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Calculus: Early Transcendentals

8th Edition
James Stewart
ISBN: 9781285741550

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BuyFindarrow_forward

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)=eαt(c1cosβt+c2sinβt) (3)

Write the expression for r .

r=α+βi (4)

Write the expression for general solution with real roots.

x(t)=c1er1t+c2er2t (5)

Write the expression for condition of under damping.

c24mk<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=(10)±(10)24(1)(100)2(1)=10±1004002=10±3002=10±1032

Simplify equation as follows.

r=5±53i (7)

Compare equation (4) and (7).

α=5β=53

Substitute 5 for α and 53 for β in equation (3),

x(t)=e5t(c1cos53t+c2sin53t) (8)

Substitute 0 for t ,

x(0)=e5(0)(c1cos53(0)+c2sin53(0))=c1cos(0)+c2sin(0)=c1

Substitute 0.1 for x(0) ,

c1=0.1

Find the value of x(0) .

Differentiate x(0) with respect to t .

x(0)=0

Differentiate equation (8) with respect to t .

x(t)=e5t(53c1sin53t+53c2cos53t)5e5t(c1cos53t+c2sin53t)

Substitute 0 for t ,

x(0)={e5(0)(53c1sin53(0)+53c2cos53(0))5e5(0)(c1cos53(0)+c2sin53(0))}=(53c1sin(0)+53c2cos(0))5(c1cos(0)+c2sin(0))=53c25c1

Substitute 0 for x(0) ,

0=53c25c1

Substitute 0.1 for c1 ,

0=53c25(0.1)0=53c2+0.553c2=0.5c2=0.553

Simplify equation as follows.

c2=1103

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

x(t)=e5t(0.1cos53t1103sin53t) (9)

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

Table 1

t e5t(0.1cos53t1103sin53t)
0 –0.1
0.2 –0.01506
0.4 0.015312
0.6 0.000229
0.8 –0.0021
1 0.000217
1.2 0.000258
1.4 –6E–05
1.5 –6.4E–05

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

Find the value of c24mk .

c24mk=1024(1)(100)=100400=300

The value of c24mk<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=(15)±(15)24(1)(100)2(1)=15±2254002=15±1752=15±572

Simplify equation as follows.

r=152±572i (10)

Compare equation (4) and (10).

α=152β=572

Substitute 152 for α and 572 for β in equation (3),

x(t)=e152t(c1cos572t+c2sin572t) (11)

Substitute 0 for t ,

x(0)=e152(0)(c1cos572(0)+c2sin572(0))=c1cos(0)+c2sin(0)=c1

Substitute 0.1 for x(0) ,

c1=0.1

Differentiate x(0) with respect to t .

x(0)=0

Differentiate equation (11) with respect to t .

x(t)=e152t(572c1sin572t+c2572cos572t)152e152t(c1cos572t+c2sin572t)

Substitute 0 for t ,

x(0)={e152(0)(572c1sin572(0)+c2572cos572(0))152e152(0)(c1cos572(0)+c2sin572(0))}=(572c1sin(0)+c2572cos(0))152(c1cos(0)+c2sin(0))=572c2152c1

Substitute 0 for x(0) ,

0=572c2152c1

Substitute 0.1 for c1 ,

0=572c2152(0.1)0=572c2+1.52572c2=1.52c2=1.52×257

Simplify equation as follows.

c2=3107

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

x(t)=e152t(0.1cos572t3107sin572t) (12)

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

Table 2

t e152t(0.1cos572t3107sin572t)
0 0
0.2 –0.01626
0.4 0.002842
0.6 –0.00189
0.8 7E–05
1 –9.1E–05
1.2 –1.2E–05
1.4 –2.6E–06
1.5 –3.1E–06

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

Find the value of c24mk .

c24mk=1524(1)(100)=225400=175

The value of c24mk<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.

c24mk=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=(20)±(20)24(1)(100)2(1)=20±4004002=20±02=10

Substitute 10 for r in equation (13),

x(t)=c1e10t+c2te10t (15)

Substitute 0 for t ,

x(0)=c1e10(0)+c2(0)e10(0)=c1

Substitute 0

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Sect-17.1 P-11ESect-17.1 P-12ESect-17.1 P-13ESect-17.1 P-14ESect-17.1 P-15ESect-17.1 P-16ESect-17.1 P-17ESect-17.1 P-18ESect-17.1 P-19ESect-17.1 P-20ESect-17.1 P-21ESect-17.1 P-22ESect-17.1 P-23ESect-17.1 P-24ESect-17.1 P-25ESect-17.1 P-26ESect-17.1 P-27ESect-17.1 P-28ESect-17.1 P-29ESect-17.1 P-30ESect-17.1 P-31ESect-17.1 P-32ESect-17.1 P-33ESect-17.1 P-34ESect-17.2 P-1ESect-17.2 P-2ESect-17.2 P-3ESect-17.2 P-4ESect-17.2 P-5ESect-17.2 P-6ESect-17.2 P-7ESect-17.2 P-8ESect-17.2 P-9ESect-17.2 P-10ESect-17.2 P-11ESect-17.2 P-12ESect-17.2 P-13ESect-17.2 P-14ESect-17.2 P-15ESect-17.2 P-16ESect-17.2 P-17ESect-17.2 P-18ESect-17.2 P-19ESect-17.2 P-20ESect-17.2 P-21ESect-17.2 P-22ESect-17.2 P-23ESect-17.2 P-24ESect-17.2 P-25ESect-17.2 P-26ESect-17.2 P-27ESect-17.2 P-28ESect-17.3 P-1ESect-17.3 P-2ESect-17.3 P-3ESect-17.3 P-4ESect-17.3 P-5ESect-17.3 P-6ESect-17.3 P-7ESect-17.3 P-8ESect-17.3 P-9ESect-17.3 P-10ESect-17.3 P-11ESect-17.3 P-12ESect-17.3 P-13ESect-17.3 P-14ESect-17.3 P-15ESect-17.3 P-16ESect-17.3 P-17ESect-17.3 P-18ESect-17.4 P-1ESect-17.4 P-2ESect-17.4 P-3ESect-17.4 P-4ESect-17.4 P-5ESect-17.4 P-6ESect-17.4 P-7ESect-17.4 P-8ESect-17.4 P-9ESect-17.4 P-10ESect-17.4 P-11ESect-17.4 P-12ECh-17 P-1RCCCh-17 P-2RCCCh-17 P-3RCCCh-17 P-4RCCCh-17 P-5RCCCh-17 P-1RQCh-17 P-2RQCh-17 P-3RQCh-17 P-4RQCh-17 P-1RECh-17 P-2RECh-17 P-3RECh-17 P-4RECh-17 P-5RECh-17 P-6RECh-17 P-7RECh-17 P-8RECh-17 P-9RECh-17 P-10RECh-17 P-11RECh-17 P-12RECh-17 P-13RECh-17 P-14RECh-17 P-15RECh-17 P-16RECh-17 P-17RECh-17 P-18RECh-17 P-19RECh-17 P-20RECh-17 P-21RE

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