Baixe Daniel J. Inman - Engineering Vibration. Solution Manual-Pearson (2007) e outras Exercícios em PDF para Dinâmica dos Sistemas, somente na Docsity!

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.

Solution:

Free-body diagram:

m

k

kx

mg

Plot of mass in kg versus displacement in m

Computation of slope from mg / x

m (kg) x (m) k (N/m)

0 1 2

10

15

20

m

x

From the free-body diagram and static

equilibrium:

kx = mg ( g = 9.81 m / s

2

k = mg / x

! k

i

n

The sample standard deviation in

computed stiffness is:

( k

i

2

i = 1

n

n " 1

Using Mathcad the plot is:

x t cos

2 t

sin

2 t

0 5 10

2

2

x t

t

1.3 Solve m

x + kx = 0 for k = 4 N/m, m = 1 kg, x

0

= 1 mm, and v

0

= 0. Plot the solution.

Solution:

This is identical to problem 2, except v

0

n

k

m

= 2 rad/s

. Calculating the

initial conditions:

x 0

= c

1

• c

2

= x

0

= 1! c

2

= 1 " c

1

v ( 0 ) =

x ( 0 ) = 2 ic

1

" 2 ic

2

= v

0

= 0! c

2

= c

1

c

2

= c

1

x t

e

2 it

e

" 2 it

cos 2 t + i sin 2 t

cos 2 t " i sin 2 t

x ( t )= cos (2 t )

The following plot is from Mathcad:

Alternately students may use equation (1.10) directly to get

x ( t ) =

2

2

2

sin( 2 t + tan

! 1

[

])

= 1 sin( 2 t +

) = cos 2 t

x t cos

2 t

0 5 10

1

1

x t

t

m! x !( t ) = mg! k ( x ( t ) + x

s

) " m! x !( t ) + kx ( t ) = mg! kx

s

" m! x !( t ) + kx ( t ) = 0 " #

n

k

m

since mg = kx

s

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.

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 θ as indicated. From the static equilibrium:! kx

s

• mg sin " = 0. Summing forces

in (b) yields:

F

! i

= m! x !( t ) " m! x !( t ) = # k ( x + x

s

) + mg sin $

" m! x !( t ) + kx = # kx

s

• mg sin $ = 0

" m! x !( t ) + kx = 0

n

k

m

rad/s

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 " f

n

= 2 " ( 10 ) = 20 " rad/s. We also have that

A = 1 mm.

For an undamped system:

x ( t ) = A ( " t + !)

n

sin

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:

= " = 1 ( 20 !) = 62.8mm/s

max n

v A

Likewise for the acceleration: a ( t ) = # A " ( " t + !)

n n

sin

2

2

= = = 3948 mm/s

2 2

max

n

a A

1.10 Using Figure 1.6, verify that equation (1.10) satisfies the initial velocity condition.

Solution: Following the lead given in Example 1.1.2, write down the general expression

of the velocity by differentiating equation (1.10):

x ( t ) = A sin(!

n

t + ") #

x ( t ) = A!

n

cos(!

n

t + ")

v (0) = A!

n

cos(!

n

0 + ") = A!

n

cos( ")

From the figure:

Figure 1.

A = x

0

2

v

0

!

n

2

, cos ( =

v

0

!

n

x

0

2

v

0

!

n

2

Substitution of these values into the expression for v (0) yields

v ( 0 ) = A!

n

cos " = x

0

2

v

0

!

n

2

n

v

0

!

n

x

0

2

v

0

!

n

2

= v

0

verifying the agreement between the figure and the initial velocity condition.

1.11 (a)A 0.5 kg mass is attached to a linear spring of stiffness 0.1 N/m. Determine the natural

frequency of the system in hertz. b) Repeat this calculation for a mass of 50 kg and a

stiffness of 10 N/m. Compare your result to that of part a.

Solution: From the definition of frequency and equation (1.12)

( a )!

n

k

m

= 0.447 rad/s

f

n

n

= 0.071 Hz

( b )!

n

= 0.447rad/s, f

n

n

= 0.071 Hz

Part (b) is the same as part (a) thus very different systems can have same natural

frequencies.

1.13 Determine the natural frequency of the two systems illustrated.

(a) (b)

Figure P1. 13

Solution:

(a) Summing forces from the free-body diagram in the x direction yields:

• k

1

x

• x

• k

2

x

Free-body diagram for part a

m

x =! k

1

x! k

2

x "

m ˙ x ˙ + k

1

x + k

2

x = 0

m

x + x k

1

• k

2

= 0, dividing by m yields :

x +

k

1

• k

2

m

x = 0

Examining the coefficient of x

yields:

!

n

k

1

• k

2

m

(b) Summing forces from the free-body diagram in the x direction yields:

• k

1

x

• k

2

x

• x

• k

3

x

Free-body diagram for part b

m! x! =! k

1

x! k

2

x! k

3

x ,"

m! x! + k

1

x + k

2

x + k

3

x = 0 "

m! x! + ( k

1

• k

2

• k

3

) x = 0 "! x! +

( k

1

• k

2

• k

3

m

x = 0

n

k

1

• k

2

• k

3

m

1.14* Plot the solution given by equation (1.10) for the case k = 1000 N/m and m = 10 kg for

two complete periods for each of the following sets of initial conditions: a) x

0

= 0 m, v

0

1 m/s, b) x

0

= 0.01 m, v

0

= 0 m/s, and c) x

0

= 0.01 m, v

0

= 1 m/s.

Solution: Here we use Mathcad:

a) all units in m, kg, s

parts b and c are plotted in the above by simply changing the initial conditions as

appropriate

m 10

x0 0.

T

"n fn

!n

x t

A sin

!n t "

A

!n

x

2

!n

2

v

2

! atan

"n x

v

0 0.5 1 1.

x t

xb t

xc t

t

k 1000

n

k

m

v0 1

1.16 A machine part is modeled as a pendulum connected to a spring as illustrated in Figure

P1.16. Ignore the mass of pendulum’s rod and derive the equation of motion. Then

following the procedure used in Example 1.1.1, linearize the equation of motion and

compute the formula for the natural frequency. Assume that the rotation is small enough

so that the spring only deflects horizontally.

Figure P1.

Solution: Consider the free body diagram of the mass displaced from equilibrium:

There are two forces acting on the system to consider, if we take moments about point O

(then we can ignore any forces at O ). This yields

M

! O

= J

O

" # m!

2

""

$ = % mg !sin $ % k !sin $ ∞! cos $

m!

2

""

$ + mg !sin $ + k!

2

sin $ cos $ = 0

Next consider the small θ approximations to that sin!!! and cos != 1. Then the

linearized equation of motion becomes:

!( t ) +

mg + k "

m "

!( t ) = 0

Thus the natural frequency is

n

mg + k!

m!

rad/s

1.17 A pendulum has length of 250 mm. What is the system’s natural frequency in Hertz?

Solution:

Given: l =250 mm

Assumptions: small angle approximation of sin

From Window 1.1, the equation of motion for the pendulum is as follows:

I

O

! + mg! = 0 , where I

O

= ml

2

g

l

The coefficient of θ yields the natural frequency as:

f

n

n

= 0.996 Hz

1.18 The pendulum in Example 1.1.1 is required to oscillate once every second. What length

should it be?

Solution:

Given: f = 1 Hz (one cycle per second)

!

n

g

l

9.8 m/s

2

0.25 m

= 6.26 rad/s

l

g

f

n

1.19 The approximation of sin θ = θ, is reasonable for θ less than 10°. If a pendulum of length

0.5 m, has an initial position of θ(0) = 0, what is the maximum value of the initial angular

velocity that can be given to the pendulum with out violating this small angle

approximation? (be sure to work in radians)

Solution: From Window 1.1, the linear equation of the pendulum is

For zero initial position, the solution is given in equation (1.10) by

since sin is always less then one. Thus if we need θ < 10°= 0.175 rad, then we need to

solve:

for v 0

which yields:

v

0

< 0.773 rad/s.

( )+ ( t )= 0

g

! t!

! ( t ) =

v

0

g

sin(

g

t ) "! #

v

0

g

0

v

Problems and Solutions for Section 1.2 and Section 1.3 (1.20 to 1.51)

Problems and Solutions Section 1.2 (Numbers 1.20 through 1.30)

1.20* Plot the solution of a linear, spring and mass system with frequency ω n

=2 rad/s,

x

0

= 1 mm and v

0

= 2.34 mm/s, for at least two periods.

Solution: From Window 1.18, the plot can be formed by computing:

A =

n

n

2

x

0

2

• v

0

2

= 1.54 mm, " = tan

1

n

x

0

v

0

!

x ( t ) = A sin(!

n

t + ")

This can be plotted in any of the codes mentioned in the text. In Mathcad the

program looks like.

In this plot the units are in mm rather than meters.