Chapter 17.3, Problem 8E

### Calculus: Early Transcendentals

8th Edition
James Stewart
ISBN: 9781285741550

Chapter
Section

### Calculus: Early Transcendentals

8th Edition
James Stewart
ISBN: 9781285741550
Textbook Problem

# A spring has a mass of 1 kg and its damping constant is c = 10. The spring starts from its equilibrium position with a velocity of 1 m/s. Graph the position function for the following values of the spring constant k: 10, 20, 25, 30, 40. What type of damping occurs in each case?

To determine

To graph: The position function for spring constants k=10,20,25,30,40 .

Explanation

For spring constant k=10 :

Write the expression for differential equation of motion of a spring with damping constant.

md2xdt2+cdxdt+kx=0 (1)

Write the expression for auxiliary equation.

mr2+cr+k=0 (2)

Write the expression for general solution with real roots.

x(t)=c1er1t+c2er2t (3)

Write the expression for condition of over damping.

c2âˆ’4mk>0 (4)

Find the expression for differential equation using equation (1).

Substitute 1 for m, 10 for c and 10 for k in equation (1),

(1)d2xdt2+(10)dxdt+10x=0d2xdt2+10dxdt+10x=0

Find the auxiliary equation using equation (2).

Substitute 1 for m, 10 for c and 10 for k in equation (2),

(1)r2+10r+10=0r2+10r+10=0

Find the value of r .

r=âˆ’(10)Â±(10)2âˆ’4(1)(10)2(1)=âˆ’10Â±100âˆ’402=âˆ’10Â±602

Consider r1 and r2 are two roots of r .

r1=âˆ’10+602=âˆ’5+15r2=âˆ’10âˆ’602=âˆ’5âˆ’15

Substitute âˆ’5+15 for r1 and âˆ’5âˆ’15 for r2 in equation (3),

x(t)=c1e(âˆ’5+15)t+c2e(âˆ’5âˆ’15)t (5)

Substitute 0 for t ,

x(0)=c1e(âˆ’5+15)(0)+c2e(âˆ’5âˆ’15)(0)=c1+c2

Substitute 0 for x(0) ,

c1+c2=0 (6)

Differentiate equation (5) with respect to t .

xâ€²(t)=(âˆ’5+15)c1e(âˆ’5+15)t+(âˆ’5âˆ’15)c2e(âˆ’5âˆ’15)t

Substitute 0 for t ,

xâ€²(0)=(âˆ’5+15)c1e(âˆ’5+15)(0)+(âˆ’5âˆ’15)c2e(âˆ’5âˆ’15)(0)=(âˆ’5+15)c1+(âˆ’5âˆ’15)c2

Substitute 1 for xâ€²(0) ,

(âˆ’5+15)c1+(âˆ’5âˆ’15)c2=1 (7)

Solve for c1 and c2 using equations (6) and (7).

âˆ’(âˆ’5+15)c1âˆ’(âˆ’5+15)c2+(âˆ’5+15)c1+(âˆ’5âˆ’15)c2=0+1(âˆ’5âˆ’15)c2âˆ’(âˆ’5+15)c2=1c2[âˆ’5âˆ’15+5âˆ’15]=1c2[âˆ’215]=1

Simplify equation as follows.

c2=1[âˆ’215]=âˆ’1215

Substitute âˆ’1215 for c2 in equation (6),

c1âˆ’1215=0c1=1215

Substitute 1215 for c1 and âˆ’1215 for c2 in equation (5),

x(t)=1215e(âˆ’5+15)tâˆ’1215e(âˆ’5âˆ’15)t (8)

Substitute different values for t in equation (8) and calculated values of x(t) are tabulated in Table 1.

Table 1

 t 1215e(âˆ’5+15)tâˆ’1215e(âˆ’5âˆ’15)t 0 0 0.2 0.078179 0.4 0.073906 0.6 0.059579 0.8 0.046596 1 0.036209 1.2 0.028098 1.4 0.021797 1.5 0.019198 2 0.010174 2.5 0.005391 3 0.002857

Draw the graph using values tabulated in Table 1 as shown in Figure 1.

Find the value of c2âˆ’4mk .

c2âˆ’4mk=102âˆ’4(1)(10)=100âˆ’40=60

The value of c2âˆ’4mk>0 , which satisfies the condition given in equation (4). So with k=10 , the motion is in over damping condition.

Thus, the position function for spring constant k=10 is drawn and motion is in over damping condition.

For spring constant k=20 :

Find the expression for differential equation using equation (1).

Substitute 1 for m, 10 for c and 20 for k in equation (1),

(1)d2xdt2+(10)dxdt+20x=0d2xdt2+10dxdt+20x=0

Find the auxiliary equation using equation (2).

Substitute 1 for m, 10 for c and 20 for k in equation (2),

(1)r2+10r+20=0r2+10r+20=0

Find the value of r .

r=âˆ’(10)Â±(10)2âˆ’4(1)(20)2(1)=âˆ’10Â±100âˆ’802=âˆ’10Â±202

Consider r1 and r2 are two roots of r .

r1=âˆ’10+202=âˆ’5+5r2=âˆ’10âˆ’202=âˆ’5âˆ’5

Substitute âˆ’5+5 for r1 and âˆ’5âˆ’5 for r2 in equation (3),

x(t)=c1e(âˆ’5+5)t+c2e(âˆ’5âˆ’5)t (9)

Substitute 0 for t ,

x(0)=c1e(âˆ’5+5)(0)+c2e(âˆ’5âˆ’5)(0)=c1+c2

Substitute 0 for x(0) ,

c1+c2=0 (10)

Differentiate equation (9) with respect to t .

xâ€²(t)=(âˆ’5+5)c1e(âˆ’5+5)t+(âˆ’5âˆ’5)c2e(âˆ’5âˆ’5)t

Substitute 0 for t ,

xâ€²(0)=(âˆ’5+5)c1e(âˆ’5+5)(0)+(âˆ’5âˆ’5)c2e(âˆ’5âˆ’5)(0)=(âˆ’5+5)c1+(âˆ’5âˆ’5)c2

Substitute 1 for xâ€²(0) ,

(âˆ’5+5)c1+(âˆ’5âˆ’5)c2=1 (11)

Solve for c1 and c2 using equations (10) and (11).

âˆ’(âˆ’5+5)c1âˆ’(âˆ’5+5)c2+(âˆ’5+5)c1+(âˆ’5âˆ’5)c2=0+1(âˆ’5âˆ’5)c2âˆ’(âˆ’5+5)c2=1c2[âˆ’5âˆ’5+5âˆ’5]=1c2[âˆ’25]=1

Simplify equation as follows.

c2=1[âˆ’25]=âˆ’125

Substitute âˆ’125 for c2 in equation (10),

c1âˆ’125=0c1=125

Substitute 125 for c1 and âˆ’125 for c2 in equation (9),

x(t)=125e(âˆ’5+5)tâˆ’125e(âˆ’5âˆ’5)t (12)

Substitute different values for t in equation (12) and calculated values of x(t) are tabulated in Table 2.

Table 2

 t 125e(âˆ’5+5)tâˆ’125e(âˆ’5âˆ’5)t 0 0 0.2 0.075884 0.4 0.061565 0.6 0.039658 0.8 0.023826 1 0.013952 1.2 0.008089 1.4 0.00467 1.5 0.003546 2 0.000893 2.5 0.000225 3 5.65Ã—10âˆ’5

Draw the graph using values tabulated in Table 2 as shown in Figure 2.

Find the value of c2âˆ’4mk .

c2âˆ’4mk=102âˆ’4(1)(20)=100âˆ’80=20

The value of c2âˆ’4mk>0 , which satisfies the condition given in equation (4). So with k=20 , the motion is in over damping condition.

Thus, the position function for spring constant k=20 is drawn and motion is in over damping condition.

For spring constant k=25 :

Write the expression for general solution with same roots.

x(t)=c1ert+c2tert (13)

Write the expression for condition of critical damping.

c2âˆ’4mk=0 (14)

Find the expression for differential equation using equation (1).

Substitute 1 for m, 10 for c and 25 for k in equation (1),

(1)d2xdt2+(10)dxdt+25x=0d2xdt2+10dxdt+25x=0

Find the auxiliary equation using equation (2).

Substitute 1 for m, 10 for c and 25 for k in equation (2),

(1)r2+10r+25=0r2+10r+25=0

Find the value of r

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