All 8 Chapters Covered
SOLUTION MANUAL
,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
2
m 15 i
= i=1
= 0.164
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:
mx + kx =0 (1)
Assume: x(t) = ae . Then:
rt x = arert
and x = ar e . Substitute into equation (1) to
2 rt
get:
mar2ert + kaert = 0
mr2 + k = 0
k
r= i
m
Thus there are two solutions:
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 1
2c1 + 2 2c1 = 5 i c1 = 5 i, and c2 = + 5 i
2 4 2 4
Therefore the solution is:
1 2it
5 2it 1 5
x= i e + + i e
2 4 2 4
Using the Euler formula to evaluate the exponential terms yields:
1 1
x= 5 i (cos 2t + i sin 2t ) + + 5 i (cos 2t i sin 2t )
2 4 2 4
3
( x(t ) = cos 2t + 5 sin 2t = sin(2t + 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
SOLUTION MANUAL
,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
2
m 15 i
= i=1
= 0.164
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:
mx + kx =0 (1)
Assume: x(t) = ae . Then:
rt x = arert
and x = ar e . Substitute into equation (1) to
2 rt
get:
mar2ert + kaert = 0
mr2 + k = 0
k
r= i
m
Thus there are two solutions:
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 1
2c1 + 2 2c1 = 5 i c1 = 5 i, and c2 = + 5 i
2 4 2 4
Therefore the solution is:
1 2it
5 2it 1 5
x= i e + + i e
2 4 2 4
Using the Euler formula to evaluate the exponential terms yields:
1 1
x= 5 i (cos 2t + i sin 2t ) + + 5 i (cos 2t i sin 2t )
2 4 2 4
3
( x(t ) = cos 2t + 5 sin 2t = sin(2t + 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
