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Please answer #41. Pic 1 shows the question and pic 2 shows section 4.4. Thank you.

Please answer #41. Pic 1 shows the question and pic 2 shows section 4.4. Thank you.

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 #41. 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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