1125a
- 3.60u = 0.1125a
a + 32u = 0
$
Since a = u , this equation becomes
$
u + 32u = 0
Comparing to that of Standard form, vn = 232 rad>s. Then
f =
vn
232
=
= 0.9003 Hz = 0.900 Hz
2p
2p
Ans:
f = 0.900 Hz
1214
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exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
22–26.
A flywheel of mass m, which has a radius of gyration about
its center of mass of kO, is suspended from a circular shaft
that has a torsional resistance of M = Cu. If the flywheel
is given a small angular displacement of u and released,
determine the natural period of oscillation.
L
SOLUTION
Equation of Motion: The mass moment of inertia of the wheel about point O is
IO = mkO 2. Referring to Fig. a,
$
a+
©MO = IO a;
-Cu = mkO 2u
$
u +
O
u
C
u = 0
mkO 2
Comparing this equation to the standard equation, the natural circular frequency of
the wheel is
vn =
1 C
C
=
kO A m
A mkO 2
Thus, the natural period of the oscillation is
t =
2p
m
= 2pkO
vn
AC
Ans.
Ans:
t = 2pkO
1215
m
AC
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exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
22–27.
k = 5 lb/ft
The 6-lb weight is attached to the rods of negligible mass.
Determine the natural frequency of vibration of the weight
when it is displaced slightly from the equilibrium position
and released.
2 ft
SOLUTION
O
TO is the equilibrium force.
TO =
3 ft
6(3)
= 9 lb
2
Thus, for small u,
c + ©MO = IO a;
6(3) - C 9 + 5(2)u D (2) = a
$
6
b a3u b(3)
32.2
Thus,
$
u + 11.926u = 0
vn = 211.926 = 3.45 rad/s
Ans.
Ans:
vn = 3.45 rad>s
1216
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exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
*22–28.
The platform AB when empty has a mass of 400 kg, center
of mass at G1, and natural period of oscillation t1 = 2.38 s.
If a car, having a mass of 1.2 Mg and center of mass at G2, is
placed on the platform, the natural period of oscillation
becomes t2 = 3.16 s. Determine the moment of inertia of
the car about an axis passing through G2.
O
1.83 m
2.50 m
G2
G1
A
B
SOLUTION
Free-body Diagram: When an object arbitrary shape having a mass m is pinned at O
and being displaced by an angular displacement of u, the tangential component of its
weight will create the restoring moment about point O.
Equation of Motion: Sum moment about point O to eliminate Ox and Oy.
(1)
- mg sin u(l) = IOa
a + ©MO = IOa :
$
d2u
= u and sin u = u if u is small, then substituting these
2
dt
values into Eq. (1), we have
Kinematics: Since a =
$
-mglu = IOu
From Eq. (2), v2n =
t =
or
mgl
$
u +
u = 0
IO
(2)
mgl
mgl
, thus, vn =
, Applying Eq. 22–12, we have
B IO
IO
IO
2p
= 2p
B mgl
vn
(3)
When the platform is empty, t = t1 = 2.38 s, m = 400 kg and l = 2.50 m.
Substituting these values into Eq. (3), we have
2.38 = 2p
(IO)p
C 400(9.81)(2.50)
(IO)p = 1407.55 kg # m2
When the car is on the platform, t = t2 = 3.16 s, m = 400 kg + 1200 kg = 1600 kg.
2.50(400) + 1.83(1200)
and
l =
= 1.9975 m
IO = (IO)C + (IO)p = (IO)C +
1600
1407.55. Substituting these values into Eq. (3), we have
3.16 = 2p
(IO)C + 1407.55
(I ) = 6522.76 kg # m2
D 1600(9.81)(1.9975) O C
Thus, the mass moment inertia of the car about its mass center is
(IG)C = (IO)C - mCd2
= 6522.76 - 1200(1.832) = 2.50(103) kg # m2
Ans.
Ans:
(IG)C = 2.50(103) kg # m2
1217
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exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
22–29.
The plate of mass m is supported by three symmetrically
placed cords of length l as shown. If the plate is given a
slight rotation about a vertical axis through its center and
released, determine the natural period of oscillation.
l
l
Solution
120
$
1
ΣMz = Iza -3(T sin f)R = mR2 u
2
l
120
R
120
sin f K f
$
6T
f = 0
u +
Rm
ΣFz = 0 3T cos f - mg = 0
f = 0, T =
$
u +
mg
R
, f = u
3
l
6 mg R
a
ba u b = 0
Rm 3
l
$
2g
u +
u = 0
l
t =
l
2p
= 2p
vn
A 2g
Ans.
Ans:
t = 2p
1218
l
A 2g
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exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
22–30.
Determine the differential equation of motion of the 3-kg block
when it is displaced slightly and released. The surface is smooth
and the springs are originally unstretched.
k
500 N/m
k
500 N/m
3 kg
SOLUTION
T + V = const.
T =
1
#
(3)x2
2
V =
1
1
(500)x 2 + (500)x2
2
2
#
T + V = 1.5x 2 + 500x 2
# $
#
1.5(2x) x + 1000xx = 0
$
3x + 1000x = 0
$
x + 333x = 0
Ans.
Ans:
$
x + 333x = 0
1219
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exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
22–31.
Determine the natural period of vibration of the pendulum.
Consider the two rods to be slender, each having a weight
of 8 lb>ft.
O
2 ft
SOLUTION
y =
1(8)(2) + 2(8)(2)
= 1.5 ft
8(2) + 8(2)
1 ft
1 ft
1
1
IO =
c (2)(8)(2)2 + 2(8)(1)2 d
32.2 12
+
1 1
c (2)(8)(2)2 + 2(8)(2)2 d = 2.8157 slug # ft2
32.2 12
h = y (1 - cos u)
T + V = const
T =
#
#
1
(2.8157)(u)2 = 1.4079 u2
2
V = 8(4)(1.5)(1 - cos u) = 48(1 - cos u)
#
T + V = 1.4079u2 + 48(1 - cos u)
# $
#
1.4079 (2u)u + 48(sin u)u = 0
For small u, sin u = u, then
$
u + 17.047u = 0
t =
2p
=
vn
2p
17.047
Ans.
= 1.52 s
Ans:
t = 1.52 s
1220
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exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
*22–32.
Determine the natural period of vibration of the 10-lb
semicircular disk.
0.5 ft
SOLUTION
Datum at initial level of center of gravity of disk.
¢ = r(1 - cos u)
E = T + V
#
1
I (u)2 + Wr(1 - cos u)
2 IC
#
#
$
E = u(IICu + Wr sin u) = 0
=
For small u,
sin u = u
$
Wr
u = 0
u +
IIC
r =
4(0.5)
= 0.212 ft
3p
IA = IG + mr2
1 10
10
(
)(0.5)2 = IG +
(0.212)2
2 32.2
32.2
IG = 0.02483 slug # ft2
IIC = IG + m(r - r)2
= 0.02483 +
10
(0.5 - 0.212)2
32.2
= 0.05056 slug # ft2
t =
IIC
2p
0.05056
= 2p
= 2p
A Wr
A 10(0.212)
vn
Ans.
t = 0.970 s
Ans:
t = 0.970 s
1221
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exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
22–33.
If the 20-kg wheel is displaced a small amount and released,
determine the natural period of vibration. The radius of
gyration of the wheel is kG = 0.36 m. The wheel rolls
without slipping.
k 500 N/m
G
0.5 m
Solution
Energy Equation. The mass moment of inertia of the wheel about its mass center
is IG = mkG = 20(0.361)2 = 2.592 kg # m2. Since the wheel rolls without slipping,
vG = vr = v(0.5). Thus,
T =
1
1
I v2 + mv2G
2G
2
1
1
(2.592)v2 + (20)[v10.52]2
2
2
#
2
= 3.796 v = 3.796u 2
=
When the disk undergoes a small angular displacement u, the spring stretches
s = u(1) = u, Fig. a. Thus, the elastic potential energy is
Ve =
Thus, the total energy is
1 2
1
ks = (500)u 2 = 250u 2
2
2
#
E = T + V = 3.796u 2 + 250u 2
Time Derivative. Taking the time derivative of the above equation,
#$
#
7.592u u + 500uu = 0
#
$
u(7.592u + 500u) = 0
#
Since u ≠ 0, then
$
7.592u + 500u = 0
$
u + 65.8588u = 0
Comparing to that of standard form, vn = 265.8588 = 8.1153 rad>s. Thus,
t =
2p
2p
=
= 0.7742 s = 0.774 s
vn
8.1153
Ans.
Ans:
t = 0.774 s
1222
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exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
22–34.
Determine the differential equation of motion of the 3-kg
spool. Assume that it does not slip at the surface
of contact as it oscillates. The radius of gyration of the spool
about its center of mass is kG = 125 mm.
k 400 N/m
200 mm
100 mm
G
SOLUTION
Kinematics: Since no slipping occurs, sG = 0.1u hence sF =
0.3
S = 0.3u. Also,
0.1 G
#
vG = 0.1u.
E = T + V
#
1
[(3)(0.125)2]u2 +
2
#
= 0.03844u2 + 18u2
#$
#
0.076875uu + 36uu = 0
# $
0.076875u(u + 468.29u)
$
u + 468u = 0
E =
1
1
(3)(0.1u)2 + (400)(0.3u)2 = const.
2
2
= 0 Since 0.076875u Z 0
Ans.
Ans:
$
u + 468u = 0
1223
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exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
22–35.
Determine the natural period of vibration of the 3-kg
sphere. Neglect the mass of the rod and the size of
the sphere.
k
500 N/m
O
300 mm
300 mm
SOLUTION
E = T + V
#
1
1
(3)(0.3u)2 + (500)(dst + 0.3u)2 - 3(9.81)(0.3u)
2
2
$
#
E = u[(3(0.3)2u + 500(dst + 0.3u)(0.3) - 3(9.81)(0.3)] = 0
=
By statics,
T(0.3) = 3(9.81)(0.3)
T = 3(9.81) N
dst =
3(9.81)
500
Thus,
$
3(0.3)2u + 500(0.3)2u = 0
$
u + 166.67u = 0
vn = 2166.67 = 12.91 rad>s
t =
2p
2p
= 0.487 s
=
vn
12.91
Ans.
Ans:
t = 0.487 s
1224
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exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
*22–36.
If the lower end of the 6-kg slender rod is displaced a small
amount and released from rest, determine the natural
frequency of vibration. Each spring has a stiffness of
k = 200 N>m and is unstretched when the rod is hanging
vertically.
O
2m
k
Solution
k
2m
Energy Equation. The mass moment of inertia of the rod about O is
1
1
I0 = ml 2 = (6) ( 42 ) = 32 kg # m2. Thus, the Kinetic energy is
3
3
#
#
1
1
T = I0v2 = (32)u 2 = 16u 2
2
2
with reference to the datum set in Fig. a, the gravitational potential energy is
Vg = mg y = 6(9.81)( - 2 cos u) = -117.72 cos u
When the rod undergoes a small angular displacement u the spring deform
x = 2 sin Ω. Thus the elastic potential energy is
1
1
Ve = 2a kx2 b = 2c (200)(2 sin u)2 d = 800 sin2 u
2
2
Thus, the total energy is
#
E = T + V = 16u 2 + 800 sin2 u - 117.72 cos u
Time Derivative. Taking the first time derivative of the above equation
#$
#
#
32u u + 1600(sin u cos u)u + 117.72(sin u)u = 0
Using the trigonometry identity sin 2u = 2 sin u cos u, we obtain
#$
#
#
32u u + 800(sin 2u)u + 117.72(sin u)u = 0
#
$
u(32u + 800 sin 2u + 117.72 sin u) = 0
#
Since u ≠ 0,
$
32u + 800 sin 2u + 117.72 sin u) = 0
Since u is small, sin 2u ≃ 2u and sin u = u. The above equation becomes
$
32u + 1717.72u = 0
$
u + 53.67875u = 0
Comparing to that of standard form, vn = 253.67875 = 7.3266 rad>s.
Thus,
f =
vn
7.3266
=
= 1.1661 Hz = 1.17 Hz
2p
2p
Ans.
Ans:
f = 1.17 Hz
1225
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exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
22–37.
The disk has a weight of 30 lb and rolls without slipping on
the horizontal surface as it oscillates about its equilibrium
position. If the disk is displaced, by rolling it counterclockwise
0.2 rad, determine the equation which describes its
oscillatory motion and the natural period when it is released.
k 80 lb/ft
0.5 ft
Solution
Energy Equation. The mass moment of inertia of the disk about its center of gravity
1
1 30
is IG = mr 2 = a
b ( 0.52 ) = 0.11646 slug # ft 2. Since the disk rolls without
2
2 32.2
slipping, vG = vr = v(0.5). Thus
T =
1
1
I v2 + mv2G
2G
2
1 30
1
(0.1146) v2 + a
b[v(0.5)]2
2
2 32.2
#
= 0.17469 v2 = 0.17469u 2
=
When the disk undergoes a small angular displacement u the spring stretches
s = ur = u(0.5), Fig. a. Thus, the elastic potential energy is
Ve =
Thus, the total energy is
1 2
1
ks = (80)[u(0.5)]2 = 10u 2
2
2
#
E = T + V = 0.17469u 2 + 10u 2
#
E = 0.175u 2 + 10u 2
Ans.
Time Derivative. Taking the time derivative of the above equation,
#$
#
0.34938u u + 20uu = 0
#
$
u(0.34938u + 20u) = 0
#
Since u ≠ 0, then
#
0.34938u + 20u = 0
$
u + 57.244u = 0
$
Ans.
u = 57.2u = 0
Comparing to that of standard form,
vn = 257.2444 = 7.5660 rad>s.
Thus
t =
2p
2p
=
= 0.8304 s = 0.830 s
vn
7.5660
Ans.
Ans:
#
E = 0.175u 2 + 10 u 2
t = 0.830 s
1226
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exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
22–38.
The machine has a mass m and is uniformly supported by
four springs, each having a stiffness k. Determine the natural
period of vertical vibration.
G
SOLUTION
k
T + V = const.
V = mgy +
T + V =
d
—
2
1
#
m(y)2
2
T =
k
d
—
2
1
(4k)(¢s - y)2
2
1
1
#
m(y)2 + m g y + (4k)(¢s - y)2
2
2
# $
#
#
m y y + m g y - 4k(¢s-y )y = 0
$
m y + m g + 4ky - 4k¢s = 0
Since ¢s =
mg
4k
Then
$
my + 4ky = 0
y +
4k
y = 0
m
vn =
t =
4k
Cm
2p
m
= p
Ck
vn
Ans.
Ans:
t = p
1227
m
Ak
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exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
22–39.
The slender rod has a weight of 4 lb>ft. If it is supported in
the horizontal plane by a ball-and-socket joint at A and a
cable at B, determine the natural frequency of vibration
when the end B is given a small horizontal displacement
and then released.
A
0.75 ft
B
SOLUTION
1.5 ft
f =
1.5umax
0.75
¢ = 0.75(1 - cos f)
0.75(1 - 1 +
= 0.75(
¢G =
Tmax =
=
f2
)
2
4u2max
)
2
1
¢ = 0.75u2max
2
1
I v2max
2 A
1 1 4(1.5)
[ (
)(1.5)2]vn2 u2max
2 3 32.2
= 0.0699 vn2 u2max
Vmax = W¢ G = 4(1.5)(0.75u2max)
Tmax = Vmax
0.0699vn2 u2max = 4.5 u2max
vn 2 = 64.40
vn = 8.025 rad>s
f =
vn
8.025
=
= 1.28 Hz
2p
2p
Ans.
Ans:
f = 1.28 Hz
1228
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exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
*22–40.
If the slender rod has a weight of 5 lb, determine the natural
frequency of vibration. The springs are originally
unstretched.
k = 4 lb/ft
1 ft
O
SOLUTION
Energy Equation: When the rod is being displaced a small angular displacement of u ,
2 ft
the compression of the spring at its ends can be approximated as x1 = 2u and
x2 = 1u. Thus, the elastic potential energy when the rod is at this position is
1
1
1
1
Ve = k1 x21 + k2 x2 = (5)(2u)2 + (4)(1u)2 = 12u2. The datum is set at the
2
2
2
2
rod’s mass center when the rod is at its original position. When the rod undergoes a
k = 5 lb/ft
small angular displacement u, its mass center is 0.5(1 - cos u) ft above the datum
hence its gravitational potential energy is Vg = 5[0.5(1 - cos u)]. Since u is small,
cos u can be approximated by the first two terms of the power series, that is,
u2
u2
cos u = 1 - . Thus, Vg = 2.5c 1 - a 1 - b d = 1.25u2
2
2
V = Ve + Vg = 12u2 + 1.25u2 = 13.25u2
5
1
The mass moment inertia of the rod about point O is IO =
a
b A 32 B
12 32.2
5
+
A 0.52 B = 0.1553 slug # ft2. The kinetic energy is
32.2
T =
#
#
1
1
IO v2 = (0.1553) u2 = 0.07764u2
2
2
The total energy of the system is
#
U = T + V = 0.07764u2 + 13.25u2
[1]
Time Derivative: Taking the time derivative of Eq.[1], we have
#$
#
0.1553 uu + 26.5uu = 0
#
$
u A 0.1553 u + 26.5u B = 0
#
Since u Z 0, then
$
0.1553u + 26.5u = 0
$
u + 170.66u = 0
[2]
From Eq.[2], p2 = 170.66, thus, p = 13.06 rad/s. Applying Eq.22–14, we have
f =
p
13.06
=
= 2.08 Hz
2p
2p
Ans.
Ans:
f = 2.08 Hz
1229
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exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
22–41.
If the block-and-spring model is subjected to the periodic
force F = F0 cos vt, show that the differential equation of
$
motion is x + (k>m)x = (F0>m) cos vt, where x is
measured from the equilibrium position of the block. What
is the general solution of this equation?
Equilibrium
position
x
k
m
F
F0 cos vt
SOLUTION
+ ©F = ma ;
:
x
x
$
F0 cos vt - kx = mx
F0
k
$
x + x =
cos vt
m
m
F0
$
cos vt
x + p2x =
m
(Q.E.D.)
Where p =
k
Cm
(1)
The general solution of the above differential equation is of the form of
x = xc + xp.
The complementary solution:
xc = A sin pt + B cos pt
The particular solution:
sp = .C cos vt
(2)
$
xP = -Cv2 cos v t
(3)
Substitute Eqs. (2) and (3) into (1) yields:
-Cv2 cos v t + p2 (C cos vt) =
F0
m
=
C = 2
p - v2
F0>k
1 - a vp b
F0
cos vt
m
2
The general solution is therefore
s = A sin pt + B cos pt +
F0>k
v 2
1 - a b
p
Ans.
cos vt
The constants A and B can be found from the initial conditions.
Ans:
x = A sin vnt + B cos vnt +
1230
Fo >k
1 - (v>vn)2
cos vt
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exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
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 ΣFy = may ; k(yst + y) - mg - F0 = - my
$
my + ky + kyst - mg = F0
However, from equilbrium kyst - mg = 0, therefore
$
my + ky = F0
F0
k
k
$
y + y = where vn =
m
m
Am
F0
$
y + v2ny = m
[1]
The general solution of the above differential equation is of the form of y = yc + yp.
yc = A sin vnt + B cos vnt
[2]
yP = C
$
yP = 0
[3]
Substitute Eqs. [2] and [3] into [1] yields :
0 + v2nC =
F0
F0
F0
C =
=
2
m
k
mp
The general solution is therefore
y = A sin vnt + B cos vnt +
F0
k
Ans.
The constants A and B can be found from the initial conditions.
Ans:
y = A sin vnt + B cos vnt +
1231
FO
k
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exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
22–43.
A 4-lb weight is attached to a spring having a stiffness
k = 10 lb>ft. The weight is drawn downward a distance of
4 in. and released from rest. If the support moves with a
vertical displacement d = 10.5 sin 4t2 in., where t is in
seconds, determine the equation which describes the
position of the weight as a function of time.
SOLUTION
y = A sin vnt + B cos vnt +
d0
1 - a v0 b
v
sin v0t
2
n
#
v = y = Avn cos vnt - Bvn sin vnt +
d0v0
1 - av b
v0
2
cos v0t
n
The initial condition when t = 0, y = y0 , and v = v0 is
y0 = 0 + B + 0
v0 = Avn - 0 +
B = y0
d0 v0
1 - a vn b
v0
2
A =
v0
vn
d0v0
vn -
v0 2
vn
Thus,
d 0 v0
d0
v0
sin v0t
y = q
2 r sin vnt + y0 cos vnt +
v
v 2
0
vn
vn 1 - a v0 b
vn
n
vn =
k
10
=
= 8.972
Am
A 4>32.2
d0
v0 2
1 - Q vn R
=
0.5>12
1 -
2
4
A 8.972
B
= 0.0520
(0.5>12)4
d0 v 0
v0
= 0 = - 0.0232
v02
42
vn
vn 8.972 - 8.972
vn
y = (-0.0232 sin 8.97t + 0.333 cos 8.97t + 0.0520 sin 4t) ft
1232
Ans.
Ans:
y = 5 -0.0232 sin 8.97 t + 0.333 cos 8.97 t + 0.0520 sin 4 t6 ft
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*22–44.
A 4-kg block is suspended from a spring that has a stiffness
of k = 600 N>m. The block is drawn downward 50 mm from
the equilibrium position and released from rest when t = 0.
If the support moves with an impressed displacement of
d = 110 sin 4t2 mm, where t is in seconds, determine the
equation that describes the vertical motion of the block.
Assume positive displacement is downward.
SOLUTION
vn =
k
600
=
= 12.25
Am
A 4
The general solution is defined by Eq. 22–23 with kd0 substituted for F0.
y = A sin vnt + B cos vnt + ±
d0
v 2
c1 - a b d
vn
≤ sin vt
d = (0.01 sin 4t)m, hence d0 = 0.01, v = 4, so that
y = A sin 12.25t + B cos 12.25t + 0.0112 sin 4t
y = 0.05 when t = 0
0.05 = 0 + B + 0;
B = 0.05 m
#
y = A(12.25) cos 12.25t - B(12.25) sin 12.25t + 0.0112(4) cos 4t
v = y = 0 when t = 0
0 = A(12.25) - 0 + 0.0112(4);
A = - 0.00366 m
Expressing the result in mm, we have
Ans.
y = ( - 3.66 sin 12.25t + 50 cos 12.25t + 11.2 sin 4t) mm
Ans:
y = ( - 3.66 sin 12.25t + 50 cos 12.25t + 11.2 sin 4t) mm
1233
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22–45.
Use a block-and-spring model like that shown in Fig. 22–14a,
but suspended from a vertical position and subjected to a
periodic support displacement d = d0 sin v0t, determine the
equation of motion for the system, and obtain its general
solution. Define the displacement y measured from the
static equilibrium position of the block when t = 0.
SOLUTION
$
k(y - d0 sin v0t + yst) - mg = -my
+ c ©Fx = max;
$
my + ky + kyst - mg = kd0 sin v0t
However, from equilibrium
kyst - mg = 0, therefore
kd0
k
$
y =
sin v t
y +
m
m
where vn =
k
Am
kd0
$
sin v t
y + vn 2 y =
m
Ans. (1)
The general solution of the above differential equation is of the form of y = yc + yp,
where
yc = A sin vnt + B cos vnt
yp = C sin v0t
(2)
$
yp = -Cv02 sin v0t
(3)
Substitute Eqs. (2) and (3) into (1) yields:
-Cv2 sin v0 t + vn2(C sin v0 t) =
kd0
m
=
C =
vn 2 - v 0 2
kd0
sin v0t
m
d0
v0 2
1 - a b
vn
The general solution is therefore
y = A sin vnt + B cos vnt +
d0
sin v t
v0 2
1 - a b
vn
Ans.
The constants A and B can be found from the initial conditions.
Ans:
y = A sin vnt + B cos vnt +
1234
d0
1 - (v>vn)2
sin vt
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exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
22–46.
A 5-kg block is suspended from a spring having a stiffness
of 300 N>m. If the block is acted upon by a vertical force
F = 17 sin 8t2 N, where t is 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 t = 0. Assume that positive displacement
is downward.
k
300 N/m
SOLUTION
The general solution is defined by:
F0
k
y = A sin vnt + B cos vnt + §
¥ sin v0t
v0 2
1 - a b
vn
F
7 sin 8t
Since
F = 7 sin 8t,
vn =
F0 = 7 N,
v0 = 8 rad>s,
k = 300 N>m
k
300
=
= 7.746 rad>s
Am
A 5
Thus,
7
300
¥ sin 8t
y = A sin 7.746t + B cos 7.746t + §
2
8
1 - a
b
7.746
y = 0.1 m when t = 0,
0.1 = 0 + B - 0;
B = 0.1 m
#
y = A(7.746) cos 7.746t - B(7.746) sin 7.746t - (0.35)(8) cos 8t
#
y = y = 0 when t = 0,
#
y = A(7.746) - 2.8 = 0;
A = 0.361
Expressing the results in mm, we have
Ans.
y = (361 sin 7.75t + 100 cos 7.75t - 350 sin 8t) mm
Ans:
y = (361 sin 7.75t + 100 cos 7.75t)
-350 sin 8t) mm
1235
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exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
22–47.
The uniform rod has a mass of m. If it is acted upon by a
periodic force of F = F0 sin vt, determine the amplitude of
the steady-state vibration.
A
L
2
k
L
2
SOLUTION
Equation of Motion: When the rod rotates through a small angle u, the springs
L
u. Thus, the force in each spring is
compress and stretch s = r AGu =
2
kL
Fsp = ks =
u. The mass moment of inertia of the rod about point A is
2
1
IA = mL2. Referring to the free-body diagram of the rod shown in Fig. a,
3
+ ©MA = IAa;
k
FO sin vt cos u(L) - mg sin u a
=
F F0 sin vt
kL
L
L
b - 2a
u bcos u a b
2
2
2
#
1
mL2u
3
Since u is small, sin u 0 and cos u 1. Thus, this equation becomes
$
1
1
mLu + (mg + kL)u = FO sin vt
3
2
$
3FO
3 g
k
u + ¢ + ≤u =
sin vt
m
2 L
mL
(1)
The particular solution of this differential equation is assumed to be in the form of
(2)
up = C sin vt
Taking the time derivative of Eq. (2) twice,
$
up = -Cv2 sin vt
(3)
Substituting Eqs. (2) and (3) into Eq. (1),
-Cv2 sin vt +
CB
3FO
3 g
k
a + ≤ (C sin vt) =
sin vt
m
2 L
mL
3FO
3 g
k
a + ≤ - v2 R sin vt =
sin vt
m
2 L
mL
C =
C =
3FO > mL
3 g
k
a + b - v2
m
2 L
3FO
3
(mg + Lk) - mLv2
2
Ans.
Ans:
C = 3
3FO
2
2 (mg + Lk) - mLv
1236
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exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
*22–48.
The 30-lb block is attached to two springs having a stiffness
of 10 lb>ft. A periodic force F = (8 cos 3t) lb, where t is in
seconds, is applied to the block. Determine the maximum
speed of the block after frictional forces cause the free
vibrations to dampen out.
k 10 lb/ft
F 8 cos 3t
k 10 lb/ft
SOLUTION
Free-body Diagram: When the block is being displaced by amount x to the right, the
restoring force that develops in both springs is Fsp = kx = 10x.
Equation of Motion:
+ ©F = 0;
:
x
- 2(10x) + 8 cos 3t =
30
a
32.2
[1]
a + 21.47x = 8.587 cos 3t
Kinematics: Since a =
d2x
$
= x, then substituting this value into Eq. [1], we have
dt2
$
x + 21.47x = 8.587 cos 3t
[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
xp = C cos 3t
Taking second time derivative and substituting into Eq. [2], we have
- 9C cos 3t + 21.47C cos 3t = 8.587 cos 3t
C = 0.6888 ft
Thus,
[3]
xp = 0.6888 cos 3t
Taking the time derivative of Eq. [3], we have
#
yp = xp = - 2.0663 sin 3t
Thus,
A yp B max = 2.07 ft>s
Ans.
Ans:
(vp)max = 2.07 ft>s
1237
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exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
22–49.
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
SOLUTION
k =
18
F
=
= 1285.71 N>m
¢y
0.014
v0 = 2 Hz = 2(2p) = 12.57 rad>s
d0 = 0.015 m
vn =
k
1285.71
=
= 17.93
4
Am
A
Using Eq. 22–22, the amplitude is
(xp)max = 3
d0
0.015
3 = 3
3
12.57 2
v0 2
1
Q 17.93 R
1 - a b
vn
Ans.
(xp)max = 0.0295 m = 29.5 mm
Ans:
(xp)max = 29.5 mm
1238
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exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
22–50.
Find the differential equation for small oscillations in terms
of u for the uniform rod of mass m. Also show that if
c 6 2mk>2, then the system remains underdamped. The
rod is in a horizontal position when it is in equilibrium.
2a
a
B
A
k
C
u
c
SOLUTION
#
Equation
of Motion: When the rod is in equilibrium, u = 0°, Fc = cyc = 0 and
$
u = 0. writing the moment equation of motion about point B by
referring to the free-body diagram of the rod, Fig. a,
a
- FA (a) - mg a b = 0
2
+ ©MB = 0;
FA =
mg
2
mg
FA
. When the rod rotates about
=
k
2k
point B through a small angle u, the spring stretches further by s1 = au. Thus, the
Thus, the initial stretch of the spring is sO =
force in the spring is FA = k(s0 + s1) = k ¢
mg
+ au ≤ . Also, the velocity of end C
2k
#
#
#
#
of the rod is vc = yc = 2au. Thus, Fc = cyc = c(2au). The mass moment of inertia of
1
a 2
the rod about B is IB =
m(3a)2 + m a b = ma2. Again, referring to Fig. a and
12
2
writing the moment equation of motion about B,
ka
©MB = IB a;
#
mg
a
+ au b cos u(a) + A 2au B cos u(2a) - mg cos ua b
2k
2
$
= - ma2u
$
#
4c
k
u +
cos uu +
(cos u)u = 0
m
m
Since u is small, cos u 1. Thus, this equation becomes
$
4c #
k
u +
u + u = 0
m
m
Ans.
Comparing this equation to that of the standard form,
vn =
k
Am
ceq = 4c
Thus,
k
= 22mk
Am
cc = 2mvn = 2m
For the system to be underdamped,
ceq 6 cc
4c 6 22mk
c 6
1
2mk
2
Ans.
1239
Ans:
$
4c #
k #
u +
u +
u = 0
m
m
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exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
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
Solution
For the steady-state vibration, the displacement is
yp =
F0 >k
1 - (v0 >vn)2
cos vt
F (100 cos 2t) N
Here F0 = 100 N, k = 800 N>m, v0 = 2 rad>s and
vn =
k
800
=
= 220 rad>s.
Am
A 40
Thus
yp =
100>800
1 - ( 2> 120 ) 2
cos 2t
yP = 0.15625 cos 2t
Taking the time derivative of this equation
#
vp = yp = -0.3125 sin 2t
(2)
vp is maximum when sin 2t = 1. Thus
Ans.
(vp)max = 0.3125 m>s
Ans:
(vp)max = 0.3125 m>s
1240
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exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
*22–52.
Use a block-and-spring model like that shown in
Fig. 22–14a but suspended from a vertical position and
subjected to a periodic support displacement of
d = d0 cos v0t, determine the equation of motion for the
system, and obtain its general solution. Define the
displacement y measured from the static equilibrium
position of the block when t = 0.
SOLUTION
$
+ T ©Fy = may; kd0 cos v0t + W - kdst - ky = my
Since W = kdst,
kd0
k
$
y + y =
cos v0t
m
m
(1)
yC = A sin vny + B cos vny (General sol.)
yP = C cos v0t (Particular sol.)
Substitute yp into Eq. (1)
C(- v02 +
C =
kd0
k
) cos v0t =
cos v0t
m
m
kd0
m
Q
k
2
m - v0 R
Thus, y = yC + yP
y = A sin vnt + B cos vnt +
Q
kd0
m
k
- v02 R
m
Ans.
cos v0t
Ans:
y = A sin vnt + B cos vnt +
1241
kd0
m
k
a - v02b
m
cos v0t
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exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
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.
V
SOLUTION
k =
vn =
25(9.81)
F
=
= 4905 N>m
¢y
0.05
k
Cm
=
4905
C 25
= 14.01 rad>s
Resonance occurs when
Ans.
v = vn = 14.0 rad>s
Ans:
v = 14.0 rad>s
1242
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exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
22–54.
In Prob. 22–53, determine the amplitude of steady-state
vibration of the fan if its angular velocity is 10 rad>s .
V
SOLUTION
k =
vn =
25(9.81)
F
=
= 4905 N>m
¢y
0.05
k
Cm
=
4905
C 25
= 14.01 rad>s
The force caused by the unbalanced rotor is
F0 = mr v2 = 3.5(0.1)(10)2 = 35 N
Using Eq. 22–22, the amplitude is
(xp)max
F0
k
4
= 4
v 2
1 -a b
p
(xp)max
35
4905
4 = 0.0146 m
= 4
10 2
1 - a
b
14.01
Ans.
(xp)max = 14.6 mm
Ans:
(xp)max = 14.6 mm
1243
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exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
22–55.
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
18 rad>s? Hint: See the first part of Example 22.8.
V
SOLUTION
k =
vn =
25(9.81)
F
=
= 4905 N>m
¢y
0.05
k
Cm
=
4905
= 14.01 rad>s
C 25
The force caused by the unbalanced rotor is
F0 = mrv2 = 3.5(0.1)(18)2 = 113.4 N
Using Eq. 22–22, the amplitude is
(xp)max = 4
(xp)max
F0
k
v 2
1 - a b
p
4
113.4
4905
4 = 0.0355 m
= 4
18 2
b
1 - a
14.01
Ans.
(xp)max = 35.5 mm
Ans:
(xp)max = 35.5 mm
1244
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exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
*22–56.
The small block at A has a mass of 4 kg and is mounted on
the bent rod having negligible mass. If the rotor at B causes
a harmonic movement dB = (0.1 cos 15t) m, where t is in
seconds, determine the steady-state amplitude of vibration
of the block.
0.6 m
A
O
1.2 m
SOLUTION
$
4(9.81)(0.6) - Fs (1.2) = 4(0.6)2u
+ ©MO = IO a;
V
k
Fs = kx = 15(x + xst - 0.1 cos 15t)
xst =
4(9.81)(0.6)
1.2(15)
15 N/m
B
Thus,
$
-15(x - 0.1 cos 15t)(1.2) = 4(0.6)2u
x = 1.2u
u + 15u = 1.25 cos 15t
Set xp = C cos 15t
-C(15)2 cos 15t + 15(C cos 15t) = 1.25 cos 15t
C =
1.25
= -0.00595 m
15 - (15)2
umax = C = 0.00595 rad
Ans.
ymax = (0.6 m)(0.00595 rad) = 0.00357 rad
Ans:
ymax = 0.00357 rad
1245
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exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
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.
V
SOLUTION
k =
F
150
=
= 1800 lb>ft
d
1>12
Resonance occurs when
vn =
k
1800
=
= 19.66
Am
A 150>32.2
Ans.
v = vn = 19.7 rad>s
Ans:
v = 19.7 rad>s
1246
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exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
22–58.
What will be the amplitude of steady-state vibration of the
motor in Prob. 22–57 if the angular velocity of the flywheel
is 20 rad>s?
V
SOLUTION
The constant value FO of the periodic force is due to the centrifugal force of the
unbalanced mass.
FO = ma n = mrv2 = a
0.25
10
b a b (20)2 = 2.588 lb
32.2
12
Hence F = 2.588 sin 20t
k =
150
F
=
= 1800 lb>ft
d
1>12
vn =
k
1800
=
= 19.657
Am
A 150>32.2
From Eq. 22–21, the amplitude of the steady state motion is
C = 4
F0>k
2.588>1800
4 = 4
4 = 0.04085 ft = 0.490 in.
2
v0 2
20
b
1 - a b
1 - a
vn
19.657
Ans.
Ans:
C = 0.490 in.
1247
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exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
22–59.
Determine the angular velocity of the flywheel in Prob. 22–57
which will produce an amplitude of vibration of 0.25 in.
V
SOLUTION
The constant value FO of the periodic force is due to the centrifugal force of the
unbalanced mass.
FO = man = mrv2 = a
10
0.25
b a bv2 = 0.006470v2
32.2
12
F = 0.006470v2 sin vt
k =
150
F
=
= 1800 lb>ft
d
1>12
vn =
k
1800
=
= 19.657
Am
A 150>32.2
From Eq. 22.21, the amplitude of the steady-state motion is
C = 4
F0>k
4
v 2
1 - a b
vn
0.006470 a
v2
b
1800
0.25
4
4
=
2
v
12
b
1 - a
19.657
Ans.
v = 19.0 rad>s
Ans:
v = 19.0 rad>s
1248
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exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
*22–60.
The 450-kg trailer is pulled with a constant speed over the
surface of a bumpy road, which may be approximated by a
cosine curve having an amplitude of 50 mm and wave length of
4 m. If the two springs s which support the trailer each have a
stiffness of 800 N>m, determine the speed v which will cause
the greatest vibration (resonance) of the trailer. Neglect the
weight of the wheels.
v
s
2m
100 mm
2m
SOLUTION
The amplitude is d0 = 50 mm = 0.05 m
The wave length is l = 4 m
k = 2(800) = 1600 N>m
vn =
t =
k
1600
=
= 1.89 rad>s
Am
A 450
2p
2p
=
= 3.33 s
vn
1.89
For maximum vibration of the trailer, resonance must occur, i.e.,
v0 = vn
Thus, the trailer must travel l = 4 m, in t = 3.33 s, so that
vR =
l
4
=
= 1.20 m.s
t
3.33
Ans.
Ans:
vR = 1.20 m.s
1249
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exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
22–61.
Determine the amplitude of vibration of the trailer in
Prob. 22–60 if the speed v = 15 km>h.
v
s
2m
100 mm
2m
SOLUTION
v = 15 km>h =
15(1000)
m>s = 4.17 m>s
3600
d0 = 0.05 m
As shown in Prob. 22–50, the velocity is inversely proportional to the period.
1
Since = f the the velocity is proportional of f, vn and v0
t
Hence, the amplitude of motion is
(xp)max = `
1 -
(xp)max = `
2
1 - (4.17
1.20 )
d0
A B
v0 2
vn
` = `
0.05
d0
1 -
A vvR B 2
`
` = 0.00453 m
Ans.
(xp)max = 4.53 mm
Ans:
(xp)max = 4.53 mm
1250
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exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
22–62.
The motor of mass M is supported by a simply supported
beam of negligible mass. If block A of mass m is clipped
onto the rotor, which is turning at constant angular velocity
of v, 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 d = PL3>48EI at this point.
Here E is Young’s modulus of elasticity, a property of the
material, and I is the moment of inertia of the beam’s crosssectional area.
r
A
L
2
L
2
SOLUTION
In this case, P = keqd. Then, keq =
frequency of the system is
48EI
P
P
=
=
. Thus, the natural
3
d
PL >48EI
L3
48EI
48EI
L3
vn =
=
=
Cm
R M
B ML3
keq
Here, FO = man = m(v2r). Thus,
Y =
FO>keq
1 - a
v 2
b
vn
m(v2r)
48EI>L3
Y =
1 Y =
v2
48EI>ML3
mrv2L3
48EI - Mv2L3
Ans.
Ans:
Y =
1251
mrv2L3
48EI - Mv2L3
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exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
22–63.
The spring system is connected to a crosshead that oscillates
vertically when the wheel rotates with a constant angular
velocity of V . If the amplitude of the steady-state vibration
is observed to be 400 mm, and the springs each have a
stiffness of k = 2500 N>m, determine the two possible
values of V at which the wheel must rotate. The block has a
mass of 50 kg.
v
SOLUTION
k
200 mm
k
In this case, keq = 2k = 2(2500) = 5000 N>m Thus, the natural circular frequency
of the system is
vn =
keq
Dm
5000
= 10 rad>s
A 50
=
Here, dO = 0.2 m and (YP)max = ; 0.4 m, so that
(YP)max =
;0.4 =
dO
1 - ¢
v 2
≤
vn
0.2
1 - ¢
v 2
≤
10
v2
= 1 ; 0.5
100
Thus,
v2
= 1.5
100
v = 12.2 rad>s
Ans.
v2
= 0.5
100
v = 7.07 rad>s
Ans.
or
Ans:
v = 12.2 rad>s
v = 7.07 rad>s
1252
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exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
*22–64.
The spring system is connected to a crosshead that oscillates
vertically when the wheel rotates with a constant angular
velocity of v = 5 rad>s. If the amplitude of the steady-state
vibration is observed to be 400 mm, determine the two
possible values of the stiffness k of the springs. The block
has a mass of 50 kg.
v
SOLUTION
k
200 mm
k
In this case, keq = 2k Thus, the natural circular frequency of the system is
vn =
keq
Dm
=
2k
= 20.04k
A 50
Here, dO = 0.2 m and (YP)max = ; 0.4 m, so that
(YP)max =
;0.4 =
dO
1 - ¢
v 2
≤
vn
0.2
1 - ¢
5
20.04k
≤
2
625
= 1 ; 0.5
k
Thus,
625
= 1.5
k
k = 417 N>m
Ans.
625
= 0.5
k
k = 1250 N>m
Ans.
or
Ans:
k = 417 N>m
k = 1250 N>m
1253
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exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
22–65.
A 7-lb block is suspended from a spring having a stiffness of
k = 75 lb>ft. The support to which the spring is attached is
given simple harmonic motion which may be expressed as
d = (0.15 sin 2t) ft, where t is in seconds. If the damping
factor is c>cc = 0.8, determine the phase angle f of forced
vibration.
SOLUTION
vn =
k
Cm
=
75
= 18.57
7
b
a
S 32.2
d = 0.15 sin 2t
d0 = 0.15, v = 2
f¿ = tan-1 §
2a
c
v
ba b
cc vn
v 2
1 - a b
vn
¥ = tan-1 §
2(0.8)a
1 - a
2
b
18.57
2
2
b
18.57
¥
Ans.
f¿ = 9.89°
Ans:
f′ = 9.89°
1254
© 2016 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently
exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
22–66.
Determine the magnification factor of the block, spring, and
dashpot combination in Prob. 22–65.
SOLUTION
vn =
k
=
Am
75
= 18.57
7
Q ¢ 32.2 ≤
d = 0.15 sin 2t
d0 = 0.15,
v = 2
1
MF =
C
B1 - a
v
v
c
b R + c2a b a b d
cn vn
vn
2 2
2
1
=
C
B1 - a
2
2
2
b R + c2(0.8) a
bR
18.57
18.57
2 2
Ans.
MF = 0.997
Ans:
MF = 0.997
1255
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exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
22–67.
A block having a mass of 7 kg is suspended from a spring
that has a stiffness k = 600 N>m. If the block is given an
upward velocity of 0.6 m>s from its equilibrium position at
t = 0, 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 F = 150 ƒ v ƒ 2 N, where v is in m>s.
SOLUTION
c = 50 N s>m
vn =
k = 600 N>m
m = 7 kg
k
600
=
= 9.258 rad>s
Am
A 7
cc = 2mvn = 2(7)(9.258) = 129.6 N # s>m
Since c 6 cz, the system is underdamped,
v d = vn
B
1 - a
c 2
50 2
b = 8.542 rad>s
b = 9.258 1 - a
cc
129.6
B
50
c
=
= 3.751
2m
2(7)
From Eq. 22-32
y = D C e- A2m Bt sin (vdt + f) S
c
c
#
v = y = D C e- A2m Bt vd cos (vdt + f) +
A-
y = De- A2m Bt C vd cos (vdt + f) c
c
c
B e- A2m Bt sin (vdt + f) S
2m
c
sin (vdt + f) D
2m
Applying the initial condition at t = 0, y = 0 and y = - 0.6 m>s.
0 = D[e - 0 sin (0 + f)]
sin f = 0
since D Z 0
f = 0°
- 0.6 = De - 0 [8.542 cos 0° - 0]
D = - 0.0702 m
y = [ -0.0702 e - 3.57t sin (8.540)] m
Ans.
1256
Ans:
y = 5 - 0.0702 e -3.57t sin (8.540)6 m
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exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
*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 c cc 0.20.
Neglect the mass of the beam.
SOLUTION
d =
2
= 0.167 ft
12
v = 100a
k =
2p
b = 10.47 rad>s
60
200
= 1200 lb>ft
2
12
FO = mrv2 = a
p =
k
=
Am
5
1
b a b (10.47)2 = 1.419 lb
32.2 12
1200
= 13.90 rad>s
200
Q 32.2
F0
k
C¿ =
C
B1 - ¢ ≤ R + B2¢ ≤ ¢ ≤ R
v
p
2
2
c
cc
v
p
2
1.419
1200
=
C
1 -
10.47
13.90
2
2
+
2(0.20)
10.47
13.90
2
= 0.00224 ft
C
Ans.
0.0269 in.
Ans:
C ′ = 0.0269 in.
1257
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exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
22–69.
Two identical dashpots are arranged parallel to each other,
as shown. Show that if the damping coefficient c 6 2mk,
then the block of mass m will vibrate as an underdamped
system.
c
c
k
SOLUTION
When the two dash pots are arranged in parallel, the piston of the dashpots have the
same velocity. Thus, the force produced is
#
#
#
F = cy + cy = 2cy
The equivalent damping coefficient ceq of a single dashpot is
#
2cy
F
ceq = # = # = 2c
y
y
For the vibration to occur (underdamped system), ceq 6 cc. However, cc = 2mvn
k
. Thus,
= 2m
Am
ceq 6 cc
k
Am
2c 6 2m
c 6 2mk
Ans.
Ans:
#
F = 2cy
k
Am
c 6 1mk
cc = 2m
1258
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exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
22–70.
The damping factor, c>cc , 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 x1 and x2 , as shown in Fig. 22–16, show that
ln x1> x2 = 2p1c>cc2> 21 - 1c> cc22. The quantity ln x1> x2
is called the logarithmic decrement.
SOLUTION
Using Eq. 22–32,
x = Dc e - A2m B t sin (vdt + f) d
c
The maximum displacement is
xmax = De - A2m B t
c
At t = t1, and t = t2
x1 = De - A2m B t1
c
x2 = De - A2m B t2
c
Hence,
x1
De - A2m Bt1
=
c
c
x2
De - A2m Bt2 = e - A2m B(t1 - t2)
c
Since vdt2 - vdt1 = 2p
then t2 - t1 =
so that ln a
2p
vd
x1
cp
b =
x2
mvd
Using Eq. 22–33, cc = 2mvn
vd = vn
B
1 - a
cc
c 2
c 2
b =
1 - a b
cc
c
2mA
r
So that,
x1
ln a b =
x2
2p a
c
b
cc
Q.E.D.
c 2
1- a b
cc
B
Ans:
x1
In a b =
x2
1259
2pa
A
c
b
cc
1 - a
c 2
b
cc
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exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
22–71.
If the amplitude of the 50-lb cylinder’s steady-state vibration
is 6 in., determine the wheel’s angular velocity v.
v
9 in.
c
k
200 lb/ft
25 lb s/ft
k
200 lb/ft
SOLUTION
In this case, Y =
Then
6
9
= 0.5 ft, dO =
= 0.75 ft, and keq = 2k = 2(200) = 400 lb>ft.
12
12
vn =
keq
Cm
=
400
= 16.05 rad>s
B (50>32.2)
cc = 2mvn = 2 ¢
50
≤ (16.05) = 49.84 lb # s>ft
32.2
c
25
= 0.5016
=
cc
49.84
dO
Y =
D
B1 - ¢
2
2(c>cc)v 2
v 2
≤ R + ¢
≤
vn
vn
0.75
0.5 =
D
B1 - ¢
2
2
2(0.5016)v 2
v
≤ R + ¢
≤
16.05
16.05
15.07(10 - 6)v4 - 3.858(10 - 3)v2 - 1.25 = 0
Solving for the positive root of this equation,
v2 = 443.16
Ans.
v = 21.1 rad>s
Ans:
v = 21.1 rad>s
1260
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exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
*22–72.
The block, having a weight of 12 lb, is immersed in a liquid
such that the damping force acting on the block has a
magnitude of F = 10.7 ƒ v ƒ 2 lb, where v is in ft>s. If the block
is pulled down 0.62 ft and released from rest, determine the
position of the block as a function of time. The spring has a
stiffness of k = 53 lb>ft. Assume that positive displacement
is downward.
k
SOLUTION
c = 0.7 lb # s>ft
vn =
k = 53 lb>ft
m =
12
= 0.3727 slug
32.2
k
53
=
= 11.925 rad>s
Am
A 0.3727
cc = 2mvn = 2(0.3727)(11.925) = 8.889 lb # s>ft
Since c 6 cc the system is underdamped.
vd = vn
B
1 - a
c 2
0.7 2
b = 11.925 1 - a
b = 11.888 rad>s
cc
B
8.889
c
0.7
=
= 0.9392
2m
2(0.3727)
From Eq. 22–32 y = D c e - A2m B t sin (vdt + f) d
c
c
c
c
#
b e- A2m B t sin (vdt + f) d
y = y = D c e- A2m B tvd cos (vdt + f) + a 2m
y = De- A2m Bt c vd cos (vdt + f) c
c
sin (vdt + f) d
2m
Appling the initial condition at t = 0, y = 0.62 ft and y = 0.
0.62 = D C e-0 sin (0 + f) D
(1)
D sin f = 0.62
0 = De -0 C 11.888 cos (0 + f) - 0.9392 sin (0 + f) D
since D Z 0
(2)
11.888 cos f - 0.9392 sin u = 0
Solving Eqs. (1) and (2) yields:
f = 85.5° = 1.49 rad
D = 0.622 ft
y = 0.622[e-0.939t sin (11.9t + 1.49)]
Ans.
Ans:
y = 0.622 [e - 0.939t sin (11.9t + 1.49)]
1261
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exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
22–73.
The bar has a weight of 6 lb. If the stiffness of the spring is
k = 8 lb>ft and the dashpot has a damping coefficient
c = 60 lb # s>ft, determine the differential equation which
describes the motion in terms of the angle u of the bar’s
rotation. Also, what should be the damping coefficient of the
dashpot if the bar is to be critically damped?
A
B
C
c
k
2 ft
3 ft
SOLUTION
$
1
6
#
6(2.5) - (60y2)(3) - 8(y1 + yst)(5) = c a
b(5)2 du
3 32.2
$
#
1.5528u + 180y2 + 40y1 + 40yst - 15 = 0
[1]
a + ©MA = IA a;
From equilibrium 40yst - 15 = 0. Also, for small u, y1 = 5u and y2 = 3u hence
#
y2 = 3u.
From Eq. [1]
$
#
1.5528u + 180(3u) + 40(5u) = 0
#
#
1.55u + 540u + 200u = 0
Ans.
By comparing the above differential equation to Eq. 22-27
m = 1.55
vn =
k = 200
£
A cd # p B c =
9 A cd # p B c
2m
200
= 11.35 rad>s
A 1.55
c = 9cd # p
2
≥ -
k
= 0
m
2
2
2km = 2200(1.55) = 3.92 lb # s>ft
9
9
Ans.
Ans:$
#
1.55u + 540u + 200u = 0
(cdp)c = 3.92 lb # s>ft
1262
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exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
22–74.
A bullet of mass m has a velocity of v0 just before it strikes
the target of mass M. If the bullet embeds in the target, and
the vibration is to be critically damped, determine the
dashpot’s critical damping coefficient, and the springs’
maximum compression. The target is free to move along the
two horizontal guides that are “nested” in the springs.
k
v0
c
k
SOLUTION
Since the springs are arranged in parallel, the equivalent stiffness of the single spring
system is keq = 2k. Also, when the bullet becomes embedded in the target,
mT = m + M. Thus, the natural frequency of the system is
vn =
keq
=
B mT
2k
Bm + M
When the system is critically damped
c = cc = 2mTvn = 2(m + M)
2k
= 28 (m + M)k
Bm + M
Ans.
The equation that describes the critically dampened system is
x = (A + Bt)e-vn t
When t = 0, x = 0. Thus,
A = 0
Then,
x = Bte-vn t
(1)
Taking the time derivative,
#
v = x = Be - vn t - Bvn te-vn t
v = Be-vn t(1 - vn t)
(2)
Since linear momentum is conserved along the horizontal during the impact, then
+ B
A;
mv0 = (m + M)v
v = a
m
bv0
m + M
Here, when t = 0, v = a
B = a
m
b v . Thus, Eq. (2) gives
m + M 0
m
b v0
m + M
And Eqs. (1) and (2) become
x = ca
m
bv0 d te-vn t
m + M
(3)
v = ca
m
bv0 d e-vn t(1 - vn t)
m + M
(4)
1263
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exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
22–74. Continued
The maximum compression of the spring occurs when the block stops. Thus,
Eq. (4) gives
0 = ca
Since a
m
b v d (1 - vn t)
m + M 0
m
b v Z 0, then
m + M 0
1 - vn t = 0
t =
1
m + M
=
vn
B 2k
Substituting this result into Eq. (3)
xmax = c a
= c
m
m + M -1
bv d a
be
m + M 0 A 2k
m
1
dv
e A 2k(m + M) 0
Ans.
1264
Ans:
cc = 28(m + M)k
m
1
xmax = c
dv
e A 2k(m + M) 0
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exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
22–75.
A bullet of mass m has a velocity v0 just before it strikes the
target of mass M. If the bullet embeds in the target, and the
dashpot’s damping coefficient is 0 6 c V cc, determine
the springs’ maximum compression. The target is free to
move along the two horizontal guides that are“nested” in
the springs.
k
v0
c
k
SOLUTION
Since the springs are arranged in parallel, the equivalent stiffness of the single spring
system is keq = 2k. Also, when the bullet becomes embedded in the target,
mT = m + M. Thus, the natural circular frequency of the system
vn =
keq
C mT
=
2k
Bm + M
The equation that describes the underdamped system is
x = Ce - (c>2mT)t sin (vdt + f)
(1)
When t = 0, x = 0. Thus, Eq. (1) gives
0 = C sin f
Since C Z 0, sin f = 0. Then f = 0. Thus, Eq. (1) becomes
x = Ce - (c>2mT)t sin vd t
(2)
Taking the time derivative of Eq. (2),
#
v = x = C B vde - (c>2mT)t cos vdt v = Ce - (c>2mT)t B vd cos vdt -
c - (c>2mT)t
e
sin vdt R
2mT
c
sin vdt R
2mT
(3)
Since linear momentum is conserved along the horizontal during the impact, then
+ B
A;
mv0 = (m + M)v
v = a
When t = 0, v = a
a
m
bv
m + M 0
m
b v . Thus, Eq. (3) gives
m + M 0
m
b v = Cvd
m + M 0
C = a
v0
m
b
m + M vd
And Eqs. (2) becomes
x = ca
v0
m
b d e-(c>2mT) t sin vdt
m + M vd
(4)
1265
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22–75. Continued
The maximum compression of the spring occurs when
sin vdt = 1
p
2
vdt =
t =
p
2vd
Substituting this result into Eq. (4),
xmax = c a
However, vd =
keq
C mT
p
v0
m
b de -[c>2(m + M)] A 2vd B
m + M vd
- a
c2
c 2
1
2k
b =
=
C m + M 4(m + M)2 2(m + M)
2mT
28k(m + M) - c2. Substituting this result into Eq. (5),
xmax =
2mv0
2
28k(m + M) - c
e
-c
pc
228k(m + M) - c2)
d
Ans.
Ans:
xmax =
1266
2mv0
28k(m + M) - c
e -pc>(228k(m + M) - c )
2
2
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exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
*22–76.
Determine the differential equation of motion for the
damped vibratory system shown. What type of motion
occurs? Take k = 100 N>m, c = 200 N # s>m, m = 25 kg.
k
SOLUTION
k
k
m
Free-body Diagram: When the block is being displaced by an amount y vertically
downward, the restoring force is developed by the three springs attached the block.
c
Equation of Motion:
+ c ©Fx = 0;
c
#
$
3ky + mg + 2cy - mg = - my
$
#
my + 2cy + 3ky = 0
(1)
Here, m = 25 kg, c = 200 N # s>m and k = 100 N>m. Substituting these values into
Eq. (1) yields
$
#
25y + 400y + 300y = 0
$
#
y + 16y + 12y = 0
Ans.
Comparing the above differential equation with Eq. 22–27, we have m = 1kg,
k
12
c = 16 N # s>m and k = 12 N>m. Thus, vn =
=
= 3.464 rad>s.
Am
A1
cc = 2mvn = 2(1)(3.464) = 6.928 N # s>m
Since c 7 cc , the system will not vibrate. Therefore it is overdamped.
Ans.
Ans:
#
$
y + 16y + 12y = 0
Since c 7 cc, the system will not vibrate.
Therefore it is overdamped.
1267
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exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
22–77.
Draw the electrical circuit that is equivalent to the
mechanical system shown. Determine the differential
equation which describes the charge q in the circuit.
k
m
F F0 cos vt
c
SOLUTION
For the block,
mx + cx + kx = F0 cos vt
Using Table 22–1,
1
Lq + Rq + ( )q = E0 cos vt
C
Ans.
Ans:
1268
1
Lq + Rq + a bq = E0 cos vt
C
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exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
22–78.
Draw the electrical circuit that is equivalent to the
mechanical system shown. What is the differential equation
which describes the charge q in the circuit?
c
k
k
SOLUTION
m
For the block,
#
$
mx + cx + 2k = 0
Using Table 22–1,
2
$
#
Lq + Rq + ( )q = 0
C
Ans.
1269
Ans:
2
$
#
Lq + Rq + a bq = 0
C
Engineering
Mechanics Dynamics 14th Edition Hibbeler Solutions Manual
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22–79.
Draw the electrical circuit that is equivalent to the
mechanical system shown. Determine the differential
equation which describes the charge q in the circuit.
k
SOLUTION
m
For the block
c
$
#
my + cy + ky = 0
Using Table 22–1
1
$
#
Lq + Rq + q = 0
C
Ans.
Ans:
1
$
#
Lq + Rq + q = 0
C
1270
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