   Chapter 17.3, Problem 7E

Chapter
Section
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 e−5t(−0.1cos53t−1103sin53t) 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 e−152t(−0.1cos572t−3107sin572t) 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

Still sussing out bartleby?

Check out a sample textbook solution.

See a sample solution

The Solution to Your Study Problems

Bartleby provides explanations to thousands of textbook problems written by our experts, many with advanced degrees!

Get Started

In Exercises 1728, use the logarithm identities to obtain the missing quantity.

Finite Mathematics and Applied Calculus (MindTap Course List)

find the real roots of each equation by factoring. 131. 4t2 + 2t 2 = 0

Applied Calculus for the Managerial, Life, and Social Sciences: A Brief Approach

2. What is the domain of ?

Mathematical Applications for the Management, Life, and Social Sciences

Evaluate 11(x2x3)dx a) 32 b) 56 c) 12 d) 23

Study Guide for Stewart's Single Variable Calculus: Early Transcendentals, 8th

In Exercises 21-26, use the laws of logic to prove the propositions. p(pq)pq

Finite Mathematics for the Managerial, Life, and Social Sciences 