Mechanical Vibration solved examples

Mechanical Vibrations: 4600-431 December 20, 2006

Example Problems

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 have mass m and the radius of the disk is r. a) Find the equations of motion for this system. c

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

k

c) If the block is displaced 18 cm to the right and released from rest find the resulting angular displacement of the disk with m = 3 kg, k = 21 N/m,

(m, r)

m

k

r = 9 cm, c = 63 N · s/m,

1

4k

Problem 2: 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.

x

k

a) Find the equations of motion.

c

(m, r)

b) With c = 16 N/(m/s), m = 2 kg, r = 0.10 m for what value of the spring stiffness k is the damping ratio of the system onehalf 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 20 cm to the right (from static equilibrium) and released from rest. Problem 3: From the figure shown to the right k2

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

r

b) what are the damping ratio and natural frequency of the system in terms of the parameters m, b, k1 , and k2 ;

r 2

m

c) can you draw an equivalent springmass-damper system?

m

b

2

k1

Problem 4: For the mechanical system shown to the right, the uniform rigid bar has mass m and pinned at point O. For this system:

ˆ k

a) find the equations of motion;

ˆı

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

m ℓ 2

c) For:

y ℓ 2

θ

m = 1.50 kg,

ℓ = 45 cm,

c = 0.125 N/(m/s),

O

k = 250 N/m,

2m x

find the angular displacement of the bar θ(t) for the following initial conditions: θ(0) = 0,

c

k

˙ θ(0) = 10 rad/s.

Assume that in the horizontal position the system is in static equilibrium and that all angles remain small. 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 FR is included, although both its magnitude and direction are unspecified. In terms of the identifed coordinates, the angular acceleration of the bar αβ/F and the F linear acceleration of the block aG are αβ/F

ˆ = θ¨ k,

F

−k y ˆ

−T ˆ

FR

T ˆ

−k x ˆ

−c x˙ ˆ

aG = x ¨ ˆ.

We can also relate the identified coordinates as ℓ x = θ, y = ℓ θ. 2 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 written as ³ ´ X F ¨ ˆ, F = m aG −→ T − k x − c x˙ ˆ = 2 m x ¶ µ 2 X ℓ ˆ ˆ = m ℓ θ¨ k. M O = I O αβ/F −→ − T − k y ℓ k 2 3 3

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

ℓ m ℓ2 ¨ −kyℓ = θ. 2 3

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

√ c ℓ2 3c beq , = q 4 = √ 5 k ℓ 2 5 m ℓ2 2 keq meq 2 50 km 2 4 6 s s r 2 5kℓ keq 3k 4 = ωn = = 5 m ℓ2 meq 2 m 6 p

c) Evaluating the damping ratio and natural frequency we find that for the given values of 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 ³ ³ p ´ ³ p ´´ θ(t) = e−ζ ωn t a sin ωn 1 − ζ 2 t + b cos ωn 1 − ζ 2 t , where a and b are arbitrary constants used to fit the initial conditions. Evaluating θ(t) ˙ at t = 0 yields and θ(t) p ˙ θ(0) = b = 0, θ(0) = −ζ ωn b + ωn 1 − ζ 2 a = 10 rad/s, so that the general solution becomes ³ ´ 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, 2 and moment of inertia IG = mr about the 2 mass center G and is assumed to roll without slip. The identified coordiante θ measures the rotation of the disk with respect to the equilibrium position.

k

g

c

a) Find the equations of motion for this system.

G (m, r)

k

b) If the disk is released from rest with θ(0) = − π2 rad, find the resulting angular displacement θ(t) for m = 3 kg, k = 36 N/m,

r = 15 cm, c = 3 N · s/m,

c) What is the force in the upper spring during this motion? Problem 6: For the mechanical system shown to the right, the uniform rigid bar has mass m and pinned at point O. For this system:

ℓ 2

k


θ

m


z ℓ 2

m

ˆ

c) For:

x

m = 2 kg, ℓ = 25 cm, c = 0.25 N/(m/s), k = 50 N/m,

k

c

ˆı

find the angular displacement of the bar θ(t) for the following initial conditions: θ(0) = 0,

˙ θ(0) = 10 rad/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 equilibrium at θ = 0, and that all angles remain small. Solution: a) We identify the coordinates x and z as shown above, which are related to the angular

5

displacement θ as: x=

ℓ θ, 2

z=

An appropriate free-body diagram is shown to the right. Applying linear momentum balance on the block yields X F F = m aG ,

ℓ θ. 2 k z ˆ −T ˆ T ˆ

(T − k x − c x) ˙ ˆ = m x ¨ ˆ.

FR

Likewise, angular momentum balance on the bar provides X MO = I O αβ/F , µ ¶ ℓ m ℓ2 ¨ ˆ ℓ ˆ −T − k z k = θ k. 2 2 12

−k x ˆ

−c x˙ ˆ

Combining these equations and eliminating the tension, the equation of motion can be written as 7m ¨ θ + c θ˙ + 2 k θ = 0. 6 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 s r √ keq 12 k 3b beq = , ζ= p . =√ ωn = meq 7m 2 keq meq 28k m Problem 7: The block shown to the right rests on a frictionless surface. Find the response of the system if the block is displaced from its static equilibrium position 15 cm to the right and released from rest. m = 4.0 kg, k1 = 1.5 N/m,

ˆ x ˆı

k1

k2 m

b = 0.25 N/(m/s), k2 = 0.50 N/(m/s).

b

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 F = m aG , (−k1 x − k2 x − b x) ˙ ˆı = m x ¨ ˆı, 6

−k1 x ˆı −k2 x ˆı −b x˙ ˆı

or, writing this in standard form mx ¨ + b x˙ + (k1 + k2 ) x = 0. Further, the system is released from rest so that the initial conditions are x(0) = x0 = 15 cm,

x(0) ˙ = 0 cm/s.

Problem 8: For the system shown to the right, the disk 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.

x

k z

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

c

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;

φ

θ

(m, r) ˆ

C

eˆ2

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

ˆı

d) what is the static equilibrium displacement of the disk?

eˆ1

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 direcˆ2 ) as tions (ˆ e1 , e ˆı = cos φ e ˆ1 + sin φ e ˆ2 , ˆ = − sin φ e ˆ1 + cos φ e ˆ2 .

ˆ1 −k z e −m g ˆ

ˆ1 −c x˙ e

ˆ1 fr e ˆ2 N e

The moment produced by gravity about point C is Mgravity

= rGC × (−m g ˆ), =

ˆ ˆ2 ) × (−m g ˆ) = −m g r sin φ k. (r e

Angular momentum balance about the contact point C yields X MC = I C αD/F , µ ¶ ³ ´ 3 m r2 ¨ ˆ ˆ = (2 r) k z + r c x˙ − m g r sin φ k θ k. 2 7

Eliminating the coordinates z and θ, we can write the equation of motion in terms of x as 3m x ¨ + c x˙ + 4 k x = m g sin φ. 2 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 = p ωn 1 − ζ 2 , where s r c beq keq 8k = √ ωn = = , ζ= p , meq 3m 2 keq meq 2 6km so that ωd =

r

8k 3m

r

1−

c2 , 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 =

8

m g sin φ . 4k

Problem 9: In the figure shown to the right, in the absence of gravity the springs are unstretched in the equilibrium position.

z2

k2

r2

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

θ ˆ

b) If the system is released from the unstretched position of the springs, what is the maximum angular velocity of the disk during the resulting motion?

r1

I

ˆı

z1

m x k1

Solution: a) We define the coordinates x, θ, z1 , and z2 as shown in the figure, which are related as x = −r1 θ,

z1 = r1 θ,

−k2 z2 ˆı

z2 = −r2 θ.

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 F = m aG ,

FR −T ˆ

k1 z1 ˆ T ˆ

(T − m g) ˆ = m x ¨ ˆ,

−m g ˆ

while angular momentum balance on the disk yields X

MO

ˆ (T r1 − k1 z1 r1 + k2 z2 r2 ) k

= I O αD/F , ˆ = I θ¨ k.

Eliminating the unknown tension T from these equations and using the coordinate relations, the equation of motion becomes ´ ´ ³ ³ I + m r12 θ¨ + k1 r12 + k2 r22 θ = m g r1 . 9

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

m g r1 . k1 r12 + k2 r22

With this, the equilibrium deflection of each spring is found to be m g r12 , k1 r12 + k2 r22 m g r1 r2 =− . k1 r12 + k2 r22

z1,eq = r1 θeq = z2,eq = −r2 θeq

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 s k1 r12 + k2 r22 . ωn = I + m r12 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 m g r1 θ(t) = θeq (1 − cos(ωn t)) = k1 r12 + k2 r22

Ã

1 − cos

Ãs

k1 r12 + k2 r22 t I + m r12

The angular velocity of the disk becomes ´ ³ ˙ = θeq ωn sin(ωn t), θ(t) which has amplitude Ω = θeq ωn = p

10

m g r1 (k1 r12

+ k2 r22 ) (I + m r12 )

!!

.

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

x

a) determine the governing equations of motion;

m 6k

b

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 = 2 N/m, b = 4 N/(m/s), and m = 3 kg, find the displacement of the mass x(t) when the system is started with the initial conditions: x(0) = 0.10 m,

x(0) ˙ = 0 m/s.

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

k


z2

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

m

ℓ 2

k

θ

m

z1

ℓ 2

ˆ

c ˆı

Solution: a) In addition to θ, we define two additional coordinates, z1 and z2 , to measure the deflection at the left and right ends of the bar.

11

These coordinates are related as: ℓ z1 = θ, 2

−k z2 ˆ

ℓ z 2 = θ = z1 . 2

A free-body diagram for this system is shown to the right. Applying angular momentum balance on the bar eliminates the appearance of the reaction force and leads to: X ˆ MG = I G θ¨ k, ³

T − k z1 − cz˙1

´ℓ ˆ k 2

=

−T ˆ T ˆ

k z1 ˆ

G FR

m ℓ2 ¨ ˆ θ k. 12

c z˙1 ˆ

Likewise, application of linear momentum balance on the block yields: X F F = m aG , ³ ´ − k z2 − T ˆ = m¨ z2 ˆ Eliminating the unknown tension T and solving for z1 in terms of z2 , the equation of motion becomes: m ℓ2 ¨ c ℓ2 ˙ k ℓ2 θ+ θ+ θ = 0. 3 4 2 b) A critically damped system occurs when ζ = 1. For this system: √ 3c . ζ=√ 32 k m Solving for ccr yields: ccr =

Problem 12: (Spring 2003) Find the response of the system shown to the right if the block is pulled down by 15 cm and released form rest. m = 2.0 kg, k1 = 0.5 N/m,

r

32 k m . 3

ˆ

k1 b

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

k2

ˆı x

m Solution: For this system, the two springs in series may be replaced by an equivalent spring, with constant: 1 k1 k2 keq = 1 1 = k +k . + 1 2 k1 k2

12

Therefore, the free-body diagram is shown to the right. Applying linear momentum balance to the block yields: X F F = m aG , ³ ´ ¨ ˆ, keq x + b x˙ ˆ = −m x

keq x ˆ

b x˙ ˆ

which can finally be written as:

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

x(0) ˙ = 0 m/s.

With this, the damping ratio and natural frequency are: r √ 1 −1 3 s , ζ= . ωn = 12 4 Therefore, the system is underdamped and the general response can be written as: ³ ´ 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: ! r !! Ã√ Ã Ã√ 13 3 −t/8 3 13 √ t + √ t x(t) = . e cos cos 20 13 8 3 8 3

Problem 13: (Spring 2003) In the figure shown to the right, in the absence of gravity the springs are unstretched in the equilibrium position.

x2 k2

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

r2 θ

b) If the system is released from the unstretched position of the springs, what is the maximum angular velocity of the disk during the resulting motion?

r1

ˆ ˆı

m x1 k1

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 unstretched position. These coordinates are related through the following transformations: x1 = −r1 θ,

x2 = −r2 θ.

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 applied about the center, yielding: X ˆ MG = I G θ¨ k, ´ ³ ˆ = I G θ¨ k. ˆ T r1 + k2 r2 x2 k

−k2 x2 ˆı

G FR −T ˆ T ˆ

Also, applying linear momentum balance to the block yields: X F F = m aG , ³ ´ T − k1 x1 − m g ˆ = m x ¨1 ˆ.

−m g ˆ −k1 x1 ˆ

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: ´ ´ ³ ³ I G + m r12 θ¨ + k1 r12 + k2 r22 θ = (m g r1 ). This equation of motion determines the equilibrium position θeq (with θëq = 0) to be: θeq =

m g r1 . k1 r12 + k2 r22

Therefore, the equilibrium displacements in each spring are: x1,eq = −r1 θeq =

m g r12 , k1 r12 + k2 r22

x2,eq = −r2 θeq =

m g r1 r2 . k1 r12 + k2 r22

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 .

14

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

= =

1 1 k1 z12 + k2 z22 , 2 2 ´ 1³ k1 r12 + k2 r22 φ2 . 2

Also, the kinetic energy becomes: T

1 G ˙2 1 I φ + m z˙12 , 2 2 ´ 1³ G I + m r12 φ˙ 2 . 2

= =

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 =

´ 1³ G I + m r12 φ˙ 2max . 2

Finally, conservation of energy implies that V0 = T1 , and solving for φ˙ max we find that: s (m g r1 )2 . φ˙ max = G (I + m r12 )(k1 r12 + k2 r22 )

Problem 14: (Spring 2003) 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.

ˆ

x 3k

z ˆı


θ

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

m

Solution: a) We define the three coordinates as shown as the figure, related as: x = −r θ,

z = −2 r θ, 15

z = 2 x.

c

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

−3k z ˆı G −k x ˆı

−c x˙ ˆı

C f ˆı

The equations of motions can be developed directly with angular momentum balance about the contact point, so that: X ˆ MC = I C θ¨ k, ³ ´ 2 ˆ = 3 m r θ¨ k. ˆ (3k z) 2r + (k x) r + (c x) ˙ r k 2 Finally, writing this equation in terms of a single coordinate, we obtain: ¶ µ ¡ ¢ 3 m r2 ¨ θ + (c r2 ) θ˙ + 13 k r2 θ = 0. 2 b) For an underdamped response, the frequency of oscillation is ωd = ωn this system, we find that: r c 26 k , ζ=√ , ωn = 3m 78 k m so that: ωd =

r

p

1 − ζ 2 . With

26 k 2 c2 . − 3m 9 m2

Problem 15: In the system shown to the right, the pulley has mass m and radius r, so that the moment 2 of inertia about the mass center is IG = mr 2 .

r

a) Find the governing equations of motion;

r 2

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

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

k k

Solution: 16

c

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

=

V

=

1 1 1 mr2 ˙2 θ , m x˙ 21 + m x˙ 22 + 2 2 2 2 1 1 k x21 + k x22 . 2 2

However, these three coordinates are dependent through the transformations: r x2 = r θ. x1 = θ, 2 Therefore, the Lagrangian reduces to: µ ¶ µ ¶ 1 7mr2 ˙2 1 5kr2 L=T −V = θ2 . θ − 2 4 2 4 Further, the generalized force resulting from the viscous damper becomes: Qθ = −

cr2 ˙ θ, 4

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: r c 5k . , ζ= √ ωn = 7m 2 35 km Therefore, the damped natural frequency becomes: r r p c2 5k 2 1− ωd = ωn 1 − ζ = . 7m 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: In the figure, the disk has mass m, radius r, 2 and moment of inertia IG = mr about the 2 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:

b G k

a) find the equations of motion for this system; b) for m = 2 kg, b = 0.5 (N · s)/m, and k = 8 N/m, find the frequency of oscillation for the system. 17

m

Solution: a) Let x denote the translational displacement of G in the ˆı direction, while θ denotes the ˆ direction. If T is the tension between the disk angular displacement of the disk in the k 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 − bx˙ + T + f ) ˆı = m¨ x ˆı, 2 ˆ ˆ = mr θ¨ k, (rf ) k 2 (−T ) ˆı = m¨ x ˆı. 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 x ¨ + bx˙ + kx = 0. 2 b) From the above equation we identify the undamped natural frequency and damping ratio as: r 2k b ωn = , ξ=√ . 5m 10km Thus, the damped natural frequency is: ωd = ωn

p

1−

ξ2

=

r

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 singledegree-of-freedom system in the form: m¨ x + bx˙ + kx = 0,

x1 = 1.00 m bc

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

x2 = 0.75 m bc

a) Determine the equivalent damping constant. b) What is the exponential decay rate of the transient solution? Solution: 18

a) We use the logarithmic decrement so that: µ ¶ ¶ µ x1 δ1 0.75 m p , δ1 = − ln ζ= = 0.288, = − ln x0 1.00 m 4π 2 + δ12 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 p , ωn 1 − ζ 2 √ p 2 m 4π + δ12 √ = 3.145 s. k

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 is pinned to the ground.

k

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

(m, l) z b

b) what is the frequency of the resulting motion; c) if the mass of the spring is taken to be mspring = 1 kg, find the new frequency of the oscillations.

Solution: a) We will define the inclination of the bar from the horizontal position as θ, so that the ˆ Therefore, using angular momentum balance about the angular acceleration is θ¨ k. point of rotation, we find: mbar l2 ¨ θ + bl z˙ + kl z = 0. 3 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: mbar z¨ + b z˙ + k z = 0. 3 For the parameter values given above, this becomes: ¶ µ ³ ´ ³ ´ 2 kg z¨ + 2 N/(m/s) z˙ + 4 N/m z = 0. 3 19

For this system, the damping ratio and natural frequency can be expressed as: r √ 3k 3b ωn = , ζ = √ mbar 2 kmbar r √ 3 . = 6 rad/s = 8 p b) The frequency of oscillation, ωd = ωn 1 − ζ 2 , reduces to: s √ µ ¶2 15 3k 3b − rad/s = 1.94 rad/s. = ωd = mbar 2mbar 2 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 the bar about the point of rotation is: IO

= = =

mbar l2 + meq l2 , 3 mbar l2 ³ mspring ´ 2 l , + 3 3 (mbar + mspring ) l2 . 3

Therefore, the new frequency of oscillation is: s ¶2 µ √ 3b 3k = 3 rad/s = 1.73 rad/s. − ωd = (mbar + mspring ) 2(mbar + mspring )

Problem 19: The non-uniform beam supports an end-mass of m = 10 kg. If the response of the system is such that: x1 = 0.25 m t1 = 1.00 s

m

x2 = 0.20 m, t2 = 4.00 s,

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

(t1 , x1 ) bc

(t2 , x2 ) bc

Solution: 20

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: p (2π)2 + δ 2 δ . ωn = , ζ=p T (2π)2 + δ 2 Therefore, the stiffness and damping constants reduce to: k = m ωn2 = m

(2π)2 + δ 2 , T2

b = m 2ζωn = m

2δ . T

For this system, we find that T = 4 s and δ = 0.223, and the stiffness and damping constant reduce to: k = 24.7 N/m,

Problem 20: For the single-degree-of-freedom mechanical system shown in the figure:

b = 1.12 N/(m/s).

2r

k

a) determine the governing equations of motion; r

b) what are the damping ratio and undamped natural frequency of this system; m

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

c

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

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 = kr 2 θ, where θ is the angular displacement of the massless pulley from static equilibrium. In addition, x we find that θ is related to the linear displacement of mass m as θ = 2r . As a result, in the ˆ direction, the equation governing the motion of the mass is: m¨ x + bx˙ +

k x = 0. 4

As a result, we find that the the equivalent spring constant is keq = k4 . b) The damping ratio and natural frequency are simply: r √ b b keq k =√ = √ . , ωn = ζ= p m 2 m 2 keq m km 21

In terms of the given parameters, we find ζ = 4.0 and ωn = 3 rad/s. We note that the damping ratio has no units. c) For an overdamped system, the general solution is: ³ ³ ´ ³ ´´ p p x(t) = exp(−ζωn t) c1 exp (ωn ζ 2 − 1)t + c2 exp −(ωn ζ 2 − 1)t , and with these initial conditions this reduces to: Ã p ´ ³ p ζ + ζ2 − 1 p exp (ωn ζ 2 − 1)t + x(t) = x0 exp (−ζωn t) 2 ζ2 − 1 ! p ³ ´ p ζ − ζ2 − 1 2 p exp −(ωn ζ − 1)t , 2 ζ2 − 1 Ã ! √ √ 4 + 15 3 √15t 4 − 15 − 3 √15t −6t √ + √ e2 e 2 = x0 e , 2 15 2 15 ¡ ¢ = x0 e−6.0t 1.016e5.81t + 0.01640e−5.81t . Problem 21: For the spring-mass-damper system shown to the right, x is measured from the static equilibrium position, and the surface is frictionless.

x

a) Determine the governing equations of motion.

m 6k

b) What is the period of each oscillation in terms of the system parameters (m, k, c)? c) For what value of c is the system critically damped? d) If the system is released with the initial conditions: 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). 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 F = (−6k x − 2c x) ˙ ˆı = m¨ x ˆı = m aG , 22

2c

and the equation of motion can be written in standard form as: x ¨ + (2ζωn ) x˙ +

(ωn2 )

x = 0,

with

ωn =

r

6k , m

c ζ=√ . 6km b) In p terms of ζ and ωn , the period of oscillation is simply T = (2π)/ωd , where ωd = ωn 1 − ζ 2 . Therefore, for this system, the period reduces to: T =r

6k m

s

2π m µ ¶2 = 2π √6km − c2 , c 1− √ 6km

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 be written as: ³ √ ´ 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 A and φ, we find: r 5 A= m, φ = 0 rad, 28 so that the general solution can be written as: r ³ √ ´ 5 −2 t e sin 2 35 t m. x(t) = 28

23

Problem 22: For the system shown to the right, x is measured 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:

x (I, r)

k m

a) find the equations of motion which determine x(t); b) what is the period of the free oscillations? m

Problem 23: 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 undamped natural frequency of this system;

k

2k

b g

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

Solution: We assume that the ˆı and ˆ directions are standard orthonormal basis in the horizontal and vertical directions respectively. 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 + bx˙ + 3kx = mg.

24

b) Dividing through by the mass m, we find: x ¨+ so that: 2ζωn =

b 3k x˙ + x = g, m m

b , m

ωn2 =

3k , m

which can be solved to yield: ωn =

r

b ζ= √ . 2 3mk

3k , m

c) In equilibrium, the system is stationary, so that x˙ eq = 0 and x ëq = 0. Substitution into the governing equations yields: 3kxeq = mg, → xeq =

mg . 3k

Problem 24: For the single-degree-of-freedom mechanical system shown in the figure: a) find the linearized governing equations of motion for small θ;

k

b) find the frequency of oscillation for the response;

G

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

(m, l) θ

4k

Solution: a) The displacement of each end of the bar in the ˆ direction is: x± = ±

l sin θ . 2

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

MG

l2 ˆ (4k sin θ + k sin θ) k, 4 5kl2 ˆ = − sin θk. 4 = −

Thus angular momentum balance about G provides: X ˆ MG = I G θ¨k, −

5kl2 ˆ sin θk 4 25

=

ml2 ¨ˆ θk. 12

For small angular displacements sin θ ∼ θ, and the governing equations of motion are therefore: 15k θ¨ + θ = 0. m b) This system is undamped. Therefore the frequency of the oscillation is equal to the undamped natural frequency: r 15k ω = ωn = . m c) This system possesses the general solution: θ(t) = c1 sin ωt + c2 cos ωt, which, for the initial conditions given above, yields the solution: r 15k t. θ(t) = θ0 cos m Problem 25: We obtain the differential equation: m¨ x + 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 vibration to be reduced to within 2% of zero? Solution: a) Written in nondimensional form, the equations of motion are: x ¨+

k b x˙ + x = 0, m m

and so we identify the damping ratio and natural frequency as: r b k ζ= √ , ωn = . m 2 km ζ = 1 corresponds to critical damping, so that: √ bcritical = 2 km = 12. b) For b = 24 we find: b ζ= √ = 2, 2 km 26

ωn =

r

k = 3. m

Therefore the eigenvalues of this system are: ³ ´ p √ λ1,2 = ωn ζ ± ζ 2 − 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 √ ∼ 5.0 −6 + 3 3

Problem 26: For the single-degree-of-freedom mechanical system shown in the figure, the bar has mass 2 m and length l (so that IO = ml3 ). If the spring is unstretched when θ = 0:

k

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

ℓ 2

ℓ 2

θ

(m, l)

b) find the frequency of oscillation for the free response;

c

Neglect gravity.

Solution: a) Using angular momentum balance about the fixed point O, we find: X MO = IO αβ/F F, µ ¶ 2 l2 ˆ ˆ = ml θ¨k, −kl2 sin θ − c θ˙ cos θ k 4 3 so that the equation of motion can be written as: ml2 ¨ l2 θ + c θ˙ cos θ + k l2 sin θ = 0. 3 4 Linearizing this equation about θ = 0, we obtain: 3c ˙ 3k θ+ θ = 0. θ¨ + 4m m b) The frequency of oscillation, that is, the damped natural frequency, is given as: p ωd = ωn 1 − ζ 2 , where ωn is the undamped natural frequency and ζ we find: r 3k , ζ= ωn = m 27

is the damping ratio. For this system √ 3c √ , 8 km

so that the damped natural frequency is: s ¶2 µ 3k 3c ωd = . − m 8m Problem 27: In the figure, the disk has mass m, radius r, 2 and moment of inertia IG = mr about the 2 mass center G.

g

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

x c

b) If the disk is released from rest with initial displacement x(0) = x0 , find the minimum value of the coefficient of friction for which the disk does not slip.

G (m, r)

k

Problem 28: For the spring-mass-damper system shown to the right, x is measured from the static equilibrium position. If the surface is assumed to be frictionless:

x


m 6k

b

b) what is the period of each oscillation; c) what value of the damping coefficient b corresponds to critical damping? d) if k = 1 N/m and m = 4 kg, find the displacement of the mass x(t) if the system is critically damped and started with the initial conditions x(0) = 0, x(0) ˙ = x˙ 0 ; Problem 29: For the system shown to the right, x is measured 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:

x (I, r)

k m

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). m

28


r


r 2

b) What is the equivalent mass of the system; m

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

k c

k

Solution: ˆ direction (clockwise), and a) Define θ as the angular displacement of the disk in the −k x1 and x2 as the displacements of the two blocks so that: x1 =

r θ, 2

x2 = rθ.

We define the tension in the left and right cable as T1 and T2 respectively. Thus, linear momentum balance on the two blocks, and angular momentum balance on the disk yield: r − T2 r 2 T1 − kx1 − cx˙ 1 T2 + kx2 T1

mr2 ¨ θ, 2 = m¨ x1 , = −m¨ x2 .

= −

Eliminating the two unknown tensions from these three equations, we find that the equation of motion (on θ) can be reduced to: ¶ ¶ µ 2¶ µ µ cr 5kr2 7mr2 ¨ ˙ θ = 0. θ+ θ+ 4 4 4 2

b) Examination of the above equation shows that meq = 7mr 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: r r p c2 5k 1− ωd = ωn 1 − ζ 2 = . 7m 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:

x

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

b m k

b) Identify the damping constant b that gives rise to critical damping; Solution: For this system the differential equation of motion can be written as: m¨ x + bx˙ + kx = 0, k b x ¨ + x˙ + x = 0, m m subject to specified initial conditions. a) With the given values for the mass, stiffness, and damping coefficient, the above equation becomes: x ¨ + (6.5 kg/s)x˙ + (9 kg/s2 )x = 0, whose characteristic equation reduces to: λ2 +

13 λ + 9 = 0, 2

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: 9 x(t) = c1 e− 2 t + 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: ´ ³ ´ ³ −1 −1 x(t) = 4 m e−(9/2 s )t − e−(2 s )t . b) For this system, the damping ratio is: b . ζ= √ 2 km 30

Thus, for a critically damped system ζ = 1, and solving for b with m = 2 kg and k = 18 N/m, we find: bcr = 12 (N · s)/m. Problem 32: In the figure, the disk has mass m, radius r, 2 and moment of inertia IG = mr about the 2 mass center G.

g


x b

b) What value of b correspond to critical damping?

G

Solution:

(m, r)

k

c) Find the displacement of the center of the disk when the system is critically damped and released from rest with x(0) = x0 .

ˆ direction as measured We define θ as the angular displacement of the disk in the k 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: x = −rθ.

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 = −bx˙ ˆı.

Using linear and angular momentum balance on the disk, we find that: X F F = (f − kx − bx) ˙ ˆı = m¨ x ˆı = m aG , 2 X ˆ = mr θ¨ k ˆ = IG αβ/F F. MG = (f r) k 2 Eliminating the unknown frictional force, and using the kinematic constraint, we find the equation of motion is: ¶ µ 3m x ¨ + bx˙ + kx = 0. 2 b) We can write this differential equation in standard form, that is: r x ¨ + (2ζωn )x˙ + (ωn2 )x = 0,

with

ωn =

ζ=√

31

2k , 3m

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 + c2 t e−ωn t , while this can be differentiated with respect to time to obtain the velocity: ¡ ¢ x(t) ˙ = (c2 − ωn c1 ) − (ωn c2 ) t e−ωn t . 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) 0 = x(0) ˙

= c1 , = c2 − ωn c1 .

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 =

p (2k)/(3m).

Problem 33: In the figure, the disk has mass m, radius r, and moment of inertia IG about the mass center, and the applied moment has a constant ˆ If the disk rolls without slip magnitude M k. (µ is sufficiently large):

g x


c G

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

(m, r)

k

c) what is the stretch in the spring when the system is in equilibrium? Solution: a) The governing equations of motion are: µ ¶ I M m+ 2 x ¨ + cx˙ + kx = − . r r b) The equivalent mass, damping, and stiffness are: meq = m +

ˆ M k

I , r2

32

ceq = c,

keq = k.

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

2

M . kr

Frictionally Damped Systems

Problem 34: The block shown to the right rests on a rough surface with coefficient of friction µ and m = 6 kg,

ˆ

k = 128 N/m.

ˆı

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

g x

z A

b) With µ = 0.50, if the block is displaced 30 cm to the right and released from rest, how long will it take the block to come to rest?

B

k k

m

µ

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

F

vB − vA

= z˙ ˆı, = (−x˙ ˆı) − (x˙ ˆı) ,

so that z˙ = −2 x. ˙ Therefore the kinematic relationship becomes z = −2 x.

−m g ˆ T ˆı

−k x ˆı

An appropriate free-body diagram for this system is shown to the right. Note that the unknown friction force is denoted as fr ˆı and the tension in the cable is T . Finally, examining the spring in the cable, the tension T and the displacement z are related as

T ˆı fr ˆı N ˆ −T ˆı

T = k z = −2 k x.

Applying linear momentum balance to the block yields X F F = (2 T − k x + fr ) ˆı + (N − m g) ˆ = m x ¨ ˆı = m aG , and in terms of x the equation of motion becomes mx ¨ + 5 k x = fr . 33

T ˆı

If the block slips then fr = −µ m g sgn(x) ˙ while is sticking occurs |fr | ≤ µ m g. a) If the block is in static equilibrium at a displacement x = xeq , then x ëq ≡ 0 and the equation of motion reduces to 5 k xeq = fr , so that equilibrium is maintained provided |fr | = |5 k xeq | ≤ µ m g. This inequality is satisfied provided |xeq | ≤

µmg . 5k

Problem 35: For the system shown to the right, the block slides on a rough surface (coefficient of friction µ) inclined at an angle of φ with respect to vertical. If the block is subject to a periodic force of the form

g k

F (t) = F0 sin(ω t), a) find the equations of motion. Do not neglect gravity;

φ

b) find the amplitude of the steady-state response using Mc when m = 1.25 kg, k = 20 N/m, φ = 30◦ , µ = 0.125, F0 = 4 N,

ω = 2.00 rad/s,

34

m

F (t)

Problem 36: [(Spring 2003)] The system shown in the figure has mass m and rests on a plane inclined at an angle φ. The coefficient of friction for the rough surface is µ and the system is released from rest at the unstretched position of the spring (with stiffness k).

x

k

a) If µ = 0, what is the equilibrium displacement of the mass (as measured from the unstretched position)?

ˆ2 ˆı2

µ ˆ

m θ

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

ˆı

c) Find the value of θ so that the system comes to rest after one full cycle exactly at the equilibrium position of the system found in part a (so that the friction force vanishes when the system comes to rest), with: m = 2 kg,

k = 32 N/m,

µ = 0.35, Solution: The unit directions ˆı2 and ˆ2 are defined to be coincident with the inclined plane and the coordinate x represents the displacement of the mass from the unstretched position of the spring, as shown in the figure. 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 displacement and rotation of the disk through the coordinate relations above.

−k x ˆı2

N ˆ2

−m g ˆ

f ˆı2

Therefore, linear momentum balance yields the following equations: X F F = m aG , ³ ´ ³ ´ ³ ´ f − k x ˆı2 + N ˆ2 − m g ˆ = m x ¨ ˆı2 , ³ ´ ³ ´ f − k x + m g sin θ ˆı2 + N − m g cos θ ˆ2 = m x ¨ ˆı2 . Therefore, this leads to the following scalar equations in the ˆı2 and ˆ2 directions: mx ¨ + 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: mx ¨ + k x = m g sin θ. The equilibrium displacement of the mass, xeq , then can be found to be: xeq =

mg sin θ. k

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 static equilibrium: mg z =x− sin θ, k 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=−

4µN . k

Therefore, if the system comes to rest at exactly the equilibrium position, then this decrease in amplitude must exactly match the initial displacement. That is: ¯ ¯ ¯ ¯ 4 µ N ¯ ¯¯ m g ¯ ¯ = ¯− |∆ A| = ¯¯− sin θ ¯ = |z(0)|. k ¯ k

Solving for θ:

tan θ = 4 µ. Problem 37: The spring mass system rests on a surface with coefficient of friction µ and x is measured from the unstretched position of the spring.. If the initial conditions of the system are chosen to be x(0) = 0 and x(0) ˙ = x˙ 0 , find the range of x˙ 0 so that the system comes to rest after exactly one cycle of motion.

x

g m

k µ 36

Problem 38: For the spring-mass system with Coulomb damping: x


m

b) what is the period of each oscillation.

k

k µ

Solution: a) We measure the displacement of the mass from the static equilibrium of the frictionless F system, i.e., µ = 0, so that the acceleration of the block is aG = x ¨ˆı. Thus linear momentum balance yields: m¨ xˆı = Fspringˆı + fµˆı + (N − mg)ˆ. The spring force is Fspring = −2kx, while the force due to sliding friction opposes the velocity and is simply: x˙ fµ = −µmg , |x| ˙ since the normal force balances the gravitational force, i.e., N = mg. If the block is stationary the magnitude of the frictional force is less than µmg. Therefore, the governing equations of motion are: m¨ x + 2kx = fµ ,

with

fµ |fµ |

˙ = −µmg |xx| |x| ˙ = 6 0, ˙ ≤ µmg x˙ = 0.

b) Coulombic damping does not effect the frequency of oscillation, which is simply: r 2k . ω= m Therefore the period of the oscillation is: T =

37

√ 2π 2π m . = √ ω 2k

Problem 39: For the spring-mass-damper system shown to the right, x is measured from the static equilibrium position. If the coefficient of friction is µ:

x


m 6k

2k

b) what is the period of each oscillation;

µ

c) if the system is released from rest with x(0) = x0 > 0, what is the minimum value of x0 so that the block slips; d) find the range of initial displacements so that the system comes to rest after one complete cycle. Problem 40: For the spring-mass system with Coulombic damping, x is measured from the unstretched position of the spring. If the coefficient of friction is µ and the gravitational constant is g:

x I=0

k m


µ

b) if the system is released from rest in the unstretched position (x(0) = 0, x(0) ˙ = 0), for what values of µ will the system move; m

c) what is the displacement (from the unstretched position) of the upper block when it first comes to rest? 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 in part c we ask for the displacement from the unstretched position of the spring, rather than from equilibrium. Therefore we include the gravitational force which will influence this result. a) With the frictional force defined as F = f ˆı, linear momentum balance on the upper and lower block yields: m¨ x + kx = f + T, m¨ x = mg − T, where the frictional force is defined as: x˙ f = −µmg |x| ˙ , |f | ≤ µmg,

38

x˙ 6= 0, x˙ = 0.

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 value of x. ˙ b) If the system is released from rest in the unstretched position, it will remain there provided the magnitude of the frictional force is less than µmg—the transition to movement 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 a work-energy analysis. Since the initial and final kinetic energy is zero, the work done by the frictional force balances out the change in potential energy from the spring and gravity. Problem 41: For the spring-mass system with Coulombic damping, x is measured from the unstretched position of the spring. If the coefficient of friction is µ and the gravitational constant is g:

x

g m


k

b) if the system is released from rest, so that x(0) ˙ = 0, for what range of initial displacements (from the unstretched position) will the block come to rest when the block first comes again to rest (x(t ˙ 1 ) = 0 for t1 > 0)?

µ

Solution: a) The equations of motion can be written as: m¨ x + kx = f,

39

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

= −µmg

|f |

≤ µmg,

x˙ , |x| ˙

x˙ 6= 0, x˙ = 0.

b) If the system is released from rest, the initial displacement must be sufficiently large so that the block slides, rather than remaining at rest. Sliding does not occur if the force due 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 amplitude of the motion decays. Consider the block sliding to the left (x˙ < 0), released from rest with initial displacement x(0) > µmg k . Thus the equation of motion becomes: m¨ x + kx = µmg, which has the general solution: 

 µ ¶ µmg k µmg   x(t) = x(0) − cos t + . k m k

Therefore, when the block comes again to rest at time t1 (unknown), the mass is at the position: µmg x(t1 ) = 2 − x(0). k At this point, the block sticks if and only if |x(t1 )| ≤ we find that: µmg x(0) ≤ 3 . k

µmg k .

Therefore, solving for x(0),

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

40

Problem 42: For the system shown to the right, x is measured from the unstretched position of the spring. Each block has mass m and the disk has moment of inertia I and radius r. The coefficient of friction between the upper block and the table is µ. If the gravitational constant is g:

x (I, r)

k m µ

a) find the equations of motion which determine x(t); b) what is the minimum value of µ so that the system slips when release from rest with x(0) = 0;

m

c) what is the period of the free oscillations? d) if the system is released from rest, what is the range of initial displacements x(0) so that the systems comes to rest after exactly one complete cycle? Solution: We begin by defining two additional coordinates, θ, which describes the rotation of ˆ direction (clockwise), and y which measures the displacement of the disk in the −k the hanging mass in the −ˆ direction. These additional coordinates are related to the displacement 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: ³ ´ ³ ´ X F F = − k x + T1 + f ˆı + N − mg ˆ = m¨ x ˆı = m aG1 , ³ ´ X ˆ = −I O θ¨k ˆ = I O αD/F F, MO = T 1 r − T 2 r k ³ ´ X F F = T2 − mg ˆ = −m¨ y ˆı = m aG2 , Notice that I O = 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 ), while using the constraint equations, we find that the equation of motion for this system reduces to: ¶ µ I ¨ + k x = f + mg, 2m + 2 x r where f the value of the frictional force in the ˆı direction, can be written as: ½ x˙ −µmg |x| x˙ 6= 0 ˙ , f= f0 , |f0 | ≤ µmg, x˙ = 0.

41

b) The minimum value for slip is simply µmin = 0. If we would like to find the range of µ 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 for this quantity and using the frictional inequality, we find: |fstatic | = |kx0 − mg| ≤ µmg. Therefore, solving for µ yields: ¯ ¯ ¯ ¯ ¯ kx0 ¯ ¯ kx0 ¯¯ µ ≥ ¯¯ , − 1¯¯ = ¯¯1 − mg mg ¯

which provides a necessary condition for sticking at x = x0 . So for sliding to occur for x0 = 0, this implies that µ < 1. c) The period of oscillation for a frictionally damped system is identical to that of an undamped system. Therefore: s 2m + rI2 2π T = = 2π . ωn k d) Let δ describe the displacement of the system from equilibrium. The amplitude of oscillation 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 µmg < |δfinal | ≤ , k k

the initial displacement δ0 from the equilibrium in the absence of friction must be in the range: µmg 3µmg 5µmg µmg − +∆= < |δ0 | ≤ = + ∆. k k k k However, the equilibrium position corresponds to xeq = mg k , and so the allowable range of x0 is: mg ¯¯ 5µmg 3µmg ¯¯ < ¯x0 − . ¯≤ k k k

3

Forced Single Degree-of-freedom Systems

42

Problem 43: 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 ˆ M (t) = M0 sin(ω t) k.

M (t)

k G

a) Find the equations of motion for this rotation of the bar. (m, ℓ)

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

k

m = 3 kg,

b = 16 N/(m/s), 1 k = 8 N/m, ℓ = m, 4 1 ω = 4 rad/s, M0 = N·m 16 Problem 44: The block of mass m = 20 kg shown to the right rests on a rigid foundation and is subject to a time-dependent load

f (t) ˆ

F (t) = f0 sin(ω t) ˆ. a) Design an undamped foundation to achieve isolation ≥ 33% for all forcing frequencies ω > 3 π rad/s;

m

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

43

b

Problem 45: 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).

k

ˆ M (t) k r

a) Find the equations of motion for the angular displacement of the disk. r/2

I

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,

m

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

k

Problem 46: 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.

k

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

r C

b) what is the damped natural frequency;

m

c) determine the steady-state amplitude of vibration when the rotor spins at an angular speed of ω = 5 rad/s with: k = 256 N/m, m = 5 kg,

b

b = 12 N/(m/s), r = 10 cm

44

b

Problem 47: In the figure, the disk has mass m, radius r, 2 and moment of inertia IG = mr about the 2 mass center G and is assumed to roll without slip. The attached plate undergoes harmonic motion of the form

g u(t) c

u(t) = u0 sin(ω t). G a) Find the equation of motion in terms of the displacement between the moving plate and the center of the disk;

(m, r)

k

b) What are the damping ratio and natural frequency for this system? c) If the system is critically damped, find the amplitude of the relative displacement of the disk for m = 3 kg, k = 36 N/m,

r = 0.10 m c = 3 N/(m/s)

u0 = 0.05 m,

ω = 5 rad/s

Problem 48: For the mechanical system shown to the right, the uniform rigid bar is massless and pinned at point O while a force is applied at A of the form

ℓ 3

k

O

f (t) = t e−σ t .

θ

A

For this system:

2ℓ 3

a) find the equations of motion; b) Identify the damping ratio and natural frequency in terms of the parameters m, c, k, and ℓ.

f (t)

c) With m = 2 kg,

ℓ = 30 cm,

c = 0.25 N/(m/s), σ = 2.00 s

k = 50 N/m, −1

B

,

find the convolution integral for the response of the system. You need not evaluate the integral. Assume that the system is in static equilibrium at θ = 0, and that all angles remain small.

45

m

k

c

Problem 49: The block shown to the right rests on a rough surface with coefficient of friction µ (assume that any cables can support compression and tension). Find the amplitude of the vibrations of the block if f (t) = f0 cos(ω t), with m = 4.0 kg,

x k

m

k = 64 N/m,

µ = 0.05, f0 = 40 N, ω = 4 rad/s. m −f (t) ˆ

Problem 50: 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.

k

b

a) find the damped natural frequency;

r

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

C

k = 108 N/m, m = 3 kg,

m

b = 9 N/(m/s), r = 7.5 cm

46

Problem 51: For the mechanical system shown to the right, the uniform rigid bar has mass m and length ℓ, and is pinned at point O. A harmonic force is applied at A. For this system: O

k


θ

B


ℓ 2

A

c) For: 4m

m = 6 kg, ℓ = 25 cm, c = 0.50 N/(m/s), k = 80 N/m, f0 = 2.00 N, ω = 10 rad/s,

f0 sin(ω t) c

find the steady-state amplitude of the displacement of the block. Assume that the system is in static equilibrium at θ = 0, and that all angles remain small. Problem 52: In the figure, the disk has mass m, radius r, 2 and moment of inertia IG = mr about the 2 mass center G and is assumed to roll without slip. The attached plate undergoes harmonic motion of the form

g u(t) c

u(t) = u0 sin(ω t).

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

k

b) What are the damping ratio and natural frequency for this system? c) If the system is critically damped, find the amplitude of the rotation of the disk for m = 3 kg, u0 = 10 cm,

k = 36 N/m, ω = 3 rad/s

47

(m, r)

Problem 53: 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 = 200 kg. When the system is operated at ω = 25 rad/s the phase φ of the response with respect to the rotation of the unbalanced disk is measured to be π/2 rad and the steadystate vibration amplitude is X = 2.00 cm. When the rotation rate of the disk is much larger than this value the amplitude reduces to X = 0.50 cm. Find the stiffness and damping constant for the foundation and the distance between the center of rotation C and the mass center G.

k

b

ε

G

C m

Problem 54: For the mechanical system shown to the right, the uniform rigid bar is massless and pinned at point O while a harmonic force is applied at A. For this system:

ℓ 3

k

a) find the equations of motion; b) Identify the damping ratio and natural frequency in terms of the parameters m, c, k, and ℓ.

A

B

O θ 2ℓ 3

m

c) For: f0 sin(ω t)

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 of the block. d) What is the magnitude of the force transmitted to the ground through the sping and damper attached to the block? (Do not include the spring attached at A.) Assume that the system is in static equilibrium at θ = 0, and that all angles remain small.

48

k

c

Problem 55: The block shown to the right rests on a rough surface with coefficient of friction µ and the block is subject to a compressive force of N = 20 N (do not include gravity, just this normal load and assume that any cables can support compression and tension).

x k

a) If f (t) = f0 = constant, find the range of initial displacements for which the block will remain stationary if released from rest (it will stick).

m

(m, r)

−f (t) ˆı

b) Find the amplitude of the vibrations of the block if f (t) = f0 cos(ω t), with m = 4.0 kg, r = 12.5 cm, f0 = 40 N,

k = 64 N/m, µ = 0.50, ω = 4 rad/s.

Problem 56: 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.

k x

a) find the equations of motion. Do not neglect gravity; b) what is the static equilibrium displacement of the disk?

φ

c) if the disk is subject to a periodic moment ˆ M (t) = M0 sin(ω t)k, find the displacement of the center of the disk if it is released from rest at the unstretched position of the spring.

49

(m, r)

M (t)

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

x

ˆ

z 2k

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

ˆı θ G

b

b) If the applied moment takes the form:

k

M (t)

(m, ρ)

M (t) = (2 N · m) sin(4 t),

C

find the steady-state amplitude of the translation of the center of the disk when: k = 16 N/m, m = 2 kg,

b = 2 N/(m/s), ρ = 0.125 m

c) Determine the steady state amplitude of the friction force. 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 x. 2

µ

ρ (k x + b x) ˙ +

ˆ M (t) k −2 k z ˆı

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 angular momentum balance about the contact point C X MC = I C αD/F ,

which yields

−m g ˆ

−b x˙ ˆı

−k x ˆı

fr ˆı N ˆ

¶ 2 3ρ ˆ ˆ = 3 m ρ θ¨ k. (2 k z) + M (t) k 2 2

Using the above coordinate transformations this equation can be written as 3m 11 k M (t) x ¨ + b x˙ + x=− . 2 2 ρ b) For the numerical values given above (with consistent units), this equation reduces to 3x ¨ + 2 x˙ + 88 x = −8 sin(4 t),

50

from which we can identify the appropriate parameters as: r r 88 6 1 ωn = , r= , ζ=√ . 3 11 264 Therefore, the amplitude of the translational oscillations becomes X=

8 F r M(r, ζ) = k 88 ¡

1 1 =√ . q ´ ³ 2 ¢ 26 6 2 6 1 1 − 11 + 2 √264 11

Likewise, the phase shift of the response is:

tan φ =

2ζ r = 1 − r2

1 2 √264

1−

q

6 11

6 11

=

1 , 5

so that φ = 0.20 rad = 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 ³ ´ ³ ´ X F ¨ ˆı = m aG . F = fr − k x − 2 k z − b x˙ ˆı + N − m g = m x 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 ³ ´ ³ ´ fr (t) = 4 k − m ω 2 X sin(ω t − φ) + b ω X cos(ω t − φ). The magnitude of the friction force is then found to be r³ ° ° ´2 ³ ´2 ° ° 4 k − m ω2 + b ω . °f ° = X

For these parameter values kf k = 6.47 N.

51

Problem 58: (Spring 2003) 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,

k

b

a) find the damped natural frequency; b) what is the steady-state amplitude of vibration when the rotor spins at an angular speed of ω = 8 π rad with: k = 32 N/m, m = 4 kg,

ε

G

O

b = 16 N/(m/s), ε = 2.5 cm

m

c) If the system is undamped (i.e., b = 0 N/(m/s)) for what range of operating speeds (ω) will the amplitude of the force transmitted to the ground FT be less than 1 N. Problem 59: (Spring 2003) 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 (in terms of the given parameters—do not substitute in numerical values yet);

f (t) = sin(ω t) ℓ 2

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

k

θ

m

z ℓ 2

ˆ

c

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

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:

52

ℓ θ 2 A free-body diagram for this system is shown to the right. Applying angular momentum balance on the bar eliminates the appearance of the reaction force and leads to: X ˆ MG = I G θ¨ k, z=

³

f (t) − k z − cz˙

´ℓ ˆ k 2

=

f (t) ˆ

k z1 ˆ

G FR c z˙1 ˆ

m ℓ2 ¨ ˆ θ k. 12

Solving for z in terms of θ, the equation of motion becomes: ℓ m ℓ2 ¨ c ℓ2 ˙ k ℓ2 θ+ θ+ θ = f (t). 12 4 4 2 b) In standard form, this equation of motion can be written as: ¶ ¶ µ ¶ µ µ 6 3k 3c θ= sin(ω t), θ˙ + θ¨ + m m mℓ so that: ωn =

r

3k , m

√ 3c , ζ= √ 2 km

M0 =

6 . mℓ

The amplitude of the moment transmitted to the ground can be written as: p 1 + (2 ζ r)2 , MT = (meq M0 ) p (1 − r2 )2 + (2 ζ r)2 q ¡ ¢2 p 1 + c kω k 2 + (c ω)2 ℓ ℓ q¡ q¡ = = . ¢ ¡ ¢ 2 2 2 2 ¢2 2 2 2 1 − m3 ωk k − m 3ω + c kω + (c ω) The amplitude of the force transmitted to the ground is then FT = MT /(ℓ/2), or: p k 2 + (c ω)2 . F T = q¡ 2 ¢2 2 + (c ω) k − m 3ω 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 1 p , 2 2 ωn2 (1 − r ) + (2 ζ r)2 1 2 2 q¡ q¡ = ¢ ¡ ¢ 2 2 2 kℓ ℓ c ω m ω + k 1− 3 k k− 53

1 ¢ m ω2 2 3

2

+ (c ω)

.

Substituting in the numerical values given in the problem statement, we find that: Θ = ¯¡ ¯ k−

8 2 ω2 3

¢¯ ¯

Therefore, if the amplitude of vibration is less than π/6: 8 ¯ ¯k −

2 ω2 ¯ 3

¯

48 π

This inequality has two solutions: k≥

π , 6 ¯ ¯ ¯ 2 ω 2 ¯¯ . ≤ ¯¯k − 3 ¯ ≤

48 2 ω 2 + , π 3

k≤

2 ω2 48 − . 3 π

Since this condition must be satisfied for all ω ≥ 20 rad/s, we take the second inequality and find that: 48 2 (20)2 − = 251. k≤ 3 π Problem 60: (Spring 2003) 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.

ˆ

x 3k z

ˆı


θ

b) if the forcing takes the form: ½ f0 , 0 ≤ t < t0 , f (t) = f0 /2, t0 ≤ t,

k

f (t)

m

find the response of the system with zero initial conditions. Solution: a) We define the three coordinates as shown as the figure, related as: x = −r θ,

z = −2 r θ,

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 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.

54

z = 2 x.

−3k z ˆı G −k x ˆı

C fr ˆı

f (t) ˆı

The equations of motions can be developed directly with angular momentum balance about the contact point, so that: X ˆ MC = I C θ¨ k, ³ ´ 2 ˆ = 3 m r θ¨ k. ˆ (3k z) 2r + (k x) r − f (t) r k 2 Finally, writing this equation in terms of a single coordinate, we obtain: µ ¶ ¢ 3 m r2 ¨ ¡ θ + 13 k r2 θ = r f (t). 2 In standard form: θ¨ +

µ

26 k 3m

¶

θ=

2 f (t) . 3mr

b) We use the convolution integral to determine the response, so that: θ(t) =

Z

0

t

F (τ ) h(t − τ ) dτ,

and for this system: f (t) 2 f (t) F (t) = = , meq 3mr

1 sin(ωn t) = h(t) = ωn

r

3m sin 26 k

Ãr

! 26 k t . 3m

Because the forcing function changes abruptly at t = t0 , the solution must be written separately for 0 < t ≤ t0 , and t > t0 : Z 2 f0 t sin(ωn (t − τ )) dτ, 0 < t ≤ t0 , x(t) = 3m 0 ωn Z Z t f0 sin(ωn (t − τ )) 2 f0 t0 sin(ωn (t − τ )) dτ + dτ, t > t0 . x(t) = 3m 0 ωn 3 m t0 ωn Evaluating these integrals, the solution becomes: x(t) = x(t) =

2 f0 26 k f0 26 k

³ ³

´ 1 − cos(ωn t) ,

´ 1 + cos(ωn (t − t0 )) − 2 cos(ωn t) ,

55

0 < t ≤ t0 , t > t0 .

Problem 61: (Spring 2003) The system shown to the right is subject to base excitation. Find the steady-state response of the system in terms of z, with: m = 2.0 kg,

ˆ m

b = 4.0 N/(m/s),

k1 = 3.00 N/m,

k1

k2 = 12.00 N/m.

ˆı

k1 z

b x

k2

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

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 2 k1

1 +

1 k2

=

m

2 k1 k2 = 4 N/m. 2 k1 + k2

keq

The new equivalent system is shown to the right. An appropriate free-body diagram is shown to the right. In terms of the identified coordinates, the acceleration of the mass center is: F aG = x ¨ ˆ = (¨ z+u ¨) ˆ,

z

u(t)

−keq z ˆ

with u ¨(t) = −(u0 ω 2 ) sin(ω t). Therefore, linear momentum balance on the mass yields: X F F = m aG , ³ ´ − keq z − b z˙ ˆ = m x ¨ ˆ 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˙ + ωn2 z = u0 ω 2 sin(ω t),

56

b

−b z˙ ˆ

with: ωn =

r

b ζ= p . 2 keq m

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: r2 Λ= p , (1 − r2 )2 + (2 ζ r)2

tan ψ =

2ζ r , 1 − r2

and r =

ω ωn

For the numerical values of this problem: ωn =

so that:

√

1 ζ=√ , 2

2,

2 Λ= √ , 5

tan ψ =

r=

√ 2,

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 = 3.03 rad. Finally: 1 z(t) = √ sin(2 t − 2.03). 5

Problem 62: In the figure, the disk has mass m, radius r, 2 about the and moment of inertia IG = mr 2 mass center G, subject to an applied moment of the form:

g x

ˆ M (t) = M0 sin(ωt)k.

M (t)

c

a) If the coefficient of friction is µ, find the condition which determines if the disk rolls with or without slip, and find the governing equations of motion when the disk rolls with and without slip;

G k

b) If µ is sufficiently large so that the disk rolls without slipping, what are the equivalent mass, stiffness, and damping of the system; c) If again the disk rolls without slipping, find the amplitude of the steady-state motion of G, that is, x(t), when the system is critically damped, with M0 = 4 N · m, m = 1 kg, r = 0.1 m, and k = 2 N/m. 57

(m, r)

Problem 63: In the figure, the disk has mass m, radius r, and moment of inertia IG = α m about the mass center G. The disk is subject to a timevarying force f (t) = 4 sin(ω t).

g x


c f (t)

G

b) After the transient solution decays, find the amplitude of the force transmitted to the ground through the springdamper element.

(m, r)

k

c) For what value of the damping ratio is this transmitted force less than twice the applied load for all values of the forcing frequency? Problem 64: The disk shown in the figure rolls without slip and is subject to a time-varying moment M (t) = sin(t).

g


x

ˆ M (t) k

b

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

G

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

k

(m, r)

d) Determine the amplitude of the frictional force during the steady-state motion. Solution: ˆ represent the standard orthonormal basis, while the translaa) We assume that (ˆı, ˆ, k) ˆ Linear and angular tional displacement of G is xˆı and the angular displacement is θk. momentum balance on the disk yield: m¨ x = −kx − bx˙ − fµ , 2 mr ¨ θ = −fµ r − M (t), 2 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: M (t) 3m x ¨ + bx˙ + kx = . 2 r b) For M (t) = 0, the frequency of oscillation is the damped natural frequency ωd = 58

ωn

p

1 − ζ 2 , where: ωn =

r

2k = 3m

r

2 , 3

ζ=√

so that the damped natural frequency is ωd =

√1 3

b 1 =√ , 2 6km

= 0.577.

c) In nondimensional form, the system is represented as: x ¨ + 2ζωn x˙ + ωn2 x = F0 sin(t), where ζ and ωn are given above, and: F0 =

2 2 = . 3m 3

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

1 (1 −

r2 )2

+ (2ζr)2

,

where r = ωωn is the frequency ratio. For this system we find r = As a result, the steady-state amplitude is: A=

q

3 2,

so that M =

√2 . 13

2 F0 M = √ = 0.555. ωn2 13

d) The response of the disk is x(t) = A sin(t + φ), where A is given above and: √ − 3 −2ζr = . tan φ = 1 − r2 − 21 From the expression for angular momentum balance, we find: fµ

where cos φ =

√1 13

M (t) m x ¨− , 2 r 1 = − √ sin(t + φ) − sin(t), 13 ¶ µ ¶ ¸ ·µ 1 1 √ sin φ cos(t) + √ cos φ + 1 sin(t) , = − 13 13 =

and sin φ =

√ 2 3 √ , 13

so that the amplitude of the frictional force is: 4 |fµ | = √ . 13

59

Problem 65: In the figure, the disk has mass m, radius r, 2 and moment of inertia IG = mr about the 2 mass center G, subject to an applied moment of the form:

g x

ˆ M (t) = M0 sin(ωt)k.

M (t)

b

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

G

a) find the equations of motion for this system;

(m, r)

k

b) for m = 2 kg, b = 0.5 (N · s)/m, and k = 8 N/m, find the steady-state amplitude of the rotation of the disk? Solution: In addition to x, defined in the figure, we define θ as the angular displacement of the ˆ direction. If the disk rolls without slip, x disk from the unstretched position in the k 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: Fspring = −kx ˆı,

Fdamper = −bx˙ ˆı.

Using linear and angular momentum balance on the disk, we find that: X F F = (f − kx − bx) ˙ ˆı = m¨ x ˆı = m aG , 2 X ˆ = IG αβ/F F. ˆ = mr θ¨ k MG = (M0 sin(ωt) + f r) k 2 Eliminating the unknown frictional force, and using the kinematic constraint, we find the equation of motion is: ¶ µ M0 3m x ¨ + bx˙ + kx = − sin(ωt). 2 r b) For the given values of the parameters, we find that: ωn2 =

8 , 3

1 ζ= √ , 8 6

F0 = −

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

= =

F0 M, ωn2 q¡ 60

8 3

0 −M 3r ¢2 ¡ ω ¢ 2 . − ω2 + 6

M0 . 3r

Problem 66: The mass m = 2 kg is supported by an elastic cantilever beam attached to a foundation which undergoes harmonic motion of the form: u(t) = 4 sin(ωt) m,

ˆ m

ˆı

x

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

z

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);

u(t) = 4 sin(ω t) m

b) what is the amplitude of the resulting motion in terms of the forcing frequency ω? Solution: a) The equivalent spring for this cantilever beam is: keq =

16 N AE = = 80 N/m. l 0.2 m F

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 + keq z z¨ + 40z

= −m¨ u, 2 = 4ω sin(ωt).

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

4ω 2 q¡ 40

=

1 1−

4ω 2 . |40 − ω 2 |

=

61

ω2 40

¢2 ,

Problem 67: 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,

ˆ k

b ˆı

a) find the damped natural frequency; b) what is the steady-state amplitude of vibration when the rotor spins at this angular speed.

ε

G

O

x m

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 = −kx − bx, ˙ m¨ x − εω sin(ωt) = −kx − bx, ˙ 2

or in more standard form: x ¨+

b k x˙ + x = εω 2 sin(ωt). m m

a) In the above system we find ωn = frequency can be written: ωd = ωn

q

p

k m

1−

and ζ =

ζ2

=

r

√b , 2 km

so that the damped natural

k b2 − . m 4m2

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

ω2 (ωn2 − ω 2 )2 + (2ζωωn )2

which, with ω = ωd , reduces the amplitude to: 1 − ζ2 A=ε p , ζ 4 − 3ζ 2 with ζ defined above.

62

,

Problem 68: The system shown in the figure is supported by a foundation that undergoes harmonic motion of the form:

ˆ

g

m ˆı

u(t) = 4 sin(ωt) m, If the mass and stiffness are m = 2 kg, and k = 8 N/m:

z

k x

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);

u(t) = 4 sin(ω t) m

b) what is the amplitude of the resulting motion in terms of the frequency ratio r; c) for what forcing frequencies is the resulting amplitude of the steady-state motion Z ≤ 8? 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: ³ ´ F aG = u ¨ + z¨ ˆ. Therefore, linear momentum balance takes the form: X F F = m aG , ³ ´ ³ ´ − kz − mg ˆ = m z¨ + u ¨ ˆ.

−m g ˆ

−k z 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 + kz z¨ + (4 s−2 )z

= −mg − m¨ u, = −9.81 m/s2 + 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 = 4ω 2 . 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)

= Up

=

r2 (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 = p

We may solve for r to yield: r

.

4r2 (1 − r2 )2

,

s

2 ± 4Z Zcr cr , 2 − 16 Zcr ) (r r Zcr Zcr . , = Zcr − 4 Zcr + 4

=

Thus for Zcr = 8, we find that Z < 8 in the range: r √ 2 0 2. 3 Problem 69: The single-degree-of-freedom system shown is subject to a harmonic force. If the natural frequency is ωn = 4 rad/s and m = 1 kg:

x

a) determine the spring and damping constants when the system is critically damped;

F (t) = sin(ω t) N m

k

b

b) determine the amplitude of the total force transmitted to the ground under steady-state oscillations when ω = 1 rad/s. Solution: The equation of motion for this system takes the form: x ¨+

b k F (t) x˙ + x = . m m m

a) In terms of the system parameters, the damping ratio and natural frequency can be written as: r b k . , ωn = ζ= √ m 2 km 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 p of the total force transmitted to the ground, which is defined as Ft , is Ft = Xωn2 1 + (2ζr)2 , where X is the amplitude of the response and r = ωωn = 14 is the frequency ratio. Thus, for this system: ! Ã p 1 F0 p · ωn2 1 + (2ζr)2 , Ft = 2 ωn (1 − r2 )2 + (2ζr)2 p √ F0 1 + (2ζr)2 8 5 = p = 1.05. = 17 (1 − r2 )2 + (2ζr)2 Problem 70: A constant force is applied to the undamped single degree-of-freedom system for a duration of t1 , at which point it is removed, that is: ½ 0 ≤ t < t1 , F0 ˆı, F (t) = 0, t ≥ t1 .

x F (t) m 4k

If the system starts with zero initial conditions, determine the resulting displacement of the mass x(t). You may either use the convolution integral or you may try to solve this explicitly. Solution: The equation of motion for this system is: x ¨+

F (t) 4k x= , m m

which has an impulse response of the form: sin ωn t h(t) = , mωn

r

ωn = 2

k . m

Therefore, using the convolution integral, the response of the system is: Z t F (ξ)h(t − τ ) dτ, x(t) = 0  Z t F0 sin(ωn (t − τ ))   dτ, 0 ≤ t < t1 ,   0 m ωn =  Z τ F sin(ω (t − τ ))  n 0   dτ, t ≥ t1 . m ωn 0  ´ F0 ³   1 − cos(ω t) , 0 ≤ t < t1 , n   mωn2 = ´  F0 ³    cos(ω (t − t )) − cos(ω t) , t ≥ t1 . n 1 n mωn2 65

Problem 71: A mass m = 2 kg is rigidly connected to a rigid massless bar of length ℓ = 40 cm, which is pinned to a wall. The system is supported by a combination of springs and damper as shown in the figure, and subjected to a timedependent moment M (t) = 12 sin(t). If k = 12 N/m and b = 6 N/(m/s):

b

ˆ

M (t) ˆı ℓ 2

a) find the undamped natural frequency and the damping ratio of the system;

z2 z1

θ

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

6k

2k

k

Problem 72: The system shown in the figure is supported by a foundation that undergoes an exponentially decaying motion of the form:

g

m

u(t) = 16e−t/4 . If the mass and stiffness are m = 5 kg, and k = 45 N/m:

z

k

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);

u(t)

b) find the resulting solution z(t) if the mass is started from rest and the spring is initially unstretched (assume g = 10 m/s2 and use the convolution integral); c) what is the amplitude of the oscillations as t increases (i.e. u(t) → 0)?

66

m

Problem 73: In the system shown at right, the disk is assumed to be massless while: m = 2 kg, b = 14 N/(m/s),

2r

k

k = 4 N/m, r = 1 cm.

r

a) Determine the governing equations of motion; b) What is the period of oscillation; m

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

b

Problem 74: In the mechanical system shown each spring is identical, with spring constant k. If m = 1 kg:

ˆı

a) determine the spring constant k and damping constant b so that the unforced system is critically damped and the exponential rate of decay is τ = 2 s−1 ;

m

b

b) with the initial conditions: x(0) = 0,

ˆ

f (t)

x(0) ˙ = 2 m/s,

k

find the resulting solution of x(t) for the unforced problem, i.e., f (t) = 0; c) with f (t) = sin(2t), determine the amplitude of the force transmitted to the supporting structure.

67

x

Problem 75: 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,

ˆ k

b ˆı

a) find the damped natural frequency; b) for:

ε m = 4 kg,

b = 0.5 (N · s)/m,

ε = 0.1 m,

O

k = 8 N/m

x m

what is the steady-state amplitude of vibration when the rotor spins at this angular speed?

2k

Problem 76: The disk shown in the figure has mass m = 4 kg, and is subject to a time-dependent moˆ If k = 2 N/m and ment M (t) = M0 sin(t) k. r = 0.2 m:

4k

IG

a) Find the steady-state amplitude of the response when: b = 1 N/(m/s),

G

b

r

M0 = 12 N · m.

b) What is the amplitude of the oscillations when the system is critically damped?

r 2

k

2k

68

Problem 77: For the system shown to the right, the block slides on a rough surface (coefficient of friction µ) inclined at an angle of φ with respect to vertical. If the block is subject to a periodic force of the form

g k

F (t) = F0 sin(ω t), a) find the equations of motion. Do not neglect gravity;

m

φ

F (t)

b) find the amplitude of the steady-state response using Mc when m = 1.25 kg, k = 20 N/m, φ = 30◦ , µ = 0.125, F0 = 4 N,

4

ω = 2.00 rad/s,

Multi Degree-of-freedom Systems

Problem 78: For the system shown to the right a) use Lagrange’s equations to determine the equations of motion; b) find the mode shapes and natural frequencies of the system;

m 2k

c) determine the modal equations for this system.

69

4m k

F (t) ˆı

) ,d (4 m

Problem 79: For the system shown to the right a) find the equations of motion; b) are your equations statically coupled, dynamically coupled, both or neither?

k

k

2 (m,

d)

f (t)

m

b

2k

Problem 80: For the system shown to the right a) use Lagrange’s equations to determine the equations of motion; k

4k

b) find the mode shapes and natural frequencies of the system. Normalize the mode shapes so that u T M u = 1.

3m k

k m

Problem 81: For the system shown to the right the bar of length ℓ is massless and the block on the right is subject to a time-dependent force of the form

k k

F (t) = F0 sin(ω t).

G

The intermediate springs are located a distance 2ℓ from the center of the bar.

k

x2 x1

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

m m

b) Determine the natural frequencies and mode shapes of this system. F (t)

c) Determine the modal equations. d) What is the steady-state amplitude of the in-phase motion? 70

Problem 82: (Spring 2003) For the mechanical system shown to the right, the uniform rigid bar is supported by identical springs. For this system: k

a) find the equations of motion in terms of x and θ (and in terms of the given parameters—do not substitute in numerical values yet);

k

x θ

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

ℓ 3

G m, ℓ

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

F0 sin(ω t) m

a) find the equations of motion (and in terms of the given parameters—do not substitute in numerical values yet);

βm

k

αk

b) for α = β = 2, find the natural frequencies and mode shapes of the system and normalize the mode shapes by M ; c) determine the equations of motion that describe the response of each mode. Problem 84: (Spring 2003) In the figure shown to the right, in the absence of gravity the springs are unstretched in the equilibrium position. Determine the equations of motion for this system.

x2

k2

IG

m r2

b

θ r1 x3 z k1

ˆ m

ˆı x1

71

k1

Solution: Because we can, we define five different coordinates to describe the dynamical behavior of this two degree-of-freedom system, leading to the following transformations: x2 = −r2 θ,

x3 = −r1 θ,

A free-body diagram for this system is shown to the right. We develop three equations of motion based on linear momentum balance on both blocks and angular momentum balance on the disk:

z = x3 − x1 .

−(k2 x2 + b x˙ 2 ) ˆı

−T ˆı

T ˆı

Block 1: ³

X

k1 z − k1 x1

Block 2: ³

´

X

T − b x˙ 2 − k2 x2

Disk:

F

F

= m a G1 , FR

ˆ = m x ¨1 ˆ F

´

X

−k1 z ˆ

F

= m a G2 ,

k1 z ˆ

ˆı = m x ¨2 ˆı

MG ³ ´ ˆ T r2 + k1 r1 z k

= I G αD/F , −k1 x1 ˆ

ˆ = I G θ¨ k

From the equations on block 2 and the disk, we eliminate the unknown tension T from the system to obtain: I G θ¨ − m r2 x ¨2 − b r2 x˙ 2 − k2 r2 x2 − k1 r1 z = 0. From this equation, and the equation on block 1, we eliminate the coordinates z and x2 , and obtain the equations of motion to be: ³

I G + m r22

´

mx ¨1 + 2 k1 x1 + k1 r1 θ ´ ³ θ¨ + b r22 θ˙ + k1 r1 x1 + k1 r12 + k2 r22 θ

Problem 85: For the system shown in the figure:

=

0,

=

0.

x

a) what is the degree-of-freedom for this system?

k

k

(I, r)

2m

b) using Lagrange’s equations, determine the differential equations that govern the motion.

m

72

Solution: a) This system contains three masses which are each allowed to move in only one direction. The upper mass slides horizontally with displacement x1 , while the disk rotates through an angle θ. Finally, the suspended mass moves vertically and its position can be described by the coordinate x2 . However, because the disk and the suspended mass are 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-degreeof-freedom system. b) We utilize the coordinates x1 , x2 , and θ, as measured from the unstretched position of the two springs. Therefore, the kinetic and potential energies are written as: T

=

V

=

1 1 1 (2m) x˙ 21 + m x˙ 22 + I θ˙2 , 2 2 2 1 1 k x21 + k (rθ − x1 )2 − m g x2 . 2 2

However, x2 and θ are related by the above relationship. Thus eliminating θ, the energies become: 1 1 1 I 2 T = (2 m) x˙ 21 + m x˙ 22 + x˙ , 2 2 2 r2 2 1 1 k x21 + k (x2 − x1 )2 − m g x2 , V = 2 2 which, using Lagrange’s equations, yields the equations of motion: 2m x ¨ + 2 k x1 − k x2 ¶1 µ I ¨2 − k x1 + k x2 m+ 2 x r Problem 86: For the system shown in the figure:

=

0,

= m g.

k2

a) find the mass and the stiffness matrix;

k1

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

m2 m1

k3

m3

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. Maintaining a unit displacement of each mass in turn requires forces of the form: (x1 , x2 , x3 ) = (x1 , x2 , x3 ) = (x1 , x2 , x3 ) =

(1, 0, 0) (0, 1, 0) (0, 0, 1)

→ → → 73

f f f

= = =

(k1 + k2 , −k2 , 0)T , (−k2 , k2 + 2k3 , −2k3 )T , (0, −2k3 , 2k3 )T .

Therefore, with these coordinates the stiffness matrix is:   k1 + k2 −k2 0 k2 + 2k3 −2k3  K =  −k2 0 −2k3 2k3 Alternatively, we can define the potential energy of the system as: V

= =

1 1 1 (k1 )(x1 )2 + k2 (x2 − x1 )2 + (2k3 )(x3 − x2 )2 , 2 2 2 1 1 1 (k1 + k2 )(x1 )2 − (2k2 )(x1 x2 ) + (k2 + 2k3 )(x2 )2 2 2 2 1 1 − (2k3 )(x2 x3 ) + (2k3 )(x3 )2 , 2 2

which leads to the same stiffness matrix. To determine the mass matrix, we could use the inertia influence coefficients, but, for variety, we determine the kinetic energy as: T =

1 1 1 m1 x˙ 21 + m2 x˙ 22 + m3 x˙ 23 . 2 2 2

Therefore, the mass matrix is: 

m1 M = 0 0

0 m2 0

 0 0  m3

b) With this choice of coordinates, the mass matrix is diagonal and the stiffness matrix contains nonzero off-diagonal terms. Thus, the system is statically coupled but dynamically uncoupled. Problem 87: For the system shown at right: k2


k1

k1 m1

b) is the system statically or dynamically coupled, or both;

m1

k2 m2

c) find the matrix M −1 K .

74

Problem 88: In the system shown to the and stiffness matrix are: · m1 0 M = 0 m2 · k1 + k2 K = −k2

right, the mass ¸

,

−k2 k2

¸

while the damping is proportional, i.e., C = βM +αK . If m1 is subject to harmonic forcing F (t) = sin t, with: m1 = 2.0 kg, k1 = 6.0 N/m

m2 = 1.0 kg, k2 = 2.0 N/m,

β = 0.1, α = 0.1

a) determine the eigenvectors of this system, normalized by the kinetic energy inner product; b) what are the damped natural frequencies of this system; c) find the steady-state amplitude of vibration of the mode of vibration with the lowest natural frequency. Problem 89: The two-degree-of-freedom system shown is subject to a harmonic force applied to the block of mass 2m, of the form: ¡ ¢ f (t) = (2 sin(t)) N ˆı.

f (t) 2m k

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

2m k

k

a) the mass, damping, and stiffness matrices; b) the forced, damped equation (singledegree-of-freedom) that describes the motion of each mode; c) the steady-state response of the system. Solution: a) Using influence coefficients, we find that the mass and stiffness matrices are: ¸ ¸ · · 2k −k 2m 0 . , K = M = −k 2k 0 2m Therefore, with proportional damping, the damping matrix becomes C = αK , so that

75

we find: M = (4 kg)

·

1 0

¸

0 1

,

K = (4 N/m)

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

·

Problem 90: In the multi-degree-of-freedom system shown in the figure, the block with mass 4m slides on a smooth, frictionless surface. If the pulley is massless:

2 −1 −1 2

·

¸

2 −1 −1 2

,

.

x1

z

2k

a) using Lagrange’s equations, determine the differential equations governing the motion, as measured from static equilibrium;

¸

r

k 4m

b) with m = 1 kg and k = 16 N/m, find the natural frequencies and mode shapes for the free vibration of this system. Normalize the mode shapes so that with respect to the mass matrix the amplitude of each mode is one;

x2 2m

c) find the general solution to these equations for the above values of m and k. 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 potential energy, the kinetic and potential energies for this system can be written as T

=

V

=

¢ ¢ 1¡ 1¡ 4 m x˙ 21 + 2 m x˙ 22 , 2 2 1¡ ¢ 2 1¡ ¢ 2 2 k x1 + k z , 2 2

In terms of x1 and x2 , the potential energy becomes V=

¢2 ¢ 1 ¡ ¢ 2 1 ¡ ¢¡ 1¡ ¢ 2 1¡ 1¡ ¢ 2 2 k x1 + k x2 − x1 = 3 k x1 + − 2 k x1 x2 + k x2 . 2 2 2 2 2

Therefore the equations of motion become 4m x ¨1 + 3 k x1 − k x2 2m x ¨2 − k x1 + k x2

= =

0, 0,

or in matrix form m

·

4 0

0 2

¸ ·

x ¨1 x ¨2

¸ 76

+k

·

3 −1

−1 1

¸ ·

x1 x2

¸

= 0.

b) The corresponding eigenvalue problem for the above system is ¸ · 3 ¡ −1 ¢ ¢ k k ¡ − 14 4 M K u =λu −→ = βu 1 m − 21 m 2 The characteristic equation is ¶µ ¶ µ 1 5 1 1 3 −β − β − = β 2 − β + = 0, 4 2 8 4 4 with the solution β=

5±3 8

−→

λ = ω2 =

½

k k , 4m m

¾

.

Returning this to the eigenvalue problem, the mode shapes are defined by the equation 3 1 ui1 + ui2 = βi ui1 , 4 4 so that ui =

·

1 4 βi − 3

¸

ci

u1 =

·

1 −2

Normalizing ui by the mass matrix implies that · £ ¤ 4m 1 (4 βi − 3) 1 = uiT M ui = c2i 0

¸

c1 ,

u2 =

¸ ·

0 2m

·

¸

1 1

c2 .

1 (4 βi − 3)

¸

.

Solving for ci ci =

s

1 2 m (2 + (4 βi − 3)2 )

Finally, the normalized eigenpairs are " # r √ 1 k 12 m , u1 = , ω1 = 2 − √12 4m m

−→

c1 =

ω2 =

r

r

1 , c2 = 12 m

k , m

u2 =

r

"

1 6m

√1 6m √1 6m

#

.

c) With the above mode shapes and natural frequencies the general solution becomes q (t) =

·

x1 (t) x2 (t)

¸

=

2 X

(Ai sin(ωi t) + Bi cos(ωi t)) ui ,

i=1

=

Ã

Ãr

! Ãr !! " # √ 1 k k 12 m A1 sin t + B1 cos t 2 − √12 4m 4m m # ! Ãr !! " Ã Ãr 1 √ k k 6m . t + B2 cos t + A2 sin √1 m m 6m

77

Problem 91: For the system shown in the figure: a) find the mass and the stiffness matrices;

k3

k1

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

m1

k2

m3

k1 m2

Solution: a) For coordinates we choose (x1 , x2 , x3 ) as the displacements of each mass with respect to inertial space. Using influence coefficients, we find that:     m1 0 0 k1 + k2 −k2 0 k1 + 2k2 −k2  M =  0 m2 0  , K =  −k2 0 −k2 k2 0 0 m3 Alternatively, if we choose coordinates (x1 , x2 , z), where z represents the stretch in the spring connecting m2 and m3 , we can determine the mass and stiffness matrices from the Lagrangian. The kinetic and potential energies are: T

=

V

=

1 1 1 m1 x˙ 21 + m2 x˙ 22 + m3 (x˙ 2 − x˙ 3 )2 , 2 2 2 1 1 1 1 2 k1 x1 + k2 (x2 − x1 )2 + k2 z 2 + k1 x22 . 2 2 2 2

Thus, with these coordinates the mass and stiffness matrices become:    k1 + k2 −k2 m1 0 0 k1 + k2 K =  −k2 M =  0 m2 + m3 −m3  , 0 0 0 −m3 m3

 0 0  k2

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

78


r

a) What is the degree-of-freedom for this system?

r 2

b) Find the governing equations of motion; k

c) If m = 1 kg and k = 4 N/m, what are the frequencies of oscillation for the motion and the corresponding mode shapes, normalized by the kinetic energy inner product?

m m k

Solution: a) Let the displacement of the left block, disk, and right block be described as (−x1 ˆ), ˆ and (x2 ˆ) respectively. Although x1 and θ are related by the following constraint: (θ k), x1 =

r θ, 2

x2 is independent from the above two coordinates. Therefore, the system has two degrees-of-freedom. b) With the above coordinates, the kinetic and potential energies can be written as: T

=

V

=

1 1 mr2 ˙2 1 mx˙ 21 + mx˙ 22 + θ , 2 2 2 2 1 2 1 kx + k(x2 − rθ)2 , 2 1 2

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

= T − V, · ¸ ¤ 1 1 £ 2 2 2 2 m 3x˙ 1 + x˙ 2 − k 5x1 − 4x1 x2 + x2 . = 2 2

Using Lagrange’s equations of motion, the governing equations are: 3m x ¨1 + 5k x1 − 2k x2 mx ¨2 − 2k x1 + k x2

= 0, = 0.

c) From the above equations, the mass and stiffness matrices can be written as: ¸ ¸ · · 5 −2 3 0 , K =k M =m −2 1 0 1 79

The characteristic matrix, A = M −1 K becomes: ¸ · 5 k − 32 3 A= , m −2 1 and the characteristic equation can be written as: β2 −

8 1 β + = 0, 3 3

k β is an eigenvalue of the characteristic where, if β is a solution to this equation, λ = m matrix A. This quadratic equations has solutions of the from: √ 4 ± 13 β= . 3

To determine the eigenvectors, we return to the characteristic matrix A, so that Au = λu. The elements of u then satisfy the equation: 2 5 u 1 − u 2 = β u1 . 3 3 Thus, if u1 = 1, this yields: u2 =

1∓

√ 13 2

for

β=

4±

√ 13 . 3

Normalizing by the kinetic energy inner product, we find that: ˆ=p u

u (u, u)M

=p

u 3u21 + u22

Problem 93: For the system shown in the figure, the surface is assumed to be frictionless. a) Using Lagrange’s equations, determine the differential equations governing the motion (measured from static equilibrium);

k 4m

(m, r)

b) Find the mode shapes and natural frequencies of the resulting motion.

k

m

80

Problem 94: The multi-degree-of-freedom system shown in the figure is subject to harmonic forcing of the form f (t) = sin(4 t). k

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

m

b) If m = 2 kg and k = 4 N/m, the natural frequencies of the system are found to be: ω1 = 0.38 rad/s,

2k

2k

ω2 = 2.62 rad/s,

ω3 = 1.41 rad/s. m Find the corresponding mode shapes and normalize them so that (ui , ui )M = 1.

m f (t)

c) Determine the forced, uncoupled equations of motion for the modal amplitudes Qi (t). Problem 95: For the system shown in the figure, the surface is assumed to be frictionless. If each block is displaced by a distance d (down and to the right), find the resulting motion of the system.

x1 (m, r)

k m

θ

z

k m

Solution: We define the coordinates x1 , x2 , θ, and z as shown in the figure, so that x1 = −r θ,

81

z = r θ − x2 .

x2

With these coordinates, the kinetic and potential energies can be written as µ ¶ 1 ¡ ¢ 2 1 m r2 ˙2 1 ¡ ¢ 2 T = θ + m x˙ 1 + m x˙ 2 , 2 2 2 2 1¡ ¢ 2 1¡ ¢ 2 k x1 + k z . V= 2 2 Expressing these only in terms of the coordinates x1 and x2 , we obtain µ ¶ 1¡ ¢ 2 1 3m x˙ 21 + m x˙ 2 , T = 2 2 2 1¡ ¢ 2 1¡ ¢ 1¡ ¢ 2 1¡ ¢ 1¡ ¢ 2 2 V= k x1 + k (−x1 − x2 ) = 2 k x1 + 2 k x1 x2 + k x2 . 2 2 2 2 2 Therefore, the mass and stiffness matrix can be identifed as · 3 ¸ · ¸ · ¸ 2 1 x1 (t) 0 M =m 2 , K =k with q (t) = 1 1 x2 (t) 0 1 and the equations of motion are: ¸ · ¸ · · 3 x ¨1 2 0 2 +k m x ¨2 1 0 1

1 1

¸ ·

x1 x2

¸

= 0.

The solution to this equation requires the solution of an eigenvalue problem of the form · 4 2 ¸ k 3 3 u = λ u, (M −1 K ) u = m 1 1 which is determined from the characteristic equation ·µ ¸ ¶ ³ ´ 4 2 k −1 = 0, − β (1 − β) − det M K − λ I = m 3 3 with λ =

k m

β. This quadratic equation has the solution ) (r r ½ ¾ 1 7±5 k 2k = , 2 −→ ω = , β= 6 3 3m m

With this, the eigenvectors are determined by returning to the original eigenvalue equation · 4 2 ¸ · · ¸ ¸ 3 βi − 4 ui1 ui1 3 3 , −→ ui2 = ui1 = βi ui2 ui2 1 1 2 In addition, normalizing the eigenvectors by the mass matrix · ¶ ¸ · ¸ µ £ ¤ 3m 0 6 + (3 βi − 4)2 ui1 2 1 = uiT M ui = ui1 ui2 . = m u2i1 ui2 0 m 4 Solving for ui1 the normalized eigenvectors are # " u1 =

√ 2 15 m 3 − √15 m

82

,

u2 =

"

√ 2 10 m √ 2 10 m

#

.

One can easily verify that both u1T M u2 = 0 and u2T M u1 = 0. The general solution is written as q (t) =

2 ³ X i=1

´ Ai sin(ωi t) + Bi cos(ωi t) ui

subject to the initial conditions q (0) =

·

d −d

Premultiplying by uiT M yields ¡ ¢ uiT M q (0) = uiT M ui Bi ,

¸

q˙ (0) =

,

·

0 0

¸

.

¡ ¢ uiT M q˙ (0) = uiT M ui (Ai ωi ).

Since uiT M ui = 1 from our normalization, the constants are directly solved to be r r m m A1 = 0, B1 = 6 d , A2 = 0, B2 = d 15 10 Finally, the solution to the specific initial conditions becomes ( Ãr !· !· Ãr ¸ ¸) · ¸ k 2k d 4 1 x1 (t) cos t t + cos . q (t) = = −6 1 x2 (t) 5 3m m

Problem 96: The two-degree-of-freedom system shown is subject to a harmonic force applied to the block of mass 2m, of the form:

f (t) m

2m

f (t) = (2 sin(t) N) ˆı.

k

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

x1

k

2k

z x2

a) the equations of motion; b) the forced, damped equation (singledegree-of-freedom) that describes the motion of each mode; Solution: a) We define the coordinates x1 , x2 , and z, which are related as z = x2 − x1 .

The equations of motion become · ¸ · ¸ · ¸ 2 0 2 −1 f (t) m +k = . 0 1 −1 3 0 83

b) The corresponding eigenvalue problem can be written as · ¶ ¸ µ ¡ −1 ¢ k k 1 − 12 M K = β u u= −1 3 m m and the characteristic equation becomes (1 − β) (3 − β) −

5 1 = β 2 − 4 β + = 0. 2 2

This quadratic equation has the solutions ( √ ) r √ √ √ ( 8 − 3) k ( 8 + 3) k 3 2 √ √ , −→ ω = β =2± . , 2 2m 2m Returning to the eigenvalue equation, the eigenvectors satisfy the equation ui1 −

1 ui2 = βi ui1 2

−→

ui2 = 2 (1 − βi ) ui1

so that s √ √ · ¸ ( 8 − 3) k 1 √ √ , u1 = ω1 = , 6−2 2m s √ √ ¸ · ( 8 + 3) k √1 √ . ω2 = , u2 = −( 6 + 2) 2m For each eigenvector the kinetic energy inner product is √ √ u1T M u1 = 14 − 4 6, u2T M u2 = 14 + 4 6, Finally, the modal equation for the first mode can be written as ¡ ¢ ¡ T ¢ ¨ 1 + u T K u1 Q1 = u T f (t), u1 M u1 Q 1 1 u1T f (t) 2 ¨ Q1 + ω Q1 = u T M u1 , 1 ³ √ √ ´1 f (t) ( 8− 3) k ¨ √ , √ Q = Q1 + 1 2m 14−4 6 while the response of the second mode is governed by ¡ T ¢ ¡ ¢ ¨ 2 + u T K u2 Q2 = u T f (t), u2 M u2 Q 2 2 T ¨ 2 + ω 2 Q2 = uT2 f (t) , Q 2 u2 M u2 ³ √ √ ´ 3) k f (t) ¨ 2 + ( 8+ √ , √ Q2 = 14+4 Q 2m 6

84

Mechanical Vibration solved examples

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