Chapter 17.3, Problem 7E

### 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 spring constant is k = 100. The spring is released at a point 0.1 m above its equilibrium position. Graph the position function for the following values of the damping constant c: 10, 15, 20, 25, 30. What type of damping occurs in each case?

To determine

To graph: The position function for spring constants c=10,15,20,25,and30 .

Explanation

For spring constant c=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 complex roots.

x(t)=eÎ±t(c1cosÎ²t+c2sinÎ²t) (3)

Write the expression for r .

r=Î±+Î²i (4)

Write the expression for general solution with real roots.

x(t)=c1er1t+c2er2t (5)

Write the expression for condition of under damping.

c2âˆ’4mk<0 (6)

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

Substitute 1 for m, 10 for c and 100 for k in equation (1),

(1)d2xdt2+(10)dxdt+100x=0d2xdt2+10dxdt+100x=0

Find the auxiliary equation using equation (2).

Substitute 1 for m, 10 for c and 100 for k in equation (2),

(1)r2+10r+100=0r2+10r+100=0

Find the value of r .

r=âˆ’(10)Â±(10)2âˆ’4(1)(100)2(1)=âˆ’10Â±100âˆ’4002=âˆ’10Â±âˆ’3002=âˆ’10Â±10âˆ’32

Simplify equation as follows.

r=âˆ’5Â±53i (7)

Compare equation (4) and (7).

Î±=âˆ’5Î²=53

Substitute âˆ’5 for Î± and 53 for Î² in equation (3),

x(t)=eâˆ’5t(c1cos53t+c2sin53t) (8)

Substitute 0 for t ,

x(0)=eâˆ’5(0)(c1cos53(0)+c2sin53(0))=c1cos(0)+c2sin(0)=c1

Substitute âˆ’0.1 for x(0) ,

c1=âˆ’0.1

Find the value of xâ€²(0) .

Differentiate x(0) with respect to t .

xâ€²(0)=0

Differentiate equation (8) with respect to t .

xâ€²(t)=eâˆ’5t(âˆ’53c1sin53t+53c2cos53t)âˆ’5eâˆ’5t(c1cos53t+c2sin53t)

Substitute 0 for t ,

xâ€²(0)={eâˆ’5(0)(âˆ’53c1sin53(0)+53c2cos53(0))âˆ’5eâˆ’5(0)(c1cos53(0)+c2sin53(0))}=(âˆ’53c1sin(0)+53c2cos(0))âˆ’5(c1cos(0)+c2sin(0))=53c2âˆ’5c1

Substitute 0 for xâ€²(0) ,

0=53c2âˆ’5c1

Substitute âˆ’0.1 for c1 ,

0=53c2âˆ’5(âˆ’0.1)0=53c2+0.553c2=âˆ’0.5c2=âˆ’0.553

Simplify equation as follows.

c2=âˆ’1103

Substitute âˆ’0.1 for c1 and âˆ’1103 for c2 in equation (8),

x(t)=eâˆ’5t(âˆ’0.1cos53tâˆ’1103sin53t) (9)

Substitute different values for t in equation (9) and calculated values of x(t) are tabulated in Table 1.

Table 1

 t eâˆ’5t(âˆ’0.1cos53tâˆ’1103sin53t) 0 â€“0.1 0.2 â€“0.01506 0.4 0.015312 0.6 0.000229 0.8 â€“0.0021 1 0.000217 1.2 0.000258 1.4 â€“6Eâ€“05 1.5 â€“6.4Eâ€“05

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)(100)=100âˆ’400=âˆ’300

The value of c2âˆ’4mk<0 , which satisfies the condition given in equation (6). So with c=10 , the motion is under damping condition.

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

For spring constant c=15 :

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

Substitute 1 for m, 15 for c and 100 for k in equation (1),

(1)d2xdt2+(15)dxdt+100x=0d2xdt2+15dxdt+100x=0

Find the auxiliary equation using equation (2).

Substitute 1 for m, 15 for c and 100 for k in equation (2),

(1)r2+15r+100=0r2+15r+100=0

Find the value of r .

r=âˆ’(15)Â±(15)2âˆ’4(1)(100)2(1)=âˆ’15Â±225âˆ’4002=âˆ’15Â±âˆ’1752=âˆ’15Â±5âˆ’72

Simplify equation as follows.

r=âˆ’152Â±572i (10)

Compare equation (4) and (10).

Î±=âˆ’152Î²=572

Substitute âˆ’152 for Î± and 572 for Î² in equation (3),

x(t)=eâˆ’152t(c1cos572t+c2sin572t) (11)

Substitute 0 for t ,

x(0)=eâˆ’152(0)(c1cos572(0)+c2sin572(0))=c1cos(0)+c2sin(0)=c1

Substitute âˆ’0.1 for x(0) ,

c1=âˆ’0.1

Differentiate x(0) with respect to t .

xâ€²(0)=0

Differentiate equation (11) with respect to t .

xâ€²(t)=eâˆ’152t(âˆ’572c1sin572t+c2572cos572t)âˆ’152eâˆ’152t(c1cos572t+c2sin572t)

Substitute 0 for t ,

xâ€²(0)={eâˆ’152(0)(âˆ’572c1sin572(0)+c2572cos572(0))âˆ’152eâˆ’152(0)(c1cos572(0)+c2sin572(0))}=(âˆ’572c1sin(0)+c2572cos(0))âˆ’152(c1cos(0)+c2sin(0))=572c2âˆ’152c1

Substitute 0 for xâ€²(0) ,

0=572c2âˆ’152c1

Substitute âˆ’0.1 for c1 ,

0=572c2âˆ’152(âˆ’0.1)0=572c2+1.52572c2=âˆ’1.52c2=âˆ’1.52Ã—257

Simplify equation as follows.

c2=âˆ’3107

Substitute âˆ’0.1 for c1 and âˆ’3107 for c2 in equation (11),

x(t)=eâˆ’152t(âˆ’0.1cos572tâˆ’3107sin572t) (12)

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

Table 2

 t eâˆ’152t(âˆ’0.1cos572tâˆ’3107sin572t) 0 0 0.2 â€“0.01626 0.4 0.002842 0.6 â€“0.00189 0.8 7Eâ€“05 1 â€“9.1Eâ€“05 1.2 â€“1.2Eâ€“05 1.4 â€“2.6Eâ€“06 1.5 â€“3.1Eâ€“06

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

Find the value of c2âˆ’4mk .

c2âˆ’4mk=152âˆ’4(1)(100)=225âˆ’400=âˆ’175

The value of c2âˆ’4mk<0 , which satisfies the condition given in equation (6). So with c=15 , the motion is under damping condition.

Thus, the position function for spring constant c=15 is drawn and motion is in under damping condition.

For spring constant c=20 :

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, 20 for c and 100 for k in equation (1),

(1)d2xdt2+(20)dxdt+100x=0d2xdt2+20dxdt+100x=0

Find the auxiliary equation using equation (2).

Substitute 1 for m, 20 for c and 100 for k in equation (2),

(1)r2+20r+100=0r2+20r+100=0

Find the value of r .

r=âˆ’(20)Â±(20)2âˆ’4(1)(100)2(1)=âˆ’20Â±400âˆ’4002=âˆ’20Â±02=âˆ’10

Substitute âˆ’10 for r in equation (13),

x(t)=c1eâˆ’10t+c2teâˆ’10t (15)

Substitute 0 for t ,

x(0)=c1eâˆ’10(0)+c2(0)eâˆ’10(0)=c1

Substitute âˆ’0

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