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38. When there is damping, but still no forcing term, the initial value problem becomes y" + 2cy' + wy = 0, y(0) = yo, y'(0) = vo, (4.17) where yo and vo are the initial displacement and velocity of the mass, respectively. Show that the Laplace transform of the solution can be written Yos + vo + 2cyo Y (s) = (4.18) (s +c)² + (@¿ – c²)' A moment's reflection will reveal that taking the inverse transform of Y (s) will depend upon the sign of w – c². In Exercises 39–41, we will examine three cases. Notice that each has a counterpart in Section 4.4. - 39. Case 1: w – c² > 0. Show that this is the underdamped case of Section 4.4. Compute the inverse Laplace trans- form of Y (s) to show that the solution in this case is given by y(t) = yoe -ct cos w – c² Yoc + vo sin c2 40. Case 2: w; – c² damped case of Section 4.4. Compute the inverse Laplace transform of Y(s) to show that the solution in this case is given by = 0. Show that this is the critically y(t) = yoe" + (vo + byo)te¬et. 41. Case 3: w – c² < 0. Show that this is the overdamped case of Section 4.4. To simplify calculations, let's set the intitial displacement as y(0) = yo = 0. Show that with this assumption the transform in (4.18) becomes vo Y (s) = %3D (s +c)² – (c² – 3)' (4.19) vo c+ Use the technique of partial fractions to decompose (4.19), and then find the solution y(t).

38. When there is damping, but still no forcing term, the initial value problem becomes y" + 2cy' + wy = 0, y(0) = yo, y'(0) = vo, (4.17) where yo and vo are the initial displacement and velocity of the mass, respectively. Show that the Laplace transform of the solution can be written Yos + vo + 2cyo Y (s) = (4.18) (s +c)² + (@¿ – c²)' A moment's reflection will reveal that taking the inverse transform of Y (s) will depend upon the sign of w – c². In Exercises 39–41, we will examine three cases. Notice that each has a counterpart in Section 4.4. - 39. Case 1: w – c² > 0. Show that this is the underdamped case of Section 4.4. Compute the inverse Laplace trans- form of Y (s) to show that the solution in this case is given by y(t) = yoe -ct cos w – c² Yoc + vo sin c2 40. Case 2: w; – c² damped case of Section 4.4. Compute the inverse Laplace transform of Y(s) to show that the solution in this case is given by = 0. Show that this is the critically y(t) = yoe" + (vo + byo)te¬et. 41. Case 3: w – c² < 0. Show that this is the overdamped case of Section 4.4. To simplify calculations, let's set the intitial displacement as y(0) = yo = 0. Show that with this assumption the transform in (4.18) becomes vo Y (s) = %3D (s +c)² – (c² – 3)' (4.19) vo c+ Use the technique of partial fractions to decompose (4.19), and then find the solution y(t).

Advanced Engineering Mathematics
Advanced Engineering Mathematics
10th Edition
ISBN:9780470458365
Author:Erwin Kreyszig
Publisher:Erwin Kreyszig
Chapter2: Second-order Linear Odes
Section: Chapter Questions
Problem 1RQ
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Please answer #39. Pic 1 shows the question and pic 2 shows section 4.4. Thank you.

38. When there is damping, but still no forcing term, the initial value problem becomes y" + 2cy' + wy = 0, y(0) = yo, y'(0) = vo, (4.17) where yo and vo are the initial displacement and velocity of the mass, respectively. Show that the Laplace transform of the solution can be written Yos + vo + 2cyo Y (s) (4.18) (s + c)² + (w – c²)' A moment's reflection will reveal that taking the inverse transform of Y (s) will depend upon the sign of w – c². In Exercises 39–41, we will examine three cases. Notice that each has a counterpart in Section 4.4. 39. Case 1: w – c² > 0. Show that this is the underdamped case of Section 4.4. Compute the inverse Laplace trans- form of Y (s) to show that the solution in this case is given by y(t) = yoe -ct cos Yoc + vo + sin 40. Case 2: w – c2 damped case of Section 4.4. Compute the inverse Laplace transform of Y (s) to show that the solution in this case is given by 0. Show that this is the critically y(t) = yoe¬et + (vo + byo)te¬ct. 41. Case 3: w – c² < 0. Show that this is the overdamped case of Section 4.4. To simplify calculations, let's set the intitial displacement as y(0) = yo = 0. Show that with this assumption the transform in (4.18) becomes vo Y (s) = (s + c)² – (c² – w;) - (4.19) vo (++c+ Use the technique of partial fractions to decompose (4.19), and then find the solution y(t).
Transcribed Image Text:38. When there is damping, but still no forcing term, the initial value problem becomes y" + 2cy' + wy = 0, y(0) = yo, y'(0) = vo, (4.17) where yo and vo are the initial displacement and velocity of the mass, respectively. Show that the Laplace transform of the solution can be written Yos + vo + 2cyo Y (s) (4.18) (s + c)² + (w – c²)' A moment's reflection will reveal that taking the inverse transform of Y (s) will depend upon the sign of w – c². In Exercises 39–41, we will examine three cases. Notice that each has a counterpart in Section 4.4. 39. Case 1: w – c² > 0. Show that this is the underdamped case of Section 4.4. Compute the inverse Laplace trans- form of Y (s) to show that the solution in this case is given by y(t) = yoe -ct cos Yoc + vo + sin 40. Case 2: w – c2 damped case of Section 4.4. Compute the inverse Laplace transform of Y (s) to show that the solution in this case is given by 0. Show that this is the critically y(t) = yoe¬et + (vo + byo)te¬ct. 41. Case 3: w – c² < 0. Show that this is the overdamped case of Section 4.4. To simplify calculations, let's set the intitial displacement as y(0) = yo = 0. Show that with this assumption the transform in (4.18) becomes vo Y (s) = (s + c)² – (c² – w;) - (4.19) vo (++c+ Use the technique of partial fractions to decompose (4.19), and then find the solution y(t).
Damped harmonic motion Now c > 0. The differential equation x" + 2cx' + wx = 0 has the characteristic equation 2? + 2cà + w = 0. The roots are - Vc2 – w3 and d2 = -c + . (4.15) = -C – We have three cases to consider depending on the sign of the discriminant c² – wz. 1. c < wo. This is the underdamped case. The roots in (4.15) are distinct complex numbers. Hence, the general solution is x(t) = e¯ [Cj cos wt + C2 sin wt], where W = Vw3 – c². 2. c > wo. This is the overdamped case. Now the roots in (4.15) are distinct and real. Further, since /c² – w < Vc² = c, we have à1 < ^2 < 0. The general solution is x(t) = Cje*1' + C2e^2' . 3. c = wo. This is the critically damped case, and in this case, the root in (4.15) is a double root, 2 = -c. The general solution is x (1) = Cje¯t + C2te¯". In each of the cases the solution decays to zero as t → ∞ due to the exponential term in the solution, and the fact that c > 0. In the critically damped case, this follows since, for c > 0, lim,→∞ t/et = 0 by l'Hôpital's rule. In the underdamped case the cosine and sine terms cause the solution to oscil- late with frequency w as it converges to zero. Notice that this frequency is always smaller than the natural frequency of the spring. In the other two cases there is no oscillation.
Transcribed Image Text:Damped harmonic motion Now c > 0. The differential equation x" + 2cx' + wx = 0 has the characteristic equation 2? + 2cà + w = 0. The roots are - Vc2 – w3 and d2 = -c + . (4.15) = -C – We have three cases to consider depending on the sign of the discriminant c² – wz. 1. c < wo. This is the underdamped case. The roots in (4.15) are distinct complex numbers. Hence, the general solution is x(t) = e¯ [Cj cos wt + C2 sin wt], where W = Vw3 – c². 2. c > wo. This is the overdamped case. Now the roots in (4.15) are distinct and real. Further, since /c² – w < Vc² = c, we have à1 < ^2 < 0. The general solution is x(t) = Cje*1' + C2e^2' . 3. c = wo. This is the critically damped case, and in this case, the root in (4.15) is a double root, 2 = -c. The general solution is x (1) = Cje¯t + C2te¯". In each of the cases the solution decays to zero as t → ∞ due to the exponential term in the solution, and the fact that c > 0. In the critically damped case, this follows since, for c > 0, lim,→∞ t/et = 0 by l'Hôpital's rule. In the underdamped case the cosine and sine terms cause the solution to oscil- late with frequency w as it converges to zero. Notice that this frequency is always smaller than the natural frequency of the spring. In the other two cases there is no oscillation.
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