System Dynamics
System Dynamics
3rd Edition
ISBN: 9780073398068
Author: III William J. Palm
Publisher: MCG
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Question
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Chapter 4, Problem 4.51P
To determine

(a)

The value of x(t) at c=680.

Expert Solution
Check Mark

Answer to Problem 4.51P

The value of x(t) at c=680 is x(t)=4.1674.81e2t+0.641e15t.

Explanation of Solution

Given information:

The value of step function is 5000 and the equation of motion is 40x¨+cx˙+12000x=f(t) the value of c is 680.

Write the Equation of motion.

40x¨+cx˙+12000x=f(t) ....... (I)

Here, acceleration is x¨, constant is c, velocity is x˙, the step function is f(t) and the displacement is x.

Calculation:

Substitute 680 for c and 5000 for c in Equation (I).

40x¨+680x˙+12000x=5000x¨+17x˙+30x=125 ...... (II)

Take Laplace transform in equation (II)

L(x¨)+17L(x˙)+30L(x)=L(125)s2X(s)sx(0)x˙(0)+17[sX(s)x(0)]+30X(s)=125s ........ (III)

Substitute 0 for x(0) and 0 for x˙(0) in Equation (III).

s2X(s)s×00+17[sX(s)0]+30X(s)=125ss2X(s)+17[sX(s)]+30X(s)=125sX(s)=125s(s+2)(s+15) ........ (IV)

Apply partial fraction on Equation (IV).

125s(s+2)(s+15)=As+Bs+2+Cs+15125=A(s+2)(s+15)+Bs(s+15)+Cs(s+2) ...... (V)

Equate constant term of equation (V).

125=30AA=4.167

Equate coefficient of s of Equation (V) and substitute 4.167 for A.

0=17(4.167)+15B+2C15B+2C=70.839 ....... (VI)

Equate coefficient of s2 of Equation (V) and substitute 4.167 for A.

0=4.167+B+CB+C=4.167 ...... (VII)

Solve equation (VI) and (VII).

B=4.81C=0.641

Substitute 4.81 for B, 0.641 for C and 4.167 for A in Equation (V).

125s(s+2)(s+15)=4.167s+4.81s+2+0.641s+15

Substitute 4.167s+4.81s+2+0.641s+15 for 125s(s+2)(s+15) in equation (IV).

X(s)=4.167s+4.81s+2+0.641s+15 ....... (VIII)

Take inverse Laplace of Equation (VIII).

L1X(s)=L1[4.167s+4.81s+2+0.641s+15]L1X(s)=L1(4.167s)+L1(4.81s+2)+L1(0.641s+15)x(t)=4.1674.81e2t+0.641e15t.

Conclusion:

The value of x(t) at c=680 is x(t)=4.1674.81e2t+0.641e15t.

To determine

(b)

The value of x(t) at c=400.

Expert Solution
Check Mark

Answer to Problem 4.51P

The value of x(t)

c=400 is x(t)=4.1674.167e5tcos5t18.635sin5t.

Explanation of Solution

Given information:

The value of step function is 5000 and the equation of motion is 40x¨+cx˙+12000x=f(t) the value of c is 400.

Write the Equation of motion.

40x¨+cx˙+12000x=f(t) ......... (IX)

Here, acceleration is x¨, constant is c, velocity is x˙, the step function is f(t) and the displacement is x.

Calculation:

Substitute 400 for c and 5000 for c in Equation (IX).

40x¨+400x˙+12000x=5000 ........ (X)

Take Laplace transform in equation (X)

L(40x¨)+400L(x˙)+1200L(x)=L(5000)40s2X(s)sx(0)x˙(0)+400[sX(s)x(0)]+1200X(s)=5000s ........ (XI)

Substitute 0 for x(0) and 0 for x˙(0) in Equation (XI).

40s2X(s)s×00+400[sX(s)0]+1200X(s)=5000ss2X(s)+10[sX(s)]+30X(s)=125sX(s)=125s(s2+10s+30) ........ (XII)

Apply partial fraction on Equation (XII).

125s(s2+10s+30)=As+Bs+Cs2+10s+30125=A(s2+10s+30)+s(Bs+C) ...... (XIII)

Equate constant term of equation (XIII).

125=30AA=4.167

Equate coefficient of s of Equation (XIII) and substitute 4.167 for A.

0+10A+CC=10(4.167)C=41.67 ....... (XIV)

Equate coefficient of s2 of Equation (XIII) and substitute 4.167 for A.

0=A+BB=4.167 ...... (XV)

Substitute 4.167 for B, 41.67 for C and 4.167 for A in Equation (XIII).

125s(s2+10s+30)=4.167s4.167s+41.67s2+10s+30

Substitute 4.167s4.167s+41.67s2+10s+30 for 125s(s2+10s+30) in equation (XIII).

X(s)=4.167s4.167s+41.67s2+10s+30 ....... (XVI)

Take inverse Laplace of Equation (XVI).

L1X(s)=L1[4.167s4.167s+41.67s2+10s+30]L1X(s)=L1(4.167s)L1(4.167s+41.67s2+10s+30)L1X(s)=4.167L1(4.167s+41.67s2+10s+30+2525)L1X(s)=4.167L1(4.167s+41.67(s+5)2+5)

Further simplifying

L1X(s)=4.167L1(4.167s+41.67(s+5)2+5)x(t)=4.167L1(4.167s(s+5)2+5)L1(41.67(s+5)2+5)x(t)=4.1674.167e5tcos5t41.675sin5tx(t)=4.1674.167e5tcos5t18.635sin5t.

Conclusion:

The value of x(t)c=400 is x(t)=4.1674.167e5tcos5t18.635sin5t.

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Chapter 4 Solutions

System Dynamics

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