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1193
Engingeering Mechanics Dynamics in SI Units 14th Edition Hibbeler
SOLUTIONS MANUAL
Full download at: //testbankreal.com/download/engingeering-mechanics-dynamics-si-units-14th-edition-hibbeler-solutions-manual/
221.
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 $
+ T Fy = may; mg - k(y + yst) = my where kyst = mg
$ k y +
m y = 0
k Hence p =
B m Where k =
8(9.81)
0.175 = 448.46 N > m
448.46
= B 8
= 7.487
$ $ 6 y + (7.487)2y = 0 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
From Eq. (2) v0 = Ap cos 0 - 0 A = p
= 7.487
= 0.2003 m
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)]
= 0.192 m Ans.
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1194
Ans:
y + 56.1 y = 0
y 0 t = 0.22 s = 0.192 m
1195
222.
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 blocks motion. Assume that positive
displacement is downward.
SOLUTION
k p =
A m =
A
800
2 = 20
2-Kg
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,
x = - 0.05 cos (20t) Ans.
Ans:
x = - 0.05 cos (20t)
1196
223.
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
F
k = y
=
15(9.81)
0.2 = 735.75 N>m
k vn =
A m =
A
735.75
15 = 7.00
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
y = 0.107 sin (7.00t) + 0.100 cos (7.00t)
f = tan - 1 a B
b = tan - 1 a 0.100
b = 43.0
Ans.
Ans.
A 0.107
Ans:
y = 0.107 sin (7.00t) + 0.100 cos (7.00t)
f = 43.0
1197
*224
When a 2-kg block is suspended from a spring, the spring is
stretched a distance of 40 mm. Determine the frequency and
the period of vibration for a 0.5-kg block attached to the same
spring.
SOLUTION
F
k = y
=
2(9.81)
0.040 = 490.5 N>m
k vn =
A m =
A
490.5
0.5 = 31.321
vn f =
2p =
31.321
2p = 4.985 = 4.98 Hz
Ans.
1 1t =
f =
4.985 = 0.201 s
Ans.
Ans:
f = 4.98 Hz
t = 0.201 s
1198
225.
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 F
k = = x
3(9.81)
0.060
= 490.5 N>m
k vn =
A m =
A
490.5
0.2 = 49.52 = 49.5 rad >s
Ans.
vn f =
2p =
49.52
2p = 7.88 Hz
1 1t =
f =
7.88 = 0.127 s
Ans.
Ans:
vn = 49.5 rad > s
t = 0.127 s
1199
Engingeering Mechanics Dynamics in SI Units 14th Edition Hibbeler
SOLUTIONS MANUAL
Full download at: //testbankreal.com/download/engingeering-mechanics-dynamics-si-units-14th-edition-hibbeler-solutions-manual/
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