Explanation For spring constant k = 10 : Write the expression for differential equation of motion of a spring with damping constant. m d 2 x d t 2 + c d x d t + k x = 0 (1) Write the expression for auxiliary equation. m r 2 + c r + k = 0 (2) Write the expression for general solution with real roots. x ( t ) = c 1 e r 1 t + c 2 e r 2 t (3) Write the expression for condition of over damping. c 2 − 4 m k > 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 ) d 2 x d t 2 + ( 10 ) d x d t + 10 x = 0 d 2 x d t 2 + 10 d x d t + 10 x = 0 Find the auxiliary equation using equation (2). Substitute 1 for m , 10 for c and 10 for k in equation (2), ( 1 ) r 2 + 10 r + 10 = 0 r 2 + 10 r + 10 = 0 Find the value of r . r = − ( 10 ) ± ( 10 ) 2 − 4 ( 1 ) ( 10 ) 2 ( 1 ) = − 10 ± 100 − 40 2 = − 10 ± 60 2 Consider r 1 and r 2 are two roots of r . r 1 = − 10 + 60 2 = − 5 + 15 r 2 = − 10 − 60 2 = − 5 − 15 Substitute − 5 + 15 for r 1 and − 5 − 15 for r 2 in equation (3), x ( t ) = c 1 e ( − 5 + 15 ) t + c 2 e ( − 5 − 15 ) t (5) Substitute 0 for t , x ( 0 ) = c 1 e ( − 5 + 15 ) ( 0 ) + c 2 e ( − 5 − 15 ) ( 0 ) = c 1 + c 2 Substitute 0 for x ( 0 ) , c 1 + c 2 = 0 (6) Differentiate equation (5) with respect to t . x ′ ( t ) = ( − 5 + 15 ) c 1 e ( − 5 + 15 ) t + ( − 5 − 15 ) c 2 e ( − 5 − 15 ) t Substitute 0 for t , x ′ ( 0 ) = ( − 5 + 15 ) c 1 e ( − 5 + 15 ) ( 0 ) + ( − 5 − 15 ) c 2 e ( − 5 − 15 ) ( 0 ) = ( − 5 + 15 ) c 1 + ( − 5 − 15 ) c 2 Substitute 1 for x ′ ( 0 ) , ( − 5 + 15 ) c 1 + ( − 5 − 15 ) c 2 = 1 (7) Solve for c 1 and c 2 using equations (6) and (7). − ( − 5 + 15 ) c 1 − ( − 5 + 15 ) c 2 + ( − 5 + 15 ) c 1 + ( − 5 − 15 ) c 2 = 0 + 1 ( − 5 − 15 ) c 2 − ( − 5 + 15 ) c 2 = 1 c 2 [ − 5 − 15 + 5 − 15 ] = 1 c 2 [ − 2 15 ] = 1 Simplify equation as follows. c 2 = 1 [ − 2 15 ] = − 1 2 15 Substitute − 1 2 15 for c 2 in equation (6), c 1 − 1 2 15 = 0 c 1 = 1 2 15 Substitute 1 2 15 for c 1 and − 1 2 15 for c 2 in equation (5), x ( t ) = 1 2 15 e ( − 5 + 15 ) t − 1 2 15 e ( − 5 − 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 1 2 15 e ( − 5 + 15 ) t − 1 2 15 e ( − 5 − 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 c 2 − 4 m k . c 2 − 4 m k = 10 2 − 4 ( 1 ) ( 10 ) = 100 − 40 = 60 The value of c 2 − 4 m k > 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 ) d 2 x d t 2 + ( 10 ) d x d t + 20 x = 0 d 2 x d t 2 + 10 d x d t + 20 x = 0 Find the auxiliary equation using equation (2). Substitute 1 for m , 10 for c and 20 for k in equation (2), ( 1 ) r 2 + 10 r + 20 = 0 r 2 + 10 r + 20 = 0 Find the value of r . r = − ( 10 ) ± ( 10 ) 2 − 4 ( 1 ) ( 20 ) 2 ( 1 ) = − 10 ± 100 − 80 2 = − 10 ± 20 2 Consider r 1 and r 2 are two roots of r . r 1 = − 10 + 20 2 = − 5 + 5 r 2 = − 10 − 20 2 = − 5 − 5 Substitute − 5 + 5 for r 1 and − 5 − 5 for r 2 in equation (3), x ( t ) = c 1 e ( − 5 + 5 ) t + c 2 e ( − 5 − 5 ) t (9) Substitute 0 for t , x ( 0 ) = c 1 e ( − 5 + 5 ) ( 0 ) + c 2 e ( − 5 − 5 ) ( 0 ) = c 1 + c 2 Substitute 0 for x ( 0 ) , c 1 + c 2 = 0 (10) Differentiate equation (9) with respect to t . x ′ ( t ) = ( − 5 + 5 ) c 1 e ( − 5 + 5 ) t + ( − 5 − 5 ) c 2 e ( − 5 − 5 ) t Substitute 0 for t , x ′ ( 0 ) = ( − 5 + 5 ) c 1 e ( − 5 + 5 ) ( 0 ) + ( − 5 − 5 ) c 2 e ( − 5 − 5 ) ( 0 ) = ( − 5 + 5 ) c 1 + ( − 5 − 5 ) c 2 Substitute 1 for x ′ ( 0 ) , ( − 5 + 5 ) c 1 + ( − 5 − 5 ) c 2 = 1 (11) Solve for c 1 and c 2 using equations (10) and (11). − ( − 5 + 5 ) c 1 − ( − 5 + 5 ) c 2 + ( − 5 + 5 ) c 1 + ( − 5 − 5 ) c 2 = 0 + 1 ( − 5 − 5 ) c 2 − ( − 5 + 5 ) c 2 = 1 c 2 [ − 5 − 5 + 5 − 5 ] = 1 c 2 [ − 2 5 ] = 1 Simplify equation as follows. c 2 = 1 [ − 2 5 ] = − 1 2 5 Substitute − 1 2 5 for c 2 in equation (10), c 1 − 1 2 5 = 0 c 1 = 1 2 5 Substitute 1 2 5 for c 1 and − 1 2 5 for c 2 in equation (9), x ( t ) = 1 2 5 e ( − 5 + 5 ) t − 1 2 5 e ( − 5 − 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 1 2 5 e ( − 5 + 5 ) t − 1 2 5 e ( − 5 − 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 5.65 × 10 − 5 Draw the graph using values tabulated in Table 2 as shown in Figure 2. Find the value of c 2 − 4 m k . c 2 − 4 m k = 10 2 − 4 ( 1 ) ( 20 ) = 100 − 80 = 20 The value of c 2 − 4 m k > 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 ) = c 1 e r t + c 2 t e r t (13) Write the expression for condition of critical damping. c 2 − 4 m k = 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 ) d 2 x d t 2 + ( 10 ) d x d t + 25 x = 0 d 2 x d t 2 + 10 d x d t + 25 x = 0 Find the auxiliary equation using equation (2). Substitute 1 for m , 10 for c and 25 for k in equation (2), ( 1 ) r 2 + 10 r + 25 = 0 r 2 + 10 r + 25 = 0 Find the value of r

8th Edition

ISBN: 9781305266643

Author: James Stewart

Publisher: Cengage Learning

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Chapter 17.3, Problem 8E

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

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Chapter 17.3, Problem 7E

Chapter 17.3, Problem 9E

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Chapter 17 Solutions

Multivariable Calculus

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Ch. 17.1 - Prob. 11E Ch. 17.1 - Prob. 12E Ch. 17.1 - Prob. 13E Ch. 17.1 - Prob. 14E Ch. 17.1 - Prob. 15E Ch. 17.1 - Prob. 16E Ch. 17.1 - Prob. 17E Ch. 17.1 - Prob. 18E Ch. 17.1 - Solve the initial-value problem. 19. 9y" + 12y' +...Ch. 17.1 - Prob. 20E Ch. 17.1 - Prob. 21E Ch. 17.1 - Prob. 22E Ch. 17.1 - Solve the initial-value problem. 23. y" y' 12y =...Ch. 17.1 - Solve the initial-value problem. 24. 4y" + 4y' +...Ch. 17.1 - Prob. 25E Ch. 17.1 - Prob. 26E Ch. 17.1 - Prob. 27E Ch. 17.1 - Prob. 28E Ch. 17.1 - Prob. 29E Ch. 17.1 - Prob. 30E Ch. 17.1 - Prob. 31E Ch. 17.1 - Solve the boundary-value problem, if possible. 32....Ch. 17.1 - Prob. 33E Ch. 17.1 - If a, b, and c are all positive constants and y(x)...Ch. 17.2 - Prob. 1E Ch. 17.2 - Prob. 2E Ch. 17.2 - Prob. 3E Ch. 17.2 - Prob. 4E Ch. 17.2 - Prob. 5E Ch. 17.2 - Prob. 6E Ch. 17.2 - Prob. 7E Ch. 17.2 - Prob. 8E Ch. 17.2 - Prob. 9E Ch. 17.2 - Prob. 10E Ch. 17.2 - Prob. 11E Ch. 17.2 - Prob. 12E Ch. 17.2 - Prob. 13E Ch. 17.2 - Prob. 14E Ch. 17.2 - Prob. 15E Ch. 17.2 - Prob. 16E Ch. 17.2 - Prob. 17E Ch. 17.2 - Prob. 18E Ch. 17.2 - Prob. 19E Ch. 17.2 - Prob. 20E Ch. 17.2 - Solve the differential equation using (a)...Ch. 17.2 - Prob. 22E Ch. 17.2 - Prob. 23E Ch. 17.2 - Solve the differential equation using the method...Ch. 17.2 - Solve the differential equation using the method...Ch. 17.2 - Solve the differential equation using the method...Ch. 17.2 - Prob. 27E Ch. 17.2 - Prob. 28E Ch. 17.3 - A spring has natural length 0.75 m and a 5-kg...Ch. 17.3 - Prob. 2E Ch. 17.3 - A spring with a mass of 2 kg has damping constant...Ch. 17.3 - Prob. 4E Ch. 17.3 - Prob. 5E Ch. 17.3 - Prob. 6E Ch. 17.3 - Prob. 7E Ch. 17.3 - Prob. 8E Ch. 17.3 - Suppose a spring has mass m and spring constant k...Ch. 17.3 - Prob. 10E Ch. 17.3 - Prob. 11E Ch. 17.3 - Prob. 12E Ch. 17.3 - Prob. 13E Ch. 17.3 - Prob. 14E Ch. 17.3 - Prob. 15E Ch. 17.3 - The battery in Exercise 14 is replaced by a...Ch. 17.3 - Prob. 17E Ch. 17.3 - Prob. 18E Ch. 17.4 - Prob. 1E Ch. 17.4 - Prob. 2E Ch. 17.4 - Prob. 3E Ch. 17.4 - Prob. 4E Ch. 17.4 - Prob. 5E Ch. 17.4 - Prob. 6E Ch. 17.4 - Prob. 7E Ch. 17.4 - Prob. 8E Ch. 17.4 - Prob. 9E Ch. 17.4 - Prob. 10E Ch. 17.4 - Prob. 11E Ch. 17.4 - Prob. 12E Ch. 17 - (a) Write the general form of a second-order...Ch. 17 - Prob. 2RCC Ch. 17 - (a) Write the general form of a second-order...Ch. 17 - Prob. 4RCC Ch. 17 - Prob. 5RCC Ch. 17 - Prob. 1RQ Ch. 17 - Prob. 2RQ Ch. 17 - Prob. 3RQ Ch. 17 - Prob. 4RQ Ch. 17 - Solve the differential equation. 1. 4y" y =0 Ch. 17 - Prob. 2RE Ch. 17 - Prob. 3RE Ch. 17 - Solve the differential equation. 4. y" + 8y' + 16y...Ch. 17 - Prob. 5RE Ch. 17 - Prob. 6RE Ch. 17 - Prob. 7RE Ch. 17 - Prob. 8RE Ch. 17 - Prob. 9RE Ch. 17 - Solve the differential equation. 10....Ch. 17 - Prob. 11RE Ch. 17 - Solve the initial-value problem. 12. y" 6y' + 25y...Ch. 17 - Solve the initial-value problem. 13. y" 5y' + 4y...Ch. 17 - Prob. 14RE Ch. 17 - Solve the boundary-value problem, if possible. 15....Ch. 17 - Prob. 16RE Ch. 17 - Prob. 17RE Ch. 17 - Prob. 18RE Ch. 17 - A series circuit contains a resistor with R = 40 ,...Ch. 17 - Prob. 20RE Ch. 17 - Assume that the earth is a solid sphere of uniform...

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

Solution Summary: The author explains the position function for spring constants k=10,20,25,30,40.

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

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Chapter 17 Solutions