System Dynamics
System Dynamics
3rd Edition
ISBN: 9780073398068
Author: III William J. Palm
Publisher: MCG
Question
Book Icon
Chapter 13, Problem 13.22P
To determine

(a)

The expression for X1(jω)F(jω).

The expression for X2(jω)F(jω).

Expert Solution
Check Mark

Answer to Problem 13.22P

The expression for X1(jω)F(jω) is 4ς2+μ2r2(2ς)2[(1(1+μ)r2)2+μ2r2(1r2)2].

The expression for X2(jω)F(jω) is 2ςω12(2ς)2[(1(1+μ)r2)2+μ2r2(1r2)2].

Explanation of Solution

Given information:

The main mass is m1.

Write the expression for force of mass 1.

f=m1x¨1+c(x˙1x˙2)+k2x1+k2x1m1x¨1=fc(x˙1x˙2)kx1..... (I)

Here, mass of body 1 is m1, damping coefficient is c, velocity of mass 1 is x˙1, velocity of mass 2 is x˙2, stiffness is k, displacement is x, acceleration of mass 1 is x¨1 and the coefficient of friction is f.

Write the expression for absorber mass.

m2x¨2=cx˙2+cx˙1..... (II)

Here, mass of the absorber is m2. and the acceleration of absorber is x¨2.

Take Laplace transform of Equation (I).

L(m1x¨1)=L(f)L[c(x˙1x˙2)]L[kx1]m1s2X1(s)+csX1(s)csX2(s)+kX1(s)=F(s)(m1s2+cs+k)X1(s)csX2(s)=F(s)..... (III)

Take Laplace transform of Equation (II).

L[m2x¨2]=L[cx˙2]+L[cx˙1]m2s2X2(s)=csX2(s)+csX1(s)csX1(s)+(m2s2+cs)X2(s)=0..... (IV)

Apply Cramer rule in Equation (III) and (IV).

(m1s2+cs+kcscsm2s2+cs)(X1(s)X2(s))=(F(s)0)..... (V)

Further soving Equation (V).

D(s)=m1m2s4+cm2s3+km2s2+m1cs3+(cs)2+ckscs2D(s)=s(m1m2s3+c(m2+m1)s2+km2s+m1cs3+ck)..... (VI)

Write the expression for parameter μ.

μ=m1m2..... (VII)

Write the expression for parameter ω12.

ω12=km1..... (VIII)

Write the expression for parameter ς.

ς=c2m1k..... (IX)

Write the expression for parameter r.

r=ωω1..... (X)

Here, angular speed of main mass is ω1.

Substitute jω for s in Equation (VI).

D(jω)=s(m1m2(jω)3+c(m2+m1)(jω)2+km2(jω)+m1c(jω)3+ck)D(jω)=|jω[(m1m2ω3jc(m2+m1)ω2+km2ωj+ck)]..... (XI)

Substitute km1 for ω12, ς for c2m1k, r for ωω1 and μ for m1m2 in Equation (XI).

D(jω)=|jω[m1(μm1)(rω1)3j(2ςm1k)(m1μm1)(rω1)2+(ω12m1)(μm1)(rω1)j+(2ςm1k)(ω12m1)]|=m1ω14r(2ς)2[(1(1+μ)r2)2+μ2r2(1r2)2]..... (XII)

Apply Cramer rule in Equation (V) to obtain D1(s).

(m1s2+cs+kcscsm2s2+cs)(X1(s)X2(s))=(F(s)0)D1(s)=F(s)×(m2s2+cs)(0×cs)D1(s)=F(s)(m2s2+cs)..... (XIII)

Apply Cramer rule in Equation (V) to obtain D2(s).

(m1s2+cs+kF(s)cs0)=0D2(s)=(cs)×F(s)D2(s)=csF(s)..... (XIV)

Write the expression for X1(s).

X1(s)=D1(s)D(s)..... (XV)

Substitute F(s)(m2s2+cs) for D1(s) and m1ω14r(2ς)2[(1(1+μ)r2)2+μ2r2(1r2)2] for D(s) in Equation (XIV).

X1(s)=F(s)(m2s2+cs)m1ω14r(2ς)2[(1(1+μ)r2)2+μ2r2(1r2)2]X1(s)F(s)=(m2s2+cs)m1ω14r(2ς)2[(1(1+μ)r2)2+μ2r2(1r2)2]..... (XV)

Substitute jω for s in Equation (XV).

X1(jω)F(jω)=(m2(jω)2+c(jω))m1ω14r(2ς)2[(1(1+μ)r2)2+μ2r2(1r2)2]=m2ω2+cωjm1ω14r(2ς)2[(1(1+μ)r2)2+μ2r2(1r2)2]..... (XVI)

Simplify Equation (XVI) by substituting km1 for ω12, ς for c2m1k, r for ωω1 in Equation (XVI).

X1(jω)F(jω)=m2ω2+cωjm1ω14r(2ς)2[(1(1+μ)r2)2+μ2r2(1r2)2]=m1rω144ς2+μ2r2

Substitute m1rω144ς2+μ2r2 for m2ω2+cωj in Equation (XVI).

X1(jω)F(jω)=m2ω2+cωjm1ω14r(2ς)2[(1(1+μ)r2)2+μ2r2(1r2)2]X1(jω)F(jω)=4ς2+μ2r2m1ω12(2ς)2[(1(1+μ)r2)2+μ2r2(1r2)2]kX1(jω)F(jω)=k4ς2+μ2r2m1ω12(2ς)2[(1(1+μ)r2)2+μ2r2(1r2)2]..... (XVII)

Write the expression for stiffness.

k=m1ω12..... (XVIII)

Substitute m1ω12 for k in Equation (XVII).

kX1(jω)F(jω)=m1ω124ς2+μ2r2m1ω12(2ς)2[(1(1+μ)r2)2+μ2r2(1r2)2]kX1(jω)F(jω)=4ς2+μ2r2(2ς)2[(1(1+μ)r2)2+μ2r2(1r2)2]..... (XIX)

Write the expression for X2(s).

X2(s)=D2(s)D(s)..... (XX)

Substitute csF(s) for D2(s) and m1ω14r(2ς)2[(1(1+μ)r2)2+μ2r2(1r2)2] for D(s) in equation (XX).

X2(s)=csF(s)m1ω14r(2ς)2[(1(1+μ)r2)2+μ2r2(1r2)2]X2(s)F(s)=csm1ω14r(2ς)2[(1(1+μ)r2)2+μ2r2(1r2)2]..... (XXI)

Substitute jω for s in Equation (XXI).

X2(jω)F(jω)=c(jω)m1ω14r(2ς)2[(1(1+μ)r2)2+μ2r2(1r2)2]..... (XXII)

Simplify Equation (XXII).

X2(jω)F(jω)=|c(jω)m1ω14r(2ς)2[(1(1+μ)r2)2+μ2r2(1r2)2]|X2(jω)F(jω)=2ςm12ω12rm1ω14r(2ς)2[(1(1+μ)r2)2+μ2r2(1r2)2]X2(jω)F(jω)=2ςω12(2ς)2[(1(1+μ)r2)2+μ2r2(1r2)2].

Conclusion:

The expression for X1(jω)F(jω) is 4ς2+μ2r2(2ς)2[(1(1+μ)r2)2+μ2r2(1r2)2].

The expression for X2(jω)F(jω) is 2ςω12(2ς)2[(1(1+μ)r2)2+μ2r2(1r2)2].

To determine

(b)

The plot between kX1F and ωkm1.

Expert Solution
Check Mark

Answer to Problem 13.22P

The plot between kX1F and ωkm1 is shown in Figure-(1).

Explanation of Solution

Given information:

The main mass is m1 and the absorber mass is m2.

To plot the graph, consider some constant values such as 10 for m1, 30 for m2, 104 for k and [0,2] for r.

Calculation:

Substitute 10 for m1 and 30 for m2 in equation (VII).

μ=1030=0.3333

Substitute 104 for k and 10 for m1 in Equation (VIII).

ω12=10410ω12=1000ω1=1000ω1=31.6

Assume 0 for ω and 0.1 for ς.

Substitute 0 for ω and 31.6 for ω1 in Equation (X).

r=031.6=0

Substitute 0 for r, 0.3333 for μ and 0.1 for ς in Equation (XVIII).

kX1(jω)F(jω)=4(0.1)2+(0.3333)2(0)2(2(0.1))2[(1(1+0.3333)(0)2)2+(0.3333)2(0)2(1(0)2)2]=0.20.141421=1.414

Calculate ωkm1

Substitute 0 for ω and 104 for k and 10 for m1 in ωkm1.

010410=0

Prepare a table for various values of parameters.

ω ς kX1(jω)F(jω) ωkm1
0 0.1 1.414 0
10 0.1 1.29 0.316
20 0.1 3.01 0.632
30 0.1 9.35 0.948
0 0.3 1.414 0
10 0.3 1.16 0.316
20 0.3 2.2 0.632
30 0.3 5.58 0.948
0 0.5 1.414 0
10 0.5 1.15 0.316
20 0.5 2.13 0.632
30 0.5 4.74 0.948
0 1 1.414 0
10 1 1.14 0.316
20 1 2.08 0.632
30 1 5.05 0.948

The plot between kX1F and ωkm1 is shown below.

System Dynamics, Chapter 13, Problem 13.22P , additional homework tip  1

Figure-(1)

Conclusion:

The plot between kX1F and ωkm1 is shown below.

System Dynamics, Chapter 13, Problem 13.22P , additional homework tip  2

Want to see more full solutions like this?

Subscribe now to access step-by-step solutions to millions of textbook problems written by subject matter experts!
Knowledge Booster
Background pattern image
Recommended textbooks for you
Text book image
Elements Of Electromagnetics
Mechanical Engineering
ISBN:9780190698614
Author:Sadiku, Matthew N. O.
Publisher:Oxford University Press
Text book image
Mechanics of Materials (10th Edition)
Mechanical Engineering
ISBN:9780134319650
Author:Russell C. Hibbeler
Publisher:PEARSON
Text book image
Thermodynamics: An Engineering Approach
Mechanical Engineering
ISBN:9781259822674
Author:Yunus A. Cengel Dr., Michael A. Boles
Publisher:McGraw-Hill Education
Text book image
Control Systems Engineering
Mechanical Engineering
ISBN:9781118170519
Author:Norman S. Nise
Publisher:WILEY
Text book image
Mechanics of Materials (MindTap Course List)
Mechanical Engineering
ISBN:9781337093347
Author:Barry J. Goodno, James M. Gere
Publisher:Cengage Learning
Text book image
Engineering Mechanics: Statics
Mechanical Engineering
ISBN:9781118807330
Author:James L. Meriam, L. G. Kraige, J. N. Bolton
Publisher:WILEY