Mechanical vibration solved examples

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Mechanical Vibrations: 4600-431 Example Problems March 1, 2010

1 Free Vibration of Single Degree-of-freedom Systems

1.1 Solved Problems

Problem 1:

For the mechanical system shown to the right, the uniform rigid bar has mass m and pinned at point O. For this system:

a) find the equations of motion;

b) Identify the damping ratio and natural frequency in terms of the parameters m, c, k, and ℓ.

c) For:

m = 1.50kg, ℓ = 45cm, c = 0.125N/(m/s), k = 250N/m, find the angular displacement of the bar θ(t) for the following initial conditions:

θ(0) = 0, ˙θ(0) = 10rad/s. Assume that in the horizontal position the system is in static equilibrium and that all angles remain small.

ℓ 2 ℓ 2 m O k y 2 m k c x θ ˆı ˆ Solution:

a) In addition to the coordinate θ identified in the original figure, we also define x and y as the displacment of the block and end of the bar respecively. The directions ˆı and ˆ are defined as shown in the figure.

A free body diagram for this system is shown to the right. Note that the tension in the cable between the bar and the block is unknown and represented with T while the reaction force FRis included, although both

its magnitude and direction are unspecified. In terms of the identifed coordinates, the angular acceleration of the bar αβ and the

linear acceleration of the block aG are

αβ= ¨θ ˆk, aG= ¨x ˆ. F_R −k y ˆ −T ˆ T ˆ −k x ˆ −c ˙x ˆ

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 balance applied to the block and angular momentum balance aout O on the bar. These can be

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written as X F = m aG−→ T − k x − c ˙xˆ = 2 m ¨x ˆ, X MO = IOαβ−→ − T ₂ℓ− k y ℓ ˆ k = m ℓ 2 3 θ ˆ¨k.

Solving the first equation for T and substituting into the second equation yields − (2 m ¨x + k x + c ˙x) ℓ

2 − k y ℓ = m ℓ2

3 θ.¨

Using the coordinate relations we can obtain the equation of motion as 5 m ℓ2 6 θ +¨ c ℓ2 4 ˙θ + 5 k ℓ2 4 θ = 0.

b) In the above equation the equivalent mass, damping, and stiffness are meq= 5 m ℓ2 6 , beq= c ℓ2 4 , keq= 5 k ℓ2 4 . From these the damping ratio and natural frequency are

ζ = beq 2pkeqmeq = c ℓ2 4 2q5 k ℓ2 4 5 m ℓ2 6 = √ 3 c 2√50 k m, ωn= s keq meq = s_{5 k ℓ}2 4 5 m ℓ2 6 =r 3 k 2 m

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

ωn= 15.8rad/s, ζ = 7.91 × 10−4.

Therefore the system is underdamped and the general solution can be written as θ(t) = e−ζ ωnt_{a sin}_ω np1 − ζ2t + b cosωnp1 − 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 ωnb + ωnp1 − ζ2a = 10rad/s,

so that the general solution becomes

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Problem 2:

c) For:

m = 2kg, ℓ = 25cm, c = 0.25N/(m/s), k = 50N/m, find the angular displacement of the bar θ(t) for the following initial conditions:

θ(0) = 0, ˙θ(0) = 10rad/s. d) for this motion, find the tension in the

cable connecting the rod and the block as a function of time.

Assume that the system is in static equilib-rium at θ = 0, and that all angles remain small. ℓ 2 ℓ 2 m k z m k c x θ ˆı ˆ Solution:

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

x = ℓ

2θ, z = ℓ 2θ. An appropriate free-body diagram is

shown to the right. Applying linear mo-mentum balance on the block yields

X

F = m aG,

(T − k x − c ˙x) ˆ = m ¨x ˆ. Likewise, angular momentum balance on the bar provides

X MO = IOαβ, −T ℓ 2− k z ℓ 2 ˆ k = m ℓ 2 12 θ ˆ¨k. FR k z ˆ −T ˆ T ˆ −c ˙x ˆ −k x ˆ

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

7 m

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

7 m

6 , beq= b, keq= 2 k, and the natural frequency and damping ratio are

ωn = s keq meq =r 12 k 7 m, ζ = beq 2pkeqmeq = √ 3 b √ 28k m. Problem 3:

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 static equilibrium position 15cm to the right and released from rest.

m = 4.0kg, b = 0.25N/(m/s), k1= 1.5N/m, k2= 0.50N/(m/s). x k1 b k2 m ˆı ˆ Solution:

An appropriate free-body diagram is shown to the right. Notice that the two springs are effectively in parallel, as the displacement across each spring is identical. Linear momentum balance on this block provides X F = m aG, (−k1x − k2x − b ˙x) ˆı = m ¨x ˆı, −k1x ˆı −k2x ˆı −b ˙x ˆı

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= 15cm, ˙x(0) = 0cm/s.

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Problem 4:

For the system shown to the right, the disk of mass m rolls without slip and x measures the displacement of the disk from the unstretched position of the spring. The surface is inclined at an angle of φ with respect to vertical.

a) find the equations of motion. Do not neglect gravity;

b) if the system is underdamped, what is the frequency of the free vibrations of this system in terms of the parameters k, c, and m;

c) for what value of the damping constant c 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 ˆı ˆ ˆe₁ ˆe₂ Solution:

a) In addition to x, the displacement of the center of the disk, we identify the coordinates z and θ, the displacement across the spring and the rotation of the disk respectively. These additional coordinates are related to x as

z = 2 x, x = −r θ.

An appropriate free-body diagram is shown to the right. We note that (ˆı, ˆ) are related to the direc-tions (ˆe1, ˆe2) as

ˆı = cos φ ˆe1+ sin φ ˆe2,

ˆ = − sin φ ˆe1+ cos φ ˆe2.

−k z ˆe1

−c ˙x ˆe1

fr ˆe1

N ˆe2

−m g ˆ

The moment produced by gravity about point C is M gravity = rGC× (−m g ˆ),

= (r ˆe2) × (−m g ˆ) = −m g r sin φ ˆk.

Angular momentum balance about the contact point C yields X MC = ICαD, (2 r) k z + r c ˙x − m g r sin φ ˆk = 3 m r 2 2 θ¨ ˆ k.

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

3 m

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

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

ωn p1 − ζ2, where ωn = s keq meq =r 8 k 3 m, ζ = beq 2pkeqmeq = c 2√6 k m, so that ωd= r 8 k 3 m r 1 − c 2 24 k m,

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

ccr= 2

√ 6 k m.

d) The system is stationary in static equilibrium, so that x ≡ x0= constant—both ˙x and

¨

x vanish, and the equation of motion reduces to 4 k x0= m g sin φ.

Solving for x0, the equilibrium displacement is

x0=

m g sin φ 4 k .

Problem 5:

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

a) Determine the deflection of each spring from its unstretched length when the system shown is in equilibrium. b) If the system is released from the

un-stretched position of the springs, what is the maximum angular velocity of the disk during the resulting motion?

r2 r1 k1 k2 m I ˆı ˆ x z1 z2 θ Solution:

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

as shown in the figure, which are related as

x = −r1θ, z1= r1θ, z2= −r2θ.

Notice that because of these coordinate definitions, a rotation with positive θ gives rise to a negative value in both x and z2. Likewise, we see that x = −z1.

Using the free-body diagram shown to the right, linear momentum balance on the block provides

X

F = m aG,

(T − m g) ˆ = m ¨x ˆ,

while angular momentum balance on the disk yields FR k1z1 ˆ −k2z2ˆı −T ˆ T ˆ −m g ˆ X MO = IOαD, (T r1− k1z1r1+ k2z2r2) ˆk = I ¨θ ˆk.

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

I + m r21 ¨θ + k1r12+ k2r22 θ = m g r1.

The equilibrium rotation of the disk thus is found to be θeq=

m g r1

k1r21+ k2r22

.

With this, the equilibrium deflection of each spring is found to be z1,eq= r1θeq= m g r2 1 k1r12+ k2r22 , z2,eq= −r2θeq= − m g r1r2 k1r21+ k2r22 .

b) The general free response of the disk can be expressed as θ(t) = θeq+ A sin(ωnt) + B cos(ωnt),

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

ωn= s k1r12+ k2r22 I + m r2 1 .

The system is released with the initial conditions: θ(0) = 0, ˙θ(0) = 0,

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so that solving for the arbitrary constants

A = 0, B = −θeq.

Therefore the solution is

θ(t) = θeq(1 − cos(ωnt)) = m g r1 k1r21+ k2r22 1 − cos s k1r21+ k2r22 I + m r2 1 t !! .

The angular velocity of the disk becomes ˙θ(t) =

θeqωn

sin(ωnt),

which has amplitude

Ω = θeqωn=

m g r1

p(k1r21+ k2r22) (I + m r21)

Problem 6:

b) what value of the damping constant c gives rise to a critically damped

sys-tem? ℓ2 ℓ 2 m k c z1 m k z2 θ ˆ ˆı Solution:

a) In addition to θ, we define two additional coordinates, z1and z2, to measure the

deflec-tion at the left and right ends of the bar. These coordinates are related as:

z1=

ℓ

2 θ, z2= ℓ

2 θ = z1. A free-body diagram for this system is shown to the right. Applying angular mo-mentum balance on the bar eliminates the appearance of the reaction force and leads to: X MG = IG θ ˆ¨k, T − k z1− c ˙z1 ℓ 2 kˆ = m ℓ2 12 θ ˆ¨k. G FR k z1ˆ c ˙z1ˆ −k z2ˆ −T ˆ T ˆ

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Likewise, application of linear momentum balance on the block yields: X F = m aG, − k z2− T ˆ = m¨z2ˆ

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

motion becomes: m ℓ2 3 θ +¨ c ℓ2 4 ˙θ + k ℓ2 2 θ = 0. b) A critically damped system occurs when ζ = 1. For this system:

ζ = √

3 c √

32 k m.

Solving for ccr yields:

ccr=

r 32 k m 3 .

Problem 7:

Find the response of the system shown to the right if the block is pulled down by 15cm and released form rest.

m = 2.0kg, b = 0.5N/(m/s), k1= 0.5N/m, k2= 0.25N/(m/s). x k1 k2 b m ˆ ˆı Solution:

For this system, the two springs in series may be replaced by an equivalent spring, with constant: keq= 1 1 k1 + 1 k2 = k1 k2 k1+ k2 .

Therefore, the free-body diagram is shown to the right. Applying linear momentum balance to the block yields:

X F = m aG, keqx + b ˙x ˆ = −m ¨x ˆ, which can finally be written as:

m ¨x + b ˙x + keqx = 0.

keqx ˆ b ˙x ˆ

With the numerical values given above, this becomes: (2kg) ¨x + 1 2 N m/s ˙x + 1 6 N m x = 0, x(0) = 3 20m, ˙x(0) = 0m/s.

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With this, the damping ratio and natural frequency are: ωn = r 1 12s −1_, _{ζ =} √ 3 4 .

Therefore, the system is underdamped and the general response can be written as: x(t) = e−ζ ωnt_{A cos(ω}

dt) + B sin(ωdt)

.

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

x(t) = 3 20e −t/8 _cos √ 13 8√3t ! +r 3 13cos √ 13 8√3t !! . Problem 8:

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

a) Determine the deflection of each spring from its unstretched length when the system shown is in equilibrium. b) If the system is released from the

un-stretched position of the springs, what is the maximum angular velocity of the disk during the resulting motion?

r2 r1 k1 k2 m x2 θ x1 ˆ ˆı 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 unstretched position. These coordinates are related through the following transformations:

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An appropriate free-body diagram for this system is shown to the right. Notice that the gravitational force must be included to determine the equilibrium deflection in the system. To eliminate the reaction force on the disk, angular momentum balance is ap-plied about the center, yielding:

X

MG = IGθ ˆ¨k,

T r1+ k2r2x2 ˆk = IGθ ˆ¨k.

Also, applying linear momentum balance to the block yields:

X F = m aG, T − k1x1− m g ˆ = m ¨x1ˆ. FR G −m g ˆ T ˆ −T ˆ −k2x2ˆı −k1 x1ˆ

Finally, eliminating the unknown tension from these equations and using the above coordinate transformations, this single-degree-of-freedom system can be modeled with the equation: IG+ m r21 ¨ θ +k1 r12+ k2r22 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 r22

.

Therefore, the equilibrium displacements in each spring are: x1,eq= −r1θeq= m g r 2 1 k1 r21+ k2r22 , x2,eq= −r2θeq= m g r1 r2 k1 r12+ k2r22 .

b) Define new coordinates z1 and z2, which measure the displacement in springs 1 and 2

with 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 equilibrium position:

φ = θ − θeq.

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

V = 1 2 k1 z 2 1+ 1 2 k2 z 2 2, = 1 2 k1 r21+ k2 r22 φ2_.

Also, the kinetic energy becomes:

T = 1 2 I G _φ_˙2₊1 2 m ˙z 2 1, = 1 2 IG+ m r21 ˙ φ2.

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If the system is released from rest at the unstretched position of the springs, then: φ(0) = −θeq= −

m g r1

k1 r12+ k2 r22

, φ(0) = 0.˙

At this initial state, the potential and kinetic energies become: T0= 0, V0=

(m g r1)2

2(k1 r12+ k2 r22)

.

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

V1= 0, T1= 1 2 IG+ m r21 ˙ φ2max.

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

˙ φmax= s (m g r1)2 (IG_{+ m r}2 1)(k1 r12+ k2 r22) . Problem 9:

For the system shown to the right, the disk of mass m rolls without slip and x measures the displacement of the disk from the unstretched position of the spring.

b) if the system is underdamped, what is the frequency of the free vibrations of this system in terms of the parameters k, c, and m; k 3k c m x z θ ˆı ˆ 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 is

shown to the right. Notice that the force in the upper spring depends on z, rather than x, while the friction force has an unknown magnitude f . Because the disk is assumed to roll without slip, we are unable to specify the value of f , but instead can relate the dis-placement and rotation of the disk through the coordinate relations above.

−k x ˆı −3k z ˆı −c ˙x ˆı f ˆı G C

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

X

MC = IC θ ˆ¨k,

(3k z) 2r + (k x) r + (c ˙x) r ˆk = 3 m r2 2 θ ˆ¨k.

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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 = ωnp1 − ζ2. With

this system, we find that: ωn = r 26 k 3 m, ζ = c √ 78 k m, so that: ωd= r 26 k 3 m − 2 c2 9 m2. Problem 10:

In the system shown to the right, the pulley has mass m and radius r, so that the moment of inertia about the mass center is IG=mr

2 2 .

a) Find the governing equations of mo-tion;

b) Find the frequency of oscillation for free vibrations of the system;

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

r r 2 m m k k c Solution:

a) We choose coordinates (x1, x2, θ), where x1measures the displacement of the first mass

in the −ˆ direction, x2measures the displacement of the second mass in the ˆ direction,

and θ measures the angular rotation of the wheel in the ˆk direction. The kinetic and potential energies for this system are:

T = 1 2 m ˙x 2 1+ 1 2 m ˙x 2 2+ 1 2 mr2 2 ˙θ 2_, V = 1 2 k x 2 1+ 1 2 k x 2 2.

However, these three coordinates are dependent through the transformations: x1=

r

2 θ, x2= r θ. Therefore, the Lagrangian reduces to:

L = T − V = 1 2 7mr2 4 ˙θ2₋1 2 5kr2 4 θ2_.

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Further, the generalized force resulting from the viscous damper becomes: Qθ= −

cr2

4 ˙θ,

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=

r 5k

7m, ζ =

c 2√35 km.

Therefore, the damped natural frequency becomes:

ωd= ωnp1 − ζ2= r 5k 7m r 1 − c 2 140 km.

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

ccr= 2

√ 35 km.

Problem 11:

In the figure, the disk has mass m, radius r, and moment of inertia IG = mr

2

2 about the

mass center G, and is attached to a block of mass m which rolls across the surface. If the disk rolls without slip (µ is sufficiently large), while the block moves without friction:

a) find the equations of motion for this system;

b) for m = 2kg, b = 0.5(N · s)/m, and k = 8N/m, find the frequency of oscillation for the system.

m b

k

G

Solution:

a) Let x denote the translational displacement of G in the ˆı direction, while θ denotes the angular displacement of the disk in the ˆkdirection. If T is the tension between the disk and the mass and f is the frictional force acting on the disk, the equations of motion on the disk and the mass are:

(−kx − b ˙x + T + f) ˆı = m¨x ˆı, (rf ) ˆk = mr

2

2 θ ˆ¨k, (−T ) ˆı = m¨x ˆı.

In addition, x and θ are related through the kinematic constraint: x = −rθ.

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Eliminating (θ, T, f ) we obtain a single degree-of-freedom system on x of the form: 5m

2 x + b ˙x + kx = 0.¨

b) From the above equation we identify the undamped natural frequency and damping ratio as: ωn = r 2k 5m, ξ = b √ 10km. Thus, the damped natural frequency is:

ωd= ωnp1 − ξ2=

r

10km − b2

25m2 .

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

Problem 12:

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

m¨x + b ˙x + kx = 0,

and experimentally measure the mass as m = kg. In free vibration we experimentally de-termine the equivalent spring constant to be k = 4N/m, and we measure the response as shown.

a) Determine the equivalent damping con-stant.

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

b c b c b c b c x1= 1.00m x2= 0.75m Solution:

a) We use the logarithmic decrement so that: ζ = δ1 p4π2_{+ δ}2 1 , δ1= − ln x1 x0 = − ln 0.75m_1.00m = 0.288,

and we find that ζ = 0.0457. With m = 1kg, the damping constant b is given as: b = 2ζ√km = 0.183N/(m/s).

We note that with m given, we can also find the period of the oscillations to be: T =2π ωd = 2π ωnp1 − ζ2 , = √_mp4π₂_{+ δ}2 1 √ k = 3.145s.

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Problem 13:

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

a) Find the linearized equations of motion in terms of z, the relative displacement between the end of the beam and the ground;

b) what is the frequency of the resulting motion;

c) if the mass of the spring is taken to be mspring = 1kg, find the new frequency

of 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 ˆk. Therefore, using angular momentum balance about the point of rotation, we find:

mbarl2

3 θ + bl ˙z + kl z = 0.¨

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

reduces to: _m

bar

3 ¨z + b ˙z + k z = 0. For the parameter values given above, this becomes:

2 3kg

¨

z +2N/(m/s) ˙z +4N/m z = 0.

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

ωn = r 3k mbar , ζ = √ 3 b 2√kmbar =√6rad/s =r 3 8.

b) The frequency of oscillation, ωd= ωnp1 − ζ2, reduces to:

ωd= s 3k mbar − 3b 2mbar 2 = √ 15 2 rad/s = 1.94rad/s.

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

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the bar about the point of rotation is: IO = mbar l 2 3 + meql 2_, = mbar l 2 3 + m_spring 3 l2, = (mbar+ mspring) l 2 3 .

Therefore, the new frequency of oscillation is:

ωd= s 3k (mbar+ mspring)− _3b 2(mbar+ mspring) 2 =√3rad/s = 1.73rad/s. Problem 14:

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

x1 = 0.25m x2 = 0.20m,

t1 = 1.00s t2 = 4.00s,

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

m b c b c b c b c (t1, x1) (t2, x2) Solution:

a) The logarithmic decrement is defined as δ = ln |x2/x1|. In terms of this quantity and

the period of oscillation T , the damping ratio and natural frequency are defined as:

ωn=

p(2π)2_{+ δ}2

T , ζ =

δ p(2π)2_{+ δ}2.

Therefore, the stiffness and damping constants reduce to: k = m ω2 n= m (2π)2_{+ δ}2 T2 , b = m 2ζωn = m 2δ T .

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

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Problem 15:

For the single-degree-of-freedom mechanical system shown in the figure:

a) determine the governing equations of motion;

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 = 1kg, b = 12(N · s)/m, and k = 9N/m. Assume the pulley is massless and neglect 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 the angular displacement of the pulley, both measured from the static equilibrium position. If T represents the tension in the cable supporting mass m, then T = kr₂θ, where θ is the 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 = _2rx. As a result, in the ˆ direction, the equation governing the motion of the mass is:

m¨x + b ˙x +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 2pkeqm =√b km, ωn= r keq m = √ k 2√m.

In terms of the given parameters, we find ζ = 4.0 and ωn = 3rad/s. We note that the

damping ratio has no units.

c) For an overdamped system, the general solution is: x(t) = exp(−ζωnt) c1exp (ωnpζ2− 1)t + c2exp −(ωnpζ2− 1)t ,

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and with these initial conditions this reduces to: x(t) = x0exp (−ζωnt) ζ + pζ2_{− 1} 2pζ2_{− 1} exp (ωnpζ2− 1)t + ζ −pζ2_{− 1} 2pζ2_{− 1} exp −(ωnpζ2− 1)t ! , = x0e−6t 4 +√15 2√15 e 3 2 √ 15t₊4 − √ 15 2√15 e −3 2 √ 15t ! , = x0e−6.0t 1.016e5.81t+ 0.01640e−5.81t . Problem 16:

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

a) Determine the governing equations of motion.

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 initial conditions:

x(0) = 0.0m, ˙x(0) = 5.0m/s, find the resulting solution x(t) if m = 2kg, k = 48N/m and c = 4N/(m/s).

x

m

6 k 2 c

Solution:

a) In terms of x, the spring and damping forces can be written as: Fspring= −6k x ˆı, Fdamper= −2c ˙x ˆı.

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

F = (−6k x − 2c ˙x)ˆı = m¨x ˆı = maG,

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

¨ x + (2ζωn) ˙x + (ωn2) x = 0, with ωn= r 6k m, ζ = √c 6km.

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

ωnp1 − ζ2. Therefore, for this system, the period reduces to:

T = 2π r 6k m s 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= 12rad/s, and ζ = 1/6. For these values, the general solution can be written

as:

x(t) = A e−2 t sin2√35 t + φ,

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

A =r 5

28m, φ = 0rad, so that the general solution can be written as:

x(t) =r 5 28 e

−2 t _sin₂√_{35 t} _m.

Problem 17:

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

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

k 2 k b

m

g

Solution:

We assume that the ˆı and ˆ directions are standard orthonormal basis in the horizontal and vertical directions respectively.

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a) The forces due to the springs in parallel and damping are: Felastic= 3k · xˆ, Fdamping= b · ˙xˆ.

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

X

F _{= m (−¨xˆ) ,} 3k · xˆ + b · ˙xˆ − mgˆ = −m¨xˆ.

Taking components in the ˆ direction, we obtain the governing equation of motion: m¨x + b ˙x + 3kx = mg.

b) Dividing through by the mass m, we find: ¨ x + b m˙x + 3k mx = g, so that: 2ζωn = b m, ω 2 n = 3k m, which can be solved to yield:

ωn =

r 3k

m, ζ =

b 2√3mk.

c) In equilibrium, the system is stationary, so that ˙xeq= 0 and ¨xeq= 0. Substitution into

the governing equations yields:

3kxeq= mg, → xeq=

mg 3k .

Problem 18:

a) find the linearized governing equations of motion for small θ;

b) find the frequency of oscillation for the response;

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

system. k 4 k (m, l) G _θ Solution:

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a) The displacement of each end of the bar in the ˆ direction is: x_± = ±l sin θ₂ .

As a result, the total moment produced by the springs about the center of mass G is: X MG = − l2 4 (4k sin θ + k sin θ) ˆk, = −5kl 2 4 sin θˆk. Thus angular momentum balance about G provides:

X MG = IGθˆ¨k, −5kl 2 4 sin θˆk = ml2 12 θˆ¨k.

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

¨ θ +15k

m θ = 0.

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

ω = ωn=r 15k

m .

c) This system possesses the general solution:

θ(t) = c1sin ωt + c2cos ωt,

which, for the initial conditions given above, yields the solution: θ(t) = θ0cos

r 15k m t.

Problem 19:

We obtain the differential equation: m¨x + b ˙x + 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 vibration to be reduced to within 2% of zero?

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a) Written in nondimensional form, the equations of motion are: ¨ x + b m˙x + k mx = 0,

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

ζ = b

2√km, ωn= r 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= r k m = 3.

Therefore the eigenvalues of this system are: λ1,2 = ωn

ζ ±pζ2_{− 1}_{= 3(−2 ±}√_3).

The dominant eigenvalue is λ = −6 + 3√3, and so the time t = τ required for the amplitude of the free vibration to be reduced to within 2% of zero is:

τ ∼ −4

−6 + 3√3 ∼ 5.0

Problem 20:

For the single-degree-of-freedom mechanical system shown in the figure, the bar has mass m and length l (so that IO = ml

2

3 ). If the

spring is unstretched when θ = 0:

a) find the linearized governing equations of motion for small θ;

b) find the frequency of oscillation for the free response; Neglect gravity. ℓ 2 ℓ 2 k c (m, l) θ Solution:

a) Using angular momentum balance about the fixed point O, we find: X MO = IOαβF, −kl2sin θ − cl 2 4 ˙θ cos θ ˆ k = ml 2 3 θˆ¨k,

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so that the equation of motion can be written as: ml2 3 θ + c¨ l2 4 ˙θ cos θ + k l 2_{sin θ = 0.}

Linearizing this equation about θ = 0, we obtain: ¨

θ + 3c 4m˙θ +

3k mθ = 0.

b) The frequency of oscillation, that is, the damped natural frequency, is given as: ωd = ωnp1 − ζ2,

where ωnis the undamped natural frequency and ζ is the damping ratio. For this system

we find: ωn =r 3k m, ζ = √ 3 c 8√km, so that the damped natural frequency is:

ωd= s 3k m − 3c 8m 2 . Problem 21:

In the system shown to the right, the pulley has mass m and radius r, so that the moment of inertia about the mass center is IG=mr

2 2 .

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

c) Find the frequency of oscillation for free vibrations of the system?

r 2 r m m k k c Solution:

a) Define θ as the angular displacement of the disk in the −ˆk direction (clockwise), and x1and 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 T1and T2 respectively. Thus, linear

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yield: T1 r 2 − T2r = − mr2 2 θ,¨ T1− kx1− c ˙x1 = m¨x1, T2+ kx2 = −m¨x2.

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

7mr2 4 ¨ θ + cr 2 4 ˙θ + 5kr2 4 θ = 0.

b) Examination of the above equation shows that meq = 7mr 2

4 . Note that this answer is

not unique, but depends on the equation of motion. For example, if we had written the equation 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= ωnp1 − ζ2= r 5k 7m r 1 − c 2 140km. Problem 22:

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

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

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

x

m k

b

Solution:

For this system the differential equation of motion can be written as: m¨x + b ˙x + kx = 0, ¨ x + b m˙x + k mx = 0, subject to specified initial conditions.

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

¨

x + (6.5kg/s) ˙x + (9kg/s2)x = 0, whose characteristic equation reduces to:

λ2+13

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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 overdamped and decays exponentially with no sustained oscillations. The general solution is given as:

x(t) = c1 e− 9 2t+ c

2 e−2t.

For the initial conditions x(0) = 0m, ˙x(0) = 10.0m/s, we find that: c1= 4m, c2= −4m,

and the general solution to this equation, subject to these initial conditions, becomes: x(t) =4m e−(9/2s−1)t− e−(2s−1)t.

b) For this system, the damping ratio is:

ζ = b 2√km.

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

bcr= 12(N · s)/m.

Problem 23:

2

2 about the

mass center G.

a) Find the equations of motion for this system assuming that the disk rolls without slip.

b) What value of b correspond to critical damping?

c) Find the displacement of the center of the disk when the system is critically damped and released from rest with x(0) = x0. (m, r) G x k b g Solution:

We define θ as the angular displacement of the disk in the ˆk direction as measured from the equilibrium position of the disk (when the spring is unstretched). Assuming that the disk rolls without slip, the rotation and translation of the disk can be related through the constraint equation:

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

Fspring= −kx ˆı, Fdamper= −b ˙x ˆı.

Using linear and angular momentum balance on the disk, we find that: X F _{= (f − kx − b ˙x)ˆı = m¨x ˆı = ma}G, X MG = (f r) ˆk = mr2 2 θ ˆ¨k = IG αβF.

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

3m 2

¨

x + b ˙x + kx = 0.

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

¨ x + (2ζωn) ˙x + (ωn2)x = 0, with ωn= r 2k 3m, ζ = √b 6km.

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+ c2t e−ωnt,

while this can be differentiated with respect to time to obtain the velocity: ˙x(t) = (c2− ωnc1) − (ωnc2) t e−ωnt.

If the system is released from rest when a known initial displacement, then the initial conditions 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−ωnt,

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Problem 24:

In the figure, the disk has mass m, radius r, and moment of inertia IGabout the mass

cen-ter, and the applied moment has a constant magnitude M ˆk. 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 when the system is in equilibrium?

(m, r) G M ˆk x k c g Solution:

a) The governing equations of motion are: m + I r2 ¨ x + c ˙x + k x = −M r .

b) The equivalent mass, damping, and stiffness are: meq= m +

I

r2, ceq= c, keq= k.

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

M kr.

1.2 Unsolved Problems

Problem 25:

In the figure, the disk has mass m and radius r. The cable wraps around the disk with no slip and is inextensible, attached to a spring of stiffness k and a damper with coefficient c. a) Find the equations of motion for this

system.

b) What are the natural frequency and damping ratio of the system?

c) If the disk is displaced 12cm down and released from rest find the resulting an-gular displacement of the disk with

m = 3kg, r = 9cm, k = 21N/m, c = 63N · s/m,

(m, r) c k

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Problem 26:

2

2 about the

mass center G and is assumed to roll without slip.

a) Find the equations of motion for this system.

b) What are the natural frequency and damping ratio for this system?

k k c (m, r) r/2 G g Problem 27:

In the spring-mass-damper system shown, the block slides on a frictionless surface, where the coordinate x measures the dis-placement of the system from the unstretched position of the spring. The mass of the block is m = 2kg, while the spring and damp-ing coefficients are k = 288N/m and b = 16(N · s)/m. Finally, the plane is inclined down by an angle φ = 20◦_.

a) Determine the static equilibrium dis-placement of the block from the un-stretched position;

b) Find the resulting solution if the sys-tem is released from the unstretched position with initial velocity ˙x(0) = −3.0m/s;

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

x m k b φ g

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Problem 28:

For the system shown to the right

a) find the equations of motion in terms of the rotation of the disk (do not neglect gravity);

b) determine the deflection of the spring from its unstretched length when the system shown is in equilibrium.

r2 r1 m k2 m I g Problem 29:

c) For:

m = 2.50kg, ℓ = 30cm, c = 1.00N/(m/s), k = 200N/m, find the angular displacement of the bar θ(t) for the following initial conditions:

θ(0) = π/6rad, ˙θ(0) = 0rad/s. Assume that in the horizontal position the system is in static equilibrium and that all angles remain small.

ℓ 2 ℓ 4 m O k 2 m c θ

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Problem 30:

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

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

c) If the block is displaced 18cm to the right and released from rest find the re-sulting angular displacement of the disk with m = 3kg, r = 9cm, k = 21N/m, c = 63N · s/m, k c m 4 k k (m, r) Problem 31:

a) Find the equations of motion. b) With

c = 16N/(m/s), m = 2kg, r = 0.10m

for what value of the spring stiffness k is the damping ratio of the system one-half of the critically damped value, so that ζ = ζcr/2?

c) With these parameter values, find the displacement of the disk if it is rolled 20cm to the right (from static equilib-rium) and released from rest.

x

c k

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Problem 32:

From the figure shown to the right

a) find the equations of motion in terms of the angular rotation of the disk; b) what are the damping ratio and natural

frequency of the system in terms of the parameters m, b, k1, and k2;

c) can you draw an equivalent spring-mass-damper system? r r 2 b k1 k2 m m Problem 33:

2

2 about the

mass center G and is assumed to roll without slip. The identified coordiante θ measures the rotation of the disk with respect to the equi-librium position.

b) If the disk is released from rest with θ(0) = −π

2rad, find the resulting

angu-lar displacement θ(t) for m = 3kg, r = 15cm, k = 36N/m, c = 3N · s/m, c) What is the force in the upper spring

during this motion?

k c k (m, r) G g

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Problem 34:

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

b) what is the frequency of oscillation of the system;

c) what value of the damping coefficient b corresponds to critical damping? d) if k = 2N/m, b = 4N/(m/s), and m =

3kg, find the displacement of the mass x(t) when the system is started with the initial conditions: x(0) = 0.10m, ˙x(0) = 0m/s. x m 6 k b Problem 35:

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 disk has moment of inertia I and radius r. If the gravitational 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 36:

2

2 about the

mass center G.

a) Find the equations of motion for this system assuming that the disk rolls without slip.

b) If the disk is released from rest with initial displacement x(0) = x0, find the

minimum value of the coefficient of fric-tion for which the disk does not slip.

(m, r) G x k c g

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Problem 37:

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

b) what is the period of each oscillation; c) what value of the damping coefficient b

corresponds to critical damping? d) if k = 1N/m and m = 4kg, find the

dis-placement of the mass x(t) if the sys-tem is critically damped and started with the initial conditions x(0) = 0,

˙x(0) = ˙x0;

x

m

6 k b

Problem 38:

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 disk has moment of inertia I. If the gravitational constant is g:

a) what is the displacement of the spring at static equilibrium;

b) find the kinetic energy of the system in terms of the coordinate x (and/or its velocity). x m m (I, r) k

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2 Forced Single Degree-of-freedom Systems

2.1 Solved Problems

Problem 39:

For the system shown to the right, the disk of mass m rolls without slip and the inner hub has radius ρ/2.

a) Find the equations of motion (in terms of the given parameters—do not substi-tute in numerical values yet);

b) If the applied moment takes the form: M (t) = (2N · m) sin(4 t), find the steady-state amplitude of the translation of the center of the disk when:

k = 16N/m, b = 2N/(m/s), m = 2kg, ρ = 0.125m c) Determine the steady state amplitude

of the friction force.

b 2 k k M (t) _{(m, ρ)} θ x z ˆ ˆı C G Solution:

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

x = −ρ θ, z = 3 2x.

An appropriate free-body diagram for this system is shown to the right. Since the disk is assumed to roll without slip, the equation of motion can be directly obtained with an-gular momentum balance about the contact point C X MC = IC αD, −b ˙x ˆı −2 k z ˆı −k x ˆı M (t) ˆk frˆı N ˆ −m g ˆ which yields ρ (k x + b ˙x) +3 ρ 2 (2 k z) + M (t) ˆ k= 3 m ρ 2 2 θ ˆ¨k. Using the above coordinate transformations this equation can be written as

3 m 2 x + b ˙x +¨ 11 k 2 x = − 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),

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from which we can identify the appropriate parameters as: ωn= r 88 3 , ζ = 1 √ 264, r = r 6 11.

Therefore, the amplitude of the translational oscillations becomes X = F kM(r, ζ) = 8 88 1 r 1 − 116 2 +2√1 264 q 6 11 2 =√1 26.

Likewise, the phase shift of the response is:

tan φ = 2 ζ r 1 − r2 = 2√1 264 q 6 11 1 − 116 = 1 5, so that φ = 0.20rad = 11.3◦.

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

X

F =fr− k x − 2 k z − b ˙x

ˆı +N − m g= m ¨x ˆı = m 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, the friction force becomes

fr(t) =

4 k − m ω2X sin(ω t − φ) +b ωX cos(ω t − φ).

The magnitude of the friction force is then found to be f = X r 4 k − m ω22₊_{b ω}2_.

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Problem 40:

a) find the equations of motion (in terms of the given parameters—do not substi-tute in numerical values yet);

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

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

ℓ 2 ℓ 2 m k c z f (t) = sin(ω t) θ ˆ ˆı Solution:

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

z = ℓ 2 θ

A free-body diagram for this system is shown to the right. Applying angular mo-mentum balance on the bar eliminates the appearance of the reaction force and leads to: X MG = IGθ ˆ¨k, f (t) − k z − c ˙z₂ℓkˆ = m ℓ 2 12 θ ˆ¨k. G FR k z1ˆ c ˙z1ˆ f (t) ˆ

Solving for z in terms of θ, the equation of motion becomes: m ℓ2 12 θ +¨ c ℓ2 4 ˙θ + k ℓ2 4 θ = 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= r 3 k m , ζ = √ 3 c 2 √k m, M0= 6 m ℓ.

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The amplitude of the moment transmitted to the ground can be written as: MT = (meqM0) p1 + (2 ζ r)2 p(1 − r2₎2_{+ (2 ζ r)}2, = ℓ 2 q 1 + c ω_k 2 q 1 −m ω3 k2 2 + c ω_k 2 = ℓ 2 pk2_{+ (c ω)}2 q k −m ω32 2 + (c ω)2 .

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

FT = pk2_{+ (c ω)}2 q k − m ω3 2 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 ω2 n 1 p(1 − r2₎2_{+ (2 ζ r)}2, = 2 k ℓ 1 q 1 − m ω2 3 k 2 + c ω k 2 = 2 ℓ 1 q k − m ω2 3 2 + (c ω)2 .

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

Θ = 8 k −2 ω 2 3 Therefore, if the amplitude of vibration is less than π/6:

8 k −2 ω 2 3 ≤ π₆, 48 π ≤ k − 2 ω2 3 .

This inequality has two solutions: k ≥ 48_π +2 ω 2 3 , k ≤ 2 ω2 3 − 48 π.

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

k ≤ 2 (20)

2

3 −

48 π = 251.

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Problem 41:

a) find the equations of motion; b) if the forcing takes the form:

f (t) =

f0, 0 ≤ t < t0,

f0/2, t0≤ t,

find the response of the system with zero initial conditions.

k 3k f (t) m x z θ ˆı ˆ 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 is shown to the right. Notice that the force in the upper spring depends on z, rather than x, while the friction force has an un-known magnitude fr. Because the disk is

assumed to roll without slip, we are unable to specify the value of fr, but instead can

relate the displacement and rotation of the disk through the coordinate relations above.

−k x ˆı −3k z ˆı f (t) ˆı frˆı G C

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

X MC = IC θ ˆ¨k, (3k z) 2r + (k x) r − f(t) r ˆk = 3 m r 2 2 θ ˆ¨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) = Z t

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and for this system: F (t) = f (t) meq = 2 f (t) 3 m r, h(t) = 1 ωn sin(ωn t) = r 3 m 26 k sin r 26 k 3 m t ! .

Because the forcing function changes abruptly at t = t0, the solution must be written

separately for 0 < t ≤ t0, and t > t0:

x(t) = 2 f0 3 m Z t 0 sin(ωn (t − τ)) ωn dτ, 0 < t ≤ t0, x(t) = 2 f0 3 m Z t0 0 sin(ωn (t − τ)) ωn dτ + f0 3 m Z t t0 sin(ωn (t − τ)) ωn dτ, 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. Problem 42:

The system shown to the right is subject to base excitation. Find the steady-state re-sponse of the system in terms of z, with:

m = 2.0kg, b = 4.0N/(m/s), k1= 3.00N/m, k2= 12.00N/m. u(t) = 0.50 sin(2 t)m z k1 k1 k2 b m x ˆ ˆı Solution:

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

x = z + u(t). Notice that the collection of springs can be replaced by a single equivalent spring, with:

keq= 1 1 2 k1+ 1 k2 = 2 k1k2 2 k1+ k2 = 4N/m.

The new equivalent system is shown to the

right. u(t)

z

keq b

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

aG= ¨x ˆ = (¨z + ¨u) ˆ,

with ¨u(t) = −(u0ω2) sin(ω t).

−keqz ˆ −b ˙z ˆ

Therefore, linear momentum balance on the mass yields: X F = m aG, − keqz − b ˙z ˆ = m ¨x ˆ

Writing this in terms of z, the equation of motion is: m ¨z + b ˙z + keqz = −m ¨u(t),

and in standard form: ¨ z + 2 ζ ωn ˙z + ωn2 z = u0 ω2 sin(ω t), with: ωn = r keq m, ζ = b 2pkeq m .

Therefore the steady state response of this system becomes: z(t) = Z sin(ω t − ψ), with Z = u0Λ(r, ζ), and: Λ = r 2 p(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 in the second quadrant, so that ψ = 3.03rad. Finally:

z(t) = √1

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Problem 43:

The disk shown in the figure rolls without slip and is subject to a time-varying moment M (t) = sin(t).

b) Find the frequency of oscillation for the unforced response, i.e. M (t) = 0; c) What is the steady-state amplitude of

the forced response?

d) Determine the amplitude of the fric-tional force during the steady-state mo-tion. (m, r) G M (t) ˆk x k b g Solution:

a) We assume that (ˆı, ˆ, ˆk) represent the standard orthonormal basis, while the transla-tional displacement of G is xˆı and the angular displacement is θˆk. Linear and angular momentum balance on the disk yield:

m¨x = −kx − b ˙x − fµ,

mr2

2 θ¨ = −fµr − M(t),

where fµ is the unknown frictional force. If the disk rolls without slipping, we find the

kinematical relation x = −rθ, and, eliminating fµ from the above balance laws, we find

the governing equation of motion can be expressed as: 3m

2 x + b ˙x + kx =¨ M (t)

r .

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

ωnp1 − ζ2, where: ωn= r 2k 3m = r 2 3, ζ = b √ 6km = 1 √ 2, so that the damped natural frequency is ωd=√1₃ = 0.577.

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

x + 2ζωn˙x + ω2nx = F0sin(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

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where r = _ωω_n is the frequency ratio. For this system we find r =q3₂, so that M =√2 13.

As a result, the steady-state amplitude is: A = F0 ω2 nM = 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 −1 2 .

From the expression for angular momentum balance, we find: fµ = m 2x −¨ M (t) r , = −√1 13sin(t + φ) − sin(t), = − ₁ √ 13sin φ cos(t) + ₁ √ 13cos φ + 1 sin(t) , where cos φ = _√1 13 and sin φ = 2_√√3

13, so that the amplitude of the frictional force is:

|fµ| =

4 √

13.

Problem 44:

2

2 about the

mass center G, subject to an applied moment of the form:

M(t) = M0sin(ωt)ˆk.

If the disk rolls without slip (µ is sufficiently large):

a) find the equations of motion for this system;

b) for m = 2kg, b = 0.5(N · s)/m, and k = 8N/m, find the steady-state amplitude 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 the disk from the unstretched position in the ˆk direction. If the disk rolls without slip, x and θ are related as:

x = −rθ.

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

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Using linear and angular momentum balance on the disk, we find that: X F _{= (f − kx − b ˙x)ˆı = m¨x ˆı = ma}G, X MG = (M0 sin(ωt) + f r) ˆk = mr2 2 θ ˆ¨k= IGαβ.

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

3m 2

¨

x + b ˙x + kx = −M_r0sin(ωt).

b) For the given values of the parameters, we find that: ωn2= 8 3, ζ = 1 8√6, F0= − M0 3r . Thus, the steady-state amplitude may be easily found as:

X = F0 ω2 nM, = − M0 3r q 8 3− ω2 2 + ω 6 2 . Problem 45:

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

u(t) = 4 sin(ωt)m,

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

a) find the equations of motion in terms of z, the relative displacement between the mass and the foundation (assume the beam has zero mass);

b) what is the amplitude of the result-ing motion in terms of the forcresult-ing fre-quency ω? u(t) = 4 sin(ω t)m z m x ˆ ˆı Solution:

a) The equivalent spring for this cantilever beam is: keq=

AE l =

16N

0.2m= 80N/m.

The acceleration of the block with respect to the ground is aG = (¨u + ¨z)ˆ, so that, in

terms of z, the equations of motion become: m¨z + keqz = −m ¨u,

¨

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b) For this undamped system, the amplitude of the resulting steady-state motion is: X = 4ω 2 40 1 q 1 −ω402 2 , = 4ω 2 |40 − ω2_|. Problem 46:

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

a) find the damped natural frequency; b) what is the steady-state amplitude of

vibration when the rotor spins at this angular speed. m ε O G k b x ˆ ˆı Solution:

We define x(t) as the vertical displacement of the geometric center of the rotor as measured from static equilibrium. As a result, the mass center G is described by the position 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 equations of motion can be written:

m ¨z = −k x − b ˙x, m ¨x − ε ω2 sin(ω t) = −k x − b ˙x, or in more standard form:

¨ x + b m˙x + k mx = εω 2_sin(ωt).

a) In the above system we find ωn =

q

k

m and ζ = b

2√km, so that the damped natural

frequency can be written:

ωd= ωnp1 − ζ2= r k m− b2 4m2.

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

Λ = ω

2

p(ω2

n− ω2)2+ (2ζωωn)2

,

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

A = ε 1 − ζ

2

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with ζ defined above.

Problem 47:

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

u(t) = 4 sin(ωt)m,

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

a) find the equations of motion in terms of z, the relative displacement between the mass and the base (assume the spring has zero unstretched length); b) what is the amplitude of the resulting

motion in terms of the frequency ratio r;

c) for what forcing frequencies is the re-sulting amplitude of the steady-state motion Z ≤ 8? u(t) = 4 sin(ω t)m z k m g x ˆ ˆı Solution:

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

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

aG=

¨

u + ¨zˆ.

Therefore, linear momentum balance takes the form: X F = m aG, − k z − m gˆ = mz + ¨¨ uˆ. −k z ˆ −m g ˆ

Substituting in the expression for u(t), and taking the component in the ˆ direction, we obtain the governing equation of motion:

m ¨z + k z = −m g − m ¨u, ¨

z + (4s−2_)z _{= −9.81m/s}2_{+ 4ω}2_sin(ωt)m,

Notice that each term has units of acceleration, that is, m/s2_{. In what follows we will}

dispense with the explicit inclusion of the units. For this system ωn= 2, ζ = 0, and the

amplitude of the forcing is F = 4ω2_{. Thus we have frequency-squared excitation.}

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the forced response can be written as z(t) = Z sin(ωt + φ), where the amplitude Z is: Z = U · Λ(r, ζ) = U r 2 p(1 − r2₎2_{+ (2ζr)}2, = 4r 2 |1 − r2_|.

c) If Z < 8 then this implies that:

Z < 4r 2 p(1 − r2₎2. Defining Zcras: Zcr= 4r2 p(1 − r2₎2,

We may solve for r to yield:

r = s Z2 cr± 4Zcr Z2 cr− 16 , = (r Zcr Zcr− 4 , r Zcr Zcr+ 4 ) .

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

0 < r <r 2

3, or r > √

2.

Problem 48:

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

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

b) determine the amplitude of the total force transmitted to the ground un-der steady-state oscillations when ω = 1rad/s. x m k _b F (t) = sin(ω t)N Solution:

The equation of motion for this system takes the form: ¨ x + b m˙x + k mx = F (t) m .

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

ζ = b

2√km, ωn= r k

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So, for a critically damped system ζ = 1, and solving for k and b we find: k = 16N/m, b = 8N · s/m.

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

Ft= Xωn2p1 + (2ζr)2, where X is the amplitude of the response and r = ωωn = 1 4 is

the frequency ratio. Thus, for this system: Ft = F0 ω2 n 1 p(1 − r2₎2_{+ (2ζr)}2 ! · ω2 np1 + (2ζr)2, = F0p1 + (2ζr) 2 p(1 − r2₎2_{+ (2ζr)}2 = 8√5 17 = 1.05. Problem 49:

A constant force is applied to the undamped single degree-of-freedom system for a dura-tion of t1, at which point it is removed, that

is:

F(t) =

F0ˆı, 0 ≤ t < t1,

0, t ≥ t1.

If the system starts with zero initial condi-tions, determine the resulting displacement of the mass x(t). You may either use the convolution integral or you may try to solve this explicitly. x m 4 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) = sin ωnt mωn

, ωn = 2

r k m.

Therefore, using the convolution integral, the response of the system is: x(t) = Z t 0 F (ξ)h(t − τ) dτ, =          Z t 0 F0 m sin(ωn(t − τ)) ωn dτ, 0 ≤ t < t1, Z τ 0 F0 m sin(ωn(t − τ)) ωn dτ, t ≥ t1. =          F0 mω2 n 1 − cos(ωnt) , 0 ≤ t < t1, F0 mω2 n cos(ωn(t − t1)) − cos(ωnt) , t ≥ t1.

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2.2 Unsolved Problems

Problem 50:

For the system shown to the right the block of mass m = 100kg is supported by springs and dampers with coefficients

k1= 1200N/m, k2= 3600N/m

b1= 100N/(m/s), b2= 120N/(m/s),

and is subject to a time dependent force F(t) = F0 sin(ω t) ˆ,

with F0= 100N and ω = 5rad/s.

a) Find the equations of motion for the displacement of the block;

b) Write these equations as a set of first-order equations (as you would for nu-merical simulation within MATLAB); c) What is the steady-state response of

the forced response;

d) What is the magnitude of the force transmitted to the ground?

k1 k2 b1 b2 m F(t) Problem 51:

The unbalanced rotor shown in the figure is pinned to a frame and supported by a spring and damper. The total mass is measured as m = 200kg. When the system is operated at ω = 25rad/s the phase φ of the response with respect to the rotation of the unbalanced disk is measured to be π/2rad and the steady-state vibration amplitude is X = 2.00cm. When the rotation rate of the disk is much larger than this value the amplitude reduces to X = 0.50cm. Find the stiffness and damp-ing coefficient for the foundation, as well as the distance ρ between the center of rotation C and the mass center G.

m ρ C G x k b

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Problem 52:

For the system shown to the right, the disk of mass m and radius r rolls without slip. If x measures the displacement of the disk from the unstretched position of the spring.

a) find the equations of motion; b) if the forcing takes the form:

f (t) =

f0t, 0 ≤ t < t0,

0, t0≤ t,

find the response of the system with zero initial conditions. If you choose to use the convolution integral you do not actually have to solve the integrand, just set it up.

b 2 k

f (t) m

Problem 53:

In the figure, the disk has mass m, radius ℓ, and moment of inertia IG = m ℓ

2 2 about

the mass center G. The disk is assumed to roll without slip and is subject to a harmonic moment of the form

M (t) = M0 sin(ω t).

b) With

m = 4kg, ℓ = 0.10m, M0= 2 N · m,

the frequency ratio of the system is r = 1 when ω = 2 π. If, in addi-tion, the amplitude of the response is Θ = π/2rad, find the damping coeffi-cient c and the stiffness k for this sys-tem. k k c M (t) (m, ℓ) ℓ/2 G g

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Problem 54:

In the figure shown to the right, the machine of mass m = 100kg rests on a regid support and is subject to a harmonic force

f (t) = F0sin(ω t).

If the support is redesigned as an elastic foun-dation, find the stiffness k and damping coef-ficient c so that the natural frequency of the new system is ωn = 10 π, and when the

sys-tem operates with a frequency ratio r = 3 the amplitude of the force transmitted to the ground is 20% of the applied forcing ampli-tude. f (t) Original c k f (t) Redesigned Problem 55:

The base of the seismograph shown to the right is subject to harmonic ground motion of the form

u(t) = u0 sin(ω t).

a) Find the equations of motion for the relative displacement z between the mass m and the base;

b) With

k = 0.25N/m, c = 0.40N/(m/s), m = 0.50kg,

Determine the steady-state amplitude of the response for u0 = 0.005m and

ω = 2 πrad/s.

k c

m

u(t) z

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Problem 56:

For the system shown to the right the bar of length ℓ has mass m and is subject to a time-dependent moment of the form

M(t) = M0 sin(ω t) ˆk.

a) Find the equations of motion for this rotation of the bar.

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

m = 3kg, b = 16N/(m/s), k = 8N/m, ℓ = 1 4m, ω = 4rad/s, M0= 1 16N · m G (m, ℓ) k b k M(t) Problem 57:

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

F(t) = f0 sin(ω t) ˆ.

a) Design an undamped foundation to achieve isolation ≥ 33% for all forcing frequencies ω > 3 πrad/s;

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

f (t) ˆ

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Problem 58:

In the figure shown to the right, the disk is subject to a time dependent moment of the form

M (t) = M0 sin(ω t).

a) Find the equations of motion for the angular displacement of the disk. b) With

k = 280N/m, b = 12N/(m/s), m = 4kg,

I = 0.40kg · m2_, _{r = 0.10m}

M0= 3N · m, ω = 5rad/s,

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

r r/2 M (t) ˆk k b k m I Problem 59:

The unbalanced rotor shown in the figure is pinned to a frame and supported by a spring and damper. The total mass of the structure is m, with a small rotating component (10% of the total mass) offset by a distance r from the center of rotation C.

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

b) what is the damped natural frequency; c) determine the steady-state amplitude of vibration when the rotor spins at an angular speed of ω = 5rad/s with:

k = 256N/m, b = 12N/(m/s), m = 5kg, r = 10cm m r C k b

Mechanical vibration solved examples

Contents

1

Free Vibration of Single Degree-of-freedom Systems

1.1

Solved Problems

1.2

Unsolved Problems

2

Forced Single Degree-of-freedom Systems

2.1

Solved Problems

2.2

Unsolved Problems