Vibrations Hw 4 556r5r

Homework Set 4 Short answer questions 𝐸𝐴

3.41 𝜔𝑛 = √𝑚𝐿 𝐽𝐺

3.42 𝜔𝑛 = √ 𝐼𝐿 3.43 𝜔𝑛 = √

48𝐸𝐼 𝑚𝐿3

Chapter Problems 2.48 Derive the differential equation governing the motion of one-degree-of-freedom system by applying the appropriate form(s) of Newton’s laws to the appropriate free-body diagrams. Use the generalized coordinate shown in Figure P2.48. Linearize nonlinear differential equations by assuming small displacements. Given: system shown Find: differential equation Solution: Free-body diagrams of the system at an arbitrary instant are shown below

Note that the force developed in the spring is proportional to the change in length of the spring. When the center of the disk is displaced a distance x from equilibrium, the end of the spring attached to the center of the disk compresses by x. When the center of the disk displaces x, the point on the disk to which the spring is attached has translated a distance x and rotated along the distance an angle . Assuming no slip between the disk and the surface,  = x/r. Hence this end of the spring has displaced 2x. The total change in length of this spring is 3x.

Summing moments about the point of between the disk and surface leads to

 M 

C ext

  M C eff

 (kx  cx )r  2k (3x)r  2k (3x)( 2r )  mx(r ) 

1 2 x mr 2 r

3 mrx  crx  19 krx  0 2

The differential equation is put into standard form by dividing by the coefficient of x leading to x 

2c 38k x  x0 3m 3m

Problem 2.48 illustrates derivation of the differential equation governing the motion of a onedegree-of-freedom system using the free-body diagram method.

2.53 Determine the differential equations governing the motion of the system by using the equivalent systems method. Use the generalized coordinates shown in Figure P2.53.

Given: x as generalized coordinate, k, m, I, r Find: differential equation Solution: Since x is measured from the system’s equilibrium position, gravity cancels with the static spring forces in the governing differential equation. Thus, for purposes of deriving the differential equation, both are ignored. It is assumed there is no slip between the cable and the pulley. Thus the angular rotation of the pulley is kinematically related to the displacement of the block by



x 2r

The equivalent systems method is used. The system is modeled by a mass-spring system of an equivalent mass and equivalent stiffness, using the generalized coordinate, x. The kinetic energy of the equivalent system at an arbitrary time is 1 T  meq x 2 2

The kinetic energy of the system at an arbitrary instant is

1 1  x  T  mx 2  I   2 2  2r 

2

1 I    m  2  x 2 2 4r 

Requiring the kinetic energy of the equivalent system to be equal to the kinetic energy of the original system at any instant leads to meq  m 

I 4r 2

The potential energy of the equivalent system at an arbitrary instant is 1 V  keq x 2 2

The potential energy of the system at hand at an arbitrary instant is 1  x v  k  2 2 1k 2 v x 24

2

Requiring the potential energies to be equal at any instant leads to keq 

k 4

The differential equation governing free vibration is meq x  k eq x  0 I  k   m  2  x  x  0 4r  4  k x  x 0 I   4 m  2  4r  

Problem 2.53 illustrates use of the equivalent system method to derive the differential equation governing free vibration of a one-degree-of-freedom system. 3.1 The mass of the pendulum bob of a cuckoo clock is 30 g. How far from the pin should the bob be placed such that its period is 1.0 sec? Given: m = 30 g, T = 1.0 sec

Find:  Solution: The pendulum is modeled as a particle of mass m attached to a massless rod. Let  be the distance between the particle and its axis of rotation. Let  be the counterclockwise displacement of the rod, measured from the vertical. Free body diagrams of the particle are shown at an arbitrary instant. 

m 2 2



m

=

mg

EXTERNAL FORCES

EFFECTIVE FORCES

Summing moments about the axis of rotation

 M 

0 ext

  M 0 eff

 mg sin   m 2 g 

  sin   0 Assuming small angular displacements

g     0  from which the natural frequency is determined as

g 

n  Requiring the period of motion to be 1 sec leads to T 1sec 

2

n

 2

 g

m g s2  2  1 4 4π 2 2 s   0.248 m 9.81

Problem 3.1 illustrates how knowledge of the period is used to determine information about the system.

3.4 When the 5.1 kg connecting rod of Figure P3.4 placed in the position shown, the spring deflects 0.5 When the end of the rod is displaced and released, resulting period of oscillation is observed as 0.15 Determine the location of the center of mass of the connecting and the centroidal mass moment of inertia of the rod.

is mm. the sec.

Given: m = 5.1 kg, ST. = 0.5 mm, T = 0.15 sec, L = 20 cm, k = 3 × 104 N/m Find: , I Solution: When the system is in equilibrium the moment of the spring force balances with the moment of the gravity force when moments are taken about the pin ,

M

0

0

 mg  k ST . L  0 

k ST . L mg

 4 N  3  10  0.0005 m 0.2 m  m   5.1 kg  9.81 m 2  sec    0.060 m  6.0 cm

Let  be the clockwise angular displacement of the rod after it is released, measured from the system’s equilibrium position. Assuming small displacements, free body diagrams of the system at an arbitrary instant are shown below

=

R mg

k (L ST ) EXTERNAL FORCES

m

:: I

EFFECTIVE FORCES

Summing moments about O, noting that the static deflection cancel with gravity

 M 

  M 0 eff .

0 ext .

 kL2  m 2  I m 2  I   kL2  0





 

(1)

kL2  0 m 2  I

The natural frequency is obtained from eq.(1) as

n 

kL2 m 2  I

(2)

The natural frequency is calculated from the natural period by

n 

2 T

(3)

Equations (2) and (3) are combined and used to solve for I as I

kL2T 2  m 2 4 2

 2 2 4 N  3  10  0.2 m  0.15 sec m 2   5.1 kg 0.060 m  2 4π  0.666 kg  m 2

Problem 3.4 illustrates the use of a free vibrations test to determine the moment of inertia of a connecting rod.

3.27 For the system shown in Figure P3.27: (a) Determine the damping ratio; (b) State whether the system is underdamped, critically damped, or overdamped; (c) Determine x(t) or for the given initial conditions Given: k = 3.2 × 104 N/m, c = 150 N ∙ s/m, cm, Ip = 0.3 kg ∙ m2, m1 = 5 kg, m2 = 40  (0)  0, (0)  2.5 rad/s Find:

(a)  (b) nature of damping (c) (t)

(t) r = 10 kg,

Solution: The differential equation is derived using the equivalent systems method. Let x1 = 3r be the downward displacement of the block of mass m1 and x2 = r be the upward displacement of the block of mass m2. The kinetic energy of the system at an arbitrary instant is 1 1 1 T  I p 2  m1 (3r) 2  m 2 (r) 2 2 2 2 1 T  I p  9r 2 m1  r 2 m 2  2 2 Since an angular displacement is chosen as the generalized coordinate the appropriate model is the torsional system. The equivalent moment of inertia is





I eq  I p  9r 2 m1  r 2 m2  1.15 kg  m 2 The potential energy of the system at an arbitrary instant is





1 1 2 k 3r   9kr 2  2 2 2

V

from which the equivalent torsional stiffness if obtained as kteq  9kr 2  2.88 103 N  m/rad

The work done by the viscous damper between two arbitrary positions is

 

W12    c r dr     cr 2 d from which the equivalent viscous damping coefficient is cteq  cr 2  1.5 N  m  s/rad

Thus the differential equation governing the motion of the system is

1.15  1.5  2.88  10 3  0   1.304  2.504  10 3  0 The natural frequency of the system is

n 

kteq

 2.504 103  50.043 rad/s

I eq

(a) The damping ratio is obtained from



c eq 2 I eq n

 0.013

(b) Since the damping ratio is less than 1 the system is underdamped.

(c) The free-vibration response of an underdamped one-degree-of-freedom system is

 (t )  Ae  t sin( d t   d ) n

where the damped natural frequency is

 d  n 1   2  d  50.0392 rad/s For the given initial conditions,

A  (0) /  d  0.050 rad

 d  tan 1 (0)  0 Noting that n  (0.013)(50.043)  0.651 , the free-vibration response is

 (t )  0.050e 0.651t sin( 50.0392t ) rad Problem 3.27 illustrates the free-vibration response of an underdamped one-degree-of-freedom system.

3.32 The amplitude of vibration of the system of Figure P3.32 decays to half of its initial value in 11 cycles with a period of sec. Determine the spring stiffness and the viscous damping coefficient.

0.3

Given: X11 = 1/2X0, T = 0.3 sec, J = 2.4 kg ∙ m2, m = 5 kg, R1 = 20 cm, R2 = 40 cm Find: c, k

Solution: Let x represent the displacement of the block, measured positive downward from the system’s equilibrium position. The equivalent system method is used to derive the differential equation governing free vibration. The angular rotation of the disk is



x R1

The change in length of the spring due to a displacement x is x2 

R2 x  2x R1

The kinetic energy of the system is

T

1 2 1 2 1  J  mx  J   m  2   2 2 2 2 R1 

Hence the system’s equivalent mass is

meq  m 

J 2.4 kg  m 2  5 kg   65 kg R12 0.2 m2

The potential energy of the system is



1 2 1 kx2  4 kx 2 2 2

V



Hence the equivalent stiffness is

keg  4 k The work done by the damping force is

W    cxdx which implies that

ceq  c Thus the differential equation governing free vibrations of the system is 65x  cx  4 kx  0 x 

c 4k x  x 0 65 65

The natural frequency is determined from the differential equation as

n 

4k 65

 

c 130 n

The damping ratio is determined as

From the information given, the logarithmic decrement is



1  x0  1 ln    ln 2  0.0630 11  x11  11

from which the damping ratio is calculated as

 

 4 2   2

 0.01

The damped natural frequency is

d 

2 2 rad   20.94 Td 0.3 sec sec

The undamped natural frequency is obtained as

n 

d 1  2

 20.94

rad sec

The spring stiffness is k  65

n2 4

 7130

N m

The damping coefficient is

c  130n  27.2

N  sec m

Problem 3.32 illustrates (a) application of the equivalent system method derive the differential equation governing free vibration of a one-degree-of-freedom system and (b) use of measured free vibration characteristics to determine system parameters. Out of book problems

1. A thin disk of mass 350 grams and radius 10 mm is attached to the end of a circular shaft of length 20 cm, radius 3 mm that is made from a material of elastic modulus 150x109 N/m2 and shear modulus 50x109 N/m2. What are the (a) natural frequency for longitudinal motion of the disk, (b) natural frequency for torsional oscillation of the disk and (c) natural frequency for transverse motion of the disk. Assume the shaft is massless in all cases.

2. A 10 kg mass is hanging from a spring of stiffness 3x105 N/m. The spring is given a displacement of 3 mm downward from equilibrium and the system released from rest. (a) What is the static deflection of the spring? (b) What is the frequency of the resulting motion in Hz? (c) What is the period of the resulting motion? (d) If the spring was given a displacement of 4 mm downward how long would it take the mass to execute one cycle of motion? (e) If the mass is increased what happens to the period: does it decrease, stay the same as before or increase? Why?

(d) If the mass is increase the natural frequency is decreased which leads to an increase of the perioid.

3. The 10 kg mass in Problem 2 is suspended from the same spring and a viscous damper of coefficient 1000 N ∙ s/m. The mass is given a displacement of 3 mm downward and released from rest. (a) What is the frequency of the resulting motion? (b) What is the period of the resulting motion? (c) What is the energy present in the system at t=0? (d) What is the logarithmic decrement? (e) What is the amplitude at the end of the second cycle of motion? (f) energy dissipated by the viscous damper during the first two cycles of motion? The damping ratio of the system is

 

c 1000 N  s/m   0.289 2mn 2 10 kg 173.2 r/s 

(a) The damped natural frequency is

d  n 1   2  173.2 r/s  1  0.2892  165.8 r/s (b) The damped period is

Td 

2

d



2  0.0379 s 165.8 r/s

(c) The energy present in the system at t=0 is the potential energy of

V





1 2 1 2 kx0  3 105 N/m  0.003 m   1.35 J 2 2

(e) The logarithmic decrement is



2 1  2



2  0.289 

 1.90

1  0.2892

(f) The amplitude at the end of two cycled of motion is given by

1 2

  ln

0   2  e2  0  e21.90  0.003 m   6.71105 m 2

(g) The energy dissipate by the viscous damper during the first two cycled is

W12  T2  V2  T1  V1  0







2 1 3 105 N/m 6.71105 m  0  1.35 J=1.349 J 2

4.

5,