System Dynamics
System Dynamics
3rd Edition
ISBN: 9780073398068
Author: III William J. Palm
Publisher: MCG
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Chapter 4, Problem 4.90P
To determine

(a)

The equations of motion of the system shown in Figure P4.90.

Expert Solution
Check Mark

Answer to Problem 4.90P

The equations of motion of the system given below,

f+9×104x16×104x2=20x1

6×104(x2x1)=60x2.

Explanation of Solution

Given:

The masses are m1=20 kg and m2=60 kg

The spring constant are k1=3×104 N/m and k2=6×104 N/m

Concept used:

Newton’s Second law of motion and Hooke’s Law are used to obtain the equation of motions.

Figure P4.90

System Dynamics, Chapter 4, Problem 4.90P , additional homework tip  1

Calculation:

Free body diagram of m1 and m2 are given below,

System Dynamics, Chapter 4, Problem 4.90P , additional homework tip  2

Applying Newton’s Second law of motion to mass m1

fk1x1k2(x1x2)=m1x1

k2(x1x2)=m2x2

Substituting values for k1,k2 and m1

f3×104x16×104(x1x2)=20x1

f9×104x1+6×104x2=20x1

Substituting values for k2 and m2

6×104(x1x2)=60x2

Therefore the equations of motion of the system given below,

f9×104x1+6×104x2=20x1 (1)

6×104(x1x2)=60x2 (2)

Where,

f = Applied force to mass m1k1,k2 = Stiffness of the spring

x1,x2 = Displacement of the spring respectively

m1,m2 = Mass of the objects.

To determine

(b)

The transfer functions for X1(s)/F(s) ,X2(s)/F(s).

Expert Solution
Check Mark

Answer to Problem 4.90P

X1(s)F(s)=s2+100020s4+11×104s2+3×107       

X2(s)F(s)=1000(20s4+11×104s2+3×107).

Explanation of Solution

Given:

From part (a),

f9×104x1+6×104x2=20x1       (1)

6×104(x1x2)=60x2         (2)

Concept used:

Laplace Transformation used to obtain the transfer functions.

Assume zero initial conditions for x1 and x2.

Calculation:

Converting the equations (1) and (2) to Laplace domain,

F(s)9×104X1(s)+6×104X2(s)=20(s2X1(s)x1(0)sx1(0))   (3)

6×104(X1(s)X2(s))=60(s2X2(s) x2(0)sx2(0))     (4)

for zero initial condition, x1(0)=0 , x2(0)=0 , x1(0)=0 , x2(0)=0

F(s)9×104X1(s)+6×104X2(s)=20s2X1(s)   (5)

6×104(X1(s)X2(s))=60s2X2(s)    (6)

from equation (6),

X2(s)=1000s2+1000X1(s)       (7)

From equation (7) and (5),

X1(s)F(s)=s2+100020s4+11×104s2+3×107       (6)

From equation (7) and (6),

X2(s)F(s)=1000(20s4+11×104s2+3×107)(7).

To determine

(c)

A plot of unit step responses of x1 for zero initial conditions.

Expert Solution
Check Mark

Answer to Problem 4.90P

System Dynamics, Chapter 4, Problem 4.90P , additional homework tip  3

Explanation of Solution

Given:

unit-step response of x1 for zero initial conditions

X1(s)F(s)=s2+100020s4+11×104s2+3×107       

Solution:

Using MATLAB,

>>sys1=tf([1,0,1000],[20,0,11*10^4,0,3*10^7]);

>> step(sys1)

System Dynamics, Chapter 4, Problem 4.90P , additional homework tip  4

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Chapter 4 Solutions

System Dynamics

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