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 (s + c)² + (w – c') Y (s) = (4.18) 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 = yoe¯# cos (t/w3 – c² YoC + vo -ct sin (t/w% – c² |w – c² - 0. Show that this is the critically 40. Case 2: aw – 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 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 + +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 (s + c)² + (w – c') Y (s) = (4.18) 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 = yoe¯# cos (t/w3 – c² YoC + vo -ct sin (t/w% – c² |w – c² - 0. Show that this is the critically 40. Case 2: aw – 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 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 + +c – Use the technique of partial fractions to decompose (4.19), and then find the solution y(t).
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 #38. Thank you.
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).
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