Explanation For spring constant c = 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 complex roots. x ( t ) = e α t ( c 1 cos β t + c 2 sin β t ) (3) Write the expression for r . r = α + β i (4) Write the expression for general solution with real roots. x ( t ) = c 1 e r 1 t + c 2 e r 2 t (5) Write the expression for condition of under damping. c 2 − 4 m k < 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 ) d 2 x d t 2 + ( 10 ) d x d t + 100 x = 0 d 2 x d t 2 + 10 d x d t + 100 x = 0 Find the auxiliary equation using equation (2). Substitute 1 for m , 10 for c and 100 for k in equation (2), ( 1 ) r 2 + 10 r + 100 = 0 r 2 + 10 r + 100 = 0 Find the value of r . r = − ( 10 ) ± ( 10 ) 2 − 4 ( 1 ) ( 100 ) 2 ( 1 ) = − 10 ± 100 − 400 2 = − 10 ± − 300 2 = − 10 ± 10 − 3 2 Simplify equation as follows. r = − 5 ± 5 3 i (7) Compare equation (4) and (7). α = − 5 β = 5 3 Substitute − 5 for α and 5 3 for β in equation (3), x ( t ) = e − 5 t ( c 1 cos 5 3 t + c 2 sin 5 3 t ) (8) Substitute 0 for t , x ( 0 ) = e − 5 ( 0 ) ( c 1 cos 5 3 ( 0 ) + c 2 sin 5 3 ( 0 ) ) = c 1 cos ( 0 ) + c 2 sin ( 0 ) = c 1 Substitute − 0.1 for x ( 0 ) , c 1 = − 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 ) = e − 5 t ( − 5 3 c 1 sin 5 3 t + 5 3 c 2 cos 5 3 t ) − 5 e − 5 t ( c 1 cos 5 3 t + c 2 sin 5 3 t ) Substitute 0 for t , x ′ ( 0 ) = { e − 5 ( 0 ) ( − 5 3 c 1 sin 5 3 ( 0 ) + 5 3 c 2 cos 5 3 ( 0 ) ) − 5 e − 5 ( 0 ) ( c 1 cos 5 3 ( 0 ) + c 2 sin 5 3 ( 0 ) ) } = ( − 5 3 c 1 sin ( 0 ) + 5 3 c 2 cos ( 0 ) ) − 5 ( c 1 cos ( 0 ) + c 2 sin ( 0 ) ) = 5 3 c 2 − 5 c 1 Substitute 0 for x ′ ( 0 ) , 0 = 5 3 c 2 − 5 c 1 Substitute − 0.1 for c 1 , 0 = 5 3 c 2 − 5 ( − 0.1 ) 0 = 5 3 c 2 + 0.5 5 3 c 2 = − 0.5 c 2 = − 0.5 5 3 Simplify equation as follows. c 2 = − 1 10 3 Substitute − 0.1 for c 1 and − 1 10 3 for c 2 in equation (8), x ( t ) = e − 5 t ( − 0.1 cos 5 3 t − 1 10 3 sin 5 3 t ) (9) Substitute different values for t in equation (9) and calculated values of x ( t ) are tabulated in Table 1. Table 1 t e − 5 t ( − 0.1 cos 5 3 t − 1 10 3 sin 5 3 t ) 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 c 2 − 4 m k . c 2 − 4 m k = 10 2 − 4 ( 1 ) ( 100 ) = 100 − 400 = − 300 The value of c 2 − 4 m k < 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 ) d 2 x d t 2 + ( 15 ) d x d t + 100 x = 0 d 2 x d t 2 + 15 d x d t + 100 x = 0 Find the auxiliary equation using equation (2). Substitute 1 for m , 15 for c and 100 for k in equation (2), ( 1 ) r 2 + 15 r + 100 = 0 r 2 + 15 r + 100 = 0 Find the value of r . r = − ( 15 ) ± ( 15 ) 2 − 4 ( 1 ) ( 100 ) 2 ( 1 ) = − 15 ± 225 − 400 2 = − 15 ± − 175 2 = − 15 ± 5 − 7 2 Simplify equation as follows. r = − 15 2 ± 5 7 2 i (10) Compare equation (4) and (10). α = − 15 2 β = 5 7 2 Substitute − 15 2 for α and 5 7 2 for β in equation (3), x ( t ) = e − 15 2 t ( c 1 cos 5 7 2 t + c 2 sin 5 7 2 t ) (11) Substitute 0 for t , x ( 0 ) = e − 15 2 ( 0 ) ( c 1 cos 5 7 2 ( 0 ) + c 2 sin 5 7 2 ( 0 ) ) = c 1 cos ( 0 ) + c 2 sin ( 0 ) = c 1 Substitute − 0.1 for x ( 0 ) , c 1 = − 0.1 Differentiate x ( 0 ) with respect to t . x ′ ( 0 ) = 0 Differentiate equation (11) with respect to t . x ′ ( t ) = e − 15 2 t ( − 5 7 2 c 1 sin 5 7 2 t + c 2 5 7 2 cos 5 7 2 t ) − 15 2 e − 15 2 t ( c 1 cos 5 7 2 t + c 2 sin 5 7 2 t ) Substitute 0 for t , x ′ ( 0 ) = { e − 15 2 ( 0 ) ( − 5 7 2 c 1 sin 5 7 2 ( 0 ) + c 2 5 7 2 cos 5 7 2 ( 0 ) ) − 15 2 e − 15 2 ( 0 ) ( c 1 cos 5 7 2 ( 0 ) + c 2 sin 5 7 2 ( 0 ) ) } = ( − 5 7 2 c 1 sin ( 0 ) + c 2 5 7 2 cos ( 0 ) ) − 15 2 ( c 1 cos ( 0 ) + c 2 sin ( 0 ) ) = 5 7 2 c 2 − 15 2 c 1 Substitute 0 for x ′ ( 0 ) , 0 = 5 7 2 c 2 − 15 2 c 1 Substitute − 0.1 for c 1 , 0 = 5 7 2 c 2 − 15 2 ( − 0.1 ) 0 = 5 7 2 c 2 + 1.5 2 5 7 2 c 2 = − 1.5 2 c 2 = − 1.5 2 × 2 5 7 Simplify equation as follows. c 2 = − 3 10 7 Substitute − 0.1 for c 1 and − 3 10 7 for c 2 in equation (11), x ( t ) = e − 15 2 t ( − 0.1 cos 5 7 2 t − 3 10 7 sin 5 7 2 t ) (12) Substitute different values for t in equation (12) and calculated values of x ( t ) are tabulated in Table 2. Table 2 t e − 15 2 t ( − 0.1 cos 5 7 2 t − 3 10 7 sin 5 7 2 t ) 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 c 2 − 4 m k . c 2 − 4 m k = 15 2 − 4 ( 1 ) ( 100 ) = 225 − 400 = − 175 The value of c 2 − 4 m k < 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 ) = 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 , 20 for c and 100 for k in equation (1), ( 1 ) d 2 x d t 2 + ( 20 ) d x d t + 100 x = 0 d 2 x d t 2 + 20 d x d t + 100 x = 0 Find the auxiliary equation using equation (2). Substitute 1 for m , 20 for c and 100 for k in equation (2), ( 1 ) r 2 + 20 r + 100 = 0 r 2 + 20 r + 100 = 0 Find the value of r . r = − ( 20 ) ± ( 20 ) 2 − 4 ( 1 ) ( 100 ) 2 ( 1 ) = − 20 ± 400 − 400 2 = − 20 ± 0 2 = − 10 Substitute − 10 for r in equation (13), x ( t ) = c 1 e − 10 t + c 2 t e − 10 t (15) Substitute 0 for t , x ( 0 ) = c 1 e − 10 ( 0 ) + c 2 ( 0 ) e − 10 ( 0 ) = c 1 Substitute − 0

8th Edition

ISBN: 9781337771474

Author: Stewart

Publisher: Cengage

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Chapter 17.3, Problem 7E

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To graph: The position function for spring constants c=10,15,20,25, and 30 .

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Chapter 17.3, Problem 6E

Chapter 17.3, Problem 8E

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Chapter 17 Solutions

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To graph: The position function for spring constants c = 10 , 15 , 20 , 25 , and 30 .

Solution Summary: The author explains the position function for spring constants c=10,15,20,25, and the expressions for auxiliary equations and general solutions.

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To graph: The position function for spring constants c=10,15,20,25, and 30 .

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Chapter 17 Solutions