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 mg / x
k
86.164
n
m mg
20 The sample standard
n
deviation in
m i1 stiffness
2
15 computed (k i ) is:
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
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,1.2 Derive the solution of mẋ˙ 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) aert . Then: x arert and x ar 2 e rt . Substitute into equation (1) to
get:
mar2ert kaert 0
mr2 k 0
k
r i
m
Thus there are two solutions:
k
k it
i
m t m
x1 c1e , and x2 c2e
where n k 2 rad/s
The sum of x1 and x2 is also a solution som
that the total solution is:
x x x c e1 2it c e 22it
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 1 5
2c1 2 2c1 5 i c1 i, and c2 i
2 4 2 4
Therefore the solution is:
1 2it 1
x 5 i e 5
2 i e 2it
2 4 4
Using the Euler formula to1evaluate the exponential terms yields:
x 5 i cos 2t i sin 2t 5 i cos 2t i sin 2t
1
2 4 2 4
5 3
x(t ) cos 2t sin 2t sin 2t 0.7297
2 2
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,Using Mathcad the plot is:
5.
2. t sin 2. t
x t cos
2
2
x t
0 5 10
2
t
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,1.3 Solve mẋ˙ kx 0 for k = 4 N/m, m = 1 kg, x0 = 1 mm, and v0 = 0. Plot the solution.
Solution:
This is identical to problem 2, except v0 0. k
initial conditions: n 2 rad/s . Calculating the
m
x 0 c1 c2 x0 1 c2 1 c1
v 0 ẋ 0 2ic1 2ic2 v0 0 c2 c1 c2 c1 0.5
1 2it 1 2it 1 1
xt e e cos2t isin 2t cos2t i sin2t
2 2 2 2
x(t)= cos (2t )
The following plot is from Mathcad:
x t cos 2. t
1
x t
0 5 10
1
t
Alternately students may use equation (1.10) directly to get
2 2 (1)2 02 21
x(t ) sin(2t tan 1[ ])
2 0
1sin(2t
) cos 2t
2
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,1.4 The amplitude of vibration of an undamped system is measured to be 1 mm. The phase shift
from t = 0 is measured to be 2 rad and the frequency is found to be 5 rad/s. Calculate the initial
conditions that caused this vibration to occur. Assume the response is of the form x(t ) A sin( nt
).
Solution:
Given: A 1mm, 2 rad, 5 rad/s . For an undamped system:
x t A sin n t 1sin 5t 2 and
v t x t A n cos nt 5 cos 5t 2
Setting t = 0 in these expressions yields:
x(0) = 1sin(2) = 0.9093 mm
v(0) = 5 cos(2) = - 2.081 mm/s
1.5 Find the equation of motion for the hanging spring-mass system of Figure P1.5, and compute
the natural frequency. In particular, using static equilibrium along with Newton’s law,
determine what effect gravity has on the equation of motion and the system’s natural
frequency.
Figure P1.5
Solution:
The free-body diagram of problem system in (a) for the static case and in (b) for the dynamic
case, where x is now measured from the static equilibrium position.
(a) (b)
From a force balance in the static case (a): mg kxs , where xs is the static deflection of the
spring. Next let the spring experience a dynamic deflection x(t) governed by summing the forces
in (b) to get
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, mx(t ) mg k(x(t ) x s ) mx(t ) kx(t ) mg kx s
k
mx(t ) kx(t ) 0 n
m
since mg kxs from static equilibrium.
1.6 Find the equation of motion for the system of Figure P1.6, and find the natural frequency. In
particular, using static equilibrium along with Newton’s law, determine what effect gravity has on
the equation of motion and the system’s natural frequency. Assume the block slides without
friction.
Figure P1.6
Solution:
Choosing a coordinate system along the plane with positive down the plane, the free- body
diagram of the system for the static case is given and (a) and for the dynamic case in (b):
In the figures, N is the normal force and the components of gravity are determined by the angle
as indicated. From the static equilibrium: kx s mg sin 0 . Summing forces in (b) yields:
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, Fi mx(t ) mx(t ) k(x x s ) mg sin
mx(t ) kx kx s mg sin 0
mx(t ) kx 0
k
rad/s
n
m
1.7 An undamped system vibrates with a frequency of 10 Hz and amplitude 1 mm. Calculate the
maximum amplitude of the system's velocity and acceleration.
Solution:
Given: First convert Hertz to rad/s: n 2 fn 2 10 20 rad/s. We also have that
A= 1 mm.
For an undamped system:
x t Asin n t
and differentiating yields the velocity: v t A n cos n t . Realizing that both the sin and cos
functions have maximum values of 1 yields:
vmax A n 1 20 62.8 mm/s
Likewise for the acceleration: at A n sin n t
2
amax A 2
1 20 2
3948 mm/s 2
n
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,1.8 Show by calculation that A sin ( nt + ) can be represented as Bsin nt + Ccos nt and calculate C
and B in terms of A and .
Solution:
This trig identity is useful: sin a b sin a cos b cos a sin b
Given: Asin n t Asin n t cos A cos n t sin
Bsin nt C cos nt
where B A cos and C A sin
1.9 Using the solution of equation (1.2) in the form x(t) Bsin nt Ccos n t
calculate the values of B and C in terms of the initial conditions x0 and v0.
Solution:
Using the solution of equation (1.2) in the form
xt Bsin nt Ccos n t
and differentiate to get:
ẋ(t) n Bcos( n t) n Csin( nt)
Now substitute the initial conditions into these expressions for the position and velocity to get:
x0 x(0) Bsin(0) C cos(0) C
v0 ẋ(0) n B cos(0) nC sin(0)
n B(1) nC(0) nB
Solving for B and C yields:
v0
B , and C x
0
n
v0
Thus x(t) sin t x
cos nt
n 0
n
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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 mg / x
k
86.164
n
m mg
20 The sample standard
n
deviation in
m i1 stiffness
2
15 computed (k i ) is:
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
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,1.2 Derive the solution of mẋ˙ 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) aert . Then: x arert and x ar 2 e rt . Substitute into equation (1) to
get:
mar2ert kaert 0
mr2 k 0
k
r i
m
Thus there are two solutions:
k
k it
i
m t m
x1 c1e , and x2 c2e
where n k 2 rad/s
The sum of x1 and x2 is also a solution som
that the total solution is:
x x x c e1 2it c e 22it
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 1 5
2c1 2 2c1 5 i c1 i, and c2 i
2 4 2 4
Therefore the solution is:
1 2it 1
x 5 i e 5
2 i e 2it
2 4 4
Using the Euler formula to1evaluate the exponential terms yields:
x 5 i cos 2t i sin 2t 5 i cos 2t i sin 2t
1
2 4 2 4
5 3
x(t ) cos 2t sin 2t sin 2t 0.7297
2 2
@
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,Using Mathcad the plot is:
5.
2. t sin 2. t
x t cos
2
2
x t
0 5 10
2
t
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,1.3 Solve mẋ˙ kx 0 for k = 4 N/m, m = 1 kg, x0 = 1 mm, and v0 = 0. Plot the solution.
Solution:
This is identical to problem 2, except v0 0. k
initial conditions: n 2 rad/s . Calculating the
m
x 0 c1 c2 x0 1 c2 1 c1
v 0 ẋ 0 2ic1 2ic2 v0 0 c2 c1 c2 c1 0.5
1 2it 1 2it 1 1
xt e e cos2t isin 2t cos2t i sin2t
2 2 2 2
x(t)= cos (2t )
The following plot is from Mathcad:
x t cos 2. t
1
x t
0 5 10
1
t
Alternately students may use equation (1.10) directly to get
2 2 (1)2 02 21
x(t ) sin(2t tan 1[ ])
2 0
1sin(2t
) cos 2t
2
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,1.4 The amplitude of vibration of an undamped system is measured to be 1 mm. The phase shift
from t = 0 is measured to be 2 rad and the frequency is found to be 5 rad/s. Calculate the initial
conditions that caused this vibration to occur. Assume the response is of the form x(t ) A sin( nt
).
Solution:
Given: A 1mm, 2 rad, 5 rad/s . For an undamped system:
x t A sin n t 1sin 5t 2 and
v t x t A n cos nt 5 cos 5t 2
Setting t = 0 in these expressions yields:
x(0) = 1sin(2) = 0.9093 mm
v(0) = 5 cos(2) = - 2.081 mm/s
1.5 Find the equation of motion for the hanging spring-mass system of Figure P1.5, and compute
the natural frequency. In particular, using static equilibrium along with Newton’s law,
determine what effect gravity has on the equation of motion and the system’s natural
frequency.
Figure P1.5
Solution:
The free-body diagram of problem system in (a) for the static case and in (b) for the dynamic
case, where x is now measured from the static equilibrium position.
(a) (b)
From a force balance in the static case (a): mg kxs , where xs is the static deflection of the
spring. Next let the spring experience a dynamic deflection x(t) governed by summing the forces
in (b) to get
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, mx(t ) mg k(x(t ) x s ) mx(t ) kx(t ) mg kx s
k
mx(t ) kx(t ) 0 n
m
since mg kxs from static equilibrium.
1.6 Find the equation of motion for the system of Figure P1.6, and find the natural frequency. In
particular, using static equilibrium along with Newton’s law, determine what effect gravity has on
the equation of motion and the system’s natural frequency. Assume the block slides without
friction.
Figure P1.6
Solution:
Choosing a coordinate system along the plane with positive down the plane, the free- body
diagram of the system for the static case is given and (a) and for the dynamic case in (b):
In the figures, N is the normal force and the components of gravity are determined by the angle
as indicated. From the static equilibrium: kx s mg sin 0 . Summing forces in (b) yields:
@
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, Fi mx(t ) mx(t ) k(x x s ) mg sin
mx(t ) kx kx s mg sin 0
mx(t ) kx 0
k
rad/s
n
m
1.7 An undamped system vibrates with a frequency of 10 Hz and amplitude 1 mm. Calculate the
maximum amplitude of the system's velocity and acceleration.
Solution:
Given: First convert Hertz to rad/s: n 2 fn 2 10 20 rad/s. We also have that
A= 1 mm.
For an undamped system:
x t Asin n t
and differentiating yields the velocity: v t A n cos n t . Realizing that both the sin and cos
functions have maximum values of 1 yields:
vmax A n 1 20 62.8 mm/s
Likewise for the acceleration: at A n sin n t
2
amax A 2
1 20 2
3948 mm/s 2
n
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,1.8 Show by calculation that A sin ( nt + ) can be represented as Bsin nt + Ccos nt and calculate C
and B in terms of A and .
Solution:
This trig identity is useful: sin a b sin a cos b cos a sin b
Given: Asin n t Asin n t cos A cos n t sin
Bsin nt C cos nt
where B A cos and C A sin
1.9 Using the solution of equation (1.2) in the form x(t) Bsin nt Ccos n t
calculate the values of B and C in terms of the initial conditions x0 and v0.
Solution:
Using the solution of equation (1.2) in the form
xt Bsin nt Ccos n t
and differentiate to get:
ẋ(t) n Bcos( n t) n Csin( nt)
Now substitute the initial conditions into these expressions for the position and velocity to get:
x0 x(0) Bsin(0) C cos(0) C
v0 ẋ(0) n B cos(0) nC sin(0)
n B(1) nC(0) nB
Solving for B and C yields:
v0
B , and C x
0
n
v0
Thus x(t) sin t x
cos nt
n 0
n
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