Solution Manual For Engineering Vibration, 5th Edition By
Daniel J. Inman All Chapters 1–8
,Problems and Solutions Section 1.1 (1.1 through 1.19)
1.1 The spring of Figure 1.2 is successively loaded with mass and the corresponding (static)
displacement is recorded below. Plot the data and calculate the spring's stiffness. Note
that the data contain some error. Also calculate the standard deviation.
m(kg) 10 11 12 13 14 15 16
x(m) 1.14 1.25 1.37 1.48 1.59 1.71 1.82
Solution:
Free-body diagram: From the free-body diagram and static
equilibrium:
kx
kx mg (g 9.81m / s 2)
k k mg / x
ki
m 86.164
n
mg
20
The sample standard deviation in
computed stiffness is:
n
m 15 (k i ) 2
i1
0.164
n 1
10
0 1 2
x
Plot of mass in kg versus displacement in m
Computation of slope from mg/x
m(kg) x(m) k(N/m)
10 1.14 86.05
11 1.25 86.33
12 1.37 85.93
13 1.48 86.17
14 1.59 86.38
15 1.71 86.05
16 1.82 86.24
,1.2 Derive the solution of m˙x˙ kx 0 and plot the result for at least two periods for the case
with n = 2 rad/s, x0 = 1 mm, and v0 = 5 mm/s.
Solution:
Given:
m˙x˙ kx 0 (1)
Assume: x(t) ae . Then:
rt
x˙ are and ˙x˙ ar e . Substitute into equation (1) to
rt 2 rt
get:
mar2ert kaert 0
mr2 k 0
k
r i
m
Thus there are two solutions:
k k
mi t
mi t
x1 c1e , and x2 c2e
k
where n 2 rad/s
m
The sum of x1 and x2 is also a solution so that the total solution is:
x x x c e2it c e2it
1 2 1 2
Substitute initial conditions: x0 = 1 mm, v0 = 5 mm/s
x 0 c1 c2 x0 1 c2 1 c1, and v 0 x˙0 2ic1 2ic2 v0 5 mm/s
2c1 2c2 5 i. Combining the two underlined expressions (2 eqs in 2 unkowns):
1 5
2c 2 2c 5 i c i, and c 1 5 i
1 1 1 2
2 4 2 4
Therefore the solution is:
1 5 2it 1 5 2it
x i e i e
2 4 2 4
Using the Euler formula to evaluate the exponential terms yields:
1 5 1 5
x i cos 2t i sin 2t i cos 2t i sin 2t
2 4 2 4
5 3
x(t ) cos 2t sin 2t sin2t 0.7297
2 2
, Using Mathcad the plot is:
5.
x t cos 2. t sin 2. t
2
2
x t
0 5 10
2
t
Daniel J. Inman All Chapters 1–8
,Problems and Solutions Section 1.1 (1.1 through 1.19)
1.1 The spring of Figure 1.2 is successively loaded with mass and the corresponding (static)
displacement is recorded below. Plot the data and calculate the spring's stiffness. Note
that the data contain some error. Also calculate the standard deviation.
m(kg) 10 11 12 13 14 15 16
x(m) 1.14 1.25 1.37 1.48 1.59 1.71 1.82
Solution:
Free-body diagram: From the free-body diagram and static
equilibrium:
kx
kx mg (g 9.81m / s 2)
k k mg / x
ki
m 86.164
n
mg
20
The sample standard deviation in
computed stiffness is:
n
m 15 (k i ) 2
i1
0.164
n 1
10
0 1 2
x
Plot of mass in kg versus displacement in m
Computation of slope from mg/x
m(kg) x(m) k(N/m)
10 1.14 86.05
11 1.25 86.33
12 1.37 85.93
13 1.48 86.17
14 1.59 86.38
15 1.71 86.05
16 1.82 86.24
,1.2 Derive the solution of m˙x˙ kx 0 and plot the result for at least two periods for the case
with n = 2 rad/s, x0 = 1 mm, and v0 = 5 mm/s.
Solution:
Given:
m˙x˙ kx 0 (1)
Assume: x(t) ae . Then:
rt
x˙ are and ˙x˙ ar e . Substitute into equation (1) to
rt 2 rt
get:
mar2ert kaert 0
mr2 k 0
k
r i
m
Thus there are two solutions:
k k
mi t
mi t
x1 c1e , and x2 c2e
k
where n 2 rad/s
m
The sum of x1 and x2 is also a solution so that the total solution is:
x x x c e2it c e2it
1 2 1 2
Substitute initial conditions: x0 = 1 mm, v0 = 5 mm/s
x 0 c1 c2 x0 1 c2 1 c1, and v 0 x˙0 2ic1 2ic2 v0 5 mm/s
2c1 2c2 5 i. Combining the two underlined expressions (2 eqs in 2 unkowns):
1 5
2c 2 2c 5 i c i, and c 1 5 i
1 1 1 2
2 4 2 4
Therefore the solution is:
1 5 2it 1 5 2it
x i e i e
2 4 2 4
Using the Euler formula to evaluate the exponential terms yields:
1 5 1 5
x i cos 2t i sin 2t i cos 2t i sin 2t
2 4 2 4
5 3
x(t ) cos 2t sin 2t sin2t 0.7297
2 2
, Using Mathcad the plot is:
5.
x t cos 2. t sin 2. t
2
2
x t
0 5 10
2
t
