Sample Problem 8/2

The 8-kg body is moved 0.2 m to the right of the equilibrium position andreleased from rest at time t � 0. Determine its displacement at time t � 2 s. Theviscous damping coefficient c is 20 and the spring stiffness k is 32 N/m.

Solution. We must first determine whether the system is underdamped, criti-cally damped, or overdamped. For that purpose, we compute the damping ratio �.

Since � � 1, the system is underdamped. The damped natural frequency is �d � � � 1.561 rad/s. The motion is given by Eq. 8/12and is

The velocity is then

Evaluating the displacement and velocity at time t � 0 gives

Solving the two equations for C and � yields C � 0.256 m and � � 0.896 rad.Therefore, the displacement in meters is

Evaluation for time t � 2 s gives x2 � �0.01616 m. Ans.

Sample Problem 8/3

The two fixed counterrotating pulleys are driven at the same angular speed�0. A round bar is placed off center on the pulleys as shown. Determine the nat-ural frequency of the resulting bar motion. The coefficient of kinetic friction be-tween the bar and pulleys is �k.

Solution. The free-body diagram of the bar is constructed for an arbitrary dis-placement x from the central position as shown. The governing equations are

Eliminating NA and NB from the first equation yields

We recognize the form of this equation as that of Eq. 8/2, so that the natural fre-quency in radians per second is �n � and the natural frequency in cy-cles per second is

Ans.ªn � 12�

�2�k g/a

�2�k g/a

x � 2�k g

a x � 0

aNB� �a2

� x�mg � 0[ΣMA � 0]

NA � NB � mg � 0[ΣFy � 0]

�k NA � �k NB � mx[ΣFx � mx]

x � 0.256e�1.25t sin (1.561t � 0.896)

x0 � C sin � � 0.2 x0 � �1.25C sin � � 1.561C cos � � 0

x � �1.25Ce�1.25t sin (1.561t � �) � 1.561Ce�1.25t cos (1.561t � �)

x � Ce���nt sin (�dt � �) � Ce�1.25t sin (1.561t � �)

2�1 � (0.625)2�n�1 � �2

�n � �k/m � �32/8 � 2 rad/s � � c2m�n

� 202(8)(2)

� 0.625

N � s/m,

610 Chapter 8 Vibrat ion and T ime Response

k

xc

8 kg

x

Equilibriumposition

mg

N

cx = 20x··

kx = 32x

a0ω 0ω

Centralposition

y

a––2

a––2x

A BG

NA NBmg

k NAµ k NBµ

�

�

�

Helpful Hint

� We note that the exponential factore�1.25t is 0.0821 at t � 2 s. Thus, � � 0.625 represents severe damp-ing, although the motion is still oscillatory.

Helpful Hints

� Because the bar is slender and doesnot rotate, the use of a momentequilibrium equation is justified.

� We note that the angular speed �0

does not enter the equation of mo-tion. The reason for this is our as-sumption that the kinetic frictionforce does not depend on the relativevelocity at the contacting surface.

c08.qxd 6/28/06 4:48 PM Page 610

Sample Problem 8/7

A simplified version of a pendulum used in impact tests is shown in the figure.Derive the equation of motion and determine the period for small oscillationsabout the pivot. The mass center G is located a distance � 0.9 m from O, and theradius of gyration about O is kO � 0.95 m. The friction of the bearing is negligible.

Solution. We draw the free-body diagram for an arbitrary, positive value of theangular-displacement variable , which is measured counterclockwise for the coor-dinate system chosen. Next we apply the governing equation of motion to obtain

or Ans.

Note that the governing equation is independent of the mass. When is small,sin � , and our equation of motion may be written as

The frequency in cycles per second and the period in seconds are

Ans.

For the given properties: Ans.

Sample Problem 8/8

The uniform bar of mass m and length l is pivoted at its center. The springof constant k at the left end is attached to a stationary surface, but the right-endspring, also of constant k, is attached to a support which undergoes a harmonicmotion given by yB � b sin �t. Determine the driving frequency �c which causesresonance.

Solution. We use the moment equation of motion about the fixed point O toobtain

Assuming small deflections and simplifying give us

The natural frequency should be recognized from the now-familiar form of theequation to be

Thus, �c � �n � will result in resonance (as well as violation of our small-angle assumption!). Ans.

�6k/m

�n � �6k/m

� 6km

� 6kbml

sin �t

��k l2

sin � l2

cos � k� l2

sin � yB� l2

cos � 112

ml2

� gr

kO

2 � 0

� gr

kO

2 sin � 0

�mgr sin � mkO

2 [ΣMO � IO ]

r

636 Chapter 8 Vibrat ion and T ime Response

G

O

Oy

Ox

θ

r–

mg

θ

r–

yB = b sin ωt

O

B

mk

l—2

l—2

k

Oy

mg

Oxθ

k )( l—2

sin θk – y

B)( l—2

sin θ

Helpful Hints

� With our choice of point O as themoment center, the bearing reac-tions Ox and Oy never enter theequation of motion.

� For large angles of oscillation, deter-mining the period for the pendulumrequires the evaluation of an ellipticintegral.

Helpful Hints

� As previously, we consider only thechanges in the forces due to a move-ment away from the equilibriumposition.

� The standard form here is � �

where M0 � and IO �

The natural frequency �n of asystem does not depend on the exter-nal disturbance.

112

ml2.

klb2

M0 sin �tIO

,

�n

2

�

�

�

�

ƒn �1

2� � gr

kO

2 � �

1ƒn

� 2��kO

2

gr

� � 2�� (0.95)2

(9.81)(0.9)� 2.01 s

c08.qxd 6/28/06 4:48 PM Page 636

Sample Problem 8/9

Derive the equation of motion for the homogeneous circular cylinder, whichrolls without slipping. If the cylinder mass is 50 kg, the cylinder radius 0.5 m,the spring constant 75 N/m, and the damping coefficient 10 determine

(a) the undamped natural frequency

(b) the damping ratio

(c) the damped natural frequency

(d) the period of the damped system.

In addition, determine x as a function of time if the cylinder is released from restat the position x � �0.2 m when t � 0.

Solution. We have a choice of motion variables in that either x or the angulardisplacement of the cylinder may be used. Since the problem statement in-volves x, we draw the free-body diagram for an arbitrary, positive value of x andwrite the two motion equations for the cylinder as

The condition of rolling with no slip is � Substitution of this conditioninto the moment equation gives F � Inserting this expression for thefriction force into the force equation for the x-direction yields

Comparing the above equation with that for the standard dampedoscillator, Eq. 8/9, allows us to state directly

(a) Ans.

(b) Ans.

Hence, the damped natural frequency and the damped period are

(c) Ans.

(d) Ans.

From Eq. 8/12, the underdamped solution to the equation of motion is

At time t � 0, x and become

The solution to the two equations in C and � gives

Thus, the motion is given by

Ans.x � �0.200e�0.0667t sin (0.998t � 1.504) m

C � �0.200 m � � 1.504 rad

x0 � �0.0667C sin � � 0.998C cos � � 0

x0 � C sin � � �0.2

x

� 0.998Ce�0.0667t cos (0.998t � �)

x � �0.0667Ce�0.0667t sin (0.998t � �)The velocity is

x � Ce���nt sin (�dt � �) � Ce�(0.0667)(1)t sin (0.998t � �)

�d � 2�/�d � 2�/0.998 � 6.30 s

�d � �n�1 � �2 � (1)�1 � (0.0667)2 � 0.998 rad/s

2��n � 23

cm

� � 13

cm�n

� 103(50)(1)

� 0.0667

�n

2 � 23

km

�cx � kx � 12

mx � mx or x � 23

cm

x � 23

km

x � 0

�12

mx.r .x

�Fr � 12

mr2 [ΣMG � I ]

�cx � kx � F � mx[ΣFx � mx]

N � s/m,

Art ic le 8/4 Vibrat ion of Rig id Bodies 637

�

�

x

mk cr

F

O

N

mg

Equilibriumposition

kx cx·

x+θ

Helpful Hints

� The angle is taken positive clock-wise to be kinematically consistentwith x.

� The friction force F may be as-sumed in either direction. We willfind that the actual direction is tothe right for x � 0 and to the leftfor x � 0; F � 0 when x � 0.

�n � �23

km

� �23

7550

� 1 rad/s

c08.qxd 6/28/06 4:48 PM Page 637

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