Mechanical Engineering Chapter 22 Homework The general solution of the above differential equation

1230

22–41.

If the block-and-spring model is subjected to the periodic

force , show that the differential equation of

motion is , where xis

measured from the equilibrium position of the block. What

is the general solution of this equation?

x

$+(k>m)x=(F

0>m) cos vt

F=F

0cos vt

F F0cos vt

k

Equilibrium

position

x

m

SOLUTION

The complementary solution:

The particular solution:

Substitute Eqs. (2) and (3) into (1) yields:

The general solution is therefore

The constants Aand Bcan be found from the initial conditions.

:

+©Fx=max;F0cos vt–kx =mx

$

n

1231

22–42.

A block which has a mass m is suspended from a spring

having a stiffness k. If an impressed downward vertical force

F=FO

acts on the weight, determine the equation which

describes the position of the block as a function of time.

SOLUTION

my

$+ky +kyst –mg =F0

However, from equilbrium

kyst –mg =0

, therefore

my

$+ky =F0

The general solution of the above differential equation is of the form of

y=y

c

+y

p.

yP=C

y

Substitute Eqs. [2] and [3] into [1] yields :

Ans:

1232

22–43.

SOLUTION

The initial condition when ,, and

Thus,

v=v0

y=y0

t=0

A4-lb weight is attached to a spring having a stiffness

The weight is drawn downward a distance of

4 in. and released from rest. If the support moves with a

vertical displacement in., where tis in

seconds, determine the equation which describes the

position of the weight as a function of time.

d=10.5 sin 4t2

k=10 lb>ft.

is

1233

*22–44.

A 4-kg block is suspended from a spring that has a stiffness

of The block is drawn downward 50 mm from

the equilibrium position and released from rest when

If the support moves with an impressed displacement of

where tis in seconds, determine the

equation that describes the vertical motion of the block.

Assume positive displacement is downward.

d=110 sin 4t2mm,

t=0.

k=600 N

>

m.

SOLUTION

The general solution is defined by Eq. 22–23 with substituted for .

F

0

kd0

1234

22–45.

Use a

bl

oc

k

-an

d

-spr

i

ng mo

d

e

l

lik

e t

h

at s

h

own

i

n F

i

g. 22–14a,

but suspended from a vertical position and subjected to a

periodic support displacement determine the

equation of motion for the system, and obtain its general

solution. Define the displacement ymeasured from the

static equilibrium position of the block when t=0.

d=d0sin v0t,

SOLUTION

However, from equilibrium

, therefore

The general solution of the above differential equation is of the form of ,

where

Substitute Eqs. (2) and (3) into (1) yields:

The general solution is therefore

y=yc+yp

kyst –mg =0

+c©Fx=max;k(y–d0sin v0t+yst)–mg =-my

$

Ans:

1235

22–46.

SOLUTION

The general solution is defined by:

Since

Thus,

y=0.1 m when t=0,

A 5-kg block is suspended from a spring having a stiffness

of If the block is acted upon by a vertical force

where tis in seconds, determine the

equation which describes the motion of the block when it is

pulled down 100 mm from the equilibrium position and

released from rest at Assume that positive displacement

is downward.

t=0.

F=17 sin 8t2N,

300 N>m.

k300 N/m

Ans:

1236

22–47.

SOLUTION

Equation of Motion: When the rod rotates through a small angle , the springs

Since is small, and .Thus, this equation becomes

The particular solution of this differential equation is assumed to be in the form of

Substituting Eqs. (2) and (3) into Eq. (1),

–Cv2sin vt+3

2ag

L+k

m

≤

(Csin vt)=3FO

mL sin vt

cos u1sin u0u

u

The uniform rod has a mass of m. If it is acted upon by a

periodic force of , determine the amplitude of

the steady-state vibration.

F=F0sin vt

kk

L

2

L

2

A

1237

*22–48.

SOLUTION

Fr

ee-body Diagram: When the block is being displaced by amount xto the right, the

Equation of Motion:

Kinematics:

Since , then substituting this value into Eq. [1], we have

T

aking second time derivative and substituting into Eq. [2], we have

T

hus,

a=d2x

$

The 30-lb block is attached to two springs having a stiffness

of

A periodic force , where tis in

seconds

, is applied to the block. Determine the maximum

speed

of the block after frictional forces cause the free

vibrations to dampen out.

F=(8 cos 3t)lb10 lb>ft.

F8 cos 3t

k10 lb/ft

1238

22–49.

SOLUTION

k=F

¢y=18

0.014 =1285.71 N>m

The light elastic rod supports a 4-kg sphere.When an 18-N

vertical force is applied to the sphere, the rod deflects

14 mm. If the wall oscillates with harmonic frequency of

2 Hz and has an amplitude of 15 mm, determine the

amplitude of vibration for the sphere.

0.75 m

1239

22–50.

SOLUTION

Equation

of Motion: When the rod is in equilibrium, , and

. writing the moment equation of motion about point Bby

referring to the free-body diagram of the rod,

Fig. a,

T

hus, the initial stretch of the spring is .When the rod rotates about

Since

is small, .Thus, this equation becomes

T

hus,

cos u1u

sO=FA

k=mg

2k

u

$

=0

Fc=cy

#

c=0u=0°

Find the differential equation for small oscillations in terms

of

for the uniform rod of mass m. Also show that if

, then the system remains underdamped. The

rod is in a horizontal position when it is in equilibrium.

c62mk>2

u

A

B

a

C

c

k

2

u

a

22–51.

The 40-kg block is attached to a spring having a stiffness of

800 N>m

. A force

F=(100 cos 2t) N

, where t is in seconds

is applied to the block. Determine the maximum speed of

the block for the steady-state vibration.

k  800 N/m

SOLUTION

For the steady-state vibration, the displacement is

Thus

Ans:

1241

*22–52.

SOLUTION

Since ,

Substitute ypinto Eq. (1)

W=kdst

Use a block-and-spring model like that shown in

Fig. 22–14abut suspended from a vertical position and

subjected to a periodic support displacement of

determine the equation of motion for the

system, and obtain its general solution. Define the

displacement ymeasured from the static equilibrium

position of the block when t=0.

d=d0cos v0t,

1242

22–53.

The fan has a mass of 25 kg and is fixed to the end of a

horizontal beam that has a negligible mass.The fan blade is

mounted eccentrically on the shaft such that it is equivalent

to an unbalanced 3.5-kg mass located 100 mm from the axis

of rotation. If the static deflection of the beam is 50 mm as a

result of the weight of the fan, determine the angular

velocity of the fan blade at which resonance will occur. Hint:

See the first part of Example 22.8.

SOLUTION

V

Ans:

1243

22–54.

SOLUTION

The force caused by the unbalanced rotor is

Using Eq. 22–22, the amplitude is

k=F

¢y=25(9.81)

0.05 =4905 N>m

10 rad>s

V

In Prob. 22–53, determine the amplitude of steady-state

vibration of the fan if its angular velocity is .

Ans:

1244

22–55.

? Hint: See the first part of Example 22.8.18 rad>s

SOLUTION

Using Eq. 22–22, the amplitude is

k=F

¢y=25(9.81)

0.05 =4905 N>m

V

What will be the amplitude of steady-state vibration of the

fan in Prob. 22–53 if the angular velocity of the fan blade is

*22–56.

SOLUTION

Thus,

–15(x–0.1 cos 15t)(1.2) =4(0.6)2u

$

The small block at Ahas a mass of 4 kg and is mounted on

the bent rod having negligible mass. If the rotor at Bcauses

a harmonic movement , where tis in

seconds, determine the steady-state amplitude of vibration

of the block.

dB=(0.1 cos 15t)m

0.6 m

1.2 m

AO

V

22–57.

The electric motor turns an eccentric flywheel which is

equivalent to an unbalanced 0.25-lb weight located 10 in.

from the axis of rotation. If the static deflection of the

beam is 1 in. due to the weight of the motor,determine

the angular velocity of the flywheel at which resonance

will occur.The motor weights 150 lb.Neglect the mass of

the beam.

SOLUTION

V

Ans:

1247

22–58.

SOLUTION

The constant value FOof the periodic force is due to the centrifugal force of the

unbalanced mass

.

W

h

at w

ill

b

e t

h

e amp

li

tu

d

e of stea

d

y-state v

ib

rat

i

on of t

h

e

motor in Prob. 22–57 if the angular velocity of the flywheel

is ?20 rad>s

V

Ans:

1248

22–59.

Determine the angular velocity of the flywheel in Prob.22–

5

7

which will produce an amplitude of vibration of 0.25 in.

SOLUTION

The constant value F

Oof the periodic force is due to the centrifugal force of the

unbalanced mass.

From Eq. 22.21, the amplitude of the steady-state motion is

V

*22–60.

The 450-kg trailer is pulled with a constant speed over the

s

urface of a bumpy road, which may be approximated by a

cosine

curve having an amplitude of 50 mm and wave length of

4m

.If the two springs swhich support the trailer each have a

stiffness

of determine the speed which will cause

t

he greatest vibration (resonance) of the trailer.Neglect the

w

eight of the wheels.

v800 N>m,

SOLUTION

T

he wave length is

F

or maximum vibration of the trailer, resonance must occur, i.e.,

l=4m

v

100mm

2m 2m

s

Ans: