978-0073380308 Chapter 9 Solution Manual Part 11

Dynamics 2e 2019

Problem 9.59

Consider a simple viscously damped harmonic oscillator governed by

Eq. (9.45), and analyze the case in which the damping coefficient

c

is negative.

Calculate the general expression for the response (without taking into account

specific initial conditions), using

mD1kg

,

cD 1Ns=m

, and

kD10 N=m

.

Comment on the system’s response.

Solution

We are told that the equation of motion for this system is given by Eq. (9.45), which is given by

of McGraw-Hill, and must be surrendered upon request of McGraw-Hill. Any duplication or distribution, either in print or electronic form, without the

permission of McGraw-Hill, is prohibited.

Problem 9.60

The

MF

for a harmonically excited spring-mass-damper system at

!0=!n1

is equal to

5

. Calculate the damping ratio of the system.

What would the damping ratio be if the

MF

were equal to

10

? Sketch

the magnification factor at

!0=!n1

as a function of the damping

ratio.

Solution

The equation for the magnification factor for this system is given by Eq. (9.73) on p. 700, which is

of McGraw-Hill, and must be surrendered upon request of McGraw-Hill. Any duplication or distribution, either in print or electronic form, without the

permission of McGraw-Hill, is prohibited.

Dynamics 2e 2021

Problem 9.61

A slider moves in the horizontal plane under the action of the harmonic forcing

F .t/ DF0sin !0t

. The slider is connected to two identical linear springs, each

of which has constant

k

. When

tD0

,

x.0/ D0

, the springs are unstretched,

D45ı

, and

LDL0

. The slider is also connected to a damper with damping

coefficient

c

. Treating

F0

,

k

,

c

, and

L0

as known quantities, neglecting friction,

and letting

Px.0/ Dvi

, (a) derive the equations of motion of the system, (b) derive

the linearized equations of motion about the initial position, and (c) determine the

amplitude of the steady-state vibrations for the linearized equations of motion.

Solution

The FBD of the slider is shown on the far right, where

of McGraw-Hill, and must be surrendered upon request of McGraw-Hill. Any duplication or distribution, either in print or electronic form, without the

permission of McGraw-Hill, is prohibited.

To linearize the equation of motion, we only need to linearize the term containing

k

since it is the only term

of McGraw-Hill, and must be surrendered upon request of McGraw-Hill. Any duplication or distribution, either in print or electronic form, without the

permission of McGraw-Hill, is prohibited.

Dynamics 2e 2023

Problem 9.62

The mechanism shown is a pendulum consisting of a pendulum bob

B

with mass

m

and a T-bar, which is pinned at

O

and has negligi-

ble mass. The horizontal portion of the T-bar is connected to two

supports, each of which has an identical spring and dashpot system,

each with spring constant

k

and damping coefficient

c

. The springs

are unstretched when

B

is vertically aligned with the pin at

O

. Mod-

eling

B

as a particle, derive the linearized equations of motion of

the system. In addition, assuming that the system is underdamped,

derive the expression for the damped natural frequency of vibration

of the system.

of McGraw-Hill, and must be surrendered upon request of McGraw-Hill. Any duplication or distribution, either in print or electronic form, without the

permission of McGraw-Hill, is prohibited.

of McGraw-Hill, and must be surrendered upon request of McGraw-Hill. Any duplication or distribution, either in print or electronic form, without the

permission of McGraw-Hill, is prohibited.

Dynamics 2e 2025

Problem 9.63

The block of mass

m

is coupled to the support

A

, which is displacing

harmonically according to

yDYsin !0t

, by the linear elastic spring

with constant kand the dashpot with constant c.

(a)

Derive its equation of motion, using

x

as the independent variable,

and explain in what way the resulting equation of motion is not in

the form of Eq. (9.65).

(b)

Next, let

´Dxy

and substitute it into the equation of motion

found in Part (a). After doing so, show that you obtain an equation

of motion in ´that is of the same form as Eq. (9.65).

(c)

Find the steady-state solution to the equation of motion found in

Part (b) and then using that, determine the steady-state solution for

x.

of McGraw-Hill, and must be surrendered upon request of McGraw-Hill. Any duplication or distribution, either in print or electronic form, without the

permission of McGraw-Hill, is prohibited.

Part (b).

Letting ´Dxyand substituting it into the first form of the equation of motion from Part (a), we obtain

Part (c).

To determine the steady-state solution for

´

, that is to the equation of motion found in Part (b), we can either

of McGraw-Hill, and must be surrendered upon request of McGraw-Hill. Any duplication or distribution, either in print or electronic form, without the

permission of McGraw-Hill, is prohibited.

Dynamics 2e 2027

Problem 9.64

The block

A

and the platform

P

of a spring scale are at rest when

the lab bench to which the scale is rigidly attached begins vibrating

sinusoidally with a frequency of

15 Hz

and amplitude of

5mm

.

The block and the platform are coupled to the base

B

of the scale

by a linear elastic spring and a viscous damper that are internal

to the scale. The combined mass of the block and platform is

mAD1:5 kg

, the spring constant is

kD50 N=m

, and the viscous

damping coefficient is cD7:5 Ns=m.

Determine the vertical motion of the platform and block as a

function of time. The base of the scale

B

is rigidly attached to the

lab bench, and the block

A

does not separate from the platform

P

during the vibration. Hint: Parts (a)–(c) of Prob. 9.63 will be

helpful.

of McGraw-Hill, and must be surrendered upon request of McGraw-Hill. Any duplication or distribution, either in print or electronic form, without the

permission of McGraw-Hill, is prohibited.

where we have used the fact that

mAgDkıst

. This equation is not in the form of Eq. (9.65) on p. 699

since the forcing term on the right hand side is of the form

Rcos !0tCSsin !0t

and not just

Rcos !0t

or

where we have used

mAD1:5 kg

,

kD50 N=m

, and

cD7:5 Ns=m

to obtain the numerical result. Since

 < 1, the system is underdamped and the complementary solution is given by

Dm!2

0Esin !0t:

Since this expression must be true for all time, we can equate the coefficients of

sin !0t

and

cos !0t

to obtain

c!0

0E)DDmA!2

02ıc!0Cc!0DmAc!3

02Cc2!2

0



Since the system is linear, it turns out that it is actually easy to solve as the superposition of a particular solution for

Rcos !0t

of McGraw-Hill, and must be surrendered upon request of McGraw-Hill. Any duplication or distribution, either in print or electronic form, without the

permission of McGraw-Hill, is prohibited.