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: 9781285741550

Author: James Stewart

Publisher: Cengage Learning

See similar textbooks

Videos

Question

Chapter 17.3, Problem 8E

To determine

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

Expert Solution & Answer

Chapter 17 Solutions

Calculus: Early Transcendentals

Knowledge Booster

$Calculus$

Learn more about

Need a deep-dive on the concept behind this application? Look no further. Learn more about this topic, calculus and related others by exploring similar questions and additional content below.

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.

Videos

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

Chapter 17 Solutions