Ejercicios Cap3, Aplic. cap4 y Laplace Civil.pdf

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problems 183

3. Vary the constants C 1, C 2, . . . , C n to make them functions of x , and obtain

n equations y (k)P , k=0,1, . . . , n−1,

y (k)

C (x ;C

1,C

2, . . . ,C

n)

C i⇒ c i(x )====⇒

i=1,2, . . . ,n y

(k)

P x ; c

1(x ), c

2(x ), . . . , c

n(x ).

4. Differentiating y (k)P with respect to x and comparing with the expressions

y (k+1)P , k=0,1, . . . ,n−2, and substituting y P and the derivatives into the

original differential equation yield n linear algebraic equations for c 1(x ),

c 2(x ), . . . , c n(x ).

5. Solve these equations for c 1(x ), c 2(x ), . . . , c n(x ).

6. Integrate to obtain the functions c 1(x ), c 2(x ), . . . , c n(x ). A particular solu-tion y P

x ; c 1(x ), c 2(x ), . . . , c n(x )

is then obtained.

Euler Differential Equations

(anx nD

n + an−1x n−1

Dn−1 + · · · + a1x D + a0) y = f (x ), D( · )≡d( · )/dx .

Letting x = ez or z = ln x , x > 0, D ( · )≡d( · )/dz , then

x nDn(·) = D ( D −1)( D −2) · · · ( D −n+1)(·), n=positive integer.

The Euler differential equation is converted to a differential equation with constant

coefficients.

Problems

Complementary Solutions

4.1 (D3 − 2D2 + D − 2) y = 0 ANS y C = A cos x + B sin x + C e2x

4.2 (D3 + D2 + 9D + 9) y = 0 ANS y C = A cos3x + B sin3x + C e−x

4.3 (D3 + D2 − D − 1) y = 0 ANS y C = (C 0+C 1x )e−x + C ex

4.4 (D3 + 8) y = 0 ANS y C = ex A cos(

√ 3x ) + B sin(

√ 3x )

+ C e−2x

4.5 (D3 − 8) y = 0 ANS y C = e−x A cos(

√ 3x ) + B sin(

√ 3x )

+ C e2x

4.6 (D4 + 4) y = 0 ANS y C = ex ( A1 cos x +B1 sin x )+e−x ( A2 cos x +B2 sin x )

4.7 (D4 + 18D2 + 81) y = 0 ANS y C = ( A0+ A1x ) cos3x +(B0+B1x ) sin3x

4.8 (D4 − 4D2 + 16) y = 0

ANS y C = e√

3x ( A1 cos x + B1 sin x ) + e−√

3x ( A2 cos x + B2 sin x



184 4 linear differential equations

4.9 (D4−2D3+2D2−2D+1) y =0 ANS y C = (C 0+C 1x )ex + A cos x +B sin x

4.10 (D4 − 5D3 + 5D2 + 5D − 6) y = 0

ANS y C =

C 1e−x

+C 2ex

+C 3e2x

+C 4e3x

4.11 (D5 − 6D4 + 9D3) y = 0 ANS y C =C 0+C 1x +C 2x 2+(D0+D1x )e3x

4.12 (D6 − 64) y = 0 ANS y C = C 1e−2x + C 2e2x

+ e−x A1 cos(

√ 3x ) + B1 sin(

√ 3x )

+ ex

A2 cos(

√ 3x ) + B2 sin(

√ 3x )

Particular Solutions — Method of Undetermined Coefficients

For the following differential equations, specify the form of a particular solution

using the method of undetermined coefficients.

4.13 (D2 + 6D + 10) y = 3x e−3x − 2e3x cos x

4.14 (D2 − 8D + 17) y = e4x (x 2 − 3x sin x )

4.15 (D2 − 2D + 2) y = (x + ex ) sin x

4.16 (D2 + 4) y = sinh x sin 2x

4.17 (D2

+2D

+2) y

=cosh x sin x

4.18 (D3 + D) y = sin x + x cos x

4.19 (D3 − 2D2 + 4D − 8) y = e2x sin2x + 2x 2

4.20 (D3 − 4D2 + 3D) y = x 2 + x e2x

4.21 (D4 + D2) y = 7x − 3cos x

4.22 (D4 + 5D2 + 4) y = sin x cos 2x

Particular Solutions — D-Operator Method

4.23 (D5 − 3D3 + 1) y = 9e2x ANS y P = e2x

4.24 (D − 1)3 y = 48x ex ANS y P = 2x 4ex

4.25 (D3 − 3D) y = 9x 2 ANS y P = −x 3 − 2x

4.26 (D5 + 4D3) y = 7 + x ANS y P = 196 x

3 (28+x )

4.27 (D2 − D − 2) y = 36x e2x ANS y P = 2e2x (3x 2−2x )

4.28 (D4 + 16) y = 64cos2x ANS y P = 2cos2x

4.29 (D4 + 4D2 − 1) y = 44sin3x ANS y P = sin3x



Problems 185

4.30 (D3 + D2 + 5D + 5) y = 5cos2x ANS y P = 2sin2x + cos2x

4.31 (D2 + 3D + 5) y = 5e−x sin2x ANS y P = −e−x (2cos2x + sin2x )

4.32 (D4

−1) y

=4e−x

ANS y P = −

x e−x

4.33 (D2 + 4) y = 8sin2x ANS y P = 1 − x sin 2x

4.34 (D3 − D2 + D − 1) y = 4sin x ANS y P = x (cos x − sin x )

4.35 (D4 − D2) y = 2ex ANS y P = x ex

General Solutions

4.36 y −

4 y +

4 y =

(1+

x )ex

+2e2x

+3e3x

ANS y = (C 0+C 1x )e2x + (x +3)ex + x 2e2x + 3e3x

4.37 (D2 − 2D + 5) y = 4ex cos2x

ANS y = ex ( A cos2x + B sin2x ) + x ex sin2x

4.38 (D2 + 4) y = 4sin2x ANS y = A cos2x + B sin2x − x cos 2x

4.39 (D2 − 1) y = 12x 2ex + 3e2x + 10cos3x

ANS y P = C 1e−

x + C 2e

x + e

x (2x

3−3x

2+3x ) + e

2x − cos3x

4.40 y + y = 2sin x − 3cos2x ANS y = A cos x +B sin x −x cos x + cos2x

4.41 y − y = ex (10 + x 2) ANS y = C 1 + C 2ex + ex

13 x

3−x 2+12x

4.42 (D2 − 4) y = 96x 2e2x + 4e−2x

ANS y = C 1e−2x + C 2e2x + e2x (8x 3 − 6x 2 + 3x ) − x e−2x

4.43 (D2

+2D

+2) y

=5cos x

+10sin2x

ANS y = e−x ( A cos x +B sin x )+ cos x +2sin x −2cos2x − sin2x

4.44 (D2 − 2D + 2) y = 4x − 2 + 2ex sin x

ANS y = ex ( A cos x + B sin x ) + 2x + 1 − x ex cos x

4.45 (D2 − 4D + 4) y = 4x e2x sin2x

ANS y = (C 0+C 1x )e2x − e2x (x sin 2x + cos2x )

4.46 (

D3

− D2

+ D − 1) y = 15sin2x ANS y = C ex + A cos x + B sin x + 2cos2x + sin2x

4.47 (D3 + 3D2 − 4) y = 40sin2x

ANS y = (C 0 + C 1x )e−2x + C 2ex + cos2x − 2sin2x



186 4 linear differential equations

4.48 y − y + y − y = 2ex + 5e2x

ANS y = A cos x + B sin x + C ex + x ex + e2x

4.49 (D3

−6D2

+11D

−6) y

=10ex sin x

ANS y = C 1ex + C 2e2x + C 3e3x + ex (3sin x − cos x )

4.50 (D3 − 2D − 4) y = 50(sin x + e2x )

ANS y = C e2x +e−x ( A cos x +B sin x )+6cos x −8sin x + 5x e2x

4.51 y − 3 y + 4 y = 12e2x + 4e3x

ANS y = (C 0 + C 1x )e2x + C 3e−x + 2x 2e2x + e3x

4.52 (D4

−8D2

+16) y

=32e2x

+16x 3

ANS y = (C 0+C 1x )e−2x + (D0+D1x )e2x + x 2e2x + x 3 + 3x

4.53 (D4 − 18D2 + 81) y = 72e3x + 729x 2

ANS y = (C 0 + C 1x )e−3x + (D0 + D1x )e3x + x 2e3x + 9x 2 + 4

Method of Variation of Parameters

4.54 y − y = x −1 − 2x −3ANS y = C 1e−x + C 2ex − x −1

4.55 y − y = 1sinh x

ANS y =C 1e−x +C 2ex −x e−x + sinh x ln1−e−2x

4.56 y − 2 y + y = ex

x ANS y =

C 0 + C 1x + x ln

x ex

4.57 y + 3 y + 2 y = sinex ANS y = C 1e−2x + C 2e−x − e−2x sinex

4.58 y − 3 y + 2 y = sine−x ANS y = C 1ex + C 2e2x − e2x sine−x

4.59 y + y = sec

3

x ANS

y = A cos x + B sin x + 1

2 sec x

4.60 y − y =

1−e2x − 1

2 ANS y =C 1e−x +C 2ex − 12 e−x sin−1ex − 1

2

1−e2x

4.61 y − y = e−2x sine−x ANS y =C 1e−x +C 2ex − sine−x −ex cose−x

4.62 y + 2 y + y = 15e−x √ x +1 ANS y = e−x

C 0 + C 1x + 4 (x +1)

52

4.63 y + 4 y = 2tan x

ANS y = A cos2x

+B sin2x

+sin2x lncos x − x cos 2x

4.64 y − 2 y + y = e2x

(ex + 1)2 ANS y = (C 0+C 1x )ex + ex ln(1+ex )

4.65 y + y = 1

1 + ex ANS y = C 1+C 2e−x − ln(e−x +1)−e−x ln(ex +1)



Problems 187

Euler Differential Equations

4.66 (x 2D2 − x D + 1) y = ln x ANS y = (C 0+C 1 ln x )x + 2 + ln x

4.67 x 2 y +

3x y +

5 y =

5

x 2 ln x

ANS y = x −1 A cos(2 ln x ) + B sin(2 ln x )

+ x −2

25 + ln x

4.68 (x 3D3 + 2x 2D2 − x D + 1) y = 9x 2 ln x

ANS y C = C 1x −1 + (C 2 + C 3 ln x )x + (3 ln x −7)x 2

4.69

(x −2)2D

2 − 3(x −2)D + 4 y = x

ANS y C = (x −2)2C 0+C 1 ln

x −2

+x − 3

2

4.70 x 3 y + 3x 2 y + x y − y = x 2

ANS y = Cx + 1√ x

A cos

√ 32

ln x + B sin

√ 32

ln x

+ x 2

7



232 5 applications of linear differential equations

On the record paper, the distance d between two adjacent peaks is measured in

length, which needs to be changed to time to yield the period T of the response.

Since the drum rotates at a speed of v rpm, i.e., it rotates an angle of 2πv in 60

seconds, hence the time T it takes to rotate an angle θ , as shown in the figure, is

given by T

60 =

θ

2πv =⇒ T =

30θ

πv .

Furthermore, since d = r θ , which is the arc length corresponding to angle θ , one

has

T =30

πv

d

r =

30d

πrv .

The frequency of vibration of tip D is

f =1

T =

πrv

30d .

Since the steady-state response and the excitation have the same frequency, one

obtains

= 2π f =π 2rv

15d .

Problems

5.1 A circular cylinder of radius r and mass m is supported by a spring of stiffness

k and partially submerges in a liquid of density γ . Suppose that, during vibration,

the cylinder does not completely submerge in the liquid. Set up the equation

of motion of the cylinder for the oscillation about the equilibrium position and

determine the period of the oscillation.

ANS m ¨ y + (k+γ πr 2) y = 0, T = 2 π

m

k+ γ πr 2

k

m

y, y, y



Problems 233

5.2 A cylinder of radius r , height h, and mass m floats with its axis vertical in a

liquid of density ρ as shown in the following figure.

☞ Archimedes' Principle: An object partially or totally submerged in a fluid

is buoyed up by a force equal to the weight of the fluid displaced.

h

Liquid Level

EquilibriumPosition

x (t )

r

1. Set up the differential equation governing the displacement x (t ), measured

relative to the equilibrium position, and determine the period of oscillation.

2. If the cylinder is set into oscillation by being pushed down a displacement x 0at t = 0 and then released, determine the response x (t ).

ANS mx + ρπr 2x = 0, T =2

r

πm

ρ; x (t ) = x 0 cos ω0t , ω0 = r

ρ π

m

5.3 A cube of mass m is immersed in a liquid as shown. The length of each side

of the cube is L. At time t = 0, the top surface of the cube is leveled with the surface

of the liquid due to buoyancy. The cube is lifted by a constant force F . Show that

the time T when the bottom surface is leveled with the liquid surface is given by

T =

L

g cos−1

1 −

mg

F

.

y, y, y m

L

L

F

t =0

Time t

t =T

F




5.4 A mass m is dropped with zero initial velocity from a height of h above a

spring of stiffness k as shown in the following figure. Determine the maximum

compression of the spring and the duration between the time when the mass con-

tacts the spring and the time when the spring reaches maximum compression.

ANS y max =

mg

k

2h+

mg

k

+

mg

k , T =

m

k

π

2 + − tan−1

mg

2hk

m

k

h

5.5 A uniform chain of length L with mass density per unit length ρ is laid on a

rough horizontal table with an initial hang of length l , i.e., y = l at t = 0 as shown

in the following figure. The coefficients of static and kinetic friction between the

chain and the surface have the same value µ. The chain is released from rest at time

t = 0 and it starts sliding off the table if (1+µ)l >µL. Show that the time T it

takes for the chain to leave the table is

T =

L

(1+µ) g cosh−1

L

(1+µ)l − µL

.

L− y

y (t )

y, y, y

5.6 A uniform chain of length L with mass density per unit length ρ is laid on a

smooth inclined surface with y = 0 at t = 0 as shown in the following figure. The

chain is released from rest at time t = 0. Show that the time T it takes for the chain

to leave the surface is

T =

L

(1− sin θ ) g cosh−1

1

sin θ

.



Problems 235

L− y

y (t ) y, y, y θ

5.7 A uniform chain of length L with mass density per unit length ρ is hung on a

small smooth pulley with y (t ) = l when t = 0, l >L/2, as shown in the Figure 5.27.

The chain is released from rest at time t = 0. Show that the time T it takes for the

chain to leave the pulley is

T =

L

2 g cosh−1

L

2 l −L

.

y, y, y

y (t )L–y (t )

A

B

O

L

r

k1

k2

a

b

Figure 5.27 Figure 5.28

5.8 A pendulum as shown in Figure 5.28 consists of a uniform solid sphere of

radius r and mass m connected by a weightless bar to hinge O. The bar is further

constrained by two linear springs of stiffnesses k1 and k2 at A and B, respectively.

It is known that the moment of inertia of a solid sphere of radius r and mass m

about its diameter is 25 mr 2. Show that the equation of motion governing the angle

of rotation of the pendulum about O and the natural period of oscillation of the

pendulum are given by

m 2

5 r 2 +L2

θ + (k1a

2 +k2b2 +mgL) θ = 0, T = 2π

m 2

5 r 2 +L2

k1a

2 +k2b2 +mgL

.




5.9 A mass m is attached to the end C of a massless rod AC as shown in the

following figure. The rod is hinged at one end A and supported by a spring of

stiffness k at the middle B. A dashpot damper having a damping coefficient c is

attached at the middle. A sinusoidal load F sin t is applied at end C .

F sint

LLx

A C B m

c

k

1. Show that the equation of motion governing displacement x (t ) of end C is

4mx + c x + k x = 4 F sin t .

2. Show that the natural circular frequency ωd of the damped free vibration of

the system is given by

ωd = ω0 1−c 2

16km

, ω0 =1

2

k

m

.

5.10 A massless rod is hinged at one end A and supported by a spring of stiffness

k at the other end D as shown in the following figure. A mass m is attached at 13 of

the length from the hinge and a dash-pot damper having a damping coefficient c is

attached at 23 of the length from the hinge. A sinusoidal load F sin t is applied at

end D.

L

F sint

Lx

A B C D

m L

c k

1. Show that the equation of motion governing displacement x (t ) of end D is

mx + 4c x + 9kx = 9F sin t .2. Show that the natural circular frequency ωd of the damped free vibration of

the system is given by

ωd = ω0

1 −

4c 2

9km , ω0 = 3

k

m.



Problems 237

5.11 A damped single degree-of-freedom system is shown in the following figure.

The displacement of the mass M is described by x (t ). The excitation is provided

by x 0(t ) = a sin t .

M

K 2

c 2

x 0(t ) x (t )

K 1

c 1

1. Show that the equation of motion governing the displacement of the mass M is given by

x + 2ζ ω0 x + ω20 x = α sin t + β cos t ,

where

ω0 =

K 1 +K 2

M , 2ζ ω0 =

c 1 +c 2 M

, α =aK 1 M

, β =ac 1 M

.

2. Determine the amplitude of the steady-state response x P (t ).

ANS

α(ω2

0 −2) +2ζ ω0β2

+

β(ω20 −2) −2ζ ω0 α

2

(ω20 −2)2 + (2ζ ω0)2

5.12 The single degree-of-freedom system shown in the following figure is sub-

jected to dynamic force F (t ) = F 0 sin t .

m k

x (t )

kc

F (t )

1. Set up the equation of motion in terms of x (t ) and determine the damped

natural circular frequency.

2. Determine the steady-state response of the system x P (t

).

ANS mx + c x + 2kx = F 0 sin t , ωd =

2k

m

1−

c 2

8km

x P (t ) =F 0

(2k−m2) sin t − c cos t

(2k−m2)2 +c 22




5.13 The single degree-of-freedom system shown in the following figure is sub-

jected to dynamic displacement x 0(t ) = a sin t at point A.

x 0(t )=a sint

x (t )

c A

k1 k2

m

1. Set up the equation of motion in terms of of x (t ).

2. If the system is lightly damped, determine the steady-state response of the

system x P (t ).

ANS mx + c x + (k1 +k2)x = ak2 sin t

x P (t ) = ak2

(k1 +k2 −m2) sin t − c cos t

(k1 +k2 −m2)2 +c 22

5.14 A precision instrument having a mass of m = 400 kg is to be mounted on a

floor. It is known that the floor vibrates vertically with a peak-to-peak amplitude

of 2 mm and frequency of 5 Hz. To reduce the effect of vibration of the floor onthe instrument, four identical springs are placed underneath the instrument. If the

peak-to-peak amplitude of vibration of the instrument is to be limited to less than

0.2 mm, determine the stiffness of each spring. Neglect damping.

ANS k = 8.97 kN/m

5.15 The single degree-of-freedom system, shown in the following figure, is

subjected to a sinusoidal load F (t ) = F 0 sin t at point A. Assume that the mass m,

the spring stiffnesses k1 and k2, and F 0 and are known. The system is at rest

when t = 0.

m A

k1 k2

F (t )=F 0 sint

x (t ) y(t )

1. Show that the differential equation governing the displacement of the mass

x (t ) is

x + ω20 x = f sin t , ω0 =

k1 k2

m(k1 +k2), f =

k2

m(k1 +k2)F 0.



Problems 239

2. For the case = ω0, determine the response of the system x (t ).

3. For the case = ω0, determine the response of the system x (t ).

ANS 2. = ω0 : x (t ) =

f

ω20 −2

−

ω0sin ω0t + sin t

;

3. = ω0 : x (t ) = − f

2ω20

− sin ω0t + ω0t cos ω0t

5.16 A vehicle is modeled by a damped single degree-of-freedom system with

mass M , spring stiffness K , and damping coefficient c as shown in the following

figure. The absolute displacement of the mass M is described by y (t ). The vehicle

is moving at a constant speed U on a wavy surface with profile y 0(x ) = µ sin x .At time t = 0, the vehicle is at x = 0.

K

U

Ox

M

c

y

y0(x )

y0

1. Show that the equation of motion governing the relative displacement of the

vehicle given by z (t ) = y (t ) − y 0(t ) is

z + 2ζ ω0 z + ω20 z = µ 2U 2 sin(Ut ), ω0 =

K

M , 2ζ ω0 =

c

M .

2. Determine the amplitude of the steady-state response z (t ), which is a partic-

ular solution of the equation of motion.

3. Assuming that the damping coefficient c = 0, determine the speed U at which

resonance occurs.

ANS

µ2U 2

(ω20 −2U 2)2 + (2ζ ω0U )2

; U =1

K

M

5.17 The landing gear of an airplane as shown in Figure 5.8 can be modeled as a

mass connected to the airplane by a spring of stiffness K and a damper of damping

coefficient c . A spring of stiffness k is used to model the forces on the tires. The

airplane lands at time t = 0 with x = 0 and moves at a constant speed U on a

wavy surface with profile y 0(x ) = µ sin x . Assuming that the airplane moves in




the horizontal direction only, determine the steady-state response of the absolute

displacement y (t ) of the mass m.

ANS y (t ) =kµ sin(Ut −ϕ)

(K +k−m2U 2)2 + (c U )2

, ϕ = tan−1 c U

K +k−m2

U 2

5.18 In Section 5.1, it is derived that the equation of motion of a single story

shear building under the base excitation x 0(t ) is given by

m ¨ y (t ) + c ˙ y (t ) + k y = −mx 0(t ),

or

¨ y (t ) + 2ζ ω0 ˙ y (t ) + ω20 y = −x 0(t ),

where

ω20 =

km

, 2ζ ω0 =c m

,

and y (t ) = x (t ) −x 0(t ) is the relative displacement between the girder and the base.

x (t )

x 0(t )=a sint

m

Rigid girder

Weightless columns k

c

For x 0(t ) = a sin t , determine the Dynamic Magnification Factor (DMF) de-

fined as

DMF =

y P (t )

max

x 0(t )max

,

where y P (t ) is the steady-state response of the relative displacement or the par-

ticular solution due to the base excitation. Plot DMF versus the frequency ratio

r = /ω0 for ζ = 0, 0.1, 0.2, and 0.3.

ANS DMF =r 2

(1−r 2)2 + (2ζ r )2, r =

ω0

5.19 Consider the undamped single degree-of-freedom system with m = 10 kg,

k = 1 kN/m. The system is subjected to a dynamic load F (t ) as shown in thefollowing figure. The system is at rest at time t = 0.

Determine the analytical expression of the displacement as a function of time up

to t = 10 sec. ANS x (t ) = 0.02(10t − sin 10t ), 0 t 5

x (t ) = 1.005 cos 10(t − 5) + 0.0007 sin 10(t − 5), 5 t 10



Problems 241

F (t ) (kN)

0

1

t (sec)5 10

m

x (t )

F (t )k

5.20 The following figure shows the configuration of a displacement meter used

for measuring the vibration of the structure that the meter is mounted on. The

structure undergoes vertical displacement a0 sin t and excites the mass-spring-

damper system of the displacement meter. The displacement of the mass is recorded

on the rotating drum. It is known that m = 1 kg, k = 1000 N/m, c = 5 N · sec/m,and the steady-state record on the rotating drum shows a sinusoidal function with

frequency of 5 Hz and peak-to-peak amplitude of 50 mm. Determine the amplitude

a0 and the frequency f = /(2π ) of the displacement of the structure.

ANS a0 = 4.0 mm, f = 5 Hz

m

c

k

a0 sint

2a

5.21 For the circuit shown in Figure 5.29(a), the switch has been at position a for

a long time prior to t = 0−. At t = 0, the switch is moved to position b. Determine

i(t ) for t > 0. ANS i(t ) = (3−9t )e−5t (A)

5.22 For the circuit shown in Figure 5.29(b), the switch has been at position a for

a long time prior to t = 0−. At t = 0, the switch is moved to position b. Show that

the differential equation governing v C (t ) for t > 0 is

d2v C

dt 2 +

R

L

dv C

dt +

1

LC v C =

V (t )

LC , v C (0+) = −RI 0,

dv C (0+)

dt = 0.

For R = 6 , C = 125 F, L = 1 H, I 0 = 1 A, V (t ) = 39sin2t (V), determine v C (t )

for t >0. ANS v C (t ) = 7e−3t (2cos4t − sin4t ) +35sin2t − 20cos2t (V)




t =0

i

a

b

4 A

2 H

0.02F 14

2

6

12V

t=0

t=0

V (t )R

a

b

L C

I 0

v C

(a) (b)

t=0

R2

R1

C 1

C 2V (t ) v C

t=0

L

C R2

R1

V 0

I (t ) v C

(c) (d)

Figure 5.29 Second-order circuits.

5.23 For the circuit shown in Figure 5.29(c), show that the differential equation

governing v C (t ) for t > 0 is

R1C 1R2C 2d2v C

dt 2 + (R1C 1 +R1C 2 +R2C 2)

dv C

dt + v C = R1C 1

dV (t )

dt ,

with the initial conditions given by v C (0+) = 0,dv C (0+)

dt =

V (0+)

R2C 2.

For R1 = 1 , R2 = 2 , C 1 = 2 F, C 2 = 1 F, V (t ) = 12e−t (V), determine v C (t ) for

t >0. ANS v C (t ) = − 83 e

− t 4 +

83 (1+ 3t )e−t (V)

5.24 For the circuit shown in Figure 5.29(d), show that the differential equation

governing v C

(t ) for t > 0 is

d2v C

dt 2 +

R1 +R2

L

dv C

dt +

1

LC v C =

V 0 +R2 I (t )

LC , v C (0+) = V 0,

dv C (0+)

dt = 0.

For R1 = R2 = 5 , C = 0.2 F, L = 5 H, V 0 = 12V, I (t ) = 2sin t (A), determine

v C (t ) for t > 0. ANS v C (t ) = 5(1+ t )e−t + 12 − 5 cos t (V)



Problems 243

5.25 Consider column AB clamped at the base and pin-supported at the top by

an elastic spring of stiffness k. Show that the buckling equation for the column is

kL3

EI tan(αL) − (αL) + (αL)3 = 0, α2 =

P

EI

.

x

y

P

A

Bk

EI, L

5.26 Consider the beam-column shown in the following figure. Determine the

lateral deflection y (x ).

ANS y (x ) = −w

P

Lx

61−

x 2

L2 +

1

α

2 x

L

−sin αx

sin αL, α2 =

P

EI

.

EI, L

x L

y

x P P

w (x )= w w



problems 291

❧ The Heaviside step function defined as

H (t −a) =

0, t <a,

1, t >a,

is very useful in describing piecewise smooth functions by combining the

following results

f (t )

1 − H (t −a)

=

f (t ), t <a,

0, t >a,

f (t ) H (t −a) − H (t −b) = 0, t <a,

f (t ), a< t <b,

0, t >b,

f (t )H (t −a) =

0, t <a,

f (t ), t >a.

❧ The Dirac delta function δ(t −a) is a mathematical idealization of impulse

functions. It is useful in modeling impulse functions, such as concentrated or

point loads.❧ Applying the Laplace transform to an nth-order linear differential equation

with constant coefficients

an y (n)(t ) + an−1 y

(n−1)(t ) + · · · + a1 y (t ) + a0 y (t ) = f (t )

converts it into a linear algebraic equation for Y (s)=L y (t )

, which can

easily be solved. The solution of the differential equation can be obtained by

determining the inverse Laplace transform y (t )=L −1Y (s).

❧ The method of Laplace transform is preferable and advantageous in solving

linear ordinary differential equations with the right-hand side functions f (t )

involving the Heaviside step function and the Dirac delta function.

Problems

Evaluate the Laplace transform of the following functions.

6.1 f (t ) = 4 t 3 − 2t 2 + 5 ANS F (s) = 24−4s+5s3

s4

6.2 f (t ) = 3sin2t − 4cos5t ANS F (s) = 6

s2 +4 −

4s

s2 + 25

6.3 f (t ) = e−2t (4cos3t + 5sin3t ) ANS F (s) = 4s+23

s2 +4s+13



292 6 the laplace transform and its applications

6.4 f (t ) = 3cosh6t + 8sinh3t ANS F (s) = 3s

s2 −36 +

24

s2 −9

6.5 f (t ) = 3t cos 2t + t 2 et ANS F (s) = 3(s2 −4)

(s2 +4)2 +

2

(s−1)3

6.6 f (t ) = t cosh 2t + t 2 sin5t + t 3 ANS F (s) = s2 + 4

(s2 −4)2+

10(3s2 − 25)

(s2 + 25)3 +

6

s4

6.7 f (t ) = 7e−5t cos2t + 9 sinh2 2t ANS F (s) = 7(s+ 5)

(s+ 5)2 + 4 +

72

s(s2 −16)

6.8 f (t ) =

0, t <π ,

sin t , t >π.ANS F (s) = −

e−πs

s2 +1

6.9 f (t ) =

0, t <1,

4t 2 +3t − 8, t >1.ANS F (s) = e−s

8

s3 +

11

s2 −

1

s

6.10 f (t ) =

0, t <1,

t 2 − 1, 1< t <2,

0, t >2.

ANS F (s)=2(s+1)e−s −(3s2 + 4s+ 2)e−2s

s3

6.11 f (t ) = sin t , t <π ,

4sin3t , t >π.ANS F (s) =

1

s2 +1 −

e−πs(11s2 + 3)

(s2 +1)(s2 + 9)

6.12 f (t ) =

2t , 0< t <2,

2 + t , 2< t <4,

10 − t , 4< t <10,

0, t >10.

ANS F (s)=2−e−2s − 2e−4s + e−10s

s2

6.13 f (t ) = t 3 δ(t −2) + 3cos5t δ(t −π) ANS F (s) = 8e−2s − 3e−πs

6.14 f (t ) = sinh 4t δ(t + 2) + e2t δ(t −1) + t 2 e−3t δ(t −2) + cosπ t δ(t −3)

ANS F (s) = e2−s+ 4e−6−2s

− e−3s

Express the following periodic functions using the Heaviside function or the Dirac

function and evaluate the Laplace transform.

6.15

t

a

T 2T 3T 4T 5T

f (t )

−a

0

ANS F (s) = a

s

1 + 2

∞n=1

(−1)n e−nTs



Problems 293

6.16

t

a

T 2T 3T 4T 5T

f (t )

0

ANS F (s) = a

Ts2 −

a

s

∞n=0

e−(n+1)Ts

6.17

t

aa sin

T πt

T 2T 3T 4T 5T

f (t )

0

ANS F (s) = aT π

T 2s2 + π2

∞n=0

e−nTs

6.18

t

f (t )

I

−I −I

2T

T

0

3T

I

−I

4T

5T

I

ANS F (s) = I ·

∞n=0

(−1)ne−nTs

Evaluate the inverse Laplace transform of the following functions.

6.19 F (s) = s

(s+1)3 ANS f (t ) = t − 1

2

t 2e−t

6.20 F (s) = 4(2s+1)

s2 −2s−3 ANS f (t ) = 7e3t

+ e−t

6.21 F (s) = 3s+ 2

s2 +6s+10 ANS f (t ) = e−3t (3cos t −7sin t )

6.22 F (s) = 3s2 +2s−1

s2 − 5s+6 ANS f (t ) = 3δ(t ) + 32e3t − 15e2t

6.23 F (s) = 30(s2 +1)(s2 − 9) ANS f (t ) = − 3sin t + sinh 3t

6.24 F (s) = 13s

(s2 −4)(s2 +9) ANS f (t ) = − cos3t + cosh 2t

6.25 F (s) = 40s

(s+1)(s+ 2)(s2 −9) ANS f (t ) = 10e−3t

− 16e−2t + 5e−t

+ e3t




6.26 F (s) = 2

s3(s2 +1) ANS f (t ) = 2cos t + t 2 − 2

6.27 F (s) = s

(s+1)(s+ 2)3 ANS f (t ) = (t 2 + t +1)e−2t − e−t

6.28 F (s) = 8

(s−1)(s+1)2(s2 +1) ANS f (t )= et −(2t + 3)e−t

+2cos t −2sin t

6.29 F (s) = 162

s3(s2 − 9) ANS f (t ) = −9t 2 − 2 + 2cosh3t

Evaluate the inverse Laplace transform of the following functions using convolution

integral .

6.30 Y (s) =

1

s(s2 +a2)2 ANS y (t ) =

1

2a4 (2− 2cos at −at sin at )

6.31 Y (s) = 1

s2(s2 +a2)2 ANS y (t ) = 1

2a5(2at +at cos at −3sin at )

6.32 Y (s) = 4

s(s2 +4s+ 4) ANS y (t ) = 1−(2t +1)e−2t

6.33 Y (s) = 16

s3(s2 +4s+ 4) ANS y (t ) = −(2t + 3)e−2t

+ 2t 2 −4 t +3

6.34 Y (s) = 6s(s2 +4s+ 3)

ANS y (t ) = − 3e−t + e−3t

+ 2

6.35 Y (s) = 5

s(s2 +4s+ 5) ANS y (t ) = −e−2t (2sin t + cos t )+1

Solve the following differential equations.

6.36 y + 4 y + 3 y = 60cos3t , y (0) = 1, y (0) = −1

ANS

y (t )=

5e

−3t −

2e

−t −

2cos3t +

4sin3t 6.37 y

+ y −2 y = 9e−2t , y (0)= 3, y (0)= − 6 ANS y (t )= et −(3t − 2)e−2t

6.38 y − y − 2 y = 2t 2 + 1, y (0) = 6, y (0) = 2

ANS y (t ) = 5e−t + 3e2t − t 2 + t − 2

6.39 y + 4 y = 8sin2t , y (0) = 1, y (0) = 4

ANS y (t ) = (−2t +1) cos2t + 3sin2t

6.40 y − 2 y + y = 4e−t + 2et , y (0) = −1, y (0) = 2

ANS y (t ) = e−t + (t 2 +5t − 2)et

6.41 y − 2 y + 2 y = 8e−t sin t , y (0) = 1, y (0) = −1

ANS y (t ) = − et sin t + e−t (cos t + sin t )



Problems 295

6.42 y − 2 y + 5 y = 8et sin2t , y (0) = 1, y (0) = −1

ANS y (t ) = −(2t −1)et cos2t

6.43 y + y − 2 y = 54t e−2t , y (0) = 6, y (0) = 0

ANS y (t ) = −(9t 2 +6t )e−2t + 6et

6.44 y − y − 2 y = 9e2t H (t −1), y (0) = 6, y (0) = 0

ANS y (t ) = 4e−t + 2e2t +

(3t −4)e2t + e3−t

H (t −1)

6.45 y + 2 y + y = 2sin tH (t −π), y (0) = 1, y (0) = 0

ANS y (t ) = (t +1)e−t −

cos t + (t +1−π)eπ−t

H (t −π)

6.46 y

+ 4 y = 8sin2tH (t −π), y (0) = 0, y

(0) = 2

ANS y (t ) = sin2t +

2(π −t ) cos2t + sin2t H (t −π)

6.47 y + 4 y = 8(t 2 +t −1)H (t −2), y (0) = 1, y (0) = 2

ANS y (t )= sin2t + cos2t +

2t 2 + 2t − 3−9cos(2t −4)− 5sin(2t −4)H (t − 2)

6.48 y − 3 y + 2 y = et H (t − 2), y (0) = 1, y (0) = 2

ANS y (t ) = e2t + (1−t )et + e2t −2H (t −2)

6.49 y − 5 y + 6 y = δ(t −2), y (0) = −1, y (0) = 1

ANS y (t ) = − 4e2t + 3e3t

+

e3(t −2) − e2(t −2)H (t − 2)

6.50 y + 4 y = 4H (t −π) + 2δ(t −π), y (0) = −1, y (0) = 2

ANS y (t ) = sin2t − cos2t + (1 + sin2t − cos2t )H (t −π)

6.51 y − y

+ 4 y − 4 y = 10e−t , y (0) = 5, y (0) = − 2, y (0) = 0

ANS

y (t ) = − e−t

+ 5et

+ cos2t − 4sin2t 6.52 y

− 5 y +4 y = 120e3t H (t −1), y (0) = 15, y (0) = − 6, y (0) = 0,

y (0) = 0 ANS y (t ) = 6et + 14e−t − 2e2t

−3e−2t + (10et +2

−5e−t +4

− 10e2t +1 +2e−2t +5 + 3e3t )H (t − 1)

6.53 y + 3 y

− 4 y = 40t 2H (t −2), y (0) = y (0) = y (0) = y (0) = 0

ANS y (t ) = −10t 2 −15+ 40et −2+8e2−t

+ 7cos(2t −4)+ 4sin(2t −4)H (t −2)

6.54 y + 4 y = (2t 2 + t + 1)δ(t −1), y (0) = 1, y (0) = −2, y (0) = 0,

y (0)= 0 ANS y (t ) = e−t cos t − sin t cosh t +

sin(t −1) cosh(t − 1)

− cos(t − 1) sinh(t −1)H (t − 1)

6.55 Determine the lateral deflection y (x ) of the beam-column as shown.




a

EI, L

y

x P P

W

ANS W 0 =W α(L−a)− sinα(L−a)

α3EI

, W 1 =W

1− cosα(L−a)

α2EI ,

y (0) =W 1 (1− cosαL)

α sinαL −W 0, y (0) = −

W 1α

sinαL,

y (x ) = y (0) + 1− cosαx

α2 y (0) +

W α(x −a)− sinα(x −a)

H (x −a)

α3EI .

For the single degree-of-freedom system shown in Figure 6.6, determine the forced vibration response x Forced(t ) due to the externally applied load f (t ) shown. The

system is assumed to be underdamped, i.e., 0<ζ <1.

6.56

t

f (t )

0 T

f 0

ANS x Forced(t ) = f 0m

η1(t ) − η1(t −T )H (t −T )

6.57

t

f (t )

2T 3T 0 T

f 0

ANS x Forced(t ) = f 0mT

η2(t ) − η2(t −T )H (t −T )

−η2(t − 2T )H (t −2T ) + η2(t − 3T )H (t −3T )

6.58

t

T 2T 3T 4T 5T

f (t )

f 0

0

ANS x Forced(t ) = f 0m

∞n=0

η1(t − 2nT )H (t − 2nT )

−η1

t −(2n+1)T

H t −(2n+1)T



Problems 297

6.59

t

T 2T 3T 4T 5T

f (t )

0

f 0

ANS x Forced(t )= f 0mT

η2(t )−2

∞n=1

η2

t −(2n−1)T

H t −(2n−1)T

−η2(t − 2nT )H (t −2nT )

6.60 For the circuit shown in Figure 6.9(a), the current source I (t ) is

I (t ) = I 0 H (−t ) + I 1(t )H (t ).

Show that the differential equation governing i(t ) is

di

dt +

L+R1R2C

R1LC i +

R1 +R2

R1LC

di

dt =

I 1(t )

LC , i(0+) =

R1I 0R1 +R2

, i(0+)

dt = 0.

For R1 = 1, R2 = 8, C = 14 F, L= 4 H, I (t )= 13sin2t

H (t )−H (t −π)

(A),

I 0=

0 A, determine i(t ) for t >0.

ANS i(t ) = 12

13 +2t

e−3t −12

13 +2(t −π)

e−3(t −π)H (t −π)

+ 113 (5sin2t −12cos2t )

1−H (t −π)

(A)

LC R1

I (t ) i

L C 2C 1

RR2 V (t )i

(a) (b)

Figure 6.9 Electric circuits.

6.61 For the circuit shown in Figure 6.9(b), the voltage source V (t ) is

V (t ) = V 0 H (− t ) + V 1(t )H (t ).

Show that the differential equation governing i(t ) is

d3i

dt 3 +

C 1 +C 2RC 1C 2

d2i

dt 2 +

1

LC 2

di

dt +

1

RLC 1C 2i =

1

L

d2V 1(t )

dt 2 +

1

RLC 1

dV 1(t )

dt ,




with the initial condtions

i(0+) = 0, di(0+)

dt =

V (0+)−V 0L

, d2i(0+)

dt 2 =

1

L

dV (0+)

dt −

V (0+)−V 0RC 2

.

For R= 8, C 1 = 14 F, C 2 =

120 F, L= 5 H, V (t )= 10H (−t )+10e−2t H (t ) (V),

determine i(t ) for t >0. ANS i(t ) = −3e−2t + (2+ cos t −3sin t )e−t (A)

6.62 Determine the deflection of a beam pinned at both ends under a uniformly

distributed load as shown.

a k

EI, L

w

b

y

x

ANS y (x ) = y (0) ·φ2(x ) + y (0) ·φ0(x )

+ w 4EI β4

1−φ3(x −a)

H (x −a) −

1−φ3(x −b)

H (x −b)

,

y (0) = α0φ2(L)−α2φ0(L)

φ22(L)+ 4β4φ2

0(L), y (0) =

α2φ2(L)+4β4α0φ0(L)

φ22(L)+ 4β4φ2

0(L),

α0 = w

4EI β4

φ3(L−a)−φ3(L−b)

, α2 = −

w

EI

φ1(L−a)−φ1(L−b)

.

6.63

Determine the deflection of a beam clamped at both ends under a concen-trated load as shown.

a k

EI, L

y

x

W

ANS y (x ) = y (0) ·φ1(x ) + y (0) ·φ0(x ) + W φ0(x −a)H (x −a), W = W EI

y (0) = α0φ1(L) − α1φ0(L)

φ21(L) − φ0(L)φ2(L)

, y (0) = α1φ1(L) − α0φ2(L)

φ21(L) − φ0(L)φ2(L)

,

α0 = − W φ0(L−a), α1 = − W φ1(L−a).



Problems 299

6.64 Determine the deflection of a free-clamped beam under a triangularly dis-

tributed load as shown.

a k

EI, L

y

x

b

w

ANS y (x ) = y (0) ·φ3(x ) + y (0) ·φ2(x )+w (x −a)−φ2(x −a)

H (x −a)

−(x −a)−φ2(x −b)−(b−a)φ3(x −b)

H (x −b)

,

w = w

4(b−a)EI β4,

y (0) = α0φ3(L) − α1φ2(L)

φ23(L) + 4β4φ0(L)φ2(L)

, y (0) = α1φ3(L) + 4β4α0φ0(L)

φ23(L) + 4β4φ0(L)φ2(L)

,

α0 = −w φ2(L−b) − φ2(L−a) + (b−a)φ3(L−b),α1 = −w

φ3(L−b) − φ3(L−a) − 4β4 (b−a)φ0(L−b)

.

6.65 Determine the deflection of a sliding-clamped beam under a triangularly

distributed load as shown.

a k

EI, L

y

x

w

ANS y (x ) = y (0) ·φ3(x ) + y (0) ·φ1(x ) − w x − a − φ2(x ) + aφ3(x )

−

(x −a) − φ2(x −a)

H (x −a)

, w =

w

4aEI β4,

y (0)= α0φ2(L)−α1φ1(L)

φ1(L)φ3(L)+4β4φ0(L)φ1(L), y (0)=

α1φ3(L)+ 4β4α0φ0(L)

φ1(L)φ3(L)+4β4φ0(L)φ1(L),

α0 = w

−φ2(L) + aφ3(L) + φ2(L−a)

,

α w

φ (L) 4aβ4φ (L) + φ (L a)

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