mechanical vibration solved examples

Mechanical Vibrations: 4600-431 Example Problems

December 20, 2006

Contents

1 Free Vibration of Single Degree-of-freedom Systems 1

2 Frictionally Damped Systems 33

3 Forced Single Degree-of-freedom Systems 42

4 Multi Degree-of-freedom Systems 69

1 Free Vibration of Single Degree-of-freedom Systems

Problem 1:

In the figure, the disk and the block havemass m and the radius of the disk is r.

a) Find the equations of motion for thissystem.

b) What are the natural frequency anddamping ratio of the system in termsof m, c, and k?

c) If the block is displaced 18 cm to theright and released from rest find the re-sulting angular displacement of the diskwith

m = 3 kg, r = 9 cm,

k = 21 N/m, c = 63 N s/m,

k

c

m

4 kk

(m, r)

1

Problem 2:

For the system shown to the right, the disk ofmass m rolls without slip and x measures thedisplacement of the disk from the unstretchedposition of the spring.

a) Find the equations of motion.

b) With

c = 16 N/(m/s), m = 2 kg,

r = 0.10 m

for what value of the spring stiffness kis the damping ratio of the system one-half of the critically damped value, sothat = cr/2?

c) With these parameter values, find thedisplacement of the disk if it is rolled20 cm to the right (from static equilib-rium) and released from rest.

x

c

k

(m, r)

Problem 3:

From the figure shown to the right

a) find the equations of motion in terms ofthe angular rotation of the disk;

b) what are the damping ratio and naturalfrequency of the system in terms of theparameters m, b, k1, and k2;

c) can you draw an equivalent spring-mass-damper system?

r

r

2

b

k1

k2

m

m

2

Problem 4:

For the mechanical system shown to theright, the uniform rigid bar has mass m andpinned at point O. For this system:

a) find the equations of motion;

b) Identify the damping ratio and naturalfrequency in terms of the parametersm, c, k, and .

c) For:

m = 1.50 kg, = 45 cm,

c = 0.125 N/(m/s), k = 250 N/m,

find the angular displacement of the bar(t) for the following initial conditions:

(0) = 0, (0) = 10 rad/s.

Assume that in the horizontal position thesystem is in static equilibrium and that allangles remain small.

2

2

m

O

k

y

2 m

k c

x

Solution:

a) In addition to the coordinate identified inthe original figure, we also define x and y asthe displacment of the block and end of thebar respecively. The directions and aredefined as shown in the figure.A free body diagram for this system isshown to the right. Note that the tension inthe cable between the bar and the block isunknown and represented with T while thereaction force FR is included, although bothits magnitude and direction are unspecified.In terms of the identifed coordinates, theangular acceleration of the bar /F and thelinear acceleration of the block

F

aG are

/F = k,F

aG = x .

FR

k y

T

T

k x c x

We can also relate the identified coordinates as

x =

2, y = .

The equations of motion for this system can be obtained with linear momentum balanceapplied to the block and angular momentum balance aout O on the bar. These can bewritten as

F = mF

aG (T k x c x

) = 2mx ,

MO = I

O /F ( T

2 k y

)k =

m2

3 k.

3

Solving the first equation for T and substituting into the second equation yields

(2mx+ k x+ c x) 2 k y = m

2

3.

Using the coordinate relations we can obtain the equation of motion as

5m2

6 +

c 2

4 +

5 k 2

4 = 0.

b) In the above equation the equivalent mass, damping, and stiffness are

meq =5m2

6, beq =

c 2

4, keq =

5 k 2

4.

From these the damping ratio and natural frequency are

=beq

2keq meq

=c 2

4

2

5 k 2

45m2

6

=

3 c

2

50 km,

n =

keqmeq

=

5 k 2

45m2

6

=

3 k

2m

c) Evaluating the damping ratio and natural frequency we find that for the given valuesof the parameters

n = 15.8 rad/s, = 7.91 104.

Therefore the system is underdamped and the general solution can be written as

(t) = e n t(a sin

(n

1 2 t)

+ b cos(n

1 2 t))

,

where a and b are arbitrary constants used to fit the initial conditions. Evaluating (t)and (t) at t = 0 yields

(0) = b = 0, (0) = n b+ n

1 2 a = 10 rad/s,

so that the general solution becomes

(t) = 0.632 e0.0125 t sin(15.8 t

).

4

Problem 5:

In the figure, the disk has mass m, radius r,

and moment of inertia IG =mr2

2 about themass center G and is assumed to roll withoutslip. The identified coordiante measures therotation of the disk with respect to the equi-librium position.

a) Find the equations of motion for thissystem.

b) If the disk is released from rest with(0) = 2 rad, find the resulting an-gular displacement (t) for

m = 3 kg, r = 15 cm,

k = 36 N/m, c = 3 N s/m,

c) What is the force in the upper springduring this motion?

k

c

k

(m, r)

G

g

Problem 6:

For the mechanical system shown to theright, the uniform rigid bar has mass m andpinned at point O. For this system:



c) For:

m = 2 kg, = 25 cm,

c = 0.25 N/(m/s), k = 50 N/m,

find the angular displacement of the bar(t) for the following initial conditions:

(0) = 0, (0) = 10 rad/s.

d) for this motion, find the tension in thecable connecting the rod and the blockas a function of time.

Assume that the system is in static equilib-rium at = 0, and that all angles remainsmall.

2

2

m

k

z

m

k cx

Solution:

a) We identify the coordinates x and z as shown above, which are related to the angular

5

displacement as:

x =

2, z =

2.

An appropriate free-body diagram isshown to the right. Applying linear mo-mentum balance on the block yields

F = mF

aG,

(T k x c x) = mx .

Likewise, angular momentum balance onthe bar provides

MO = IO /F ,(

T 2 k z

2

)k =

m2

12 k.

FR

k z

T

T

c x k x

Combining these equations and eliminating the tension, the equation of motion can bewritten as

7m

6 + c + 2 k = 0.

b) For the above equation the equivalent mass, damping, and stiffness are

meq =7m

6, beq = b, keq = 2 k,

and the natural frequency and damping ratio are

n =

keqmeq

=

12 k

7m, =

beq

2keq meq

=

3 b

28km.

Problem 7:

The block shown to the right rests on a fric-tionless surface. Find the response of the sys-tem if the block is displaced from its staticequilibrium position 15 cm to the right andreleased from rest.

m = 4.0 kg, b = 0.25 N/(m/s),

k1 = 1.5 N/m, k2 = 0.50 N/(m/s).

x

k1

b

k2m

Solution:

An appropriate free-body diagram is shown to theright. Notice that the two springs are effectively inparallel, as the displacement across each spring isidentical. Linear momentum balance on this blockprovides

F = mF

aG,

(k1 x k2 x b x) = mx ,

k1 x k2 x

b x

6

or, writing this in standard form

mx+ b x+ (k1 + k2)x = 0.

Further, the system is released from rest so that the initial conditions are

x(0) = x0 = 15 cm, x(0) = 0 cm/s.

Problem 8:

For the system shown to the right, the disk ofmass m rolls without slip and x measures thedisplacement of the disk from the unstretchedposition of the spring. The surface is inclinedat an angle of with respect to vertical.

a) find the equations of motion. Do notneglect gravity;

b) if the system is underdamped, what isthe frequency of the free vibrations ofthis system in terms of the parametersk, c, and m;

c) for what value of the damping constantc is the system critically damped;

d) what is the static equilibrium displace-ment of the disk?

x

c

k

(m, r)

z

C

e1

e2

Solution:

a) In addition to x, the displacement of the center of the disk, we identify the coordinatesz and , the displacement across the spring and the rotation of the disk respectively.

These additional coordinates are related to x as

z = 2x, x = r .

An appropriate free-body diagram is shown to theright. We note that (, ) are related to the direc-tions (e1, e2) as

= cos e1 + sin e2,

= sin e1 + cos e2.

k z e1

c x e1

fr e1N e2

mg

The moment produced by gravity about point C is

Mgravity = rGC (mg ),= (r e2) (mg ) = mg r sin k.

Angular momentum balance about the contact point C yieldsMC = I

C D/F ,((2 r) k z + r c xmg r sin

)k =

(3mr2

2

)k.

7

Eliminating the coordinates z and , we can write the equation of motion in terms of xas

3m

2x+ c x+ 4 k x = mg sin.

Since the gravitational force has been included in the development of this equation ofmotion, the coordinates are measured with respect to the unstretched position of thespring.

b) Assuming the system is underdamped, the frequency of the free vibrations is d =

n

1 2, where

n =

keqmeq

=

8 k

3m, =

beq

2keq meq

=c

2

6 km,

so that

d =

8 k

3m

1 c

2

24 km,

c) The system is critically damped when = 1, which corresponds to a damping coefficientof

ccr = 2

6 km.

d) The system is stationary in static equilibrium, so that x x0 = constantboth x andx vanish, and the equation of motion reduces to

4 k x0 = mg sin.

Solving for x0, the equilibrium displacement is

x0 =mg sin

4 k.

8

Problem 9:

In the figure shown to the right, in the ab-sence of gravity the springs are unstretchedin the equilibrium position.

a) Determine the deflection of each springfrom its unstretched length when thesystem shown is in equilibrium.

b) If the system is released from the un-stretched position of the springs, whatis the maximum angular velocity of thedisk during the resulting motion?

r2

r1

k1

k2

m

I

x

z1

z2

Solution:

a) We define the coordinates x, , z1, and z2as shown in the figure, which are relatedas

x = r1 , z1 = r1 , z2 = r2 .

Notice that because of these coordinatedefinitions, a rotation with positive gives rise to a negative value in both xand z2. Likewise, we see that x = z1.Using the free-body diagram shown tothe right, linear momentum balance onthe block provides

F = mF

aG,

(T mg) = mx ,

while angular momentum balance on thedisk yields

FRk1 z1

k2 z2

T

T

mg

MO = I

O D/F ,

(T r1 k1 z1 r1 + k2 z2 r2) k = I k.

Eliminating the unknown tension T from these equations and using the coordinaterelations, the equation of motion becomes(

I +mr21

) +

(k1 r

21 + k2 r

22

) = mg r1.

9

The equilibrium rotation of the disk thus is found to be

eq =mg r1

k1 r21 + k2 r22

.

With this, the equilibrium deflection of each spring is found to be

z1,eq = r1 eq =mg r21

k1 r21 + k2 r22

,

z2,eq = r2 eq = mg r1 r2k1 r21 + k2 r

22

.

b) The general free response of the disk can be expressed as

(t) = eq +A sin(n t) +B cos(n t),

where eq is given above, A and B are arbitrary constants, and

n =

k1 r21 + k2 r

22

I +mr21.

The system is released with the initial conditions:

(0) = 0, (0) = 0,

so that solving for the arbitrary constants

A = 0, B = eq.

Therefore the solution is

(t) = eq (1 cos(n t)) = mg r1k1 r21 + k2 r

22

(1 cos

(k1 r21 + k2 r

22

I +mr21t

)).

The angular velocity of the disk becomes

(t) =(eq n

)sin(n t),

which has amplitude

= eq n =mg r1

(k1 r21 + k2 r22) (I +mr

21)

10

Problem 10: (Spring 2003)For the spring-mass-damper system shown tothe right, x is measured from the static equi-librium position. If the surface is assumed tobe frictionless:

a) determine the governing equations ofmotion;

b) what is the frequency of oscillation ofthe system;

c) what value of the damping coefficient bcorresponds to critical damping?

d) if k = 2 N/m, b = 4 N/(m/s), and m =3 kg, find the displacement of the massx(t) when the system is started with theinitial conditions:

x(0) = 0.10 m, x(0) = 0 m/s.

x

m6 k b

Problem 11: (Spring 2003)For the mechanical system shown to theright, the uniform rigid bar has mass m andpinned at point O. For this system:


b) what value of the damping constant cgives rise to a critically damped sys-tem?

2

2

m

k

c

z1

m

k

z2

Solution:

a) In addition to , we define two additional coordinates, z1 and z2, to measure the deflec-tion at the left and right ends of the bar.

11

These coordinates are related as:

z1 =

2, z2 =

2 = z1.

A free-body diagram for this system isshown to the right. Applying angular mo-mentum balance on the bar eliminates theappearance of the reaction force and leadsto:

MG = IG k,(

T k z1 cz1)

2k =

m 2

12 k.

G

FR

k z1

c z1

k z2

T T

Likewise, application of linear momentum balance on the block yields:F = m

F

aG,( k z2 T

) = mz2

Eliminating the unknown tension T and solving for z1 in terms of z2, the equation ofmotion becomes:

m 2

3 +

c 2

4 +

k 2

2 = 0.

b) A critically damped system occurs when = 1. For this system:

=

3 c

32 k m.

Solving for ccr yields:

ccr =

32 k m

3.

Problem 12: (Spring 2003)Find the response of the system shown to theright if the block is pulled down by 15 cm andreleased form rest.

m = 2.0 kg, b = 0.5 N/(m/s),

k1 = 0.5 N/m, k2 = 0.25 N/(m/s). x

k1

k2

b

m

Solution:

For this system, the two springs in series may be replaced by an equivalent spring, withconstant:

keq =1

1k1

+ 1k2=

k1 k2k1 + k2

.

12

Therefore, the free-body diagram is shownto the right. Applying linear momentumbalance to the block yields:

F = mF

aG,(keq x+ b x

) = m x ,

which can finally be written as:

m x+ b x+ keq x = 0.

keq x b x

With the numerical values given above, this becomes:

(2 kg) x+

(1

2

N

m/s

)x+

(1

6

N

m

)x = 0, x(0) =

3

20m, x(0) = 0 m/s.

With this, the damping ratio and natural frequency are:

n =

1

12s1, =

3

4.

Therefore, the system is underdamped and the general response can be written as:

x(t) = e n t(A cos(d t) +B sin(d t)

).

Using the initial conditions to solve for A and B, we find:

x(t) =3

20et/8

(cos

(13

8

3t

)+

3

13cos

(13

8

3t

)).

Problem 13: (Spring 2003)In the figure shown to the right, in the ab-sence of gravity the springs are unstretchedin the equilibrium position.

a) Determine the deflection of each springfrom its unstretched length when thesystem shown is in equilibrium.

b) If the system is released from the un-stretched position of the springs, whatis the maximum angular velocity of thedisk during the resulting motion?

r2

r1

k1

k2

m

x2

x1

13

Solution:

a) We define , x1, and x2 as indicated in the above figure. In particular, x1 and x2represent the displacement in their respective springs as measured from their unstretchedposition. These coordinates are related through the following transformations:

x1 = r1 , x2 = r2 .

An appropriate free-body diagram for thissystem is shown to the right. Notice thatthe gravitational force must be included todetermine the equilibrium deflection in thesystem. To eliminate the reaction force onthe disk, angular momentum balance is ap-plied about the center, yielding:

MG = IG k,(

T r1 + k2 r2 x2

)k = IG k.

Also, applying linear momentum balance tothe block yields:

F = mF

aG,(T k1 x1 m g

) = m x1 .

FR

G

m g

T

T

k2 x2

k1 x1

Finally, eliminating the unknown tension from these equations and using the abovecoordinate transformations, this single-degree-of-freedom system can be modeled withthe equation: (

IG +m r21

) +

(k1 r

21 + k2 r

22

) = (m g r1).

This equation of motion determines the equilibrium position eq (with eq = 0) to be:

eq =m g r1

k1 r21 + k2 r22

.

Therefore, the equilibrium displacements in each spring are:

x1,eq = r1 eq = m g r21

k1 r21 + k2 r22

, x2,eq = r2 eq = m g r1 r2k1 r21 + k2 r

22

.

b) Define new coordinates z1 and z2, which measure the displacement in springs 1 and 2with respect to the static equilibrium position, that is:

z1 = x1 x1,eq, z2 = x2 x2,eq.

Likewise, let represent the angular displacement of the disk from the static equilibriumposition:

= eq.

14

Therefore, the potential energy of this system can be written as:

V = 12k1 z

21 +

1

2k2 z

22 ,

=1

2

(k1 r

21 + k2 r

22

)2.

Also, the kinetic energy becomes:

T = 12IG 2 +

1

2m z21 ,

=1

2

(IG +m r21

)2.

If the system is released from rest at the unstretched position of the springs, then:

(0) = eq = m g r1k1 r21 + k2 r

22

, (0) = 0.

At this initial state, the potential and kinetic energies become:

T0 = 0, V0 = (m g r1)2

2(k1 r21 + k2 r22).

Because this system is conservative, the total energy, E = T + V remains constant.Therefore, when the kinetic energy is maximal, the potential energy is minimal, that is:

V1 = 0, T1 = 12

(IG +m r21

)2max.

Finally, conservation of energy implies that V0 = T1, and solving for max we find that:

max =

(m g r1)2

(IG +m r21)(k1 r21 + k2 r

22).

Problem 14: (Spring 2003)For the system shown to the right, the disk ofmass m rolls without slip and x measures thedisplacement of the disk from the unstretchedposition of the spring.


b) if the system is underdamped, what isthe frequency of the free vibrations ofthis system in terms of the parametersk, c, and m;

k

3k

cm

x

z

Solution:

a) We define the three coordinates as shown as the figure, related as:

x = r , z = 2 r , z = 2 x.

15

A free-body diagram for this system isshown to the right. Notice that the force inthe upper spring depends on z, rather thanx, while the friction force has an unknownmagnitude f . Because the disk is assumedto roll without slip, we are unable to specifythe value of f , but instead can relate the dis-placement and rotation of the disk throughthe coordinate relations above.

k x

3k z

c x

f

G

C

The equations of motions can be developed directly with angular momentum balanceabout the contact point, so that:

MC = IC k,(

(3k z) 2r + (k x) r + (c x) r)k =

3 m r2

2 k.

Finally, writing this equation in terms of a single coordinate, we obtain:(3 m r2

2

) + (c r2) +

(13 k r2

) = 0.

b) For an underdamped response, the frequency of oscillation is d = n

1 2. Withthis system, we find that:

n =

26 k

3 m, =

c78 k m

,

so that:

d =

26 k

3 m 2 c

2

9 m2.

Problem 15:

In the system shown to the right, the pulleyhas mass m and radius r, so that the moment

of inertia about the mass center is IG =mr2

2 .

a) Find the governing equations of mo-tion;

b) Find the frequency of oscillation for freevibrations of the system;

c) For what value of the damping constantis the system critically damped?

r

r2

m

m

k

kc

Solution:

16

a) We choose coordinates (x1, x2, ), where x1 measures the displacement of the first massin the direction, x2 measures the displacement of the second mass in the direction,and measures the angular rotation of the wheel in the k direction. The kinetic andpotential energies for this system are:

T = 12m x21 +

1

2m x22 +

1

2

mr2

22,

V = 12k x21 +

1

2k x22.

However, these three coordinates are dependent through the transformations:

x1 =r

2, x2 = r .

Therefore, the Lagrangian reduces to:

L = T V = 12

(7mr2

4

)2 1

2

(5kr2

4

)2.

Further, the generalized force resulting from the viscous damper becomes:

Q = cr2

4,

so that the equation of motion for this system can be reduced to:

(7m) + c + (5k) = 0.

b) For this system the natural frequency and damping ratio are:

n =

5k

7m, =

c

2

35 km.

Therefore, the damped natural frequency becomes:

d = n

1 2 =

5k

7m

1 c

2

140 km.

c) For a critically damped system, = 1, so that we may solve for c = ccr to yield:

ccr = 2

35 km.

Problem 16:



2 about themass center G, and is attached to a block ofmass m which rolls across the surface. If thedisk rolls without slip ( is sufficiently large),while the block moves without friction:

a) find the equations of motion for thissystem;

b) for m = 2 kg, b = 0.5 (N s)/m, andk = 8 N/m, find the frequency of oscil-lation for the system.

m

b

k

G

17

Solution:

a) Let x denote the translational displacement of G in the direction, while denotes the

angular displacement of the disk in the k direction. If T is the tension between the diskand the mass and f is the frictional force acting on the disk, the equations of motionon the disk and the mass are:

(kx bx+ T + f) = mx ,

(rf) k =mr2

2 k,

(T ) = mx .

In addition, x and are related through the kinematic constraint:

x = r.

Eliminating (, T, f) we obtain a single degree-of-freedom system on x of the form:

5m

2x+ bx+ kx = 0.

b) From the above equation we identify the undamped natural frequency and dampingratio as:

n =

2k

5m, =

b10km

.

Thus, the damped natural frequency is:

d = n

1 2 =

10km b225m2

.

For the given values of the parameters, this reduces to d = 1.264 rad/s.

Problem 17:

We model a nonuniform beam as a single-degree-of-freedom system in the form:

mx+ bx+ kx = 0,

and experimentally measure the mass as m =kg. In free vibration we experimentally de-

termine the equivalent spring constant to bek = 4 N/m, and we measure the response asshown.

a) Determine the equivalent damping con-stant.

b) What is the exponential decay rate ofthe transient solution?

bcbc

bcbc

x1 = 1.00 m

x2 = 0.75 m

Solution:

18

a) We use the logarithmic decrement so that:

=1

42 + 21, 1 = ln

(x1x0

)= ln

(0.75 m

1.00 m

)= 0.288,

and we find that = 0.0457. With m = 1 kg, the damping constant b is given as:

b = 2km = 0.183 N/(m/s).

We note that with m given, we can also find the period of the oscillations to be:

T =2

d=

2

n

1 2 ,

=

m

42 + 21k

= 3.145 s.

b) The exponential decay rate is = n, which is found to be = 0.0914 s1.

Problem 18:

The rigid beam (mass m = 2 kg, length l =1.5 m) is supported by an elastic spring (k =4 N/m) and damper (b = 2 N/(m/s)), and ispinned to the ground.

a) Find the linearized equations of motionin terms of z, the relative displacementbetween the end of the beam and theground;

b) what is the frequency of the resultingmotion;

c) if the mass of the spring is taken to bemspring = 1 kg, find the new frequencyof the oscillations.

z

k

b

(m, l)

Solution:

a) We will define the inclination of the bar from the horizontal position as , so that the

angular acceleration is k. Therefore, using angular momentum balance about thepoint of rotation, we find:

mbar l2

3 + bl z + kl z = 0.

We assume that the coordinates and z are related by z = l sin , which for smallrotations reduces to z = l. Using this constraint to eliminate , the equation of motionreduces to:

mbar3

z + b z + k z = 0.

For the parameter values given above, this becomes:(2

3kg

)z +

(2 N/(m/s)

)z +

(4 N/m

)z = 0.

19

For this system, the damping ratio and natural frequency can be expressed as:

n =

3k

mbar, =

3 b

2kmbar

=

6 rad/s =

3

8.

b) The frequency of oscillation, d = n

1 2, reduces to:

d =

3k

mbar(

3b

2mbar

)2=

15

2rad/s = 1.94 rad/s.

c) If the mass of the spring is considered, it is treated as an additional equivalent massmeq = mspring/3 located at the end of the bar. Therefore the new moment of inertia ofthe bar about the point of rotation is:

IO =mbar l

2

3+meq l

2,

=mbar l

2

3+(mspring

3

)l2,

=(mbar +mspring) l

2

3.

Therefore, the new frequency of oscillation is:

d =

3k

(mbar +mspring)(

3b

2(mbar +mspring)

)2=

3 rad/s = 1.73 rad/s.

Problem 19:

The non-uniform beam supports an end-massof m = 10 kg. If the response of the systemis such that:

x1 = 0.25 m x2 = 0.20 m,t1 = 1.00 s t2 = 4.00 s,

as shown in the figure, find the equivalentstiffness and equivalent damping of the beam(assume that the beam is massless).

m

bcbc

bcbc

(t1, x1)

(t2, x2)

Solution:

20

a) The logarithmic decrement is defined as = ln |x2/x1|. In terms of this quantity andthe period of oscillation T , the damping ratio and natural frequency are defined as:

n =

(2)2 + 2

T, =

(2)2 + 2

.

Therefore, the stiffness and damping constants reduce to:

k = m 2n = m(2)2 + 2

T 2, b = m 2n = m

2

T.

For this system, we find that T = 4 s and = 0.223, and the stiffness and dampingconstant reduce to:

k = 24.7 N/m, b = 1.12 N/(m/s).

Problem 20:

For the single-degree-of-freedom mechanicalsystem shown in the figure:


b) what are the damping ratio and un-damped natural frequency of this sys-tem;

c) find the response of the system x(t)subject to the initial conditions x(0) =x0, x(0) = 0,

when m = 1 kg, b = 12 (N s)/m, and k =9 N/m. Assume the pulley is massless andneglect the effects of gravity.

2 r

r

m

k

c

Solution:

a) Let x represent the displacement of mass m in the vertical direction and measure theangular displacement of the pulley, both measured from the static equilibrium position.If T represents the tension in the cable supporting mass m, then T = kr2 , where isthe angular displacement of the massless pulley from static equilibrium. In addition,we find that is related to the linear displacement of mass m as = x2r . As a result, inthe direction, the equation governing the motion of the mass is:

mx+ bx+k

4x = 0.

As a result, we find that the the equivalent spring constant is keq =k4 .

b) The damping ratio and natural frequency are simply:

=b

2keqm

=bkm

, n =

keqm

=

k

2m.

21

In terms of the given parameters, we find = 4.0 and n = 3 rad/s. We note that thedamping ratio has no units.

c) For an overdamped system, the general solution is:

x(t) = exp(nt)(c1 exp

((n

2 1)t

)+ c2 exp

((n

2 1)t

)),

and with these initial conditions this reduces to:

x(t) = x0 exp (nt)( +

2 1

22 1 exp

((n

2 1)t

)+

2 1

22 1 exp

((n

2 1)t

)),

= x0e6t

(4 +

15

2

15e

3

2

15t +

4152

15e

3

2

15t

),

= x0e6.0t (1.016e5.81t + 0.01640e5.81t) .

Problem 21:

For the spring-mass-damper system shown tothe right, x is measured from the static equi-librium position, and the surface is friction-less.

a) Determine the governing equations ofmotion.

b) What is the period of each oscilla-tion in terms of the system parameters(m, k, c)?

c) For what value of c is the system criti-cally damped?

d) If the system is released with the initialconditions:

x(0) = 0.0 m, x(0) = 5.0 m/s,

find the resulting solution x(t) if m =2 kg, k = 48 N/m and c = 4 N/(m/s).

x

m6 k 2 c

Solution:

a) In terms of x, the spring and damping forces can be written as:

Fspring = 6k x , Fdamper = 2c x .

Using linear momentum balance on the block, we find that:F = (6k x 2c x) = mx = mFaG,

22

and the equation of motion can be written in standard form as:

x+ (2n) x+ (2n) x = 0, with n =

6k

m,

=c6km

.

b) In terms of and n, the period of oscillation is simply T = (2)/d, where d =

n

1 2. Therefore, for this system, the period reduces to:

T =2

6k

m

1

(c6km

)2 = 2 m6km c2 ,

provided c2 < 6km.

c) For a critically damped system, = 1, and solving for c, yields:

ccritical =

6km.

d) With these parameter values, the equation of motion reduces to:

x+ 4 x+ 144 x = 0,

so that n = 12 rad/s, and = 1/6. For these values, the general solution can bewritten as:

x(t) = A e2 t sin(2

35 t+ ),

where A and are arbitrary constants used to fit the initial conditions. Solving for Aand , we find:

A =

5

28m, = 0 rad,

so that the general solution can be written as:

x(t) =

5

28e2 t sin

(2

35 t)

m.

23

Problem 22:

For the system shown to the right, x is mea-sured from the unstretched position of thespring. Each block has mass m and the diskhas moment of inertia I and radius r. If thegravitational constant is g:

a) find the equations of motion which de-termine x(t);

b) what is the period of the free oscilla-tions?

x

m

m

(I, r)k

Problem 23:



b) what are the damping ratio and un-damped natural frequency of this sys-tem;

c) what is the stretch in the spring whenthe system is in equilibrium?

k 2 k b

m

g

Solution:

We assume that the and directions are standard orthonormal basis in the horizontaland vertical directions respectively.

a) The forces due to the springs in parallel and damping are:

Felastic = 3k x, Fdamping = b x.

With the inclusion of the gravitational force, the equation of motion for this system canbe written:

F = m (x) ,3k x+ b xmg = mx.

Taking components in the direction, we obtain the governing equation of motion:

mx+ bx+ 3kx = mg.

24

b) Dividing through by the mass m, we find:

x+b

mx+

3k

mx = g,

so that:

2n =b

m, 2n =

3k

m,

which can be solved to yield:

n =

3k

m, =

b

2

3mk.

c) In equilibrium, the system is stationary, so that xeq = 0 and xeq = 0. Substitution intothe governing equations yields:

3kxeq = mg, xeq = mg3k

.

Problem 24:


a) find the linearized governing equationsof motion for small ;

b) find the frequency of oscillation for theresponse;

c) if the initial velocity is zero, and (0) =0, determine the time response of thesystem.

k

4 k

(m, l)G

Solution:

a) The displacement of each end of the bar in the direction is:

x = l sin 2

.

As a result, the total moment produced by the springs about the center of mass G is:

MG = l

2

4(4k sin + k sin ) k,

= 5kl2

4sin k.

Thus angular momentum balance about G provides:MG = I

Gk,

5kl2

4sin k =

ml2

12k.

25

For small angular displacements sin , and the governing equations of motion aretherefore:

+15k

m = 0.

b) This system is undamped. Therefore the frequency of the oscillation is equal to theundamped natural frequency:

= n =

15k

m.

c) This system possesses the general solution:

(t) = c1 sint+ c2 cost,

which, for the initial conditions given above, yields the solution:

(t) = 0 cos

15k

mt.

Problem 25:

We obtain the differential equation:mx+ bx+ kx = 0,

as a model for a spring-mass-damper system with:

m = 2, k = 18.

a) Identify the damping constant b that gives rise to critical damping;

b) If, instead, b = 24, approximately how long will it take for the the amplitude of free vibrationto be reduced to within 2% of zero?

Solution:

a) Written in nondimensional form, the equations of motion are:

x+b

mx+

k

mx = 0,

and so we identify the damping ratio and natural frequency as:

=b

2km

, n =

k

m.

= 1 corresponds to critical damping, so that:

bcritical = 2km = 12.

b) For b = 24 we find:

=b

2km

= 2, n =

k

m= 3.

26

Therefore the eigenvalues of this system are:

1,2 = n

(

2 1

)= 3(2

3).

The dominant eigenvalue is = 6 + 33, and so the time t = required for theamplitude of the free vibration to be reduced to within 2% of zero is:

46 + 33 5.0

Problem 26:

For the single-degree-of-freedom mechanicalsystem shown in the figure, the bar has mass

m and length l (so that IO =ml2

3 ). If thespring is unstretched when = 0:

a) find the linearized governing equationsof motion for small ;

b) find the frequency of oscillation for thefree response;

Neglect gravity.

2

2

k

c

(m, l)

Solution:

a) Using angular momentum balance about the fixed point O, we find:MO = IO/FF,(

kl2 sin c l2

4 cos

)k =

ml2

3k,

so that the equation of motion can be written as:

ml2

3 + c

l2

4 cos + k l2 sin = 0.

Linearizing this equation about = 0, we obtain:

+3c

4m +

3k

m = 0.

b) The frequency of oscillation, that is, the damped natural frequency, is given as:

d = n

1 2,

where n is the undamped natural frequency and is the damping ratio. For this systemwe find:

n =

3k

m, =

3 c

8km

,

27

so that the damped natural frequency is:

d =

3k

m(

3c

8m

)2.

Problem 27:



2 about themass center G.

a) Find the equations of motion for thissystem assuming that the disk rollswithout slip.

b) If the disk is released from rest withinitial displacement x(0) = x0, find theminimum value of the coefficient of fric-tion for which the disk does not slip.

(m, r)

G

x

k

c

g

Problem 28:

For the spring-mass-damper system shown tothe right, x is measured from the static equi-librium position. If the surface is assumed tobe frictionless:


b) what is the period of each oscillation;

c) what value of the damping coefficient bcorresponds to critical damping?

d) if k = 1 N/m and m = 4 kg, find thedisplacement of the mass x(t) if the sys-tem is critically damped and startedwith the initial conditions x(0) = 0,x(0) = x0;

x

m6 k b

Problem 29:

For the system shown to the right, x is mea-sured from the unstretched position of thespring. Each block has mass m and the diskhas moment of inertia I. If the gravitationalconstant is g:

a) what is the displacement of the springat static equilibrium;

b) find the kinetic energy of the system interms of the coordinate x (and/or itsvelocity).

x

m

m

(I, r)k

28

Problem 30:

In the system shown to the right, the pulleyhas mass m and radius r, so that the moment

of inertia about the mass center is IG =mr2

2 .


b) What is the equivalent mass of the sys-tem;

c) Find the frequency of oscillation for freevibrations of the system?

r2

r

m

m

k

kc

Solution:

a) Define as the angular displacement of the disk in the k direction (clockwise), andx1 and x2 as the displacements of the two blocks so that:

x1 =r

2, x2 = r.

We define the tension in the left and right cable as T1 and T2 respectively. Thus, linearmomentum balance on the two blocks, and angular momentum balance on the diskyield:

T1r

2 T2r = mr

2

2,

T1 kx1 cx1 = mx1,T2 + kx2 = mx2.

Eliminating the two unknown tensions from these three equations, we find that theequation of motion (on ) can be reduced to:(

7mr2

4

) +

(cr2

4

) +

(5kr2

4

) = 0.

b) Examination of the above equation shows that meq =7mr2

4 . Note that this answer isnot unique, but depends on the equation of motion. For example, if we had written theequation of motion as 7m + c + k = 0, the equivalent mass would be meq = 7m.

c) The frequency of oscillation is given by d, which reduces to:

d = n

1 2 =

5k

7m

1 c

2

140km.

29

Problem 31:

In the spring-mass-damper system shown,the block slides with no friction. With m =2 kg, k = 18 N/m, b = 13 (N s)/m:

a) Find the resulting solution if the sys-tem is released from the unstretchedposition with initial velocity x(0) =10.0 m/s

b) Identify the damping constant b thatgives rise to critical damping;

x

m

k

b

Solution:

For this system the differential equation of motion can be written as:

mx+ bx+ kx = 0,

x+b

mx+

k

mx = 0,

subject to specified initial conditions.

a) With the given values for the mass, stiffness, and damping coefficient, the above equationbecomes:

x+ (6.5 kg/s)x+ (9 kg/s2)x = 0,

whose characteristic equation reduces to:

2 +13

2+ 9 = 0,

which has two real solutions of the form:

=13 5

4.

Thus the system has two purely real eigenvalues and the resulting system is overdampedand decays exponentially with no sustained oscillations. The general solution is givenas:

x(t) = c1 e 9

2t + c2 e

2t.

For the initial conditions x(0) = 0 m, x(0) = 10.0 m/s, we find that:

c1 = 4 m, c2 = 4 m,

and the general solution to this equation, subject to these initial conditions, becomes:

x(t) =(4 m

) (e(9/2 s

1)t e(2 s1)t).

b) For this system, the damping ratio is:

=b

2km

.

30

Thus, for a critically damped system = 1, and solving for b with m = 2 kg andk = 18 N/m, we find:

bcr = 12 (N s)/m.

Problem 32:



2 about themass center G.


b) What value of b correspond to criticaldamping?

c) Find the displacement of the center ofthe disk when the system is criticallydamped and released from rest withx(0) = x0.

(m, r)

G

x

k

b

g

Solution:

We define as the angular displacement of the disk in the k direction as measuredfrom the equilibrium position of the disk (when the spring is unstretched). Assumingthat the disk rolls without slip, the rotation and translation of the disk can be relatedthrough the constraint equation:

x = r.

a) The frictional force, which is unknown, is defined as f = f , while the forces due to thespring and damper are:

Fspring = kx , Fdamper = bx .

Using linear and angular momentum balance on the disk, we find that:F = (f kx bx) = mx = mFaG,

MG = (fr) k =mr2

2 k = IG /FF.

Eliminating the unknown frictional force, and using the kinematic constraint, we findthe equation of motion is: (

3m

2

)x+ bx+ kx = 0.

b) We can write this differential equation in standard form, that is:

x+ (2n)x+ (2n)x = 0, with n =

2k

3m,

=b

6km.

31

If the system is critically damped, then this implies that the damping ratio is unity.Therefore, solving for b when = 1, we find:

bcritical =

6km.

c) When the system is critically damped, the general solution takes the form:

x(t) =(c1 + c2 t

)en t,

while this can be differentiated with respect to time to obtain the velocity:

x(t) =((c2 nc1) (nc2) t

)en t.

If the system is released from rest when a known initial displacement, then the initialconditions are (x(0), x) = (x0, 0). Thus, returning these to the general solution, we find:

x0 = x(0) = c1,

0 = x(0) = c2 nc1.

Solving for c1 and c2, the solution to these initial conditions becomes:

x(t) = x0(1 + n t

)en t,

where recall that n =

(2k)/(3m).

Problem 33:

In the figure, the disk has mass m, radius r,and moment of inertia IG about the mass cen-ter, and the applied moment has a constantmagnitude M k. If the disk rolls without slip( is sufficiently large):


b) what are the equivalent mass, stiffness,and damping of the system;

c) what is the stretch in the spring whenthe system is in equilibrium?

(m, r)

G

M kx

k

c

g

Solution:

a) The governing equations of motion are:(m+

I

r2

)x+ cx+ kx = M

r.

b) The equivalent mass, damping, and stiffness are:

meq = m+I

r2, ceq = c, keq = k.

32

c) When the system is in equilibrium, the displacement of the disk is:

xeq = Mkr

.

2 Frictionally Damped Systems

Problem 34:

The block shown to the right rests on a roughsurface with coefficient of friction and

m = 6 kg, k = 128 N/m.

a) If the block is displaced 3 cm to theright and released, for what values of will the block remain in that position?

b) With = 0.50, if the block is displaced30 cm to the right and released fromrest, how long will it take the block tocome to rest?

x

kk

m

g

z

A B

Solution:

In addition to the variable x identified in the problem statement, we also define z tobe the stretch in the spring parallel with the cable system. As a one degree-of-freedomsystem, the variables x and z are directly related. The relative velocity across the springcan be identified as

F

vB FvA = z ,= (x ) (x ) ,

so that z = 2 x. Therefore the kinematicrelationship becomes

z = 2x.

An appropriate free-body diagram for thissystem is shown to the right. Note that theunknown friction force is denoted as fr andthe tension in the cable is T . Finally, exam-ining the spring in the cable, the tension Tand the displacement z are related as

T = k z = 2 k x.

T

T k x

fr

m g

N

T T

Applying linear momentum balance to the block yieldsF = (2T k x+ fr) + (N mg) = mx = m FaG,

and in terms of x the equation of motion becomes

mx+ 5 k x = fr.

33

If the block slips then fr = mg sgn(x) while is sticking occurs |fr| mg.

a) If the block is in static equilibrium at a displacement x = xeq, then xeq 0 and theequation of motion reduces to

5 k xeq = fr,

so that equilibrium is maintained provided

|fr| = |5 k xeq| mg.

This inequality is satisfied provided

|xeq| mg5 k

.

Problem 35:

For the system shown to the right, the blockslides on a rough surface (coefficient of fric-tion ) inclined at an angle of with respectto vertical. If the block is subject to a peri-odic force of the form

F (t) = F0 sin( t),


b) find the amplitude of the steady-stateresponse using Mc when

m = 1.25 kg, k = 20 N/m,

= 30, = 0.125,

F0 = 4 N, = 2.00 rad/s,

k

mF (t)

g

34

Problem 36:

[(Spring 2003)] The system shown in the fig-ure has mass m and rests on a plane inclinedat an angle . The coefficient of friction forthe rough surface is and the system is re-leased from rest at the unstretched positionof the spring (with stiffness k).

a) If = 0, what is the equilibrium dis-placement of the mass (as measuredfrom the unstretched position)?

b) For > 0, at what angle does theblock begin to slip?

c) Find the value of so that the systemcomes to rest after one full cycle exactlyat the equilibrium position of the sys-tem found in part a (so that the frictionforce vanishes when the system comesto rest), with:

m = 2 kg, k = 32 N/m,

= 0.35,

m

k

x 2

2

Solution:

The unit directions 2 and 2 are defined to be coincident with the inclined plane andthe coordinate x represents the displacement of the mass from the unstretched positionof the spring, as shown in the figure.

A free-body diagram for this system isshown to the right. Notice that the force inthe upper spring depends on z, rather thanx, while the friction force has an unknownmagnitude f . Because the disk is assumedto roll without slip, we are unable to specifythe value of f , but instead can relate the dis-placement and rotation of the disk throughthe coordinate relations above.

k x 2

f 2N 2

m g

Therefore, linear momentum balance yields the following equations:F = m

F

aG,(f k x

)2 +

(N)2

(m g

) = m x 2,(

f k x+m g sin )2 +

(N m g cos

)2 = m x 2.

Therefore, this leads to the following scalar equations in the 2 and 2 directions:

m x+ k x = f +m g sin ,

N = m g cos .

a) If = 0, then the friction forces vanishes and the first of the above equations reduces

35

to:m x+ k x = m g sin .

The equilibrium displacement of the mass, xeq, then can be found to be:

xeq =m g

ksin .

b) With 6= 0, an equilibrium state is maintained provided:f N,k xm g sin m g cos .Therefore, if the system is released from x = 0, the block begins to slide when:

tan = .

c) Define z to be a new coordinate measuring the displacement of the system from staticequilibrium:

z = x m gk

sin ,

so that the equations of motion become:

m z + k z = f, N = m g cos .

with initial displacement z(0) = (m g sin )/k. Over one complete cycle of motion,for a frictionally damped system the amplitude decreases by a value:

A = 4 Nk

.

Therefore, if the system comes to rest at exactly the equilibrium position, then thisdecrease in amplitude must exactly match the initial displacement. That is:

| A| =4 Nk

=m g

ksin

= |z(0)|.Solving for :

tan = 4 .

Problem 37:

The spring mass system rests on a surfacewith coefficient of friction and x is mea-sured from the unstretched position of thespring.. If the initial conditions of the sys-tem are chosen to be x(0) = 0 and x(0) = x0,find the range of x0 so that the system comesto rest after exactly one cycle of motion.

gx

mk

36

Problem 38:

For the spring-mass system with Coulombdamping:


b) what is the period of each oscillation.

x

mk k

Solution:

a) We measure the displacement of the mass from the static equilibrium of the frictionlesssystem, i.e., = 0, so that the acceleration of the block is

F

aG = x. Thus linearmomentum balance yields:

mx = Fspring + f+ (N mg).

The spring force is Fspring = 2kx, while the force due to sliding friction opposes thevelocity and is simply:

f = mg x|x| ,

since the normal force balances the gravitational force, i.e., N = mg. If the blockis stationary the magnitude of the frictional force is less than mg. Therefore, thegoverning equations of motion are:

mx+ 2kx = f, withf = mg x|x| |x| 6= 0,|f| mg x = 0.

b) Coulombic damping does not effect the frequency of oscillation, which is simply:

=

2k

m.

Therefore the period of the oscillation is:

T =2

=

2m

2k.

37

Problem 39:

For the spring-mass-damper system shown tothe right, x is measured from the static equi-librium position. If the coefficient of frictionis :


b) what is the period of each oscillation;

c) if the system is released from rest withx(0) = x0 > 0, what is the minimumvalue of x0 so that the block slips;

d) find the range of initial displacementsso that the system comes to rest afterone complete cycle.

x

m6 k 2 k

Problem 40:

For the spring-mass system with Coulombicdamping, x is measured from the unstretchedposition of the spring. If the coefficient offriction is and the gravitational constant isg:


b) if the system is released from rest in theunstretched position (x(0) = 0, x(0) =0), for what values of will the systemmove;

c) what is the displacement (from the un-stretched position) of the upper blockwhen it first comes to rest?

x

m

m

I = 0k

Solution:

Notice that x describes the displacement of both masses and, since the pulley is massless,the tension in the string connecting the masses is constant, say T . Also, notice that inpart c we ask for the displacement from the unstretched position of the spring, ratherthan from equilibrium. Therefore we include the gravitational force which will influencethis result.

a) With the frictional force defined as F = f , linear momentum balance on the upperand lower block yields:

mx+ kx = f + T,

mx = mg T,

where the frictional force is defined as:

f = mg x|x| , x 6= 0,|f | mg, x = 0.

38

Eliminating the unknown tension T , the equation of motion is given as:

2m x+ kx = f +mg,

where f is defined as above and depends on the motion of the system, that is, the valueof x.

b) If the system is released from rest in the unstretched position, it will remain thereprovided the magnitude of the frictional force is less than mgthe transition to move-ment occurs when |f | = mg. Thus the system does not move is |f | mg and x = 0.Substituting these conditions into the equations of motion we find:

|f | = |kxmg| mg,

which, solving for with x(0) = 0, implies that the system does not move if 1.Therefore, the system does move when:

< 1.

c) The displacement of the upper block when it first comes to rest is:

x1 =2(1 )mg

k.

This can be found by either solving the equations of motion explicitly, or through awork-energy analysis. Since the initial and final kinetic energy is zero, the work doneby the frictional force balances out the change in potential energy from the spring andgravity.

Problem 41:

For the spring-mass system with Coulombicdamping, x is measured from the unstretchedposition of the spring. If the coefficient offriction is and the gravitational constant isg:


b) if the system is released from rest, sothat x(0) = 0, for what range of initialdisplacements (from the unstretchedposition) will the block come to restwhen the block first comes again to rest(x(t1) = 0 for t1 > 0)?

gx

mk

Solution:

a) The equations of motion can be written as:

mx+ kx = f,

39

where f is the force due to friction, modeled by Coulombs law of friction as:

f = mg x|x| , x 6= 0,

|f | mg, x = 0.

b) If the system is released from rest, the initial displacement must be sufficiently large sothat the block slides, rather than remaining at rest. Sliding does not occur if the forcedue to friction is sufficient to balance the elastic force, that is, mg f = kx(0). Thus,solving for x(0) we find, that for sliding to occur:

|x(0)| > mgk

.

However, if |x(0)| is too large, the system will undergo multiple reversals as the ampli-tude of the motion decays. Consider the block sliding to the left (x < 0), released fromrest with initial displacement x(0) > mgk . Thus the equation of motion becomes:

mx+ kx = mg,

which has the general solution:

x(t) =

x(0) mg

k

cos( k

mt

)+mg

k.

Therefore, when the block comes again to rest at time t1 (unknown), the mass is at theposition:

x(t1) = 2mg

k x(0).

At this point, the block sticks if and only if |x(t1)| mgk . Therefore, solving for x(0),we find that:

x(0) 3mgk

.

So for a block with x(0) > 0, the allowable range for x(0) is mgk < x(0) 3mgk .Together with an identical argument for x(0) < 0 yields the total allowable range as:

mg

k< |x(0)| 3mg

k.

40

Problem 42:

For the system shown to the right, x is mea-sured from the unstretched position of thespring. Each block has mass m and the diskhas moment of inertia I and radius r. Thecoefficient of friction between the upper blockand the table is . If the gravitational con-stant is g:

a) find the equations of motion which de-termine x(t);

b) what is the minimum value of so thatthe system slips when release from restwith x(0) = 0;

c) what is the period of the free oscilla-tions?

d) if the system is released from rest, whatis the range of initial displacementsx(0) so that the systems comes to restafter exactly one complete cycle?

x

m

m

(I, r)k

Solution:

We begin by defining two additional coordinates, , which describes the rotation ofthe disk in the k direction (clockwise), and y which measures the displacement ofthe hanging mass in the direction. These additional coordinates are related to thedisplacement of the upper mass by the constraint equations:

y = x, =x

r.

a) On each mass the equations of motion can be written as:

F =

( k x+ T1 + f

)+

(N mg

) = mx = m

F

aG1 ,MO =

(T1r T2r

)k = IO k = IOD/FF,

F =(T2 mg

) = my = mFaG2 ,

Notice that IO = I 6= 0, so that provided 6= 0 the tensions T1 and T2 are not equal.Taking the components of these equations and eliminating the unknowns (T1, T2), whileusing the constraint equations, we find that the equation of motion for this systemreduces to: (

2m+I

r2

)x+ k x = f +mg,

where f the value of the frictional force in the direction, can be written as:

f =

{ mg x|x| , x 6= 0f0, |f0| mg, x = 0.

41

b) The minimum value for slip is simply min = 0. If we would like to find the range of for which slip occurs, we resort to the value of f at static equilibrium. Assuming(x, x) = (0, 0), the equations of motion reduce to:

kx = fstatic +mg,

where fstatic represents the force required to maintain static equilibrium. Solving forthis quantity and using the frictional inequality, we find:

|fstatic| = |kx0 mg| mg.

Therefore, solving for yields:

kx0mg 1

=1 kx0mg

,which provides a necessary condition for sticking at x = x0. So for sliding to occur forx0 = 0, this implies that < 1.

c) The period of oscillation for a frictionally damped system is identical to that of anundamped system. Therefore:

T =2

n= 2

2m+ Ir2

k.

d) Let describe the displacement of the system from equilibrium. The amplitude ofoscillation will decay by a value of = 4mg/k over one cycle of motion. Therefore,Since the system will come to rest within the range:

mgk

< |final| mgk

,

the initial displacement 0 from the equilibrium in the absence of friction must be inthe range:

mgk

+ =3mg

k< |0| 5mg

k=

mg

k+ .

However, the equilibrium position corresponds to xeq =mgk , and so the allowable range

of x0 is:3mg

k 3 rad/s;

b) If, for the isolator that you designed,the damping ratio was measured to be = 0.125 (rather than = 0 as as-sumed above), what is the minimumisolation achieved over this frequencyrange?

f(t)

m

43

Problem 45:

In the figure shown to the right, the disk issubject to a time dependent moment of theform

M(t) = M0 sin( t).

a) Find the equations of motion for theangular displacement of the disk.

b) With

k = 280 N/m, b = 12 N/(m/s),

m = 4 kg,

I = 0.40 kg m2, r = 0.10 mM0 = 3 N m, = 5 rad/s,

Determine the steady-state response ofdisk as a function of time.

r

r/2

M(t) k

k b

k

m

I

Problem 46:

The unbalanced rotor shown in the figure ispinned to a frame and supported by a springand damper. The total mass of the structureis m, with a small rotating component (10%of the total mass) offset by a distance r fromthe center of rotation C.

a) Find the distance between the centerof mass of the system and the center ofrotation;

b) what is the damped natural frequency;

c) determine the steady-state amplitudeof vibration when the rotor spins at anangular speed of = 5 rad/s with:

k = 256 N/m, b = 12 N/(m/s),

m = 5 kg, r = 10 cm

m

r

C

k b

44

Problem 47:



2 about themass center G and is assumed to roll withoutslip. The attached plate undergoes harmonicmotion of the form

u(t) = u0 sin( t).

a) Find the equation of motion in terms ofthe displacement between the movingplate and the center of the disk;

b) What are the damping ratio and natu-ral frequency for this system?

c) If the system is critically damped, findthe amplitude of the relative displace-ment of the disk for

m = 3 kg, r = 0.10 m

k = 36 N/m, c = 3 N/(m/s)

u0 = 0.05 m, = 5 rad/s

(m, r)

G

u(t)

k

c

g

Problem 48:

For the mechanical system shown to theright, the uniform rigid bar is massless andpinned at point O while a force is applied atA of the form

f(t) = t e t.

For this system:



c) With

m = 2 kg, = 30 cm,

c = 0.25 N/(m/s), k = 50 N/m,

= 2.00 s1,

find the convolution integral for the re-sponse of the system. You need notevaluate the integral.


O

3

2 3

k

f(t)

A

m

k c

B

45

Problem 49:

The block shown to the right rests on a roughsurface with coefficient of friction (assumethat any cables can support compression andtension). Find the amplitude of the vibra-tions of the block if f(t) = f0 cos( t), with

m = 4.0 kg, k = 64 N/m,

= 0.05,

f0 = 40 N, = 4 rad/s.

f(t)

x

km

m

Problem 50:

The unbalanced rotor shown in the figure ispinned to a frame and supported by a springand damper. The total mass of the structureis m, with a small rotating component (10%of the total mass) offset by a distance r fromthe center of rotation C.

a) find the damped natural frequency;

b) what is the steady-state amplitude ofvibration when the rotor spins at an an-gular speed of = 5 rad/s with:

k = 108 N/m, b = 9 N/(m/s),

m = 3 kg, r = 7.5 cm

m

r

C

k b

46

Problem 51:

For the mechanical system shown to theright, the uniform rigid bar has mass m andlength , and is pinned at point O. A har-monic force is applied at A. For this system:



c) For:

m = 6 kg, = 25 cm,

c = 0.50 N/(m/s), k = 80 N/m,

f0 = 2.00 N, = 10 rad/s,

find the steady-state amplitude of thedisplacement of the block.


O

2

k

f0 sin( t)

A

4m

c

B

Problem 52:



2 about themass center G and is assumed to roll withoutslip. The attached plate undergoes harmonicmotion of the form

u(t) = u0 sin( t).

a) Find the equation of motion in terms ofthe angular rotation of the disk;

b) What are the damping ratio and natu-ral frequency for this system?

c) If the system is critically damped, findthe amplitude of the rotation of the diskfor

m = 3 kg, k = 36 N/m,

u0 = 10 cm, = 3 rad/s

(m, r)

G

u(t)

k

c

g

47

Problem 53:

The unbalanced rotor shown in the figure ispinned to a frame and supported by a springand damper. The total mass is measured asm = 200 kg. When the system is operated at = 25 rad/s the phase of the response withrespect to the rotation of the unbalanced diskis measured to be /2 rad and the steady-state vibration amplitude is X = 2.00 cm.When the rotation rate of the disk is muchlarger than this value the amplitude reducesto X = 0.50 cm.Find the stiffness and damping constant forthe foundation and the distance between thecenter of rotation C and the mass center G.

m

C

G

k b

Problem 54:

For the mechanical system shown to theright, the uniform rigid bar is massless andpinned at point O while a harmonic force isapplied at A. For this system:



c) For:

m = 2 kg, = 30 cm,

c = 0.25 N/(m/s), k = 50 N/m,

f0 = 2.00 N, = 10 rad/s,

find the steady-state displacement ofthe block.

d) What is the magnitude of the forcetransmitted to the ground through thesping and damper attached to theblock? (Do not include the spring at-tached at A.)


O

3

2 3

k

f0 sin( t)

A

m

k c

B

48

Problem 55:

The block shown to the right rests on a roughsurface with coefficient of friction and theblock is subject to a compressive force of N =20 N (do not include gravity, just this normalload and assume that any cables can supportcompression and tension).

a) If f(t) = f0 = constant, find the rangeof initial displacements for which theblock will remain stationary if releasedfrom rest (it will stick).

b) Find the amplitude of the vibrations ofthe block if f(t) = f0 cos( t), with

m = 4.0 kg, k = 64 N/m,

r = 12.5 cm, = 0.50,

f0 = 40 N, = 4 rad/s.

f(t)

(m, r)

x

km

Problem 56:

For the system shown to the right, the disk ofmass m rolls without slip and x measures thedisplacement of the disk from the unstretchedposition of the spring. The surface is inclinedat an angle of with respect to vertical.


b) what is the static equilibrium displace-ment of the disk?

c) if the disk is subject to a periodic mo-ment

M (t) = M0 sin( t)k,

find the displacement of the center ofthe disk if it is released from rest at theunstretched position of the spring.

xk

(m, r) M (t)

49

Problem 57: (Spring 2003)For the system shown to the right, the disk ofmass m rolls without slip and the inner hubhas radius /2.

a) Find the equations of motion (in termsof the given parametersdo not substi-tute in numerical values yet);

b) If the applied moment takes the form:

M(t) = (2 N m) sin(4 t),

find the steady-state amplitude of thetranslation of the center of the diskwhen:

k = 16 N/m, b = 2 N/(m/s),

m = 2 kg, = 0.125 m

c) Determine the steady state amplitudeof the friction force.

x

z

C

Gb

2 k

kM(t) (m, )

Solution:

a) We identify the three coordinates x, z, and as shown in the figure above. These arerelated as:

x = , z = 32x.

An appropriate free-body diagram for thissystem is shown to the right. Since the diskis assumed to roll without slip, the equationof motion can be directly obtained with an-gular momentum balance about the contactpoint C

MC = IC D/F ,

b x

2 k z

k x

M(t) k

fr

N

mg

which yields ( (k x+ b x) +

3

2(2 k z) +M(t)

)k =

3m2

2 k.

Using the above coordinate transformations this equation can be written as

3m

2x+ b x+

11 k

2x = M(t)

.

b) For the numerical values given above (with consistent units), this equation reduces to

3 x+ 2 x+ 88 x = 8 sin(4 t),

50

from which we can identify the appropriate parameters as:

n =

88

3, =

1264

, r =

6

11.

Therefore, the amplitude of the translational oscillations becomes

X =F

kM(r, ) = 8

88

1(1 611

)2+(2 1

264

611

)2 = 126 .

Likewise, the phase shift of the response is:

tan =2 r

1 r2 =2 1

264

611

1 611=

1

5,

so that = 0.20 rad = 11.3.

c) In the development of the equation of motion, the friction force was eliminated bysumming moments about C. Using linear momentum balance we can reintroduce thefriction force as

F =(fr k x 2 k z b x

)+

(N mg

)= mx = m

F

aG.

Therefore, solving for fr we find that

fr = mx+ b x+ 4 k x.

With x(t) represented as x(t) = X sin( t ), where X and are found above, thefriction force becomes

fr(t) =(4 k m2

)X sin( t ) +

(b )X cos( t ).

The magnitude of the friction force is then found to be

f = X(

4 k m2)2

+(b )2.

For these parameter values f = 6.47 N.

51

Problem 58: (Spring 2003)The unbalanced rotor shown in the figure ispinned to a frame and supported by a springand damper. If the total mass is m while themass center G is located at an eccentricity of from the the center of rotation O,


b) what is the steady-state amplitude ofvibration when the rotor spins at an an-gular speed of = 8 rad with:

k = 32 N/m, b = 16 N/(m/s),

m = 4 kg, = 2.5 cm

c) If the system is undamped (i.e., b =0 N/(m/s)) for what range of operat-ing speeds () will the amplitude of theforce transmitted to the ground FT beless than 1 N.

m

O

G

k b

Problem 59: (Spring 2003)For the mechanical system shown to theright, the uniform rigid bar has mass m andpinned at point O. For this system:

a) find the equations of motion (in termsof the given parametersdo not substi-tute in numerical values yet);

b) if c = 0.25 N/(m/s), k = 32 N/m,m = 2 kg, and = 0.25 m, find theamplitude of the force transmitted tothe ground through combination of thespring and damper when = 4 rad/s.

c) if c = 0, m = 2 kg, and = 0.25 m,find the value of the stiffness k so thatthe bars amplitude of oscillation is lessthan /6 rad for all forcing frequenciesgreater than 20 rad/s.

2

2

mk

c

z

f(t) = sin( t)

Solution:

a) In addition to , we define the additional coordinate z, which measures the deflectionat the left end of the bar, with and z related as:

52

z =

2

A free-body diagram for this system isshown to the right. Applying angular mo-mentum balance on the bar eliminates theappearance of the reaction force and leadsto:

MG = IG k,(

f(t) k z cz)

2k =

m 2

12 k.

G

FR

k z1

c z1

f(t)

Solving for z in terms of , the equation of motion becomes:

m 2

12 +

c 2

4 +

k 2

4 =

2f(t).

b) In standard form, this equation of motion can be written as:

+

(3 c

m

) +

(3 k

m

) =

(6

m

)sin( t),

so that:

n =

3 k

m, =

3 c

2k m

, M0 =6

m .

The amplitude of the moment transmitted to the ground can be written as:

MT = (meq M0)

1 + (2 r)2

(1 r2)2 + (2 r)2 ,

=

2

1 +

(c k

)2(1 m 23 k

)2+(c k

)2 = 2k2 + (c )2(

k m 23)2

+ (c )2.

The amplitude of the force transmitted to the ground is then FT = MT /(/2), or:

FT =

k2 + (c )2(

k m 23)2

+ (c )2.

Substituting in the numerical values given in the problem statement, we find that:

FT = 1.50 N

c) The amplitude of the steady-state vibrations can be written as:

=M02n

1(1 r2)2 + (2 r)2 ,

=2

k

1(1 m 23 k

)2+(c k

)2 = 2 1(k m 23

)2+ (c )

2.

53

Substituting in the numerical values given in the problem statement, we find that:

=8(k 2 23 )

Therefore, if the amplitude of vibration is less than /6:

8k 2 23

6,

48

k 2 23 .

This inequality has two solutions:

k 48

+2 2

3, k 2

2

3 48

.

Since this condition must be satisfied for all 20 rad/s, we take the second inequalityand find that:

k 2 (20)2

3 48

= 251.

Problem 60: (Spring 2003)For the system shown to the right, the disk ofmass m rolls without slip and x measures thedisplacement of the disk from the unstretchedposition of the spring.


b) if the forcing takes the form:

f(t) =

{f0, 0 t < t0,f0/2, t0 t,

find the response of the system withzero initial conditions.

k

3k

f(t)

m

x

z

Solution:

a) We define the three coordinates as shown as the figure, related as:

x = r , z = 2 r , z = 2 x.

A free-body diagram for this system isshown to the right. Notice that the forcein the upper spring depends on z, ratherthan x, while the friction force has an un-known magnitude fr. Because the disk isassumed to roll without slip, we are unableto specify the value of fr, but instead canrelate the displacement and rotation of thedisk through the coordinate relations above.

k x

3k z

f(t)

fr

G

C

54

The equations of motions can be developed directly with angular momentum balanceabout the contact point, so that:

MC = IC k,(

(3k z) 2r + (k x) r f(t) r)k =

3 m r2

2 k.

Finally, writing this equation in terms of a single coordinate, we obtain:(3 m r2

2

) +

(13 k r2

) = r f(t).

In standard form:

+

(26 k

3 m

) =

2 f(t)

3 m r.

b) We use the convolution integral to determine the response, so that:

(t) =

t0

F () h(t ) d,

and for this system:

F (t) =f(t)

meq=

2 f(t)

3 m r, h(t) =

1

nsin(n t) =

3 m

26 ksin

(26 k

3 mt

).

Because the forcing function changes abruptly at t = t0, the solution must be writtenseparately for 0 < t t0, and t > t0:

x(t) =2 f03 m

t0

sin(n (t ))n

d, 0 < t t0,

x(t) =2 f03 m

t00

sin(n (t ))n

d +f0

3 m

tt0

sin(n (t ))n

d, t > t0.

Evaluating these integrals, the solution becomes:

x(t) =2 f026 k

(1 cos(n t)

), 0 < t t0,

x(t) =f0

26 k

(1 + cos(n(t t0)) 2 cos(n t)

), t > t0.

55

Problem 61: (Spring 2003)The system shown to the right is subject tobase excitation. Find the steady-state re-sponse of the system in terms of z, with:

m = 2.0 kg, b = 4.0 N/(m/s),

k1 = 3.00 N/m, k2 = 12.00 N/m.

u(t) = 0.50 sin(2 t) m

z

k1 k1

k2

b

m

x

Solution:

a) We define the addition coordinate x which measures the absolute displacement of themass with respect to the ground, so that:

x = z + u(t).

Notice that the collection of springs can bereplaced by a single equivalent spring, with:

keq =1

12 k1

+ 1k2=

2 k1 k22 k1 + k2

= 4 N/m.

The new equivalent system is shown to theright. u(t)

zkeq b

m

An appropriate free-body diagram is shownto the right. In terms of the identified coor-dinates, the acceleration of the mass centeris:

F

aG = x = (z + u) ,

with u(t) = (u0 2) sin( t).

keq z b z

Therefore, linear momentum balance on the mass yields:F = m

F

aG,( keq z b z

) = m x

Writing this in terms of z, the equation of motion is:

m z + b z + keq z = m u(t),

and in standard form:

z + 2 n z + 2n z = u0

2 sin( t),

56

with:

n =

keqm

, =b

2keq m

.

Therefore the steady state response of this system becomes:

z(t) = Z sin( t ),

with Z = u0 (r, ), and:

=r2

(1 r2)2 + (2 r)2 , tan =2 r

1 r2 , and r =

n

For the numerical values of this problem:

n =

2, =12, r =

2,

so that:

=25, tan =

2

1

Recall that the phase shift must be positive, so that is tan is negative, then is inthe second quadrant, so that = 3.03 rad. Finally:

z(t) =15

sin(2 t 2.03).

Problem 62:



2 about themass center G, subject to an applied momentof the form:

M (t) = M0 sin(t)k.

a) If the coefficient of friction is , find thecondition which determines if the diskrolls with or without slip, and find thegoverning equations of motion when thedisk rolls with and without slip;

b) If is sufficiently large so that thedisk rolls without slipping, what are theequivalent mass, stiffness, and dampingof the system;

c) If again the disk rolls without slipping,find the amplitude of the steady-statemotion of G, that is, x(t), when the sys-tem is critically damped, with M0 =4 N m, m = 1 kg, r = 0.1 m, andk = 2 N/m.

(m, r)

G

M (t)x

k

c

g

57

Problem 63:

In the figure, the disk has mass m, radius r,and moment of inertia IG = m about themass center G. The disk is subject to a time-varying force f(t) = 4 sin( t).


b) After the transient solution decays, findthe amplitude of the force transmit-ted to the ground through the spring-damper element.

c) For what value of the damping ratio isthis transmitted force less than twicethe applied load for all values of theforcing frequency?

(m, r)G

f(t)

x

k

c

g

Problem 64:

The disk shown in the figure rolls withoutslip and is subject to a time-varying momentM(t) = sin(t).


b) Find the frequency of oscillation for theunforced response, i.e. M(t) = 0;

c) What is the steady-state amplitude ofthe forced response?

d) Determine the amplitude of the fric-tional force during the steady-state mo-tion.

(m, r)

G

M(t) kx

k

b

g

Solution:

a) We assume that (, , k) represent the standard orthonormal basis, while the transla-

tional displacement of G is x and the angular displacement is k. Linear and angularmomentum balance on the disk yield:

mx = kx bx f,mr2

2 = fr M(t),

where f is the unknown frictional force. If the disk rolls without slipping, we find thekinematical relation x = r, and, eliminating f from the above balance laws, we findthe governing equation of motion can be expressed as:

3m

2x+ bx+ kx =

M(t)

r.

b) For M(t) = 0, the frequency of oscillation is the damped natural frequency d =

58

n

1 2, where:

n =

2k

3m=

2

3, =

b6km

=12,

so that the damped natural frequency is d =13

= 0.577.

c) In nondimensional form, the system is represented as:

x+ 2nx+ 2nx = F0 sin(t),

where and n are given above, and:

F0 =2

3m=

2

3.

Recall that the magnification factor for a harmonically driven system is:

M = 1(1 r2)2 + (2r)2 ,

where r = n is the frequency ratio. For this system we find r =

32 , so that M = 213 .

As a result, the steady-state amplitude is:

A =F02nM = 2

13= 0.555.

d) The response of the disk is x(t) = A sin(t+ ), where A is given above and:

tan =2r1 r2 =

3 12

.

From the expression for angular momentum balance, we find:

f =m

2x M(t)

r,

= 113

sin(t+ ) sin(t),

= [(

113

sin

)cos(t) +

(113

cos+ 1

)sin(t)

],

where cos = 113

and sin = 2

313

, so that the amplitude of the frictional force is:

|f| = 413.

59

Problem 65:



2 about themass center G, subject to an applied momentof the form:

M (t) = M0 sin(t)k.

If the disk rolls without slip ( is sufficientlylarge):

a) find the equations of motion for thissystem;

b) for m = 2 kg, b = 0.5 (N s)/m, andk = 8 N/m, find the steady-state am-plitude of the rotation of the disk?

(m, r)

G

M (t)x

k

b

g

Solution:

In addition to x, defined in the figure, we define as the angular displacement of thedisk from the unstretched position in the k direction. If the disk rolls without slip, xand are related as:

x = r.

a) The frictional force, which is unknown, is defined as f = f , while the forces due to thespring and damper are:

Fspring = kx , Fdamper = bx .

Using linear and angular momentum balance on the disk, we find that:F = (f kx bx) = mx = mFaG,

MG = (M0 sin(t) + fr) k =mr2

2 k = IG /FF.

Eliminating the unknown frictional force, and using the kinematic constraint, we findthe equation of motion is:(

3m

2

)x+ bx+ kx = M0

rsin(t).

b) For the given values of the parameters, we find that:

2n =8

3, =

1

8

6, F0 = M0

3r.

Thus, the steady-state amplitude may be easily found as:

X =F02nM,

=M03r(

83 2

)2+(6

)2 .

60

Problem 66:

The mass m = 2 kg is supported by an elas-tic cantilever beam attached to a founda-tion which undergoes harmonic motion of theform:

u(t) = 4 sin(t) m,

If the beam has length l = 20 cm, whileAE = 16 N:

a) find the equations of motion in termsof z, the relative displacement betweenthe mass and the foundation (assumethe beam has zero mass);

b) what is the amplitude of the result-ing motion in terms of the forcing fre-quency ?

u(t) = 4 sin( t) m

z

m

x

Solution:

a) The equivalent spring for this cantilever beam is:

keq =AE

l=

16 N

0.2 m= 80 N/m.

The acceleration of the block with respect to the ground isF

aG = (u + z), so that, interms of z, the equations of motion become:

mz + keq z = mu,z + 40z = 42 sin(t).

b) For this undamped system, the amplitude of the resulting steady-state motion is:

X =42

40

1(1 240

)2 ,=

42

|40 2| .

61

Problem 67:

The unbalanced rotor shown in the figure ispinned to a frame and supported by a springand damper. If the total mass is m while themass center G is located at an eccentricity of from the the center of rotation O,


b) what is the steady-state amplitude ofvibration when the rotor spins at thisangular speed.

m

O

G

k b

x

Solution:

We define x(t) as the vertical displacement of the geometric center of the rotor asmeasured from static equilibrium. As a result, the mass center G is described by theposition z(t) = x(t) + sin(t). Note that measures the eccentricity of the masscenter, not the location of the mass imbalance. Consequently, the governing equationsof motion can be written:

mz = kx bx,mx 2 sin(t) = kx bx,

or in more standard form:

x+b

mx+

k

mx = 2 sin(t).

a) In the above system we find n =

km and =

b2km

, so that the damped natural

frequency can be written:

d = n

1 2 =

k

m b

2

4m2.

b) For an arbitrary forcing frequency the amplitude of oscillation is A = , where:

=2

(2n 2)2 + (2n)2,

which, with = d, reduces the amplitude to:

A = 1 2

4 32 ,

with defined above.

62

Problem 68:

The system shown in the figure is supportedby a foundation that undergoes harmonic mo-tion of the form:

u(t) = 4 sin(t) m,

If the mass and stiffness are m = 2 kg, andk = 8 N/m:

a) find the equations of motion in termsof z, the relative displacement betweenthe mass and the base (assume thespring has zero unstretched length);

b) what is the amplitude of the resultingmotion in terms of the frequency ratior;

c) for what forcing frequencies is the re-sulting amplitude of the steady-statemotion Z 8?

u(t) = 4 sin( t) m

zk

mg

x

Solution:

a) We construct a free-body diagram as shown. The only forces acting on the mass arisefrom the gravitational force and the spring force.

However, the acceleration of the mass with respect tothe ground is written as:

F

aG =(u+ z

).

Therefore, linear momentum balance takes the form:F = m

F

aG,( kz mg

) = m

(z + u

).

k z

m g

Substituting in the expression for u(t), and taking the component in the direction, weobtain the governing equation of motion:

mz + kz = mg mu,z + (4 s2)z = 9.81 m/s2 + 42 sin(t) m,

Notice that each term has units of acceleration, that is, m/s2. In what follows we willdispense with the explicit inclusion of the units. For this system n = 2, = 0, and theamplitude of the forcing is F = 42. Thus we have frequency-squared excitation.

b) Using the above values for the natural frequency and the damping ratio, we find that

63

the forced response can be written as z(t) = Z sin(t+ ), where the amplitude Z is:

Z = U (r, ) = U r2

(1 r2)2 + (2r)2 ,

=4r2

|1 r2| .

c) If Z < 8 then this implies that:

Z

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