33726671 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 ζ = ζcr/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 ℓ.

c) For:

m = 1.50 kg, ℓ = 45 cm,

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

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

θ(0) = 0, θ(0) = 10 rad/s.

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

ℓ

2

ℓ

2

m

O

k

y

2 m

k c

x

θ

ı

Solution:

a) In addition to the coordinate θ identified inthe original figure, we also define x and y asthe displacment of the block and end of thebar respecively. The directions ı 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 αβ/F and thelinear acceleration of the block

F

aG are

αβ/F = θ k,F

aG = x .

FR

−k y

−T

T

−k x −c x

We can also relate the identified coordinates as

x =ℓ

2θ, y = ℓ θ.

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)

= 2m x ,

∑

MO = IO αβ/F −→(

− Tℓ

2− k y ℓ

)

k =mℓ2

3θ k.

3

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

− (2m x + k x + c x)ℓ

2− k y ℓ =

mℓ2

3θ.

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

5mℓ2

6θ +

c ℓ2

4θ +

5 k ℓ2

4θ = 0.

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

meq =5mℓ2

6, beq =

c ℓ2

4, keq =

5 k ℓ2

4.

From these the damping ratio and natural frequency are

ζ =beq

2√

keq meq

=c ℓ2

4

2√

5 k ℓ2

45 m ℓ2

6

=

√3 c

2√

50 k m,

ωn =

√

keq

meq=

√

5 k ℓ2

45 m ℓ2

6

=

√

3 k

2m

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

ωn = 15.8 rad/s, ζ = 7.91 × 10−4.

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

θ(t) = e−ζ ωn t(

a sin(

ωn

√

1 − ζ2 t)

+ b cos(

ωn

√

1 − ζ2 t))

,

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

θ(0) = b = 0, θ(0) = −ζ ωn b + ωn

√

1 − ζ2 a = 10 rad/s,

so that the general solution becomes

θ(t) = 0.632 e−0.0125 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 θ 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 withθ(0) = −π

2 rad, find the resulting an-gular displacement θ(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, ℓ = 25 cm,

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

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

θ(0) = 0, θ(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 θ = 0, and that all angles remainsmall.

ℓ2

ℓ2

m

k

z

m

k cx

θ

ı

Solution:

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

5

displacement θ as:

x =ℓ

2θ, z =

ℓ

2θ.

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) = m x .

Likewise, angular momentum balance onthe bar provides

∑

MO = IO αβ/F ,(

−Tℓ

2− k z

ℓ

2

)

k =mℓ2

12θ 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

6θ + c θ + 2 k θ = 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

ωn =

√

keq

meq=

√

12 k

7m, ζ =

beq

2√

keq meq

=

√3 b√

28k m.

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

k2

m

ı

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) ı = m x ı,

−k1 x ı

−k2 x ı

−b x ı

6

or, writing this in standard form

m x + 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 φ 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)φ

z

θ

C

ı

e1

e2

Solution:

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

These additional coordinates are related to x as

z = 2x, x = −r θ.

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

ı = cos φ e1 + sinφ e2,

= − sin φ e1 + cos φ e2.

−k z e1

−c x e1

fr e1

N e2

−mg

The moment produced by gravity about point C is

Mgravity = rGC × (−mg ),

= (r e2) × (−mg ) = −mg r sin φ k.

Angular momentum balance about the contact point C yields∑

MC = IC αD/F ,

(

(2 r) k z + r c x − mg r sinφ)

k =

(

3mr2

2θ

)

k.

7

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

3m

2x + c x + 4 k x = mg sin φ.

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 ωd =

ωn

√

1 − ζ2, where

ωn =

√

keq

meq=

√

8 k

3m, ζ =

beq

2√

keq meq

=c

2√

6 k m,

so that

ωd =

√

8 k

3m

√

1 − c2

24 k m,

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

ccr = 2√

6 k m.

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

4 k x0 = mg sinφ.

Solving for x0, the equilibrium displacement is

x0 =mg sinφ

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

ı

x

z1

z2

θ

Solution:

a) We define the coordinates x, θ, z1, and z2

as shown in the figure, which are relatedas

x = −r1 θ, z1 = r1 θ, z2 = −r2 θ.

Notice that because of these coordinatedefinitions, a rotation with positive θgives 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) = m x ,

while angular momentum balance on thedisk yields

FR

k1 z1

−k2 z2 ı

−T

T

−mg

∑

MO = IO αD/F ,

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

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

(

I + mr21

)

θ +(

k1 r21 + k2 r2

2

)

θ = mg r1.

9

The equilibrium rotation of the disk thus is found to be

θeq =mg r1

k1 r21 + k2 r2

2

.

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

z1,eq = r1 θeq =mg r2

1

k1 r21 + k2 r2

2

,

z2,eq = −r2 θeq = − mg r1 r2

k1 r21 + k2 r2

2

.

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

θ(t) = θeq + A sin(ωn t) + B cos(ωn t),

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

ωn =

√

k1 r21 + k2 r2

2

I + mr21

.

The system is released with the initial conditions:

θ(0) = 0, θ(0) = 0,

so that solving for the arbitrary constants

A = 0, B = −θeq.

Therefore the solution is

θ(t) = θeq (1 − cos(ωn t)) =mg r1

k1 r21 + k2 r2

2

(

1 − cos

(

√

k1 r21 + k2 r2

2

I + mr21

t

))

.

The angular velocity of the disk becomes

θ(t) =(

θeq ωn

)

sin(ωn t),

which has amplitude

Ω = θeq ωn =mg r1

√

(k1 r21 + k2 r2

2) (I + mr21)

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?

ℓ2

ℓ2

m

k

c

z1

m

k

z2

θ

ı

Solution:

a) In addition to θ, 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 =ℓ

2θ, z2 =

ℓ

2θ = 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 θ k,

(

T − k z1 − cz1

) ℓ

2k =

m ℓ2

12θ k.

G

FR

k z1

c z1

−k z2

−T

T

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

∑

F = mF

aG,(

− k z2 − T)

= mz2

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

m ℓ2

3θ +

c ℓ2

4θ +

k ℓ2

2θ = 0.

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

ζ =

√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

ı

Solution:

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

keq =1

1k1

+ 1k2

=k1 k2

k1 + 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:

ωn =

√

1

12s−1, ζ =

√3

4.

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

x(t) = e−ζ ωn t(

A cos(ωd t) + B sin(ωd t))

.

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

x(t) =3

20e−t/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

θ

x1

ı

13

Solution:

a) We define θ, x1, and x2 as indicated in the above figure. In particular, x1 and x2

represent the displacement in their respective springs as measured from their unstretchedposition. These coordinates are related through the following transformations:

x1 = −r1 θ, x2 = −r2 θ.

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 θ k,(

T r1 + k2 r2 x2

)

k = IG θ 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 ı

−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

)

θ +(

k1 r21 + k2 r2

2

)

θ = (m g r1).

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

θeq =m g r1

k1 r21 + k2 r2

2

.

Therefore, the equilibrium displacements in each spring are:

x1,eq = −r1 θeq =m g r2

1

k1 r21 + k2 r2

2

, x2,eq = −r2 θeq =m g r1 r2

k1 r21 + k2 r2

2

.

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 φ represent the angular displacement of the disk from the static equilibriumposition:

φ = θ − θeq.

14

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

V =1

2k1 z2

1 +1

2k2 z2

2 ,

=1

2

(

k1 r21 + k2 r2

2

)

φ2.

Also, the kinetic energy becomes:

T =1

2IG φ2 +

1

2m z2

1 ,

=1

2

(

IG + m r21

)

φ2.

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

φ(0) = −θeq = − m g r1

k1 r21 + k2 r2

2

, φ(0) = 0.

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

T0 = 0, V0 =(m g r1)

2

2(k1 r21 + k2 r2

2).

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 =1

2

(

IG + m r21

)

φ2max.

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

φmax =

√

(m g r1)2

(IG + m r21)(k1 r2

1 + k2 r22)

.

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

θ

ı

Solution:

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

x = −r θ, z = −2 r θ, 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 ı

−3k z ı

−c x ı

f ı

G

C

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

∑

MC = IC θ k,

(

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

k =3 m r2

2θ k.

Finally, writing this equation in terms of a single coordinate, we obtain:

(

3 m r2

2

)

θ + (c r2) θ +(

13 k r2)

θ = 0.

b) For an underdamped response, the frequency of oscillation is ωd = ωn

√

1 − ζ2. Withthis system, we find that:

ωn =

√

26 k

3 m, ζ =

c√78 k m

,

so that:

ωd =

√

26 k

3 m− 2 c2

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

T =1

2m x2

1 +1

2m x2

2 +1

2

mr2

2θ2,

V =1

2k x2

1 +1

2k x2

2.

However, these three coordinates are dependent through the transformations:

x1 =r

2θ, x2 = r θ.

Therefore, the Lagrangian reduces to:

L = T − V =1

2

(

7mr2

4

)

θ2 − 1

2

(

5kr2

4

)

θ2.

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

Qθ = −cr2

4θ,

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

(7m) θ + c θ + (5k) θ = 0.

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

ωn =

√

5k

7m, ζ =

c

2√

35 km.

Therefore, the damped natural frequency becomes:

ωd = ωn

√

1 − ζ2 =

√

5k

7m

√

1 − c2

140 km.

c) For a critically damped system, ζ = 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 (µ 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 ı direction, while θ 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) ı = mx ı,

(rf) k =mr2

2θ k,

(−T ) ı = mx ı.

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

x = −rθ.

Eliminating (θ, 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:

ωn =

√

2k

5m, ξ =

b√10km

.

Thus, the damped natural frequency is:

ωd = ωn

√

1 − ξ2 =

√

10km − b2

25m2.

For the given values of the parameters, this reduces to ωd = 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:

ζ =δ1

√

4π2 + δ21

, δ1 = − ln

(

x1

x0

)

= − ln

(

0.75 m

1.00 m

)

= 0.288,

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

b = 2ζ√

km = 0.183 N/(m/s).

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

T =2π

ωd=

2π

ωn

√

1 − ζ2,

=

√m

√

4π2 + δ21√

k= 3.145 s.

b) The exponential decay rate is σ = ζωn, which is found to be σ = 0.0914 s−1.

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 θ, so that the

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

mbar l2

3θ + bl z + kl z = 0.

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

mbar

3z + 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:

ωn =

√

3k

mbar, ζ =

√3 b

2√

kmbar

=√

6 rad/s =

√

3

8.

b) The frequency of oscillation, ωd = ωn

√

1 − ζ2, reduces to:

ωd =

√

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 l2

3+ meq l2,

=mbar l2

3+

(mspring

3

)

l2,

=(mbar + mspring) l2

3.

Therefore, the new frequency of oscillation is:

ωd =

√

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 δ = ln |x2/x1|. In terms of this quantity andthe period of oscillation T , the damping ratio and natural frequency are defined as:

ωn =

√

(2π)2 + δ2

T, ζ =

δ√

(2π)2 + δ2.

Therefore, the stiffness and damping constants reduce to:

k = m ω2n = m

(2π)2 + δ2

T 2, b = m 2ζωn = m

2δ

T.

For this system, we find that T = 4 s and δ = 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 θ 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 = kr

2 θ, where θ isthe angular displacement of the massless pulley from static equilibrium. In addition,we find that θ is related to the linear displacement of mass m as θ = x

2r . 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:

ζ =b

2√

keqm=

b√km

, ωn =

√

keq

m=

√k

2√

m.

21

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

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

x(t) = exp(−ζωnt)(

c1 exp(

(ωn

√

ζ2 − 1)t)

+ c2 exp(

−(ωn

√

ζ2 − 1)t))

,

and with these initial conditions this reduces to:

x(t) = x0 exp (−ζωnt)

(

ζ +√

ζ2 − 1

2√

ζ2 − 1exp

(

(ωn

√

ζ2 − 1)t)

+

ζ −√

ζ2 − 1

2√

ζ2 − 1exp

(

−(ωn

√

ζ2 − 1)t)

)

,

= x0e−6t

(

4 +√

15

2√

15e

3

2

√15t +

4 −√

15

2√

15e−

3

2

√15t

)

,

= x0e−6.0t

(

1.016e5.81t + 0.01640e−5.81t)

.

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 ı, Fdamper = −2c x ı.

Using linear momentum balance on the block, we find that:

∑

F = (−6k x − 2c x) ı = mx ı = mF

aG,

22

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

x + (2ζωn) x + (ω2n) x = 0, with ωn =

√

6k

m,

ζ =c√6km

.

b) In terms of ζ and ωn, the period of oscillation is simply T = (2π)/ωd, where ωd =

ωn

√

1 − ζ2. Therefore, for this system, the period reduces to:

T =2π

√

6k

m

√

1 −(

c√6km

)2= 2π

m√6km − c2

,

provided c2 < 6km.

c) For a critically damped system, ζ = 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 ωn = 12 rad/s, and ζ = 1/6. For these values, the general solution can bewritten as:

x(t) = A e−2 t sin(

2√

35 t + φ)

,

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

A =

√

5

28m, φ = 0 rad,

so that the general solution can be written as:

x(t) =

√

5

28e−2 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 ı 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 · x − mg = −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:

2ζωn =b

m, ω2

n =3k

m,

which can be solved to yield:

ωn =

√

3k

m, ζ =

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 =mg

3k.

Problem 24:


a) find the linearized governing equationsof motion for small θ;

b) find the frequency of oscillation for theresponse;

c) if the initial velocity is zero, and θ(0) =θ0, determine the time response of thesystem.

k

4 k

(m, l)G

θ

Solution:

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

x± = ± l sin θ

2.

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

∑

MG = − l2

4(4k sin θ + k sin θ) k,

= −5kl2

4sin θk.

Thus angular momentum balance about G provides:

∑

MG = IGθk,

−5kl2

4sin θk =

ml2

12θk.

25

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

θ +15k

mθ = 0.

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

ω = ωn =

√

15k

m.

c) This system possesses the general solution:

θ(t) = c1 sinωt + c2 cos ωt,

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

θ(t) = θ0 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:

ζ =b

2√

km, ωn =

√

k

m.

ζ = 1 corresponds to critical damping, so that:

bcritical = 2√

km = 12.

b) For b = 24 we find:

ζ =b

2√

km= 2, ωn =

√

k

m= 3.

26

Therefore the eigenvalues of this system are:

λ1,2 = ωn

(

ζ ±√

ζ2 − 1)

= 3(−2 ±√

3).

The dominant eigenvalue is λ = −6 + 3√

3, and so the time t = τ required for theamplitude of the free vibration to be reduced to within 2% of zero is:

τ ∼ −4

−6 + 3√

3∼ 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 θ = 0:

a) find the linearized governing equationsof motion for small θ;

b) find the frequency of oscillation for thefree response;

Neglect gravity.

ℓ2

ℓ2

k

c

(m, l)

θ

Solution:

a) Using angular momentum balance about the fixed point O, we find:

∑

MO = IOαβ/FF,(

−kl2 sin θ − cl2

4θ cos θ

)

k =ml2

3θk,

so that the equation of motion can be written as:

ml2

3θ + c

l2

4θ cos θ + k l2 sin θ = 0.

Linearizing this equation about θ = 0, we obtain:

θ +3c

4mθ +

3k

mθ = 0.

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

ωd = ωn

√

1 − ζ2,

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

ωn =

√

3k

m, ζ =

√3 c

8√

km,

27

so that the damped natural frequency is:

ωd =

√

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:



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 θ 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

2θ, x2 = rθ.

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 = −mr2

2θ,

T1 − kx1 − cx1 = mx1,

T2 + kx2 = −mx2.

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

(

7mr2

4

)

θ +

(

cr2

4

)

θ +

(

5kr2

4

)

θ = 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 θ + c θ + k θ = 0, the equivalent mass would be meq = 7m.

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

ωd = ωn

√

1 − ζ2 =

√

5k

7m

√

1 − c2

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:

λ2 +13

2λ + 9 = 0,

which has two real solutions of the form:

λ =−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 s−1)t)

.

b) For this system, the damping ratio is:

ζ =b

2√

km.

30

Thus, for a critically damped system ζ = 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 θ 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 = −rθ.

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

Fspring = −kx ı, Fdamper = −bx ı.

Using linear and angular momentum balance on the disk, we find that:

∑

F = (f − kx − bx) ı = mx ı = mF

aG,

∑

MG = (fr) k =mr2

2θ k = IG αβ/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 + (2ζωn)x + (ω2n)x = 0, with ωn =

√

2k

3m,

ζ =b√

6km.

31

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

bcritical =√

6km.

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

x(t) =(

c1 + c2 t)

e−ωn t,

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

x(t) =(

(c2 − ωnc1) − (ωnc2) t)

e−ωn 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 − ωnc1.

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

x(t) = x0

(

1 + ωn t)

e−ωn t,

where recall that ωn =√

(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(µ 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 = −M

kr.

2 Frictionally Damped Systems

Problem 34:

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

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

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

b) With µ = 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

µ

g

z

A B

ı

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 − F

vA = z ı,

= (−x ı) − (x ı) ,

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 ı 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 ı

T ı−k x ı

fr ı

−m g

N

−T ı T ı

Applying linear momentum balance to the block yields∑

F = (2T − k x + fr) ı + (N − mg) = m x ı = mF

aG,

and in terms of x the equation of motion becomes

m x + 5 k x = fr.

33

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

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| ≤ µmg.

This inequality is satisfied provided

|xeq| ≤µmg

5 k.

Problem 35:

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

F (t) = F0 sin(ω t),


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

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

φ = 30, µ = 0.125,

F0 = 4 N, ω = 2.00 rad/s,

k

mF (t)

φ

g

34

Problem 36:

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

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

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

c) Find the value of θ so that the systemcomes to rest after one full cycle exactly

at 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,

µ = 0.35,

θ

m

k

µ

x 2

ı2

ı

Solution:

The unit directions ı2 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 ı2

f ı2N 2

−m g

Therefore, linear momentum balance yields the following equations:∑

F = mF

aG,(

f − k x)

ı2 +(

N)

2 −(

m g)

= m x ı2,(

f − k x + m g sin θ)

ı2 +(

N − m g cos θ)

2 = m x ı2.

Therefore, this leads to the following scalar equations in the ı2 and 2 directions:

m x + k x = f + m g sin θ,

N = m g cos θ.

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

35

to:m x + k x = m g sin θ.

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

xeq =m g

ksin θ.

b) With µ 6= 0, an equilibrium state is maintained provided:∣

∣

∣f∣

∣

∣≤ µ N,

∣

∣

∣k x − m g sin θ

∣

∣

∣≤ µ m g cos θ.

Therefore, if the system is released from x = 0, the block begins to slide when:

tan θ = µ.

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

z = x − m g

ksin θ,

so that the equations of motion become:

m z + k z = f, N = m g cos θ.

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

∆ A = −4 µ N

k.

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 µ N

k

∣

∣

∣

∣

=∣

∣

∣−m g

ksin θ

∣

∣

∣= |z(0)|.

Solving for θ:tan θ = 4 µ.

Problem 37:

The spring mass system rests on a surfacewith coefficient of friction µ 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

µ

36

Problem 38:

For the spring-mass system with Coulombdamping:


b) what is the period of each oscillation.

x

mk k

µ

Solution:

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

F

aG = xı. Thus linearmomentum balance yields:

mxı = Fspring ı + fµı + (N − mg).

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

fµ = −µmgx

|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 µmg. Therefore, thegoverning equations of motion are:

mx + 2kx = fµ, withfµ = −µmg x

|x| |x| 6= 0,

|fµ| ≤ µmg x = 0.

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

ω =

√

2k

m.

Therefore the period of the oscillation is:

T =2π

ω=

2π√

m√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 µ:


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

µ

Problem 40:

For the spring-mass system with Coulombicdamping, x is measured from the unstretchedposition of the spring. If the coefficient offriction is µ 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 µ 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

µ

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 ı, linear momentum balance on the upperand lower block yields:

mx + kx = f + T,

mx = mg − T,

where the frictional force is defined as:

f = −µmg x|x| , x 6= 0,

|f | ≤ µmg, 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 µmg—the transition to move-ment occurs when |f | = µmg. Thus the system does not move is |f | ≤ µmg and x = 0.Substituting these conditions into the equations of motion we find:

|f | = |kx − mg| ≤ µmg,

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

µ < 1.

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

x1 =2(1 − µ)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 µ 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

µ

Solution:

a) The equations of motion can be written as:

mx + kx = f,

39

where f is the force due to friction, modeled by Coulomb’s law of friction as:

f = −µmgx

|x| , x 6= 0,

|f | ≤ µmg, 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, µmg ≥ f = kx(0). Thus,solving for x(0) we find, that for sliding to occur:

|x(0)| >µmg

k.

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) > µmg

k . Thus the equation of motion becomes:

mx + kx = µmg,

which has the general solution:

x(t) =

x(0) − µmg

k

cos

(

k

mt

)

+µmg

k.

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

x(t1) = 2µmg

k− x(0).

At this point, the block sticks if and only if |x(t1)| ≤ µmgk . Therefore, solving for x(0),

we find that:x(0) ≤ 3

µmg

k.

So for a block with x(0) > 0, the allowable range for x(0) is µmgk < x(0) ≤ 3µmg

k .Together with an identical argument for x(0) < 0 yields the total allowable range as:

µmg

k< |x(0)| ≤ 3

µmg

k.

40

Problem 42:

For the system shown to the right, x is mea-sured from the unstretched position of the

spring. 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 µ. 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 µ 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

µ

Solution:

We begin by defining two additional coordinates, θ, 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, θ =x

r.

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

∑

F =(

− k x + T1 + f)

ı +(

N − mg)

= mx ı = mF

aG1,

∑

MO =(

T1r − T2r)

k = −IO θk = IOαD/FF,

∑

F =(

T2 − mg)

= −my ı = mF

aG2,

Notice that IO = I 6= 0, so that provided θ 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 ı direction, can be written as:

f =

−µmg x|x| , x 6= 0

f0, |f0| ≤ µmg, x = 0.

41

b) The minimum value for slip is simply µmin = 0. If we would like to find the range ofµ 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| ≤ µmg.

Therefore, solving for µ yields:

µ ≥∣

∣

∣

∣

kx0

mg− 1

∣

∣

∣

∣

=

∣

∣

∣

∣

1 − kx0

mg

∣

∣

∣

∣

,

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

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

T =2π

ωn= 2π

√

2m + Ir2

k.

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

−µmg

k< |δfinal| ≤

µmg

k,

the initial displacement δ0 from the equilibrium in the absence of friction must be inthe range:

−µmg

k+ ∆ =

3µmg

k< |δ0| ≤

5µmg

k=

µmg

k+ ∆.

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

of x0 is:3µmg

k<

∣

∣

∣x0 −

mg

k

∣

∣

∣≤ 5µmg

k.

3 Forced Single Degree-of-freedom Systems

42

Problem 43:

For the system shown to the right the barof length ℓ has mass m and is subject to atime-dependent moment of the form

M (t) = M0 sin(ω t) k.

a) Find the equations of motion for thisrotation of the bar.

b) What is the steady-state amplitude ofthe forced response, with

m = 3 kg, b = 16 N/(m/s),

k = 8 N/m, ℓ =1

4m,

ω = 4 rad/s, M0 =1

16N · m

G

(m, ℓ)k b

k M (t)

Problem 44:

The block of mass m = 20 kg shown to theright rests on a rigid foundation and is sub-ject to a time-dependent load

F (t) = f0 sin(ω t) .

a) Design an undamped foundation toachieve isolation ≥ 33% for all forcingfrequencies ω > 3π rad/s;

b) If, for the isolator that you designed,the damping ratio was measured to beζ = 0.125 (rather than ζ = 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(ω 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 m

M0 = 3 N · m, ω = 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 ω = 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(ω 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, ω = 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−σ t.

For this system:



c) With

m = 2 kg, ℓ = 30 cm,

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

σ = 2.00 s−1,

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


O

ℓ3

2 ℓ3

k

f(t)

A

m

k c

B

θ

45

Problem 49:

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

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

µ = 0.05,

f0 = 40 N, ω = 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 ω = 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 ℓ, and is pinned at point O. A har-monic force is applied at A. For this system:



c) For:

m = 6 kg, ℓ = 25 cm,

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

f0 = 2.00 N, ω = 10 rad/s,

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


O

ℓ2

k

f0 sin(ω t)

A

4m

c

B θ

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(ω 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, ω = 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 atω = 25 rad/s the phase φ of the response withrespect to the rotation of the unbalanced diskis measured to be π/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

ε

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, ℓ = 30 cm,

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

f0 = 2.00 N, ω = 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

ℓ3

2 ℓ3

k

f0 sin(ω t)

A

m

k c

B

θ

48

Problem 55:

The block shown to the right rests on a roughsurface with coefficient of friction µ 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(ω t), with

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

r = 12.5 cm, µ = 0.50,

f0 = 40 N, ω = 4 rad/s.

−f(t) ı

(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 φ 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(ω 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)φ

49

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

a) Find the equations of motion (in termsof the given parameters—do 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, ρ = 0.125 m

c) Determine the steady state amplitudeof the friction force.

θ

x

z

ı

C

Gb

2 k

kM(t) (m, ρ)

Solution:

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

x = −ρ θ, z =3

2x.

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 αD/F ,

−b x ı

−2 k z ı

−k x ı

M(t) k

fr ı

N

−mg

which yields(

ρ (k x + b x) +3 ρ

2(2 k z) + M(t)

)

k =3mρ2

2θ k.

Using the above coordinate transformations this equation can be written as

3m

2x + b x +

11 k

2x = −M(t)

ρ.

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:

ωn =

√

88

3, ζ =

1√264

, r =

√

6

11.

Therefore, the amplitude of the translational oscillations becomes

X =F

kM(r, ζ) =

8

88

1√

(

1 − 611

)2+

(

2 1√264

√

611

)2=

1√26

.

Likewise, the phase shift of the response is:

tan φ =2 ζ r

1 − r2=

2 1√264

√

611

1 − 611

=1

5,

so that φ = 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)

ı +(

N − mg)

= m x ı = mF

aG.

Therefore, solving for fr we find that

fr = m x + b x + 4 k x.

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

fr(t) =(

4 k − mω2)

X sin(ω t − φ) +(

b ω)

X cos(ω t − φ).

The magnitude of the friction force is then found to be

∥

∥

∥f∥

∥

∥= X

√

(

4 k − mω2)2

+(

b ω)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 ofε 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 ω = 8 π rad with:

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

m = 4 kg, ε = 2.5 cm

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

m

ε

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 parameters—do not substi-

tute in numerical values yet);

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

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

ℓ2

ℓ2

mk

c

z

f(t) = sin(ω t)

θ

ı

Solution:

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

52

z =ℓ

2θ

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 θ k,

(

f(t) − k z − cz) ℓ

2k =

m ℓ2

12θ k.

G

FR

k z1

c z1

f(t)

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

m ℓ2

12θ +

c ℓ2

4θ +

k ℓ2

4θ =

ℓ

2f(t).

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

θ +

(

3 c

m

)

θ +

(

3 k

m

)

θ =

(

6

m ℓ

)

sin(ω t),

so that:

ωn =

√

3 k

m, ζ =

√3 c

2√

k m, M0 =

6

m ℓ.

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

MT = (meq M0)

√

1 + (2 ζ r)2√

(1 − r2)2 + (2 ζ r)2,

=ℓ

2

√

1 +(

c ωk

)2

√

(

1 − m ω2

3 k

)2+

(

c ωk

)2=

ℓ

2

√

k2 + (c ω)2√

(

k − m ω2

3

)2+ (c ω)

2.

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

FT =

√

k2 + (c ω)2√

(

k − m ω2

3

)2+ (c ω)

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:

Θ =M0

ω2n

1√

(1 − r2)2 + (2 ζ r)2,

=2

k ℓ

1√

(

1 − m ω2

3 k

)2+

(

c ωk

)2=

2

ℓ

1√

(

k − m ω2

3

)2+ (c ω)

2.

53

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

Θ =8

∣

∣

(

k − 2 ω2

3

)∣

∣

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

8∣

∣k − 2 ω2

3

∣

∣

≤ π

6,

48

π≤

∣

∣

∣

∣

k − 2 ω2

3

∣

∣

∣

∣

.

This inequality has two solutions:

k ≥ 48

π+

2 ω2

3, k ≤ 2 ω2

3− 48

π.

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

k ≤ 2 (20)2

3− 48

π= 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

θ

ı

Solution:

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

x = −r θ, z = −2 r θ, 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 ı

−3k z ı

f(t) ı

fr ı

G

C

54

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

∑

MC = IC θ k,

(

(3k z) 2r + (k x) r − f(t) r)

k =3 m r2

2θ k.

Finally, writing this equation in terms of a single coordinate, we obtain:

(

3 m r2

2

)

θ +(

13 k r2)

θ = r f(t).

In standard form:

θ +

(

26 k

3 m

)

θ =2 f(t)

3 m r.

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

θ(t) =

∫ t

0

F (τ) h(t − τ) dτ,

and for this system:

F (t) =f(t)

meq=

2 f(t)

3 m r, h(t) =

1

ωnsin(ωn 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 f0

3 m

∫ t

0

sin(ωn (t − τ))

ωndτ, 0 < t ≤ t0,

x(t) =2 f0

3 m

∫ t0

0

sin(ωn (t − τ))

ωndτ +

f0

3 m

∫ t

t0

sin(ωn (t − τ))

ωndτ, t > t0.

Evaluating these integrals, the solution becomes:

x(t) =2 f0

26 k

(

1 − cos(ωn t))

, 0 < t ≤ t0,

x(t) =f0

26 k

(

1 + cos(ωn(t − t0)) − 2 cos(ωn 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

ı

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 k2

2 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 ω2) sin(ω t).

−keq z −b z

Therefore, linear momentum balance on the mass yields:

∑

F = mF

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 ζ ωn z + ω2n z = u0 ω2 sin(ω t),

56

with:

ωn =

√

keq

m, ζ =

b

2√

keq m.

Therefore the steady state response of this system becomes:

z(t) = Z sin(ω t − ψ),

with Z = u0 Λ(r, ζ), and:

Λ =r2

√

(1 − r2)2 + (2 ζ r)2, tan ψ =

2 ζ r

1 − r2, and r =

ω

ωn

For the numerical values of this problem:

ωn =√

2, ζ =1√2, r =

√2,

so that:

Λ =2√5, tan ψ =

2

−1

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

z(t) =1√5

sin(2 t − 2.03).

Problem 62:



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

M (t) = M0 sin(ωt)k.

a) If the coefficient of friction is µ, 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 µ 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 = α m about themass center G. The disk is subject to a time-varying force f(t) = 4 sin(ω 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 (ı, , k) represent the standard orthonormal basis, while the transla-

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

mx = −kx − bx − fµ,

mr2

2θ = −fµr − M(t),

where fµ is the unknown frictional force. If the disk rolls without slipping, we find thekinematical relation x = −rθ, and, eliminating fµ 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 ωd =

58

ωn

√

1 − ζ2, where:

ωn =

√

2k

3m=

√

2

3, ζ =

b√6km

=1√2,

so that the damped natural frequency is ωd = 1√3

= 0.577.

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

x + 2ζωnx + ω2nx = F0 sin(t),

where ζ and ωn 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 + (2ζr)2,

where r = ωωn

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

32 , so that M = 2√

13.

As a result, the steady-state amplitude is:

A =F0

ω2n

M =2√13

= 0.555.

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

tan φ =−2ζr

1 − r2=

−√

3

− 12

.

From the expression for angular momentum balance, we find:

fµ =m

2x − M(t)

r,

= − 1√13

sin(t + φ) − sin(t),

= −[(

1√13

sin φ

)

cos(t) +

(

1√13

cos φ + 1

)

sin(t)

]

,

where cos φ = 1√13

and sinφ = 2√

3√13

, so that the amplitude of the frictional force is:

|fµ| =4√13

.

59

Problem 65:



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

M (t) = M0 sin(ωt)k.

If the disk rolls without slip (µ 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 θ as the angular displacement of thedisk from the unstretched position in the k direction. If the disk rolls without slip, xand θ are related as:

x = −rθ.

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

Fspring = −kx ı, Fdamper = −bx ı.

Using linear and angular momentum balance on the disk, we find that:∑

F = (f − kx − bx) ı = mx ı = mF

aG,

∑

MG = (M0 sin(ωt) + fr) k =mr2

2θ k = IG αβ/FF.

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

(

3m

2

)

x + bx + kx = −M0

rsin(ωt).

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

ω2n =

8

3, ζ =

1

8√

6, F0 = −M0

3r.

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

X =F0

ω2n

M,

=−M0

3r√

(

83 − ω2

)2+

(

ω6

)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(ωt) 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 ω?

u(t) = 4 sin(ω t) m

z

m

x

ı

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 = 4ω2 sin(ωt).

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

X =4ω2

40

1√

(

1 − ω2

40

)2,

=4ω2

|40 − ω2| .

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 ofε from the the center of rotation O,


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

m

ε

O

G

k b

x

ı

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) + ε sin(ωt). Note that ε measures the eccentricity of the mass

center, not the location of the mass imbalance. Consequently, the governing equationsof motion can be written:

mz = −kx − bx,

mx − εω2 sin(ωt) = −kx − bx,

or in more standard form:

x +b

mx +

k

mx = εω2 sin(ωt).

a) In the above system we find ωn =√

km and ζ = b

2√

km, so that the damped natural

frequency can be written:

ωd = ωn

√

1 − ζ2 =

√

k

m− b2

4m2.

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

Λ =ω2

√

(ω2n − ω2)2 + (2ζωωn)2

,

which, with ω = ωd, reduces the amplitude to:

A = ε1 − ζ2

ζ√

4 − 3ζ2,

with ζ 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(ωt) 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(ω t) m

zk

mg

x

ı

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 = mF

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 s−2)z = −9.81 m/s2 + 4ω2 sin(ωt) 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 ωn = 2, ζ = 0, and theamplitude of the forcing is F = 4ω2. 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(ωt + φ), where the amplitude Z is:

Z = U · Λ(r, ζ) = Ur2

√

(1 − r2)2 + (2ζr)2,

=4r2

|1 − r2| .

c) If Z < 8 then this implies that:

Z <4r2

√

(1 − r2)2.

Defining Zcr as:

Zcr =4r2

√

(1 − r2)2,

We may solve for r to yield:

r =

√

Z2cr ± 4Zcr

Z2cr − 16

,

=

√

Zcr

Zcr − 4,

√

Zcr

Zcr + 4

.

Thus for Zcr = 8, we find that Z < 8 in the range:

0 < r <

√

2

3, or r >

√2.

Problem 69:

The single-degree-of-freedom system shown issubject to a harmonic force. If the naturalfrequency is ωn = 4 rad/s and m = 1 kg:

a) determine the spring and damping con-stants when the system is criticallydamped;

b) determine the amplitude of the totalforce transmitted to the ground un-der steady-state oscillations when ω =1 rad/s.

x

mk b

F (t) = sin(ω t) N

Solution:

The equation of motion for this system takes the form:

x +b

mx +

k

mx =

F (t)

m.

a) In terms of the system parameters, the damping ratio and natural frequency can bewritten as:

ζ =b

2√

km, ωn =

√

k

m.

64

So, for a critically damped system ζ = 1, and solving for k and b we find:

k = 16 N/m, b = 8 N · s/m.

b) The amplitude of the total force transmitted to the ground, which is defined as Ft, isFt = Xω2

n

√

1 + (2ζr)2, where X is the amplitude of the response and r = ωωn

= 14 is

the frequency ratio. Thus, for this system:

Ft =

(

F0

ω2n

1√

(1 − r2)2 + (2ζr)2

)

· ω2n

√

1 + (2ζr)2,

=F0

√

1 + (2ζr)2√

(1 − r2)2 + (2ζr)2=

8√

5

17= 1.05.

Problem 70:

A constant force is applied to the undampedsingle degree-of-freedom system for a dura-tion of t1, at which point it is removed, thatis:

F (t) =

F0 ı, 0 ≤ t < t1,0 , t ≥ t1.

If the system starts with zero initial condi-tions, determine the resulting displacementof the mass x(t). You may either use theconvolution integral or you may try to solvethis explicitly.

x

m4 k

F (t)

Solution:

The equation of motion for this system is:

x +4k

mx =

F (t)

m,

which has an impulse response of the form:

h(t) =sinωnt

mωn, ωn = 2

√

k

m.

Therefore, using the convolution integral, the response of the system is:

x(t) =

∫ t

0

F (ξ)h(t − τ) dτ,

=

∫ t

0

F0

m

sin(ωn(t − τ))

ωndτ, 0 ≤ t < t1,

∫ τ

0

F0

m

sin(ωn(t − τ))

ωndτ, t ≥ t1.

=

F0

mω2n

(

1 − cos(ωnt))

, 0 ≤ t < t1,

F0

mω2n

(

cos(ωn(t − t1)) − cos(ωnt))

, t ≥ t1.

65

Problem 71:

A mass m = 2 kg is rigidly connected to arigid massless bar of length ℓ = 40 cm, whichis pinned to a wall. The system is supportedby a combination of springs and damper asshown in the figure, and subjected to a time-dependent moment M(t) = 12 sin(t). If k =12 N/m and b = 6 N/(m/s):

a) find the undamped natural frequencyand the damping ratio of the system;

b) what are the steady-state amplitudeand phase of the forced response?

m

ℓ2

M(t)

6 k

k 2 k

z1

b

z2

θ

ı

Problem 72:

The system shown in the figure is supportedby a foundation that undergoes an exponen-tially decaying motion of the form:

u(t) = 16e−t/4.

If the mass and stiffness are m = 5 kg, andk = 45 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) find the resulting solution z(t) if themass is started from rest and the springis initially unstretched (assume g =10 m/s2 and use the convolution inte-gral);

c) what is the amplitude of the oscillationsas t increases (i.e. u(t) → 0)?

u(t)

zk

mg

66

Problem 73:

In the system shown at right, the disk is as-sumed to be massless while:

m = 2 kg, k = 4 N/m,b = 1

4 N/(m/s), r = 1 cm.

a) Determine the governing equations ofmotion;

b) What is the period of oscillation;

c) If f(t) = sin t, find the amplitude andphase of the resulting motion as t → ∞.

2 r

r

bf(t)

k

m

Problem 74:

In the mechanical system shown each springis identical, with spring constant k. If m =1 kg:

a) determine the spring constant k anddamping constant b so that the un-forced system is critically damped andthe exponential rate of decay is τ =2 s−1;

b) with the initial conditions:

x(0) = 0, x(0) = 2 m/s,

find the resulting solution of x(t) for theunforced problem, i.e., f(t) = 0;

c) with f(t) = sin(2t), determine the am-plitude of the force transmitted to thesupporting structure.

f(t)

k

b

m

x

ı

67

Problem 75:

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 ofε from the the center of rotation O,


b) for:

m = 4 kg, ε = 0.1 m,

b = 0.5 (N · s)/m, k = 8 N/m

what is the steady-state amplitude ofvibration when the rotor spins at thisangular speed?

m

ε

O

G

k

2 k

b

x

ı

Problem 76:

The disk shown in the figure has mass m =4 kg, and is subject to a time-dependent mo-ment M (t) = M0 sin(t) k. If k = 2 N/m andr = 0.2 m:

a) Find the steady-state amplitude of theresponse when:

b = 1 N/(m/s), M0 = 12 N · m.

b) What is the amplitude of the oscil-lations when the system is criticallydamped?

r

r2

IG

2 k

k

4 k

b

68

Problem 77:

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

F (t) = F0 sin(ω t),


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

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

φ = 30, µ = 0.125,

F0 = 4 N, ω = 2.00 rad/s,

k

mF (t)

φ

g

4 Multi Degree-of-freedom Systems

Problem 78:

For the system shown to the right

a) use Lagrange’s equations to determinethe equations of motion;

b) find the mode shapes and natural fre-quencies of the system;

c) determine the modal equations for thissystem.

m 4m

2 k k F (t) ı

69

Problem 79:



b) are your equations statically coupled,dynamically coupled, both or neither?

(4m, d

)

f(t)

(m, 2 d)k

2 k

k

b

m

Problem 80:


a) use Lagrange’s equations to determinethe equations of motion;

b) find the mode shapes and natural fre-quencies of the system. Normalize themode shapes so that uT M u = 1.

m

3m

k k

4 k k

Problem 81:

For the system shown to the right the barof length ℓ is massless and the block on theright is subject to a time-dependent force ofthe form

F (t) = F0 sin(ω t).

The intermediate springs are located a dis-tance ℓ

2 from the center of the bar.

a) Find the equations of motion in termsof x1 and x2.

b) Determine the natural frequencies andmode shapes of this system.

c) Determine the modal equations.

d) What is the steady-state amplitude ofthe in-phase motion?

G

k

m

F (t)

x1

m

x2

k

k

70

Problem 82: (Spring 2003)For the mechanical system shown to theright, the uniform rigid bar is supported byidentical springs. For this system:

a) find the equations of motion in termsof x and θ (and in terms of the givenparameters—do not substitute in nu-

merical values yet);

b) if k = 64 N/m, m = 2 kg, and ℓ =0.25 m, find the natural frequencies andmode shapes for this system;

m, ℓ

Gℓ3

k

k

θ

x

Problem 83: (Spring 2003)For the system shown to the right:

a) find the equations of motion (and interms of the given parameters—do not

substitute in numerical values yet);

b) for α = β = 2, find the natural frequen-cies and mode shapes of the system andnormalize the mode shapes by M ;

c) determine the equations of motion thatdescribe the response of each mode.

k α k

m β m

F0 sin(ω t)

Problem 84: (Spring 2003)In the figure shown to the right, in the ab-sence of gravity the springs are unstretchedin the equilibrium position. Determine theequations of motion for this system.

r2

r1

IG

k1

k1

k2

b

m

m

x1

z

x3

x2

θ

ı

71

Solution:

Because we can, we define five different coordinates to describe the dynamical behaviorof this two degree-of-freedom system, leading to the following transformations:

x2 = −r2 θ, x3 = −r1 θ, z = x3 − x1.

A free-body diagram for this system isshown to the right. We develop three equa-tions of motion based on linear momentumbalance on both blocks and angular momen-tum balance on the disk:

Block 1:∑

F = mF

aG1,

(

k1 z − k1 x1

)

= m x1

Block 2:∑

F = mF

aG2,

(

T − b x2 − k2 x2

)

ı = m x2 ı

Disk:∑

MG = IG αD/F ,(

T r2 + k1 r1 z)

k = IG θ k

FR

−k1 x1

k1 z

−k1 z

−T ı

T ı

−(k2 x2 + b x2) ı

From the equations on block 2 and the disk, we eliminate the unknown tension T fromthe system to obtain:

IG θ − m r2 x2 − b r2 x2 − k2 r2 x2 − k1 r1 z = 0.

From this equation, and the equation on block 1, we eliminate the coordinates z andx2, and obtain the equations of motion to be:

m x1 + 2 k1 x1 + k1 r1 θ = 0,(

IG + m r22

)

θ + b r22 θ + k1 r1 x1 +

(

k1 r21 + k2 r2

2

)

θ = 0.

Problem 85:

For the system shown in the figure:

a) what is the degree-of-freedom for thissystem?

b) using Lagrange’s equations, determinethe differential equations that governthe motion.

x

2 m

m

(I, r)k k

72

Solution:

a) This system contains three masses which are each allowed to move in only one direc-tion. The upper mass slides horizontally with displacement x1, while the disk rotatesthrough an angle θ. Finally, the suspended mass moves vertically and its position canbe described by the coordinate x2. However, because the disk and the suspended massare connected by an inextensible string, their motion can be related by:

x2 = rθ.

So this system has only two independent coordinates and therefore it is a two-degree-of-freedom system.

b) We utilize the coordinates x1, x2, and θ, as measured from the unstretched position ofthe two springs. Therefore, the kinetic and potential energies are written as:

T =1

2(2m) x2

1 +1

2mx2

2 +1

2I θ2,

V =1

2k x2

1 +1

2k (rθ − x1)

2 − mg x2.

However, x2 and θ are related by the above relationship. Thus eliminating θ, the energiesbecome:

T =1

2(2m) x2

1 +1

2mx2

2 +1

2

I

r2x2

2,

V =1

2k x2

1 +1

2k (x2 − x1)

2 − mg x2,

which, using Lagrange’s equations, yields the equations of motion:

2m x1 + 2 k x1 − k x2 = 0,(

m +I

r2

)

x2 − k x1 + k x2 = mg.

Problem 86:


a) find the mass and the stiffness matrix;

b) is your system of equations dynamicallycoupled, statically coupled, or both? m1

k1 m2

m3

k2

k3 k3

Solution:

We choose the coordinates (x1, x2, x3), which represent the positions of the three masses.

a) To determine the stiffness matrix, we use the stiffness influence coefficients. Maintaininga unit displacement of each mass in turn requires forces of the form:

(x1, x2, x3) = (1, 0, 0) → f = (k1 + k2,−k2, 0)T ,(x1, x2, x3) = (0, 1, 0) → f = (−k2, k2 + 2k3,−2k3)

T ,(x1, x2, x3) = (0, 0, 1) → f = (0,−2k3, 2k3)

T .

73

Therefore, with these coordinates the stiffness matrix is:

K =

k1 + k2 −k2 0−k2 k2 + 2k3 −2k3

0 −2k3 2k3

Alternatively, we can define the potential energy of the system as:

V =1

2(k1)(x1)

2 +1

2k2(x2 − x1)

2 +1

2(2k3)(x3 − x2)

2,

=1

2(k1 + k2)(x1)

2 − 1

2(2k2)(x1x2) +

1

2(k2 + 2k3)(x2)

2

−1

2(2k3)(x2x3) +

1

2(2k3)(x3)

2,

which leads to the same stiffness matrix.

To determine the mass matrix, we could use the inertia influence coefficients, but, forvariety, we determine the kinetic energy as:

T =1

2m1x

21 +

1

2m2x

22 +

1

2m3x

23.

Therefore, the mass matrix is:

M =

m1 0 00 m2 00 0 m3

b) With this choice of coordinates, the mass matrix is diagonal and the stiffness matrixcontains nonzero off-diagonal terms. Thus, the system is statically coupled but dynam-ically uncoupled.

Problem 87:

For the system shown at right:


b) is the system statically or dynamicallycoupled, or both;

c) find the matrix M−1K .

m1

m2

k1

k2 m1

k2

k1

74

Problem 88:

In the system shown to the right, the massand stiffness matrix are:

M =

[

m1 00 m2

]

,

K =

[

k1 + k2 −k2

−k2 k2

]

while the damping is proportional, i.e., C =βM +αK . If m1 is subject to harmonic forc-ing F (t) = sin t, with:

m1 = 2.0 kg, m2 = 1.0 kg, β = 0.1,k1 = 6.0 N/m k2 = 2.0 N/m, α = 0.1

a) determine the eigenvectors of this sys-tem, normalized by the kinetic energyinner product;

b) what are the damped natural frequen-cies of this system;

c) find the steady-state amplitude of vi-bration of the mode of vibration withthe lowest natural frequency.

Problem 89:

The two-degree-of-freedom system shown issubject to a harmonic force applied to theblock of mass 2m, of the form:

f (t) =(

(2 sin(t)) N)

ı.

If the system is subject to proportional damp-ing with α = 0.25, m = 2 kg, and k =4 N/m, find:

a) the mass, damping, and stiffness matri-ces;

b) the forced, damped equation (single-degree-of-freedom) that describes themotion of each mode;

c) the steady-state response of the system.

k k k

2 m 2 m

f (t)

Solution:

a) Using influence coefficients, we find that the mass and stiffness matrices are:

M =

[

2m 00 2m

]

, K =

[

2k −k−k 2k

]

.

Therefore, with proportional damping, the damping matrix becomes C = αK , so that

75

we find:

M = (4 kg)

[

1 00 1

]

, K = (4 N/m)

[

2 −1−1 2

]

,

C = (1 N/(m/s))

[

2 −1−1 2

]

.

Problem 90:

In the multi-degree-of-freedom system shownin the figure, the block with mass 4m slideson a smooth, frictionless surface. If the pulleyis massless:

a) using Lagrange’s equations, determinethe differential equations governing themotion, as measured from static equi-librium;

b) with m = 1 kg and k = 16 N/m,find the natural frequencies and modeshapes for the free vibration of this sys-tem. Normalize the mode shapes sothat with respect to the mass matrixthe amplitude of each mode is one;

c) find the general solution to these equa-tions for the above values of m and k.

4m

2 k

2m

r

x2

k

x1z

Solution:

a) We identify the three coordiantes x1, x2, and z, with

z = x2 − x1.

Measuring the response from static equilibria and neglecting the gravitational potentialenergy, the kinetic and potential energies for this system can be written as

T =1

2

(

4m)

x21 +

1

2

(

2m)

x22,

V =1

2

(

2 k)

x21 +

1

2

(

k)

z2,

In terms of x1 and x2, the potential energy becomes

V =1

2

(

2 k)

x21 +

1

2

(

k) (

x2 − x1

)2=

1

2

(

3 k)

x21 +

1

2

(

− 2 k)

x1 x2 +1

2

(

k)

x22.

Therefore the equations of motion become

4m x1 + 3 k x1 − k x2 = 0,

2m x2 − k x1 + k x2 = 0,

or in matrix form

m

[

4 00 2

] [

x1

x2

]

+ k

[

3 −1−1 1

] [

x1

x2

]

= 0 .

76

b) The corresponding eigenvalue problem for the above system is

(

M−1 K)

u = λ u −→ k

m

[

34 − 1

4− 1

212

]

=k

m

(

β u)

The characteristic equation is

(

3

4− β

) (

1

2− β

)

− 1

8= β2 − 5

4β +

1

4= 0,

with the solution

β =5 ± 3

8−→ λ = ω2 =

k

4m,

k

m

.

Returning this to the eigenvalue problem, the mode shapes are defined by the equation

3

4ui1 +

1

4ui2 = βi ui1,

so that

ui =

[

14βi − 3

]

ci u1 =

[

1−2

]

c1, u2 =

[

11

]

c2.

Normalizing ui by the mass matrix implies that

1 = uTi M ui = c2

i

[

1 (4βi − 3)]

[

4m 00 2m

] [

1(4βi − 3)

]

.

Solving for ci

ci =

√

1

2m (2 + (4βi − 3)2)−→ c1 =

√

1

12m, c2 =

√

1

6m

Finally, the normalized eigenpairs are

ω1 =

√

k

4m, u1 =

[

1√12 m

− 2√12 m

]

, ω2 =

√

k

m, u2 =

[

1√6 m1√6 m

]

.

c) With the above mode shapes and natural frequencies the general solution becomes

q(t) =

[

x1(t)x2(t)

]

=

2∑

i=1

(Ai sin(ωi t) + Bi cos(ωi t)) ui,

=

(

A1 sin

(

√

k

4mt

)

+ B1 cos

(

√

k

4mt

))[

1√12 m

− 2√12 m

]

+

(

A2 sin

(

√

k

mt

)

+ B2 cos

(

√

k

mt

)) [

1√6 m1√6 m

]

.

77

Problem 91:


a) find the mass and the stiffness matrices;

b) is your system of equations dynamicallycoupled, statically coupled, or both? m1

k1 m3

m2

k2 k1

k3

Solution:

a) For coordinates we choose (x1, x2, x3) as the displacements of each mass with respectto inertial space. Using influence coefficients, we find that:

M =

m1 0 00 m2 00 0 m3

, K =

k1 + k2 −k2 0−k2 k1 + 2k2 −k2

0 −k2 k2

Alternatively, if we choose coordinates (x1, x2, z), where z represents the stretch in thespring connecting m2 and m3, we can determine the mass and stiffness matrices fromthe Lagrangian. The kinetic and potential energies are:

T =1

2m1x

21 +

1

2m2x

22 +

1

2m3(x2 − x3)

2,

V =1

2k1x

21 +

1

2k2(x2 − x1)

2 +1

2k2z

2 +1

2k1x

22.

Thus, with these coordinates the mass and stiffness matrices become:

M =

m1 0 00 m2 + m3 −m3

0 −m3 m3

, K =

k1 + k2 −k2 0−k2 k1 + k2 00 0 k2

b) With the first choice of coordinates, the mass matrix is diagonal while the stiffnessmatrix is not, the system is statically coupled but dynamically uncoupled. With thelatter coordinates neither matrix is diagonal so that the system is both statically anddynamically coupled.

78

Problem 92:



2 .

a) What is the degree-of-freedom for thissystem?

b) Find the governing equations of mo-tion;

c) If m = 1 kg and k = 4 N/m, whatare the frequencies of oscillation forthe motion and the corresponding modeshapes, normalized by the kinetic en-ergy inner product?

r

r2

m

m

k

k

Solution:

a) Let the displacement of the left block, disk, and right block be described as (−x1 ),

(θ k), and (x2 ) respectively. Although x1 and θ are related by the following constraint:

x1 =r

2θ,

x2 is independent from the above two coordinates. Therefore, the system has twodegrees-of-freedom.

b) With the above coordinates, the kinetic and potential energies can be written as:

T =1

2mx2

1 +1

2mx2

2 +1

2

mr2

2θ2,

V =1

2kx2

1 +1

2k(x2 − rθ)2,

Thus, using the above kinematic constraint to eliminate θ, the Lagrangian becomes:

L = T − V,

=1

2m

[

3x21 + x2

2

]

− 1

2k

[

5x21 − 4x1x2 + x2

2

]

.

Using Lagrange’s equations of motion, the governing equations are:

3m x1 + 5k x1 − 2k x2 = 0,

m x2 − 2k x1 + k x2 = 0.

c) From the above equations, the mass and stiffness matrices can be written as:

M = m

[

3 00 1

]

, K = k

[

5 −2−2 1

]

79

The characteristic matrix, A = M−1K becomes:

A =k

m

[

53 − 2

3−2 1

]

,

and the characteristic equation can be written as:

β2 − 8

3β +

1

3= 0,

where, if β is a solution to this equation, λ = kmβ is an eigenvalue of the characteristic

matrix A. This quadratic equations has solutions of the from:

β =4 ±

√13

3.

To determine the eigenvectors, we return to the characteristic matrix A, so that Au =λu . The elements of u then satisfy the equation:

5

3u1 −

2

3u2 = β u1.

Thus, if u1 = 1, this yields:

u2 =1 ∓

√13

2for β =

4 ±√

13

3.

Normalizing by the kinetic energy inner product, we find that:

u =u

√

(u ,u)M=

u√

3u21 + u2

2

Problem 93:

For the system shown in the figure, the sur-face is assumed to be frictionless.

a) Using Lagrange’s equations, determinethe differential equations governing themotion (measured from static equilib-rium);

b) Find the mode shapes and natural fre-quencies of the resulting motion.

4 m

m

(m, r)

k

k

80

Problem 94:

The multi-degree-of-freedom system shown inthe figure is subject to harmonic forcing ofthe form f(t) = sin(4 t).

a) Find the equations of motion in termsof the coordinates x1, x2, and x3, andidentify the mass and stiffness matrices.

b) If m = 2 kg and k = 4 N/m, the natu-ral frequencies of the system are foundto be:

ω1 = 0.38 rad/s, ω2 = 2.62 rad/s,

ω3 = 1.41 rad/s.

Find the corresponding mode shapesand normalize them so that (ui,ui)M =1.

c) Determine the forced, uncoupled equa-tions of motion for the modal ampli-tudes Qi(t).

m

k

m

2 k

m

2 k

f(t)

Problem 95:

For the system shown in the figure, the sur-face is assumed to be frictionless. If eachblock is displaced by a distance d (down andto the right), find the resulting motion of thesystem.

m

m

(m, r)k

k

x1

x2

z

θ

Solution:

We define the coordinates x1, x2, θ, and z as shown in the figure, so that

x1 = −r θ, z = r θ − x2.

81

With these coordinates, the kinetic and potential energies can be written as

T =1

2

(

m)

x21 +

1

2

(

mr2

2

)

θ2 +1

2

(

m)

x22,

V =1

2

(

k)

x21 +

1

2

(

k)

z2.

Expressing these only in terms of the coordinates x1 and x2, we obtain

T =1

2

(

3m

2

)

x21 +

1

2

(

m)

x22,

V =1

2

(

k)

x21 +

1

2

(

k)

(−x1 − x2)2

=1

2

(

2 k)

x21 +

1

2

(

2 k)

x1 x2 +1

2

(

k)

x22.

Therefore, the mass and stiffness matrix can be identifed as

M = m

[

32 00 1

]

, K = k

[

2 11 1

]

with q(t) =

[

x1(t)x2(t)

]

and the equations of motion are:

m

[

32 00 1

] [

x1

x2

]

+ k

[

2 11 1

] [

x1

x2

]

= 0 .

The solution to this equation requires the solution of an eigenvalue problem of theform

(M−1 K ) u =k

m

[

43

23

1 1

]

u = λ u ,

which is determined from the characteristic equation

det(

M−1 K − λ I)

=k

m

[(

4

3− β

)

(1 − β) − 2

3

]

= 0,

with λ = km β. This quadratic equation has the solution

β =7 ± 5

6=

1

3, 2

−→ ω =

√

k

3m,

√

2 k

m

With this, the eigenvectors are determined by returning to the original eigenvalue equa-tion

[

43

23

1 1

] [

ui1

ui2

]

= βi

[

ui1

ui2

]

, −→ ui2 =3βi − 4

2ui1

In addition, normalizing the eigenvectors by the mass matrix

1 = uTi M ui =

[

ui1 ui2

]

[

3 m2 00 m

] [

ui1

ui2

]

= mu2i1

(

6 + (3βi − 4)2

4

)

.

Solving for ui1 the normalized eigenvectors are

u1 =

[

2√15 m

− 3√15 m

]

, u2 =

[

2√10 m2√

10 m

]

.

82

One can easily verify that both uT1 M u2 = 0 and uT

2 M u1 = 0.

The general solution is written as

q(t) =

2∑

i=1

(

Ai sin(ωi t) + Bi cos(ωi t))

ui

subject to the initial conditions

q(0) =

[

d−d

]

, q(0) =

[

00

]

.

Premultiplying by uTi M yields

uTi M q(0) =

(

uTi M ui

)

Bi, uTi M q(0) =

(

uTi M ui

)

(Ai ωi).

Since uTi M ui = 1 from our normalization, the constants are directly solved to be

A1 = 0, B1 = 6 d

√

m

15, A2 = 0, B2 = d

√

m

10

Finally, the solution to the specific initial conditions becomes

q(t) =

[

x1(t)x2(t)

]

=d

5

cos

(

√

k

3mt

)

[

4−6

]

+ cos

(

√

2 k

mt

)

[

11

]

.

Problem 96:

The two-degree-of-freedom system shown issubject to a harmonic force applied to theblock of mass 2m, of the form:

f(t) = (2 sin(t) N) ı.

If the mass and stiffness of the system areassumed to be m = 2 kg, and k = 4 N/m,find:

a) the equations of motion;

b) the forced, damped equation (single-degree-of-freedom) that describes themotion of each mode;

k k 2 k2m m

f(t)

x1

x2

z

Solution:

a) We define the coordinates x1, x2, and z, which are related as

z = x2 − x1.

The equations of motion become

m

[

2 00 1

]

+ k

[

2 −1−1 3

]

=

[

f(t)0

]

.

83

b) The corresponding eigenvalue problem can be written as

(

M−1 K)

=k

m

[

1 − 12

−1 3

]

u =

(

k

mβ

)

u

and the characteristic equation becomes

(1 − β) (3 − β) − 1

2= β2 − 4β +

5

2= 0.

This quadratic equation has the solutions

β = 2 ±√

3

2, −→ ω2 =

(√

8 −√

3) k√2 m

,(√

8 +√

3) k√2 m

.

Returning to the eigenvalue equation, the eigenvectors satisfy the equation

ui1 −1

2ui2 = βi ui1 −→ ui2 = 2 (1 − βi) ui1

so that

ω1 =

√

(√

8 −√

3) k√2 m

, u1 =

[

1√6 − 2

]

,

ω2 =

√

(√

8 +√

3) k√2 m

, u2 =

[

1

−(√

6 + 2)

]

.

For each eigenvector the kinetic energy inner product is

uT1 M u1 = 14 − 4

√6, uT

2 M u2 = 14 + 4√

6,

Finally, the modal equation for the first mode can be written as

(

uT1 M u1

)

Q1 +(

uT1 K u1

)

Q1 = uT1 f (t),

Q1 + ω21 Q1 =

uT1

f (t)

uT1

M u1

,

Q1 +(

(√

8−√

3) k√2 m

)

Q1 = f(t)

14−4√

6,

while the response of the second mode is governed by

(

uT2 M u2

)

Q2 +(

uT2 K u2

)

Q2 = uT2 f (t),

Q2 + ω22 Q2 =

uT2

f (t)

uT2

M u2

,

Q2 +(

(√

8+√

3) k√2 m

)

Q2 = f(t)

14+4√

6,

84

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