Advanced Engineering Mathematics
10th Edition
ISBN: 9780470458365
Author: Erwin Kreyszig
Publisher: Wiley, John & Sons, Incorporated
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Please answer #41. Pic 1 shows the question and pic 2 shows section 4.4. 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).
![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.](https://content.bartleby.com/qna-images/question/1b1372ae-e668-4971-89ee-29da7ac7c466/db8f3f34-552e-4342-95b4-3cd17f2fa0a2/31zc6z_processed.jpeg)
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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