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Chapter 19, Solution 2.

Eq. 19.15: 2m m n m m nv x a xω ω= =

Given data 20.2 m/s 4 m/sm mv a= =

: 0.2 m/sm m n m mv x xω ω= = (1)

2 2 2: 4 m/sm m n m ma x xω ω= = (2)

Divide Equ. (2) by Equ. (1): 24 m/s

20 rad/s0.2 m/snω = =

Eq. (1): ( )0.2 m/s 20 rad/smx=

0.01 mmx = 10 mmmx = !

Frequency 20 rad/s

2 2n

nfωπ π

= = 3.18 Hznf = !

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Chapter 19, Solution 3.

cycle 2 rad

sin , 6s cyclem n nx x t

πω ω= = ×

sin 12mx x tπ=

12 cos12mx x tπ π=&

2144 sin12mx x tπ π= −&&

12 4 ft/smxπ =

4

0.1061 ft12mx

π= =

1.273 in.mx = !

( )2 2Max Acc. 144 0.1061 150.8 ft/sπ= = !

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Chapter 19, Solution 4.

Simple Harmonic Motion

20 lb

0.2222 in.90 lb/in.s

W

kδ = = =

( )( )90 12

41.699 rad/s 220

32.2

nk

fm

ω π= = = =

(a) Amplitude 0.222 in.s mxδ= = =

0.222 in.mx = !

41.699 rad/s

6.63662

fπ

= =

6.64 Hzf = !

(b) ( )( )41.699 rad/s 0.2222 in. 9.2655 in./sm n mv xω= = =

9.27 in./smv = !

( ) ( )22 241.699 rad/s 0.2222 in. 386.36 in./sm n ma xω= = =

232.197 ft/s=

232.2 ft/sma = !

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Chapter 19, Solution 5.

Simple Harmonic Motion

(a) ( )sinm nx x tω φ= +

( )( )2

9000 lb/ft

70 lb 32.2 lb/sn

k

mω = =

64.343 rad/s=

2

0.90765 snn

πτω

= =

0.0977 snτ = !

1

10.240 Hznn

fτ

= =

10.24 Hznf = !

(b) At 0 0 00: 0, 10 ft/st x x v= = = =&

( )( )0 0 sin 0 0m nx x ω φ φ= = + ⇒ =

( )( )0 0 cos 0m n n m nx v x xω ω φ ω= = + =&

Substituting 10 ft/s 64.343 rad/smx=

or 0.1554 ft 1.865 in.mx = =

1.865 in.mx = !

( )( )22 0.15542 ft 64.343 rad/sm m na x ω= =

2643.4 ft/s=

2643 ft/sma = !

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Vector Mechanics for Engineers: Statics and Dynamics, 8/e, Ferdinand P. Beer, E. Russell Johnston, Jr.,

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Chapter 19, Solution 6.

In Simple Harmonic Motion

(a) 2

m m na x ω=

Substituting ( )2 250 m/s 0.058 m

nω=

or ( )22862.07 rad/s

nω =

29.361 rad/sn

ω =

Now 29.361 rad/s

4.6729 Hz2 2

n

nf

ωπ π

= = =

Then ( )( )1 cycle 1

in Hz Hz1 min 60 s/min 60

f = =

So ( ) 11

6060

Hz 4.6729 Hz280.37 r/min

Hz

f = =

and 280 rpm�

(b) ( )( )0.058 m 29.361 rad/s 1.7029 m/sm m nv x ω= = =

1.703 m/smv = �

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Chapter 19, Solution 7.

Simple Harmonic Motion

(a) ( )sinm ntθ θ ω φ= +

( )2 2

1.35 snn

π πωτ

= =

4.833 rad/s=

( )cosm n ntθ θ ω ω φ= +&

m m nθ θ ω=&

m m m nv l lθ θ ω= =&

Thus, mm

n

v

lθ

ω= (1)

For a simple pendulum

ng

lω =

Thus,

( )2

2 2

9.81 m/s

4.833 rad/sn

gl

ω= =

0.420 m=

From (1)

( )( )0.4 m/s

0.42 m 4.833 rad/sm

mn

v

lθ

ω= =

0.197 rad=

or 11.287°

11.29mθ = °!

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(b) Now

ta lθ= &&

Hence, the maximum tangential acceleration occurs when θ&& is

maximum.

( )2 sinm n ntθ θ ω ω φ= − +&&

2m m nθ θ ω=&&

( ) 2t m nm

a lθ ω=

or ( ) ( )( )( )20.42 m 0.197 rad 4.833 rad/st m

a =

21.9326 m/s=

( ) 21.933 m/st ma = !

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Vector Mechanics for Engineers: Statics and Dynamics, 8/e, Ferdinand P. Beer, E. Russell Johnston, Jr., Elliot R. Eisenberg, William E. Clausen, David Mazurek, Phillip J. Cornwell © 2007 The McGraw-Hill Companies.

Chapter 19, Solution 8.

Simple Harmonic Motion:

2

60 lb/ft6.2161 rad/s

50 lb

32.2 ft/s

nk

mω = = =

6.2161

Hz2 2

nnf

ωπ π

= =

(a) 2.4 in. 0.2 ftmx = = 0.2 ftmx = !

0.989 Hznf = !

( )( )22 0.2 ft 6.2161 rad/sm m na x ω= =

27.728 ft/s=

(b) f m sF ma mgµ= =

or 2

2

7.728 ft/s0.240

32.2 ft/sm

sa

gµ = = = !

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Chapter 19, Solution 9.

6 lb/in. 72 lb/ft, 55 in./s, 4 lb.mk v W= = = =

: 0k

F ma kx mx x xm

= − = + =&& &&

Thus: 2 72579.6 24.025 rad/s

432.2

k

mω ω= = = =

Eq. (19.15): m mv x ω=

( )55 in./s 24.025 rad/smx=

2.2845 in.mx = 2.28 in.mx = !

( )( )2 2 22.2845 in. 579.6 rad /sm ma x ω= =

21324.1 in./sma = 2110.3 ft/sma = !

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Chapter 19, Solution 10.

( )60cos 10 45sin 103

x t tππ π = + −

( )60cos 10 45 sin10 cos cos10 sin3 3

t t tπ ππ π π = + −

22.5sin10 21.02886cos10t tπ π= + (1)

Now

sin(10 ) sin10 cos cos10 sinm m mx t x t x tπ φ π φ π φ+ = + (2)

Comparing (1) and (2) gives

22.5 cos , 21.02866 sinm mx xφ φ= =

(a) 2 2

0.2 s10n

n

π πτω π

= = = !

(b) 2 2 2(22.5) (21.02866)mx = +

30.8 mmmx = !

(c) 21.02866

tan22.5

φ = 0.7516 rad 43.1φ = = °!

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Chapter 19, Solution 11.

At both 600 rpm and 1200 rpm, the maximum acceleration is just equal to g.

(a) 600 rpm 62.832 rad/sω = =

Eq. (19.15): ( )

2

262.832

m m m

ga x xω= =

SI: ( )

3

2

9.812.4849 10 m

62.832mx

−= = × 2.48 mmmx = �

US: ( )2

32.20.008156 ft

62.832mx = = 0.0979 in.

mx = �

(b) 1200 rpm 125.664 rad/sω = =

Eq. (19.15): ( )

2

2125.664

m m m

ga x xω= =

SI: ( )

6

2

9.81621.2 10 m

125.664mx

−= = × 0.621 mmmx = �

US: ( )2

32.20.002039 ft

125.664mx = = 0.0245 in.

mx = �

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Chapter 19, Solution 12.

Simple Harmonic Motion, thus

( )sinm nx x tω φ= +

400 N/m16.903 rad/s

1.4 kgnk

mω = = =

Now (0) 0 sin(0 ) 0mx x φ φ= = + ⇒ =

Then

(0) cos(0 0)m nx x ω= +&

or 2.5 m/s ( )m(16.903 rad/s) 0.14790 mm mx x= ⇒ =

Then ( ) ( )0.14790 m sin 16.903 rad/sx t =

(a)

At ( )0.06 m: 0.06 m (0.14790 m)sin 16.903 rad/sx t = =

or

1 0.06 msin

0.14790 m0.02471 s

16.903 rad/s

− = =t

0.0247 st = !

(b) Now

( )cosm n nx x tω ω=&

( )2 sinm n nx x tω ω= −&&

Then, for 0.024713 st =

( )( ) ( )( )0.1479 m 16.903 rad/s cos 16.903 rad/s 0.024713 sx = &

2.285 m/s= 2.29 m/sx =& !

And

( )( ) ( )( )20.1479 m 16.903 rad/s sin 16.903 rad/s 0.024713 sx = − &&

217.143 m/s= − 217.14 m/sx =&& !

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Vector Mechanics for Engineers: Statics and Dynamics, 8/e, Ferdinand P. Beer, E. Russell Johnston, Jr., Elliot R. Eisenberg, William E. Clausen, David Mazurek, Phillip J. Cornwell © 2007 The McGraw-Hill Companies.

Chapter 19, Solution 13.

Referring to the figure of Problem 19.12

( )sinm nx x tω φ= +

( )cosm n nx x tω ω φ= +&

( )2 sinm n nx x tω ω φ= − +&&

Using the data from Problem 19.13: 0, 0.1479 m, 16.903 rad/sm nxφ ω= = =

, , arex x x& &&

And ( ) ( )0.14790 m sin 16.903 rad/sx t =

So, at 0.9 s,t =

( ) ( )( )0.1479 m sin 16.903 rad/s 0.9 sx =

0.0703 m= 70.3 mmx = !

( )( ) ( )( )0.1479 m 16.903 rad/s cos 16.903 rad/s 0.9 sx = &

2.19957 m/s= − 2.20 m/sx =& !

( )( ) ( )( )20.1479 m 16.903 rad/s sin 16.903 rad/s 0.9 sx = &&

220.083 m/s= − 220.1 m/sx =&& !

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Chapter 19, Solution 14.

(a)

sin( )m nx x tω φ= +

( )( )2

9000 lb/ft70 lb

32.2 lb/s

ω = =nk

m

64.343 rad/s=

20.9765 sn

n

πτω

= =

With the initial conditions: (0) 15 in. 1.25 ft, (0) 0x x= = =&

( )1.25 ft sin 0mx φ= +

(0) 0 cos(0 )2m nx xπω φ φ= = + ⇒ =&

1.25 ftmx =

Then

( ) (1.25 ft)sin 64.3432

x t tπ = +

(1.5) (1.25 ft)sin 64.343(1.5 s) 0.80137 ft2

xπ = + = −

( )(1.5) (1.25 ft)(64.343)cos 64.343 1.5 s 61.726 ft/s2

xπ = + = −

&

In 1.5 s, the block completes

1.5 s15.361 cycles

0.09765 s/cycle=

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So, in one cycle, the block travels

4(1.25 ft) 5 ft=

Fifteen cycles take

15(0.09765 s/cycle) 1.46477 s=

Thus, the total distance traveled is

15(5 ft) 1.25 ft (1.25 0.80137)ft 77.1 ft+ + − =

Total 77.1 ft= !

(b)

2(1.5) (1.25 ft)(64.343 rad/s) sin (64.343 rad/s)(1.5 s)2

xπ = − +

&&

23317.68 ft/s= 23320 ft/sx =&& !

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Chapter 19, Solution 15.

2

210 lb 0.31056 lb s /ft

32.2 ft/sm = = ⋅

With the given properties:

250 lb/ft 12.6886 rad/s

0.31056 lb s /ftnkm

ω = = =⋅

From free fall of the collar

( )0 2 2 1.5 ft 3 9.82853 ft/sv gh g g= = = =

The free-fall time is thus:

( )1

2 1.5 ft2 3 0.30523 sytg g g

= = = =

Now to simplify the analysis we measure the displacement from the position of static displacement of the spring, under the weight of the collar: Note that the static deflection is:

10 lb 0.2 ft50 lb/ftst

Wk

δ = = =

Then 0,mx kx+ = where x is measured positively up from the position of static deflection. The solution is:

( )sin ,m nx x tω φ= + with velocity

( )cosm n nx x tω ω φ= +

Now to determine and ,mx φ impose the conditions at impact and count the time from there. Thus: At impact:

0, 0.2 ft and 9.82853 ft/sstt x vδ= = = = − (down)

or 0.2 ft sinmx φ=

( )9.82853 ft/s 12.6886 rad/s cosmx φ− =

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Solving for and mx φ 0.800 ftmx = −

0.25268 rad φ = −

So, from time of impact, the ‘time of flight’ is the time necessary for the collar to come to rest on its downward motion. Thus, is the time such 2tthat

( )2 20 12.68862

x t t πφ= ⇒ + =

or 212.6886 0.252682

t π− =

Hence, 2 0.14371 st =(a) Thus, the period of the motion is

( ) ( )1 22 2 0.30523 s 0.14371 st tτ = + = +

0.89788 s= 0.898 sτ =

(b) After 0.4 seconds, the velocity is

( ) ( )10.4 cos 0.4m n nx x tω ω φ⎡ ⎤= − +⎣ ⎦

( )( ) ( )( )12.6886 rad/s 0.8 ft cos 12.6886 rad/s 0.4 0.30523 s⎡= − −⎣

]0.25268 rad−

5.91 ft/s= − ( )0.4 5.91 ft/sv =

Vector Mechanics for Engineers: Statics and Dynamics, 8/e, Ferdinand P. Beer, E. Russell Johnston, Jr., Elliot R. Eisenberg, William E. Clausen, David Mazurek, Phillip J. Cornwell © 2007 The McGraw-Hill Companies.

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Chapter 19, Solution 16.

sin , cos ,m n m n n ng

t tl

θ θ ω θ θ ω ω ω= = =&

9.81

2.8592 rad/s,1.2nω = = 0.18

0:1.2 m nt θ θ ω= = =&

0.052462 radiansmθ∴ =

At 1.5 s,t = ( )( )0.052462 sin 2.8592 1.5θ =

(a) 0.047826θ = − radians 2.74= − ° !

(b) ( )( ) ( )( )1.2 0.052462 2.8592 cos 2.8592 1.5v =

74.0 mm/sv = !

( )( ) ( )( )21.2 0.052462 2.8592 sin 2.8592 1.5a =

2469 mm/sa = !

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Vector Mechanics for Engineers: Statics and Dynamics, 8/e, Ferdinand P. Beer, E. Russell Johnston, Jr., Elliot R. Eisenberg, William E. Clausen, David Mazurek, Phillip J. Cornwell © 2007 The McGraw-Hill Companies.

Chapter 19, Solution 17.

(a)

( )sinm nx x tω φ= +

( )0 sin 0 0.75 ftmx x φ= + =

( )0 0 cos 0 ,2m nx xπω φ φ= = + ⇒ =&

0.75 ftmx∴ =

When the collar just leaves the spring, its acceleration is g (downward) and 0.v = Now

( )0.75 ft cos2n nx tπω ω = +

&

( ) ( )0 0 0.75 ft cos ,2 2 2n n nx v t tπ π πω ω ω = = = + ⇒ + =

&

And

( ) 20.75 ft sin2n na g tπω ω = − = − +

or ( ) 20.75 ft ng ω− = −

232.2 ft/s6.5524 rad/s

0.75 ftnω = =

Then

( )222

10 lb, 6.5524 rad/s

32.2 ft/sn n

kk m

mω ω= ⇒ = =

13.333 lb/ft=

13.33 lb/ftk = !

(b) 6.5524 rad/snω =

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At 1.6 s:t =

( ) ( )( )0.75 ft sin 6.5524 rad/s 1.6 s 0.36727 ft2

xπ = + = −

0.367 ftx = − above equilibrium !

( )( ) ( )( )0.75 ft 6.5524 rad/s cos 6.5524 rad/s 1.6 s 4.2848 ft/s2

v xπ = = + =

&

4.28 ft/sv = !

( )( ) ( )( )2 20.75 ft 6.5524 rad/s sin 6.5524 rad/s 1.6 s 15.768 ft/s2

a xπ = = − + =

&&

215.77 ft/sa = !

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Chapter 19, Solution 18.

Determine the constant k of a single spring equivalent to the three springs shown.

Springs 1 and 2:

1 1 1

1 2

1 2

, andP P P

k k kδ δ δ= + = +

′

Hence

1 2

1 2

k kk

k k′ =

+

Where k′ is the spring constant of a single spring equivalent of springs 1 and 2.

Springs k′ and 3 Deflection in each spring is the same

So 1 2 1 2 3

, and , ,P P P P k P k P kδ δ δ′= + = = =

Now 3

k k kδ δ δ′= +

1 2

3 3

1 2

k kk k k k

k k′= + = +

+

or ( )( )

( ) ( )3

3.5 kN/m 2.1 kN/m2.8 kN/m 4.11 kN/m 4.11 10 N/m

3.5 kN/m 2.1 kN/mk = + = = ×

+

(a) 3

2 20.361 s

4.11 10 N/m

13.6 kg

n

k

m

π πτ = = =×

�

1 1

2.77 Hz0.3614 s

n

n

fτ

= = = �

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(b) ( )sinm n

x x tω φ= +

And 44 mm 0.044 mmx = =

( )( )2 2 2.77 Hz 17.384 rad/sn n

fω π π= = =

And

( ) ( )0.044 m sin 17.4 rad/sx t φ = +

( )( ) ( )0.044 m 17.4 rad/s cos 17.4 rad/sx t φ = + &

Then ( )( )max

0.044 m 17.4 rad/s 0.766 m/sv = =

( )( ) ( )20.044 m 17.4 rad/s sin 17.4 rad/sx t φ = − + &&

Then ( )( )2 2

max0.044 m 17.4 rad/s 13.3 m/sa = =

max

0.765 m/sv = �

2

max13.31 m/sa = �

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Vector Mechanics for Engineers: Statics and Dynamics, 8/e, Ferdinand P. Beer, E. Russell Johnston, Jr., Elliot R. Eisenberg, William E. Clausen, David Mazurek, Phillip J. Cornwell © 2007 The McGraw-Hill Companies.

Chapter 19, Solution 19.

(a) First, calculate the spring constant

( ) ( ) ( ) ( )24 kN/m 12 kN/m 12 kN/m 48 kN/mP δ δ δ δ= + + =

48 kN/mk∴ =

Then

340 10 N/m30.984 rad/s

50 kgnk

mω ×= = =

2

0.20279 snn

πτω

= =

14.9312 Hzn

n

fτ

= =

0.203 snτ = !

4.93 Hznf = !

(b) Now

( )sinm nx x tω φ= +

And, since 0 0.060 mx =

( ) ( )0.060 m sin 30.984 rad/sx t φ = +

( )( ) ( )0.060 m 30.984 rad/s cos 30.984 rad/sx t φ = + &

( )( ) ( )20.060 m 30.984 rad/s sin 30.984 rad/sx t φ = − + &&

Hence max 1.859 m/sv = !

2max 57.6 m/sa = !

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Chapter 19, Solution 20.

Equivalent spring constant

2 2 4k k k k′ = + =

(Deflection of each spring is the same.)

( )1

2

26.8 s

10 lb

32.2 ft/s

πτ = =n

k

22

10 6.832.2

k π ⇒ =

0.2651 lb/ftk =

( )2

2

2

4

6 lb

32.2 ft/s

πτ =n

k

( )( )22

4 0.2651 lb/ft 32.2 ft/s

6 lb

π=

2.633 s=

2.63 sn

τ = �

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© 2007 The McGraw-Hill Companies.

Chapter 19, Solution 21.

Equivalent Springs

Series: 1 2

1 2

s

k kk

k k=

+

Parallel: 1 2p

k k k= +

2 2 2 2;

s p

s ps pk k

m m

π π π πτ τω ω

= = = =

( )( )

( )2 22

1 21 2

1 2 1 2

1 2

5

2

ττ

++ = = = = +

ps

p s

k k kk k

k kk k k

k k

( )( ) 2 2

1 2 1 1 2 26.25 2k k k k k k= + +

( ) ( )2 2 2

2 2 2

1

4.25 4.25 4

2

k k kk

−=

m

1

2

2.125 3.516k

k= m

1

2

14 or

4

k

k= �

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Vector Mechanics for Engineers: Statics and Dynamics, 8/e, Ferdinand P. Beer, E. Russell Johnston, Jr., Elliot R. Eisenberg, William E. Clausen, David Mazurek, Phillip J. Cornwell © 2007 The McGraw-Hill Companies.

Chapter 19, Solution 22.

For a static load P the total elongation of the spring is:

1 2 3

P P P

k k kδ = + +

1 1 1

8 kN/m 12 kN/m 16 kN/mP

= + +

6 4 3 13

48 48δ + + = =

P P

48

3.6923 kN/m13

Pk

δ= = =

33.6923 10 N/m( ) 12.153 rad/s

25 kg

ka

mω ×= = =

2 2

0.517 s12.153

π πτω

= = = 0.517 sτ = !

12.153

1.9342 Hz2 2

fωπ π

= = = 1.934 Hzf = !

(b) For 30 mm = 0.03 mmx =

( )( )0.03 m 12.153 rad/sm mv x ω= = 0.365 m/smv = !

( )( )22 0.03 m 12.153 rad/sm ma x ω= = 24.43 m/sma = !

COSMOS: Complete Online Solutions Manual Organization System

Vector Mechanics for Engineers: Statics and Dynamics, 8/e, Ferdinand P. Beer, E. Russell Johnston, Jr., Elliot R. Eisenberg, William E. Clausen, David Mazurek, Phillip J. Cornwell © 2007 The McGraw-Hill Companies.

Chapter 19, Solution 23.

For equilibrium

Equal stretch 12 N 16 N

,A C Bk kδ δ δ= = =

12 N ,A CT T kx= = + 16 NBT kx= +

( ) ( ) ( )4.08 9.81 2 12 16mx kx kx= × − + − +&&

3 0mx kx+ =&&

( ) 30.0125 cos

kx t

m=

(a) Then ( )( )16 0.0125 1 0BT k= + − =

1.280 kN/mk = !

(b) ( )

( )2 2 23 12803

941.76 rad /s40 9.81n

k

mω = = =

30.688 rad/s 4.88 Hzn fω = ⇒ = !

(c) ( ) 312 1280 0.0125 cosA

kT t

m= +

3

Minimum when cos 1k

tm

= = −

( ) ( )min12 1280 0.0125 4AT = − = −

Max Comp: 4 N !

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Vector Mechanics for Engineers: Statics and Dynamics, 8/e, Ferdinand P. Beer, E. Russell Johnston, Jr., Elliot R. Eisenberg, William E. Clausen, David Mazurek, Phillip J. Cornwell © 2007 The McGraw-Hill Companies.

Chapter 19, Solution 24.

Springs in parallel, eq 1 22k k k= +

12 16,000 2 210

0.2nk

m

π πω πτ

+= = = =

After removal

12 28

0.25nk

m

πω π= = =

Elimination: 2 12 16,000100

k

mπ +=

2 1264

k

mπ =

(a) 2 2 16,000100 64 ,

mπ π∴ = × 45.0 kgm = !

(b) 2 11

264 , 14,222 N m

45.0316

kkπ = =

1 14.22 kN/mk = !

COSMOS: Complete Online Solutions Manual Organization System

Vector Mechanics for Engineers: Statics and Dynamics, 8/e, Ferdinand P. Beer, E. Russell Johnston, Jr.,

Elliot R. Eisenberg, William E. Clausen, David Mazurek, Phillip J. Cornwell

© 2007 The McGraw-Hill Companies.

Chapter 19, Solution 25.

The equivalent spring constant for springs in series is:

1 2

1 2

e

k kk

k k=

+

1 2For andk k

( )1 2

1 2

2 2π πτ = =

+e

A A

k k k

m m k k

1For alonek

1

2πτ′ =

A

k

m

(a)

2

2 1 2k k k

ττ

= + ′

0.2 s0.16667

0.12 s

ττ

= =′

And with 2

3.5 kN/mk =

( )( )2 13.5 kN/m 1.6667 3.5 kN/mk= +

or 1

6.22 kN/mk = �

(b) 1

2πτ′ =

A

k

m

A

A

Wm

g=

( )( )

2

1 1

22

A

km

τπ

=

And with 3

16.22 kN/m 6.22 10 N/mk = = ×

( ) ( )( )

2 3

2

2

0.12 s 6.22 10 N/m2.2688 Ns /m 2.2688 kg

2A

m

π

×= = =

2.27 kgA

m = �

COSMOS: Complete Online Solutions Manual Organization System

Vector Mechanics for Engineers: Statics and Dynamics, 8/e, Ferdinand P. Beer, E. Russell Johnston, Jr., Elliot R. Eisenberg, William E. Clausen, David Mazurek, Phillip J. Cornwell © 2007 The McGraw-Hill Companies.

Chapter 19, Solution 26.

The equivalent spring constant, accounting for the springs in series, is

eq 2

kk k= +

( )( ) ( )( )eq

22

2 20.4 s

3 /2

100 lb 32.2 ft/s100 lb 32.2 ft/s

k k

π πτ = = =

It follows that 510.849 lb/ftk =

511 lb/ftk = !

After the changes

eq

20.4 s

120/32.2

k

πτ = =

Thus eq 919.528 lb/ftk =

Since

eqA

A

kkk k

k k= +

+

or

( ) ( )( )

eq

eq

510.849 919.528 510.849

2 2 510.849 919.528A

k k kk

k k

− −= =

− −

2043.4 lb/ft=

2040 lb/ftAk = !

COSMOS: Complete Online Solutions Manual Organization System

Vector Mechanics for Engineers: Statics and Dynamics, 8/e, Ferdinand P. Beer, E. Russell Johnston, Jr., Elliot R. Eisenberg, William E. Clausen, David Mazurek, Phillip J. Cornwell © 2007 The McGraw-Hill Companies.

Chapter 19, Solution 27.

Initially 2

1.6 s

A

km

πτ = =

After the 14 lb weight is added to A,

22.1 s

A B

km m

πτ′ = =

+

(a) A B

A

m m

m

ττ′ +=

22.1

1.6A B

A

m m

m

+ =

( )( )1.7227 A A Bm m m= + Note 2

14 lb

32.2 ft/sBm =

2

14 lb1.3838 1.3838

32.2 ft/sA Bm m

= =

( )22

14 lb32.2 ft/s 1.3838 19.373 lb

32.2 ft/sA AW m

= = =

19.37 lbAW = !

(b) 2

A

km

πτ =

( ) ( )( )

222 Am

k πτ

=

( )( )

22

2

19.373 lb

32.2 ft/s2

1.6 sk π

=

9.278 lb/ftk =

9.28 lb/ftk = !

COSMOS: Complete Online Solutions Manual Organization System

Vector Mechanics for Engineers: Statics and Dynamics, 8/e, Ferdinand P. Beer, E. Russell Johnston, Jr.,

Elliot R. Eisenberg, William E. Clausen, David Mazurek, Phillip J. Cornwell

© 2007 The McGraw-Hill Companies.

Chapter 19, Solution 28.

, 2 2 2s s

d dx x F kx k x

h h= = =

( ) ( )eff: 2

A AM M Fd mgx mx hΣ = Σ − = − &&

2d

k x d mg x mxhh

− = −

&&

2

2

20

kd gx x

hmh

+ − =

&&

22

2 2

2

2 2,

kd g k d g

h m h hmhω ω

= − = −

(1)

Data: 0.3 m; 0.7 m; 20 kgd h m= = =

(a) For 1 s:τ = 2

;πτ

ω= 2 22

1 s = ; 4π ω π

ω=

Eq. (1):

2

2 2 0.3 9.814

20 0.7 0.7

kπ = −

2912.4 N/mk = 2.91 kN/mk = �

(b) For Infinite:τ = 2

;πτ

ω= 0ω =

Eq. (1):

2

2 2 0.3 9.810

20 0.7 0.7

kω = = −

763.0 N/mk = 763 N/mk = �

COSMOS: Complete Online Solutions Manual Organization System

Vector Mechanics for Engineers: Statics and Dynamics, 8/e, Ferdinand P. Beer, E. Russell Johnston, Jr.,

Elliot R. Eisenberg, William E. Clausen, David Mazurek, Phillip J. Cornwell

© 2007 The McGraw-Hill Companies.

Chapter 19, Solution 29.

(a) The equation of motion is:

( )( )( ) ( )( ): 0.75 0.75 2 1.2 m 1.2A

M kθ θΣ − = &&

1.125 1.44k mθ θ− = &&

1.44 1.125 0m kθ θ+ =&&

( )31.125 1.35 10

01.44m

θ θ×

+ =&&

1054.690

m

θ θ+ =&&

Then 2 1054.69 2, and 4

n n

n

s

m

πω τω

= = =

22 1054.69 1054.69 2

or 4 4m m

π π = =

427.45 kgm =

427 kgm = �

(b)

( )sinm n

tθ θ ω φ= +

At 0t =

0.060.03636 rad

1.65

m

m

θ = =

COSMOS: Complete Online Solutions Manual Organization System

Vector Mechanics for Engineers: Statics and Dynamics, 8/e, Ferdinand P. Beer, E. Russell Johnston, Jr.,

Elliot R. Eisenberg, William E. Clausen, David Mazurek, Phillip J. Cornwell

© 2007 The McGraw-Hill Companies.

And 0θ =&&

Then 0.03636 sinm

θ φ=

( )0 cosm n n

tθ ω ω φ= +

Thus , 0.0363 rad2

m

πφ θ= =

Now

0.060.03636 rad

1.65

m

m

θ = =

( )cosm n n

tθ θ ω ω φ= +&

1054.691.5708 rad/s

427.45n

ω = =

Then

0.060.03636 rad

1.65

m

m

θ = =

( )( ) ( )0.03636 rad 1.5708 rad/s cos 1.5708 rad/s2

tπθ = +

&

( )( )max

0.03636 rad 1.5708 rad/s 0.05712 rad/sθ = =&

( ) ( )( )max

1.2 0.05712 rad/s 0.06854 m/scv m= =

( )max

68.5 mm/scv = �

COSMOS: Complete Online Solutions Manual Organization System

Vector Mechanics for Engineers: Statics and Dynamics, 8/e, Ferdinand P. Beer, E. Russell Johnston, Jr.,

Elliot R. Eisenberg, William E. Clausen, David Mazurek, Phillip J. Cornwell

© 2007 The McGraw-Hill Companies.

Chapter 19, Solution 30.

(a) ( )( )( )

( )

6 2 2 2 3 4

3 3

48 30 10 lb/in. 144 in. /ft 2 10 ft48

15 ft

P EIk

Lδ

−× ×= = =

122880 lb/ft= 51.229 10 lb/ftk = × �

(b) ( )2122880 lb/ft

51.360 rad/s

1500 lb 32.2 ft/sn

k

mω = = =

or 1 51.36

8.17 Hz2 2 2

n

n

kf

m

ωπ π π

= = = = �

COSMOS: Complete Online Solutions Manual Organization System

Vector Mechanics for Engineers: Statics and Dynamics, 8/e, Ferdinand P. Beer, E. Russell Johnston, Jr., Elliot R. Eisenberg, William E. Clausen, David Mazurek, Phillip J. Cornwell © 2007 The McGraw-Hill Companies.

Chapter 19, Solution 31.

(a) , , ePL AE

P k PAE L

δ δ δ = = =

( )26

3

0.32 in.29 10 psi

4129.57 10 lb/in.

18 in.eAE

kL

π × = = = ×

3129.6 10 lb/in.ek = × !

(b)

( )( )

( )

3

2

129.57 10 lb/in.

16 lb

32.2 ft/s81.272 Hz

2 2

e

n

kmfπ π

×

= = =

81.3 Hznf = !

COSMOS: Complete Online Solutions Manual Organization System

Vector Mechanics for Engineers: Statics and Dynamics, 8/e, Ferdinand P. Beer, E. Russell Johnston, Jr., Elliot R. Eisenberg, William E. Clausen, David Mazurek, Phillip J. Cornwell © 2007 The McGraw-Hill Companies.

Chapter 19, Solution 32.

st

Wk

δ=

Wm

g=

2 st

W

n Wstg

k g

mδω

δ= = =

1

2 2n

nst

gF

ωπ π δ

= = !

COSMOS: Complete Online Solutions Manual Organization System

Vector Mechanics for Engineers: Statics and Dynamics, 8/e, Ferdinand P. Beer, E. Russell Johnston, Jr., Elliot R. Eisenberg, William E. Clausen, David Mazurek, Phillip J. Cornwell © 2007 The McGraw-Hill Companies.

Chapter 19, Solution 33.

(a) ( )( )20.10 kg 9.81 m/s 0.981 Nmg − =

12

2

000.981 N

44

F mg x x = = ⇒ =

0.06015 m=

0 60.1 mmx = !

(b) At 0, x ( )1 12 2

0

04

2 0.060152x

dFx

dx

− − = =

8.1549 N/m=

0

8.1549 N/mex

dFk

dx = =

8.1549 N/m0.10 kg

1.4372 Hz2 2π π

= = =e

n

kmF

1.437 HznF = !

COSMOS: Complete Online Solutions Manual Organization System

Vector Mechanics for Engineers: Statics and Dynamics, 8/e, Ferdinand P. Beer, E. Russell Johnston, Jr., Elliot R. Eisenberg, William E. Clausen, David Mazurek, Phillip J. Cornwell © 2007 The McGraw-Hill Companies.

Chapter 19, Solution 34.

Using the Binomial Theorem we write

12

2

2 2

11 sin sin

21 sin sin

2

m

m

θ φθ φ

− = −

−

2 211 sin sin

2 2mθ φ= + + ⋅ ⋅ ⋅ ⋅

Neglecting terms of order higher than 2 and setting ( )2 1sin 1 cos 2 ,

2φ φ= − we have

( )2 20

1 14 1 sin 1 cos 2

2 2 2m

nl

dg

π θτ φ φ = + − ∫

2 2 20

1 14 1 sin sin cos 2

4 2 4 2m ml

dg

π θ θ φ φ = + −

∫

2

2 2

0

1 14 sin sin sin 2

4 2 8 2m ml

g

πθ θφ φ φ = + −

214 sin 0

2 4 2 2ml

g

π θ π = + +

212 1 sin

4 2m

nl

g

θτ π = +

!

COSMOS: Complete Online Solutions Manual Organization System

Vector Mechanics for Engineers: Statics and Dynamics, 8/e, Ferdinand P. Beer, E. Russell Johnston, Jr., Elliot R. Eisenberg, William E. Clausen, David Mazurek, Phillip J. Cornwell © 2007 The McGraw-Hill Companies.

Chapter 19, Solution 35.

For small oscillations: 0 2l

gτ π=

We want ( )01.01 1.01 2l

gτ τ π= =

Using the formula of Prob. 19.34, we write

20 0

11 sin 1.01

4 2mθτ τ τ = + =

( )2sin 4 1.01 1 0.04; sin 0.2; 11.542 2 2m m mθ θ θ= − = = = °

23.1mθ = °!

COSMOS: Complete Online Solutions Manual Organization System

Vector Mechanics for Engineers: Statics and Dynamics, 8/e, Ferdinand P. Beer, E. Russell Johnston, Jr., Elliot R. Eisenberg, William E. Clausen, David Mazurek, Phillip J. Cornwell © 2007 The McGraw-Hill Companies.

Chapter 19, Solution 36.

Eq. (19.20) 2

2k l

gτ π

π

=

For 0.8 m:l =

2

0.8 m2 2 1.7943 s

9.81 m/s

l

gπ π= =

Thus: ( )21.7943 s

kτπ

=

Using Table 19.1:

(a) For small oscillations: 2

0; 1mkθ

π= =

1.7943 sτ = !

(b) For 2

30 : 1.017mkθ

π= ° =

( )( )1.017 1.7943 sτ = 1.825 sτ = !

(c) For 2

90 : 1.180mkθ

π= ° =

( )( )1.180 1.7943 sτ = 2.12 sτ = !

COSMOS: Complete Online Solutions Manual Organization System

Vector Mechanics for Engineers: Statics and Dynamics, 8/e, Ferdinand P. Beer, E. Russell Johnston, Jr., Elliot R. Eisenberg, William E. Clausen, David Mazurek, Phillip J. Cornwell © 2007 The McGraw-Hill Companies.

Chapter 19, Solution 37.

From Equation 19.20:

22n

k l

gτ π

π =

For 60 , 1.686m kθ = ° =

( ) ( )( )( )3 s 2 1.686 232.2

l=

30.44484 0.19788

32.2 6.744 32.2

l l= = ⇒ =

Thus 6.3718 ft 76.462 in.l = =

76.5 in.l = !

COSMOS: Complete Online Solutions Manual Organization System

Vector Mechanics for Engineers: Statics and Dynamics, 8/e, Ferdinand P. Beer, E. Russell Johnston, Jr.,

Elliot R. Eisenberg, William E. Clausen, David Mazurek, Phillip J. Cornwell

© 2007 The McGraw-Hill Companies.

Chapter 19, Solution 38.

( ) ( )0.6 , 0.4A st B stA B

F k l F k lθ δ θ δ = + = +

(a) ( )eff:

C CM MΣ = Σ

( ) ( )0.6 0.6 0.1 0.4 0.4 0.1st st tA B

lk l lmg lk l I lmaθ δ θ δ α − + + − + = + (1)

Equilibrium position: 0θ =

( ) ( )0.6 0.1 0.4 0st stA B

lk lmg lkδ δ− + − = (2)

Substitute (2) into (1)

( ) ( )2 20.1 0.6 0.4 0

tI lma l k l kα θ θ+ + + =

or 20.1 0.52 0

tI lma l kα θ+ + =

But 2

, 0.1 , ,12

t

mlI a lα α θ= = = &&

So ( )2

2 20.1 0.52 0

12

mll m l kθ θ

+ + =

&&

or 20.09333 0.52 0l kθ θ+ =&&

25.5714 0l kθ θ+ =&&

( ) 850 N/m5.5714 0

9 kgθ θ

+ =

&&

2526.19 s 0θ θ−+ =&&

COSMOS: Complete Online Solutions Manual Organization System

Vector Mechanics for Engineers: Statics and Dynamics, 8/e, Ferdinand P. Beer, E. Russell Johnston, Jr.,

Elliot R. Eisenberg, William E. Clausen, David Mazurek, Phillip J. Cornwell

© 2007 The McGraw-Hill Companies.

Thus 2526.19 s 22.939 rad/s, 3.6508 Hz

2

n

n nf

ωωπ

−= = = =

3.65 Hznf = �

(b) ( ) ( )sin , cosm n m n n

t tθ θ ω φ θ θ ω ω φ= + = +&

m m nθ θ ω=&

( ) ( )0.6A mmx l θ= &&

( )( )( )0.0011 m/s 0.6 0.6 m 22.939 rad/sm

θ=

0.0001332 rad 0.0076m

θ = = °

0.0076m

θ = °�

COSMOS: Complete Online Solutions Manual Organization System

Vector Mechanics for Engineers: Statics and Dynamics, 8/e, Ferdinand P. Beer, E. Russell Johnston, Jr., Elliot R. Eisenberg, William E. Clausen, David Mazurek, Phillip J. Cornwell © 2007 The McGraw-Hill Companies.

Chapter 19, Solution 39.

( )eff:B BM MΣ = Σ

Diskcos2 2 2ABL L L

mg kxr I m Iθ α α α − = + +

(1)

where where from statics 2st stL

x r mg k rθ δ δ= + =

into (1) and assuming small angles ( )cos 1θ ≈

2

Lmg θ 2

stkr krθ δ− −2

Disk2ABL

I m I α = + +

so 2

2Disk 0

4ABmL

I I krθ θ

+ + + =

&&

so 2 2

22 2

2 2Disk Disk

1 14 12 4 2

n

AB

kr kr

mL mLI I mL M r

ω = =+ + + +

(2)

Disk

: 0.9 m

7.5 kg

0.25 m into (2)

6 kg

5 kN/m

Data L

m

r

M

k

= = = = =

COSMOS: Complete Online Solutions Manual Organization System

Vector Mechanics for Engineers: Statics and Dynamics, 8/e, Ferdinand P. Beer, E. Russell Johnston, Jr., Elliot R. Eisenberg, William E. Clausen, David Mazurek, Phillip J. Cornwell © 2007 The McGraw-Hill Companies.

( )( )

( )( ) ( )( ) ( )( )2

2

2 2 2

5000 N/m 0.25 m1 1 1

7.5 kg 0.9 m 7.5 kg 0.9 m 6 kg 0.25 m12 4 2

nω =+ +

2 2141.243 rad/snω =

11.8846 rad/snω =

(a) so 2 2

11.8846n

π πτω

= = 0.529 sτ = !

(b) 0.020 mmx =

( )0.02 11.8846 0.2377 m/sm m nv x ω= = =

max 238 mm/sv = !

COSMOS: Complete Online Solutions Manual Organization System

Vector Mechanics for Engineers: Statics and Dynamics, 8/e, Ferdinand P. Beer, E. Russell Johnston, Jr., Elliot R. Eisenberg, William E. Clausen, David Mazurek, Phillip J. Cornwell © 2007 The McGraw-Hill Companies.

Chapter 19, Solution 40.

At equilibrium, ( )spring tension 20 lb sin15= °

( )0

20 lb sin15stretch

50 lb/ftx

°= =

20 sin15PMΣ = °( ) 04 50

12x− 4

12x +

2

412

1 20 4 20 4 16.667 0.3105592 32.2 12 32.2 12

x x x x = + − =

!! !! !!

0.310559 16.667 0, 7.326 rad/snx x ω+ = =!!

(a) 0.858 sτ = !

(b) sin , cosn n nx A t x A tω ω ω= =!

( )0.5 7.326 3.66 in./snAω = = !

COSMOS: Complete Online Solutions Manual Organization System

Vector Mechanics for Engineers: Statics and Dynamics, 8/e, Ferdinand P. Beer, E. Russell Johnston, Jr., Elliot R. Eisenberg, William E. Clausen, David Mazurek, Phillip J. Cornwell © 2007 The McGraw-Hill Companies.

Chapter 19, Solution 41.

Kinematics

2 cos sinNon-linear terms

: sin2

A

x A

a l l

mlF N ma

θ θ θ θ

θ θ

= + +

Σ = −

& &&

&&

2 2

sin cos sin sin2 3 2A A

kl ml lM Nl ma

θθ θ θ θΣ = − + = −&&

or, letting cos 1,θ = sin cos , sin 0, since 1θ θ θ θ θ= = <<

2 22 3

0, 3 2 2n

ml kl k

mθ θ ω+ = =&&

For 1.5 ft,l = 2

6 lb, 23 lb/in. 276 lb/ft

32.2 ft/sm k= = =

(a) 1 3

7.50 Hz2 2

kf

mπ= = !

(b) 2 2 maxmax maxn n

x

Lθ ω θ ω= − = −&&

( )

( )2max

0.83 276

3 1298.747 rad/s

62 2 1.532.2

k x

m L

− = − = =

2max 98.7 rad/sθ =&& !

COSMOS: Complete Online Solutions Manual Organization System

Vector Mechanics for Engineers: Statics and Dynamics, 8/e, Ferdinand P. Beer, E. Russell Johnston, Jr.,

Elliot R. Eisenberg, William E. Clausen, David Mazurek, Phillip J. Cornwell

© 2007 The McGraw-Hill Companies.

Chapter 19, Solution 42.

(a)

( )eff:

A AM MΣ = Σ

( ) ( ) ( )1.2 0.825 2 0.75 0.825 1.2R S R t tG C

mg m g F I m a m aα+ − = + + (1)

Where ( )0.75S st

F k θ δ= +

At equilibrium, 0θ =

And

( )0: 1.2 0.825 2 0.75 0A R st

M mg m g kδΣ = + − = (2)

Substitute (2) into (1)

( ) ( ) ( ) ( )0.825 1.2 2 0.75 0.75R t tG C

I m a m a kα θ+ + = −

( ) ( ), 0.825 , 1.2t tG Ca aα θ α α= = =&&

( )( )22

21.2 kg 1.65 m

0.2725 kg m12 12

mlI = = = −

( ) ( ) ( ) ( ) ( )2 2 20.2725 0.825 1.2 1.2 2 0.75 450 0m θ θ + + + =

&&

[ ]1.08925 1.44 506.25 0m θ θ+ + =&&

506.250

1.08925 1.44mθ θ+ =

+&&

COSMOS: Complete Online Solutions Manual Organization System

Vector Mechanics for Engineers: Statics and Dynamics, 8/e, Ferdinand P. Beer, E. Russell Johnston, Jr.,

Elliot R. Eisenberg, William E. Clausen, David Mazurek, Phillip J. Cornwell

© 2007 The McGraw-Hill Companies.

Now ( ) ( )( )

2 2

22

2

2506.25 22 109.66

1.08925 1.44 0.6n n

n

fm

π πω πτ

= = = = = +

( )506.25 109.66 1.08925 1.44m= +

2.4495 kgm =

2.45 kgm = �

(b) ( ) ( )max max

0.061.2 , 0.03636 rad

1.65 1.65θ θ= = = =& B

C m

yv

( )( )1

220.03636 rad 109.66 s 0.38075 rad/s

m m nθ θ ω −= = =&

( ) ( )max

1.2 0.38075 rad/s 0.45691 m/sCv = =

( )max

457 mm/sCv = �

COSMOS: Complete Online Solutions Manual Organization System

Vector Mechanics for Engineers: Statics and Dynamics, 8/e, Ferdinand P. Beer, E. Russell Johnston, Jr., Elliot R. Eisenberg, William E. Clausen, David Mazurek, Phillip J. Cornwell © 2007 The McGraw-Hill Companies.

Chapter 19, Solution 43.

2mx kx P= −&& Mx P F= −&&

21

2Mr Frθ =&&

Eliminate , and P F θ

32 0

2

Mm x kx + + =

&&

( )2 2 5000 N/m2769.23

3 4 kg 9 kg2

nk

Mm

ω = = =++

27.735 rad/snω =

(a) 2

0.22654 snn

πτω

= =

0.227 snτ = !

(b) ( )max 12 mm 333 mm/snv ω= = !

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Chapter 19, Solution 44.

From Problem 19.43

( )2 2 3500 N/m2538.46; 23.2048 rad/s

3 4 kg 9 kg2

n nk

Mm

ω ω= = = =++

( )( ) ( )( )2 26 kg 9.81 m/s 4 kg 9.81 m/s 98.1 NN W Mg mg= = + = + =

2 21 1

2 2

xFr Mr Mr

rθ = =

&&&&

( )122 6 kg

max , ; Amplitude98.1 N

ω= = =&& &&m m nF

x x A AN

( )( )2

16 kg

20.5 23.2048 rad/s98.1 Ns Aµ = =

or 0.03036 mA =

30.4 mmA = !

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Chapter 19, Solution 45.

2

22

42

2 30.021440 m

162

r rr

br

r

ππ

π

− = =−

( )2 16 0.22546 kg2

m tr tπρ ρ = − =

( )24 4 2

4 8 44 2

6 4 9 2 3ot r r r r

I r t rρ π πρ

π π = − − + −

( )4 136 90.009342

3 4tr t

πρ ρ = − =

(a) 20, o no

tmgbI mgb

I

ρθ θ ω+ = = =&& ( )( )( )0.22546 9.81 0.021440

tρ ( )0.009342

2 5.07602, 2.2530 rad/sn nω ω= =

22.79 s

n

πτω

= = !

(b) 2

22 2 2

9.81 m/s, 1.9326 m

5.07602 rad /sn

n

g gl

lω

ω= = = =

1.933 ml = !

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Chapter 19, Solution 46.

Kinematics: / ,G O G Oa a a GO d= + =

0.3 m, 4 2 /3 0.180063 mr d r π= = =

( )2PM mgd I m r dθ θ θ Σ = − = + −

&& &&

Where 2 2

2

32

2 9

mr mrI

π= −

2n

mω = gd

m 2 322r m−

2

29

rm

π+ ( )2

65.515

r d

=

−

2

0.776264 sn

πτω

= = 0.776 sτ = !

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Chapter 19, Solution 47.

21

2AI mr=

(a) 2 4

02 3

mr rmgθ θ

π + =

&&

or 8

03

g

rθ θ

π+ =&&

8

3ng

rω

π=

2 26.82

83

nn

r

ggr

π πτω

π

= = = !

(b) 2

2 2 4

3

rb r

π = +

2 22 2 21 4 4

2 3 3B Gr r

I I mb mr m mr mmπ

= + = − + +

23

2mr=

230

2mr mgbθ θ+ =&&

or 2

2

3

gb

rθ =&&

2

2

3n

gb

rω =

2

2 27.38

162 1

93

nn

r

gg

r

π πτω

π

= = =

+

!

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Chapter 19, Solution 48.

( ) ( )2

22 1.61 1.62 0.6

3 3.6 3.6 12A

bm mI b b

= + +

211

25mb=

( )2 0.3

3.6 6

b b by

b= =

130.6

6 30

bAG b b= − =

22 211 13

0.6425 30 36B

bI mb m b m b

= − + +

223

25mb=

24.046

bGB =

(a) 211 13 2

0, 6.3325 30 n

n

mb b bmg

g

πθ θ τω

+ = = =&& !

(b) 223 24.04 2

0, 6.6725 6 n

n

mb bmg b

g

πθ θ τω

+ = = =

&& !

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Chapter 19, Solution 49.

( )2 21

12I m b c= +

1

2ta r cα θ= = &&

( )eff

1 1: sin

2 2A Α tM M mg c I ma cθ α Σ = Σ − = +

( )2

2 21 1sin

2 12 2

mmgc b c m cθ θ θ − = + +

&& &&

2 2 sin 02

3 12

g c

c bθ θ

+ = +

&&

22 2 2

6

24 13 12

cg c g b

bc b cb

ω

= =

+ +

(1)

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(a) Since 2

,πτ

ω= minimum τ occurs when 2ω is maximum. We consider

c

b

as the variable.

( )

2

2

22

4 1 86

4 1

c c cd b b bg

c bd cb

b

ω + − = =

+

2 2

4 1 8 0c c

b b + − =

2

1

4

c

b =

0.5c

b= !

(b) For a simple pendulum of length c we have 2 8.

cω = We equate the value of 2ω to that of Eq. (1).

2

6

4 1

cg gbb cc

b

=

+

2 2 2

14 1;

2 = + =

c c cb

b b b 0.707

c

b= !

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Chapter 19, Solution 50.

Consider general pendulum of centroidal radius of gyration k.

( ) ( ) 20 eff

: sinM Μ mgr mr r mkθ θ θ0Σ = Σ = +&& &&

2 2 sin 0gr

r kθ θ

+ = +

&&

For small oscillations, sin ,θ θ≈ we have

2 2

22 2 2 2

20 ; ; 2

gr gr r k

grr k r k

πθ θ ω τ πω

++ = = = = + + &&

Connecting rod suspended at A: 1.03 sτ =

160 mm 0.16 mar r= = =

Thus

( )( )( )

2 20.161.03 2

9.81 0.16

kπ

+=

( ) ( )( )2

2 2 1.030.16 9.81 0.16 0.04218

2k

π + = =

2 0.04218 0.0256 0.01658 ; 0.12876 mk k= − = = 128.8 mmk = !

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Chapter 19, Solution 51.

2

21 1 10m 0.334722 m

2 2 12I mr

= = =

α θ= &&

(a) The disk is free to rotate and is in curvilinear translation.

Thus 0Iα =

Then ( )effB B

M MΣ = Σ

sin , sint

mgl lmaθ θ θ− = ≈

20ml mglθ θ+ =&&

And 2

2 2

26

12

32.2 ft/s14.861 s

ftn

g

lω −= = =

13.8551 s

nω −=

21.6299 s

n

n

πτω

= =

1.630 sn

τ = �

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(b) When the disk is riveted at A, it rotates at an angular acceleration α

( )effB B

M MΣ = Σ

21sin , , sin

2t

mgl I lma I mrθ α θ θ− = + = ≈

2 210

2mr ml mglθ θ + + =

&&

Then

( )22 2

2 2

2

2

2632.2 ft/s ft

1213.838 s

10ft

2 2612ft

2 12

ω −

= = =

+ +

n

gl

rl

13.7199 s

nω −=

21.6891 s

n

n

πτω

= =

1.689 sn

τ = �

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Chapter 19, Solution 52.

( )0 0 eff: sin tM M W r I ma rθ αΣ = Σ − = +

2 2sinmgr mk mrθ θ θ− = +&& &&

2 2

sin 0gr

r kθ θ+ =

+&& (1)

For a simple pendulum of length OA l=

0g

lθ θ+ =&& (2)

Comparing Equations (1) and (2)

2 2r kl

r

+=

2k

GA l rr

= − = (Q.E.D) !

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Chapter 19, Solution 53.

See Solution to Problem 19.52 for derivation of

2 2sin 0

gr

r kθ θ+ =

+&&

For small oscillations sinθ θ≈ and

2 2 22 22n

n

r k kr

gF rg

π πτ πω

+= = = +

For smallest nτ we must have 2k

rr

+ a minimum

2

2

21 0

kd r

r k

dr r

+

= − =

2 2r k=

r k= (Q.E.D) !

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Chapter 19, Solution 54.

Same derivation as in Problem 19.52 with r replaced by .R

Thus 2 0gR

R kθ θ+ =

+&&

Length of the equivalent simple pendulum is

2 2 2R k kL R

R R

+= = +

( )2

2= − + =kL l r l

kr

Thus the length of the equivalent simple pendulum is the same as in Problem 19.52. It follows that the period is the same and that the new center of oscillation is at O (Q.E.D)

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Chapter 19, Solution 55.

2 2

2 2 54 4

l ld l= + =

Ignore static terms: 2

2Cl

M mg klθ θΣ = − −

2

2 2 2 52

12 4 4

mlmgl kl ml ml

θ θθ θ θ

− − = + +

&& &&&&

2

32

253

mgk

lm

ω+

=

1 6 9

2 5 10nk g

fm lπ

= + !

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Chapter 19, Solution 56.

( )eff:

4 4 2D DkL r mrL mLr

M M mgrθθ θ θΣ = Σ − − = +&& &&

30

4 2

mLr kLrmgrθ θ + + =

&&

2 4 2

3 3ng k

L mω = +

1 2 4

Hz2 3 3n

k gf

m Lπ= + !

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Chapter 19, Solution 57.

2

22

1 16 lb 1 12.5 in.12 12 12 in./ft32.2 ft/s

I ml

= = ⋅

20.04493 lb ft s= ⋅ ⋅

α θ= !!

2 0.1666712ta α θ = =

!!

sinθ θ≈

( )effC CM MΣ = Σ

( ) ( )2 20.6875 0.16667 0.16667k mg I mθ θ θ θ− + = +!! !!

( )( )160.04493 0.02778 0.47266 0.16667 16 032.2

kθ θ + + − = !!

[ ]0.05693 0.47266 2.6667 0kθ θ+ − =!! (1)

(a) For 50 lb/ft,k = 368.28 0θ θ+ =!!

368.28 3.007 Hz2 2

nnf

ωπ π

= = =

3.01 Hznf = !

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(b) For ,nτ → ∞ 0nω → oscillations will not occur

From Equation (1)

2 0.47266 2.6667 00.05873n

kω −= =

5.642 lb/ftk =

5.64 lb/ftk = !

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Chapter 19, Solution 58.

2 3

33 2 3

a ab

= =

4 4

23 3

96 96Oa a

I t mbρ

= + +

23 5,

2 2 12Ot a

m a I maρ

= =

With 220 lb s 16

, ft, 14 lb/ft32.2 ft 12

m a k⋅ = = =

So with sinθ θ≈ for small ,θ ( )effO ΟΜ ΜΣ = Σ yields

2

252 0

12 3

ma mgakaθ θ + + =

&&

22

122 0

5 3

mgaka

maθ θ + + =

&&

2 22

12 12 242

55 3 5 3n

mga g kka

a mmaω θ = + = +

2 212 32.2 24 14141.655 s

16 2055 312 32.2

nω −

= + =

11.90 rad/snω =

1.894 Hz2

nfωπ

= = !

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Chapter 19, Solution 59.

Ignore static terms

( ) ( ) ( ) ( )( ) ( )( )2 2 2 21 150 0.6 50 0.9 1.2 0.3 1.2 0.6 1.8 0.9

3 3B

M gθ θ θ θ Σ = − − + = +

&&

54.96840.63 54.9684 0, 9.3408 rad/s

0.63n

θ θ ω+ = = =&&

0

15cos cos

900t tθ θ ω ω = =

At 2 2

00.7 s, cos 0.7 1.407 rad/s

n nt θ ω θ ω= = − = −&&

21.407 rad/sα = �

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Chapter 19, Solution 60.

Determine location of the centroid G.

Let mass per unit lengthρ =

Then total mass ( ) ( )2 2m r r rρ π ρ π= + = +

About C 22

0 2r

mgc r g r gπ ρ ρπ

= + =

2ry

π=

( ) 22 2r c rρ π ρ+ =

( )2

2

r

c

π=

+

( )0 0 eff

tM M a c cα θ α θΣ = Σ = = =&& &&

sin sinn

mgc I mcaθ α θ θ− = + ≈

( )2 00 0I mc mgc I mgcθ θ θ θ+ + = + =&& &&

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But 2

0I mc I+ =

( ) ( ) ( )2

2

0 0 0 s-circ linesemi circ line

2

12

rI I I m r m= + = +

( )s-circ line 2

2

m

m r m r

r

ρπ ρ ρπ

= = =+

( )2 2

2

0

2 2

3 2 3

r r mrI r rρ π π

π ⋅ = ⋅ + = + +

( ) ( )2

2 20

2 3 2

mr rmgπ θ θ

π π + + = + +

&&

( )( )

( )( )

2

2

2 2

3 3

2 9.81 m/s2

0.32 mn

g

r

ωπ π

= =+ +

2 216.0999 s 4.0125 rad/s

n nω ω−= =

( )sin m n B

t y rθ θ ω φ θ= + =

( ) ( ) ( )sin sinB m n B nmy r t y tθ ω φ ω φ= + = +

At 0t = 0.03 m 0B By y= =&

( ) ( ) ( )0 0 cos 0 2

B B mt y y

πφ φ= = = + =&

( ) ( )0.03 sin 0 0.03 m2

B B Bm my y y

π = = + =

10.03sin 4.0125 s

2B n ny t

πω ω − = + =

( )( ) ( )0.03 cos 0.03 sin2

B n n n ny t t

πω ω ω ω = + = −

&

( )( )2 20.03 sin 0.03 cos

2B n n n ny t t

πω ω ω ω = − + =

&&

At 10 s:t = ( ) ( ) ( )( )2 20.03 4.0125 cos 4.0125 10 0.36437 m/s

B Bta y= = = −&&

( ) ( ) ( )( )0.03 4.0125 sin 4.0125 10 0.07902 m/sB Bv y= = =&

( ) ( ) ( )11

2 22 2222 2 2

0.079020.36437 0.365 m/s

0.32

B

B B t

v

a a

r

= + = − + =

�

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Chapter 19, Solution 61.

( )221 10.250

2 2 32

mI mr m α θ= = = = &&

0.650ta lα θ= = &&

(a) The disk is free to rotate and is in curvilinear translation.

Thus 0Iα =

( )effB B

M MΣ = Σ

sint

mgl lmaθ− =

2 20

n

gml mgl

lθ θ ω+ = =&&

From 19.17, the solution to this Equation is

( )sinm n

tθ θ ω φ= +

At 0,t = 2 rad, 0180 90

π πθ θ= ⋅ = =&

( )cosm n n

tθ θ ω ω φ= +&

0 0 cos2

m nt

πθ ω φ φ= = =

sin 090 2

m

π πθ = +

rad90

m

πθ =

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Thus sin90 2

nt

π πθ ω = +

( )max

maxA m nv l lθ θ ω= =&

With 26 in. 2.1667 ft, rad,90

m n

gl

l

πθ ω= = = =

( ) ( ) ( )2max

32.2 ft/s

2.1667 ft rad90 2.1667 ft

Av

π =

0.29156 ft/s=

3.4987 in./s=

( )max

3.50 in./sAv = �

(b) For disk riveted at A ( Iα included)

( )effB B

M MΣ = Σ sint

mgl I lmaθ α− = +

2 210

2mr ml mglθ θ + + =

&

2

2

2

2

nr

gl

lω =

+

sin90 2

nt

π πθ ω = +

(See (a))

( )max

maxA m nv l lθ θ ω= =&

( ) ( )( )2

2 2

32.2 ft/s 2.1667 ft

2.1667 ft90 1 10 in. 26 in.

2 12 in./ft 12 in./ft

radπ =

+

0.28135 ft/s=

3.376 in./s=

( )max

3.38 in./sAv = �

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Chapter 19, Solution 62.

( )18 0.3

8 in. = ft, 2 ,12 32.2

r m rπ = =

( )20.3

32.2m r

=

2

1 2

2 0.045767 ft

rm

bm m

= =+

2 2 21 2

10.58813 lb s ft

3= + = ⋅ ⋅AI m r m r

( )1 2 0AI m m gbθ θ+ + =&&

( )1 22

nA

m m gb

Iω

+=

1.9105 rad/s,nω = 0 cos ntθ θ ω=

2

2 rad,180

πθ° = ° = 0 sinn ntω θ ω ω= −

2

0 cosn ntα θ ω ω= −

At 5 s,t = 0.00849 rad/sω = !

0.1264 rad/sα = !

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Chapter 19, Solution 63.

( )effM MΣ = Σ

K Iθ θ− = &&

0K

Iθ θ+ =&&

2 ,nK

Iω = 2

I

Kτ π= (1)

For 50-mm-diameter sphere, 0.025 mr =

2 3 2 52 2 4 8

5 5 3 15SI mr r r rπ ρ π ρ = = =

( ) ( )5 3 6 280.025 7850 kg/m 128.45 10 kg m

15π −= = × ⋅

Solve Eq. (1) For :I

22

;4

KI τ

π=

2

R R

S S

I

I

ττ

=

(2)

Data: 4.1 s, 6.2 s, 0.45 kgR S Rmτ τ= = =

Eq. (2): 2

4.1 s0.4373

6.2 sR

S

I

I

= =

( )6 20.4373 0.4373 128.45 10 kg m−= = × ⋅R SI I

6 256.17 10 kg m−= × ⋅I

6 2

2 56.17 10 kg m

0.45 kgR

RR

Ik

m

−× ⋅= =

2 6 2124.82 10 mk −= ×

311.173 10 mk −= × 11.17 mmRk = !

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Chapter 19, Solution 64.

( )effG GM MΣ = Σ

K Iθ θ− = &&

0K

Iθ θ+ =&&

2n

K

Iω = (1)

2I

Kτ π= (2)

Data: 1.95 N m/radK = ⋅

3 kgm =

( )( )22 3 21 13 kg 0.5 m 62.5 10 kg m

12 12I ml −= = = × ⋅

(a) Eq. (2) 3 262.5 10 kg m

21.95 N m/rad

τ π−× ⋅=⋅

1.125 sτ = !

(b) Max velocity

Eq. (1) 23 2

1.95 N m/rad31.2

62.5 10 kg mω −

⋅= = =× ⋅n

K

I

5.586 rad/snω =

Simple harmonic motion

m m nω θ ω= 180 radmθ π= °=

( )( )rad 5.586 rad/s 17.548 rad/smω π= =

( ) ( ) ( )( )0.25 m 17.548 rad/s 4.387 m/sA mmv AG ω= = =

( ) 4.39 m/sA mv = !

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Chapter 19, Solution 65.

Equivalent torsional spring constant

( )2 1 1 1, , eT K T K T Kθ θ θ θ= = − =

( )2 1 2 1K K Kθ θ= +

21

1 2

K

K Kθ θ=

+

1eT K Kθ θ= =

21

1 2e

KK K

K Kθ θ=

+

1 2

1 2e

K KK

K K=

+

Newton’s Law

( )effc cM MΣ = Σ

eK Jθ θ− = &&

21

2J mr=

210

2 emr Kθ θ+ =&&

( )( )

2

2

4 402 2 32.2 12

23 1.52

23 1.5

nn eK

mr

π πτ πω

= = = +

5.2197 s=

5.22 snτ = !

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Chapter 19, Solution 66.

2 21

4ABI mr mb= +

[ ]2 0ABI k mgbθ θ+ + =&&

22

2

2

4

nk mgb

rm b

ω +=

+

2

120 lb 8, ft

1232.2 ft/sm r

= =

6.4

ft,12

b =

150 lb ft/radk = ⋅

(a) ( ) ( )

22

2

120 6.42 150 32.2

32.2 12246.93

8120 6.41232.2 4 12

nω

+ = = +

15.7139 rad/snω =

2 2

0.3998 s15.7139n

π πτω

= = =

0.400 sτ = !

(b) 0 0 max 0sin ; cos ;n n n nt tθ θ ω θ θ ω ω θ θ ω= = =& &

So max max 0 nv b bθ θ ω= =&

( )max6.4 2

ft 15.7139 rad/s 0.2925 ft/s12 180

vπ = =

max 3.51 in./sv = !

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Chapter 19, Solution 67.

( )effG GM MΣ = Σ 0K J J Kθ θ θ θ− = + =&& &&

Empty platform centroidal of platformPJ J=

( )( )22

2 2

27 N m/rad 2.2 s2 2

4 4n

n Pn

P

KJ

KJ

π π ττω π π

⋅= = = =

23.31 N m sPJ = ⋅ ⋅

Platform with object A centroidal ofAJ J A=

( )

( )2

2

2 2

4n

n P An

P A

KJ J

K

J J

τπ πτω π

′′ = = + =

′+

( )( )2

22

27 N m/rad 3.8 s3.31 N m s

4AJ

π⋅

= − ⋅ ⋅

29.88 3.31 6.57 N m sAJ = − = ⋅ ⋅

26.57 N m sAJ = ⋅ ⋅ !

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Chapter 19, Solution 68.

Geometry

3

b

l

θφ =

3 3

mg bF

l

θ=

Then 3G G

bM I Fθ θ Σ = = −

&&

or 2 2

012 3

mb mgb

Lθ θ+ =&&

(a) 2 4 2, n

n

g l

l g

πω τ πω

= = = !

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(b)

3mx F= −&&

mgx

l= −

2n

g

lω =

2l

gτ π= !

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Chapter 19, Solution 69.

( )2 2 21

2T m b c θ= + &

2 21

2V kc θ=

( )2

22 2nkc

m b cω =

+

720 N/m, 1.5 kgk m= =

17.5271 rad/snω =

00.015 m

0.033333 rad0.45 m

θ = =

0max 0.351 m/sD nv cω θ= =

max 0.351 m/sDv = !

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Chapter 19, Solution 70.

From Problem 19.69

17.5271 rad/snω =

( )0

0.25 m/s0.5555 rad/s

0.45 mθ = =&

0max 0.01902 mDn

cx

θω

= =&

max 19.02 mmDx = !

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Chapter 19, Solution 71.

22 2 21 1

2 2 2Bl

T I ml mθ θ = = +

& &

21

2 2

lV k

θ =

2

22

85

8

n

kl

mlω =

5

k

m=

1

Hz2 5n

kf

mπ= !

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Chapter 19, Solution 72.

Equilibrium:

0

0

0

22

20 lb1 ft9

k yT

T k y

y

= + =∴ =

Vibration: 22 2

1 20 ,2

T vg

=

( ) ( )

22 1

1 11 1144 1442 2 2

xV x = +

1 2:V T= 2 21

1 202 32.2 n xω

( ) ( ) 21

1 1 1442 8

x = +

2 289.8, 17.0235 rad/sn nω ω= =

(a) 2 0.36909 s,n

πτω

= = 0.369 sτ = !

(b) 1 ft 1.8915 ft/s9nmv ω = =

1.892 ft/smv = !

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Chapter 19, Solution 73.

Datum at 1

Position 1

21 0 1

1 0

2 mT J Vθ= =&

Position 2

2 20 T V mgh= =

( ) 21 cos 2sin2m

mh r rθθ= − =

2

2mrθ≈

2

2 2mV mgr

θ=

Conservation of energy 2

21 1 2 2 0

1 0 0

2 2m

mT V T V J mgrθθ+ = + + = +&

2 2 20 m n m n m mJ mgrθ ω θ ω θ θ= =&

( )2202 2

20

44 n n

n

Jmgr

J mgr

ππω τω

= = =

( )( )22 2

0 2

4

n mgrJ J mr J mr

τ

π= + + =

( )( )2

224

n mgrJ mr

τ

π= −

( ) ( )( )( )

( )( )2 2

22

1.26 s 38 kg 9.81 m/s 0.175 m38 kg 0.175 m

4π= −

2 2 22.62345 N m s 1.16375 N m s 1.4597 N m s J = ⋅ ⋅ − ⋅ ⋅ = ⋅ ⋅

21.460 kg mJ = ⋅ !

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Chapter 19, Solution 74.

Find nω as a function of c.

Datum at 2

Position 1 1 10 T V mgh= =

( )1 1 cos mV mgc θ= −

221 cos 2sin

2 2m m

mθ θθ− = ≈

2

1 2mV mgc

θ=

Position 2

22

1

2 C mT I θ= &

2 2 21

12CI I mc ml mc= + = +

22 2

2 21

02 12 m

lT m c Vθ

= + =

&

2 2 22

1 1 2 2 0 02 12 2m ml

T V T V mgc m cθ θ

+ = + + = + +

&

m n mθ ω θ=&

22 2

12 nl

gc m c ω

= +

22

2

12

ngc

lc

ω =

+

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Maximum c, when

22 2

2

22

212

0 0

12

n

lg c c g

d

dc lc

ω

+ − = = =

+

2

2 0 12

lc− =

12

lc = !

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Chapter 19, Solution 75.

Consider a general pendulum of centroidal radius of gyration .k Datum at 1 Position 1

21 0

12 mT J θ= !

1 0V =

Position 2

2 20 T V mgh= =

( ) 21 cos 2sin2m

mh r F θθ= − =

2 2

2 2 2m mV mg Fθ θ= ≈

1 1 2 2T V T V+ = +

2 20

1 1 0 02 2m mJ mgrθ θ+ = +!

m n mθ ω θ=! 2 2 2

0 n m mJ mgrω θ θ=

2 0

0

2 2n nn

mgr JJ mgr

πω τ πω

= = =

2 2 20J J mr mk mr= + = +

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(a) 2 2

2nk r

grτ π +=

For a rod suspended at A:

2 2

0.895 2 ,an a

a

k r r rgr

τ π += = = (1)

For a rod suspended at B:

2 2

0.805 2 ,bn b

b

k r r rgr

τ π += = = (2)

10.5 in. 0.875 fta br r+ = = (3)

From (1) and (2)

( )22 22 0.895

4a

agrk rπ

+ = ( )1′

( )22 22 0.805

4b

bgrk rπ

+ = ( )2′

Taking the difference ( ) ( )1 2′ ′− :

[ ]2 22 0.80125 0.648025

4a b a bgr r r rπ

− = −

( )( ) [ ]2 0.80125 0.6480254a b a b a b

gr r r r r rπ

− + = −

( ) ( ) [ ]232.2 0.80125 0.648025

4 0.875a b a br r r rπ

− = −

( ) 0.746685 0.60406a b a br r r r− = −

0.25332 0.39594 0.63979a b b ar r r r= ⇒ =

0.63979 0.875 0.533605 fta a ar r r+ = ⇒ =

6.4033 in.= 6.40 in.ar = !

(b) 2

2 20.8952a ak gr rπ

= −

( )( ) ( ) ( )2

222

0.895 s32.2 ft/s 0.533605 ft 0.533605 ft

2π= −

2 2 20.34863 ft 0.28473 ft 0.0639 ft= − =

0.25278 ft 3.0334 in.k = =

3.03 in.k = !

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Chapter 19, Solution 76.

2

2Pmr

I >

2 224 4

1 0.715 0.715 ft3 3

rd r r

π π = + − = =

2224 2

2 3Omr r

I m mdπ

= − +

0.65117 m=

2

21,

2 2OT I V mgdθθ= =&

2 2 35.35640.65117

2

n

dmg

mω = =

5.9461 rad/snω =

21.057 s

n

πτω

= = !

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Chapter 19, Solution 77.

Since vibration takes place about the position of equilibrium, we shall neglect the effect of weight and the static deflection.

Position of Max Displacement

Spring Elongations

2B ml

x θ= C mx lθ=

Position 1: 1 0T =

( )2

22 21

1 1 1

2 2 2 2B B C C B m C ml

V k x k x k k lθ θ = + = +

2 21

1 1

2 4 B C mV k k l θ = +

Position 2: 2 0V =

2

2 2 2 22

1 1 1 1 1

2 2 2 12 2 2m m m ml

T I mv ml mω ω ω = + = +

2 22

1

6 mT ml ω=

Conservation of Energy

2 2 2 21 1 2 2

1 1 1: 0

2 4 6B C m mT V T V k k l mlθ ω + = + + + =

For simple harmonic motion, ( )22 2 21 1 1;

2 4 6m n m B C m n mk k l mlω ω θ θ ω θ = + =

2 3 1

4n B Ck km

ω = +

(1)

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Note: Result is independent of length of the rod.

Data

6 lb,W = 3 lb/in. 36 lb/ft,k = = 5 lb/in. = 60 lb/ft.Ck =

2

2

3 36 lb/ft60 lb/ft 1110.9

46 lb

32.2 ft/s

nω = + =

33.33 rad/snω = 33.33

2 2nf

ωπ π

= = 5.30 Hzf = !

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Chapter 19, Solution 78.

2 21

2 2 4

l lV k mg

θ θ = +

2 2 2 2 2

2

1 1

2 4 2 2 4

L mr LT m

r

θ θ = +

& &

2 23

16

mL θ=&

Then

2

22

8 43

16

n

kL mgL

mLω

+=

2 2

3

k g

m L = +

1 2 4Hz

2 3 3nk g

fm Lπ

= + !

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Chapter 19, Solution 79.

(a) Position 1

( )21 1

10 2 st mT V k rδ θ= = +

Position 2

( )22 22 2

1 1 1 2 2 2m m stT J mv V mgh kθ δ= + = +!

1 1 2 2T V T V+ = +

( ) ( )2 22 21 1 1 1 02 2 2 2st m m m stk r J mv mgh kδ θ θ δ+ + = + + +!

2stkδ 2 2 2 2 22 2st m m m m stk r kv J mv mgh kδ θ θ θ δ+ + = + + +! (1)

When the disk is in equilibrium

0 sinC stM mg r k rβ δΣ = = − Also sin mh r βθ=

Thus 0stmgh k rδ− = (2)

Substitute (2) into (1) 2 2 2 2

m m mkr J mvθ θ= +!

m n m m m n mv r rθ ω θ θ ω θ= = =! !

( )2 2 2 2 2m m nkr J mrθ θ ω= +

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2

22n

krJ mr

ω =+

212

J mr=

22

2 2

21 32

nkr k

mmr mrω = =

+

( )2 2 0.71983 s

800 N/m23 7 kg

nn

π πτω

= = =

0.720 snτ = !

(b)

m mv rθ= !

m m nθ θ ω=!

0.01 mm m n mv r rθ ω θ= =

( ) 20.01 m 0.0873 m/s0.720 smv π

= =

!

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Chapter 19, Solution 80.

sinθ θ≈ 2

21 cos 2sin2 2m m

mθ θθ− = ≈

Position1 2

21

1 12

2 2 2m ml

T J mθ θ = +

& &

1 0V =

Position 2

( )2

2 22

0 1 cos2 2 2m ml l

T V W kθ θ = = − − +

2 2

22 2 2 4

mm

Wl klV

θ θ= − +

Conservation of Energy

1 1 2 2T V T V+ = +

( )2 2 2

2 2 21 12 0 0

2 2 4 2 2 4m

m m ml wl kl

J mθθ θ θ+ + = − +& &

21

12m n mW

J lg

θ ω θ= =&

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22 2 2 2

6 4 2 2n m mW W Wl kl

lg g

ω θ θ −+ = +

2 62 255

12

n

W klg k

WlWl

gg

ω

− + − = = +

26 9.81 m/s 120 N/m

5 0.160 m 0.6 kg

−= +

2166.43 s−=

12.901 rad/snω =

12.0532 s2

nnf

ωπ

−= =

2.05 Hznf = !

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Chapter 19, Solution 81.

( )2

21

2 4

lV k l mg

θθ

= +

2

2 21 1

2 12 2

lT ml m θ

= +

&

2

22

2 4

6

n

kl mgl

mlω

+=

2 3 3

2nk g

m lω = +

With 1500 N/m, 0.6 m, 10 kgk l m= = =

2 2 2474.525 rad /snω =

20.288 s

n

πτω

= = !

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Chapter 19, Solution 82.

Let m be the mass of rod and Cm be mass of each collar

Then 2 2 2 2

2 4 2Ckl mgl l

V m gθ θ θ

= + +

( )2 221 1

2 3 2 Cl

T m m lθ θ

= +

& &

2

222

2 4 2

6 2

C

nC

mgl m glkl

m lmlω

+ +=

+

2 4 2

6 2

C

C

mg m gkl l

mm

+ +=

+

( )( )( )

( )( )( )

2 2

2

5 kg 9.81 m/s 2.5 kg 9.81 m/s1500 N/m2 4 0.6 m 2 0.6 m

5 kg 2.5 k6 2

nω

+ + =

+

2397.62 s−=

19.4838 rad/snω =

2

0.322 sn

πτω

= = !

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Chapter 19, Solution 83.

2 2

1 20, 22 2 2

lV V mgl mg

θ θ = = +

2mglθ=

22 2 2

2 11 1

0, 22 2 3

mlT T ml θ θ

= = +

& &

2 25

6

ml θ=&

22 2 2 2 25 6

: , 6 5n n n

ml gmgl

lθ ω θ ω ω= = =&

( )( )

2 21.586 s

9.81 m/s60.75 m5

n

π πτω

= = = !

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Chapter 19, Solution 84.

6.87 NA AW m g= =

4.91 NC CW m g= =

9.81 NAC ACW m g= =

Position 1 ( ) ( ) ( )2 2 2 21

1 1 1 10.1 0.16 0.03

2 2 2 2A m C m AC m AC mT m m m Iθ θ θ θ= + + +& & & &

( )210.26

12AC ACI m=

So ( ) ( ) ( ) ( )( )2 2 2 2 21

1 10.7 0.1 0.5 0.16 1 0.03 1 0.26

2 12 mT θ = + + + &

( )2 210.02633 kg m

2 mθ= ⋅ &

1 0V =

Position 2

( )( ) ( )( ) ( )( )2 20, 0.1 1 cos 0.16 1 cos 0.03 1 cosA m C m AC mT V W W Wθ θ θ= = − − + − + −

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With 2

21 cos 2sin2 2m m

mθ θθ− = "

( )( ) ( )( ) ( )( )2 2

2 6.87 0.1 4.91 0.16 9.81 0.03 0.39292 2m mV

θ θ = − + + =

( ) ( )2

21 1 2 2

1 1: 0.02633 0 0 0.3929

2 2 2m

mT V T Vθθ+ = + + = +&

m n mθ ω θ=&

So 2 20.392914.922 s

0.02633nω −= =

2 21.6266 s

14.922n

n

π πτω

= = =

1.627 snτ = !

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Chapter 19, Solution 85.

Position 1

( ) ( ) ( )( )2

22 2 21 0

1 1 1 1 22

2 2 2 2 2 4A C m mm m

m lT m v m v I θ θ

= + + +

& &

( ) ( )21

12 2G A C mm m

mI l v v lθ= = = &

2 2

2 2 21

7

24 8 6m m mml ml

T ml mlθ θ θ

= + + =

& & &

15

2 cos cos cos2 2

lV mgl mg mglβ β β= − − = −

Position 2

2 0T =

( ) ( )2 cos cos2 2m mm l

V mgl gβ θ β θ= − − − −

( ) ( )cos cos2 2m mm l

mgl β θ β θ− + − +

[ ]25

cos cos sin sin cos cos sin sin4 m m m mV mgl β θ β θ β θ β θ= − + + −

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25

cos cos2 mV mgl β θ= −

( )2

cos 1 small angles2m

mθθ ≈ −

2

25

cos 12 2

mV mglθβ

= − −

1 1 2 2T V T V+ = +

22 27 5 5

cos 0 cos 16 2 2 2

mmml mgl mgl

θθ θ β

− = − −

&

m n mθ ω θ=&

2 2 27 5cos

6 4n m ml gω θ βθ= ⋅

2 15cos

14ng

lω β=

22 15 32.2 ft/s

cos 4025 in.14

12 in./ft

nω

= °

212.686 s−=

13.5617 snω −=

10.56686 s2

nnf

ωπ

−= =

0.567 Hznf = !

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Chapter 19, Solution 86.

D = Disk R = Rod

2ABl r=

For small oscillations: ( ) 211 cos

2m mh r rθ θ= − =

Position 1: 1 0T =

21

1

2R R mV W h m grθ= =

Position 2: 2 0V =

( )22 22

1 1 1

2 2 2D m D D R mmT I m v Iω ω= + +

( )22 2 2 2 21 1 1 1 12

2 2 2 2 12D m D m R mm r m r m rω ω ω = + +

2 21 3 1

2 2 3D R mm m r ω = +

Conservation of Energy.

1 1 2 2 :T V T V+ = +

2 2 21 1 3 10

2 2 2 3R m D R mm g m m rθ ω + = +

But for simple harmonic motion: m n mω ω θ=

( )22 21 1 3 1

2 2 2 3R m D R n mm gr m m rθ ω θ = +

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2

3 12 3

Rn

D R

m g

rm mω =

+

or 2

3 12 3

Rn

D R

W g

rW Wω =

+ (1)

Data: 4

3 lb, 5 lb, ft12R DW W r= = =

( ) ( )

2 3 32.234.094

3 1 45 3

2 3 12

nω

= = +

5.839 rad/snω =

2 2

5.839n

π πτω

= = 1.076 sτ = !

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Chapter 19, Solution 87.

Position 1: 1 0,T = 21

1

2 mV kx=

Position 2: 2 0,V = 2 2 22

1 1 12

2 2 2AB m m Disk mT m v I m vω = + +

2

2 2 22

1 1

2 2m

AB m Disk Disk mv

T m v m r m vr

= + +

( ) 22

13

2 AB Disk mT m m v= +

Conservation of Energy

1 1 2 2:T V T V+ = + ( )2 21 10 3

2 2m AB Disk mkx m m v+ = +

But for simple harmonic motion, :m n mv xω=

( )( )221 13

2 2m AB Disk n mkx m m xω= +

2

3nAB Disk

k

m mω =

+ Note: Result is independent of r

Data: 5 kN/m, 9 kg, 6 kgAB Diskk m m= = =

( )2 5000 N/m

185.1859 kg 3 6 kgnω = =

+ 13.608 rad/snω =

13.608

2 2nf

ωπ π

= = 2.17 Hzf = !

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Chapter 19, Solution 88.

( )1

41 cos

3

rV mgh mg θ

π = = −

2

1 cos2

mθθ− ≈

2

12

3

mV mgrθπ

=

2 2 2 2

2

1 1 1 1

2 2 2 2A A B B

T I I mrω ω ω = + +

Where

2 21

2 8 4 256A B

m r mrI I

= = =

and 4A B

ω ω ω= =

2 2 2 2

2

2

5

16 4 16

mr mr mrT

ωω

∴ = + =

1 2

2,

mV T=

2

mgrθ 5

3

m

π=

2 2 2

n mr ω θ16

2 32

,15

n

g

rω

π=

1 32

2 15n

gf

rπ π =

�

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Chapter 19, Solution 89.

Kinematics:

Data

0.15 , 0.05m m m AB m mv r v bω ω ω ω= = = = 10kg

ABm =

4kgD

m =

Conservation of Energy

1 1 2 2T V T V+ = + (1)

Where

( )1

1

0Position 1, Max. Displacement

cosAB m

T

V m g r b θ=

= − −

( )0.10 cos 9.81cosAB m m

m g θ θ= − = −

2 2 2

2

1 1 12

2 2 2D m D m AB AB

T m v I m vω = + +

( )( ) ( )( ) ( )( )2 22 21 1 1 12 4 0.15 4 0.15 10 0.05

2 2 2 2m m m

ω ω ω = + +

20.1475

mω=

( ) ( )20.10 9.81

AB ABV m g r b m g= − − = − = −

Into (1)

20 9.81cos 0.1475 9.81

m mθ ω− = −

( )20.1475 9.81 1 cos

m mω θ= −

But 211 cos

2m m

θ θ− ≈

2 20.1475 4.905

m mω θ= (2)

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m n m

ω ω θ=

Into (2)

( )2 2 20.1475 4.905

n m mω θ θ=

So

2 4.90533.254

0.1475n

ω = =

5.7666 rad/sn

ω =

0.9178 Hz2

nfωπ

= = 0.918 Hznf = �

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Chapter 19, Solution 90.

22 21 1

2 72 2 6

xT mx my m

= + =

&& &

Equilibrium of BC:

30

2 2 2Dmg l l

M ku

Σ = = ⋅ −

Where 0, natural length2 3

mgu l l u

k= = = +

( ) ( )2 20

12 2

2 2kk

V k l x l x u = − − = +

,gV mgy= − where

222 3

2 2

l ll x y

= − + +

And ( ) ( )2 2

2 23 3 4

3 0, 2

l l x lxy ly x lx y

− + − −+ + − = =

So 21 1 8 H.O.T.

23 3 3y x x

l

= + − +

Then 2

2 42 2 constant

3 3 3

x mgxV kx kux mg

l= + + − +

2

42

3 376

n

mgk

lm

ω+

=

1 12 8 Hz

2 7 7 3n

k gf

m lπ= + !

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Chapter 19, Solution 91.

Position 1

( )22 2 2 21

1 1 12 2

2 2 2D m D m r mT I m r c m rθ θ θ = + − +

& & &

For one disk ( )2 2 2

2 2 22

1 4 160.31987

2 3 2 9D D D D D D DA

r r rI I m c m r m m m r

π π = − = − = − =

( )2

2 2 241 0.3313

3D D Dm r c m r m rπ

− = − =

( ) 2 2 21 0.3199 0.3313 0.5D r mT m r m r θ = + +

&

( ) 2 20.6512 0.5D r mm m r θ= + &

But , D rD r

W Wm m

g g= =

So ( ) ( )2

2 21

1 60.65117 6 0.5 4 0.045866

32.2 12 m mT θ θ = + = & &

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Position 2 ( )2

2 22

0, 2 1 cos , , 1 cos 12 3 2m m

D m mr

T V m gc cθ θθ θ

π = = − = − ≅

"

( )2

2 2 22

4 4 4 62 6 1.27324

3 2 3 3 12m

D D m m mr

V m g W rθ θ θ θ

π π π = = = =

&

2 21 1 2 2 : 0.045866 0 0 1.27324m mT V T V θ θ+ = + + = +&

2 1.27324: 27.747

0.045866m n m nθ ω θ ω= = =&

2

5.2677, 1.192 sn nn

πω τω

= = =

1.192 snτ = !

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Chapter 19, Solution 92.

With 2 2

4 lb 6 lb 40, , 20 lb/ft, ft

1232.2 ft/s 32.2 ft/sm M k l

= = = =

2 2 21

2 2 2 2 2

l lV k mg Mgl

θ θ θ = + +

And for small :θ

( )2

22 2 21 1 1 1 1

2 3 2 2 2

lT ml M l Mr

r

θθ θ = + +

&& &

So

2

22 2

8 4 2 23.05935

36 4

n

kl l lmg Mg

l lm M

ω+ +

= =+

Then 2

1.308 sn

πτω

= = !

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Chapter 19, Solution 93.

2 2

2

1 1

2 12T ml θ =

&

Ignoring static terms:

2 2

1

1 1

2 2 2 2

a aV k k

θ θ = +

2

2

4ka

θ=

2

2 2 2 2

2 1

1:

24 4n m m

aT V ml kω θ θ

= =

2

2

2

6

n

ka

mlω =

6

2n

a kf

l mπ= �

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Chapter 19, Solution 94.

2 2m m m ma a

AA BB ll

θ α α θ′ ′= = = =

Position 1

( )1 10 1 coscT V mgy mgl α= = = −

For small angles 2 2

221 cos 2sin

2 2 8m m

m ma

l

α αα θ− = ≈ =

22

1 28m

aV mgl

lθ

=

Position 2 2 2 22 2

1 1 1 0

2 2 12m mT I ma Vθ θ = = =

& &

m n mθ ω θ=&

1 1 2 2T V T V+ = +

22 2 2

2

10

248n m

amgl ma

lω θ

+ +

2 3n

g

lω =

1 3

2ng

flπ

= !

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Chapter 19, Solution 95.

( ) 21 cos2m mn r rθ θ1= − = ( )m mv r r ω= −

Position 1: 21 1

10,

2 mT V Wh mgrθ= = =

Position 2: 2 0V =

( )22 2 2 2 22

1 1 1 1

2 2 2 2m m m mT I mv mk m r rω ω ω= + = + −

Conservation of energy

1 1 2 2T V T V+ = +

( )22 2 21 10

2 2m mmgr m k r rθ ω + = + −

But for simple harmonic motion, m n mω ω θ=

( ) ( )222 21 1

2 2m n mmgr m k r rθ ω θ = + −

( )

222n

rg

k r rω =

+ − (1)

For half section of pipe

2 2

1r

r r r rπ π

= − = −

Parallel-Axis Theorem: 20I I mr= +

2

2 20

2;

rI mr mr I m

π = = +

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22

41I mr

π = −

2 22

41k r

π = −

Eq. (1): 22

2 22

2

4 21 1

n

r

g

r r

πω

ππ

= − + −

2

22 2

2 2

44 4 4 21 1n

rg

grr

π πω

πππ π

= = −− + − +

( )2

21

2 4 2 4 2ng g g g

r r rπω

π π ππ

= = =− − −

2

n

πτω

= ( )2

2r

g

πτ π

−= !

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Chapter 19, Solution 96.

( ) 2sin sinm m mr rθ θ θ≈

( )2

1 cos2m

mr rθθ− ≈

Position 1 maximum deflection 1 0T =

2

1 2m

mV Wy mgrθ= =

Position 2 ( )0θ =

2 2 22

1 1 1

2 2 12m mT I mlθ θ = =

& &

m n mθ ω θ=&

2 2 22

1 1

2 12 n mT ml ω θ =

1 1 2 2T V T V+ = +

2 2 2 21 1 10

2 2 12m n mmgr mlθ ω θ + =

22

2

12 2 2

12n nn

gr l

grl

πω τ πω

= = =

3

nl

gr

πτ = !

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Chapter 19, Solution 97.

This is not a damped vibration. However, the kinetic energy of the fluid must be included

(a) Position 2 2 0T =

22

1

2 mV kx=

Position 1

2 21 spere fluid

1 1

2 4s m mr

T T T m v Vvg

= + = +

1 0V =

2 2 21 1 2 2

1 1 1 0 0

2 4 2s m m mr

T V T V m v Vv kxg

+ = + + + = +

m m m nv x x ω= =&

2 2 2 21 1 1,

2 2 2ω ω

+ = = +

s m n m n

s

r km V x kx

g rm V

g

32

2

1 1 62.4 lb/ft 4 4 in.0.15032 lb s /ft

2 2 3 12 in./ft32.2 ft/s

rV

gπ

= = ⋅

2 2

22

40 lb/ft220.54 s

1 lb0.15032 lb s /ft

32.2 ft/s

nω −= =

+ ⋅

114.850 snω −=

2

0.4231 snn

πτω

= =

0.423 snτ = !

(b) The acceleration does not change mass

0.423 snτ∴ = !

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Chapter 19, Solution 98.

Eq. (19.33)

2

1

m

m

f

n

Pkxωω

=

−

1450 N/m10.607 s

4 kgnk

mω −= = =

13 N0.28889 m

450 N/mmP

k= =

2

0.28889 m

110.607

m

f

xω

=

−

(a) 2

0.28889 m5: 0.03714 m

51

10.607

f mxω = = = −

( )In Phase 37.1 mmmx = !

(b) 2

0.28889 m10: 0.25984 m

101

10.607

f mxω = = = −

( )In Phase 260 mmmx = !

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Chapter 19, Solution 99.

Eq. (19.33)

22 ,

1

m

m n

f

n

Pkkxm

ωωω

= =

−

( ) 22, or m m

mm ff

P Px k

x mk m ωω= =

+−

(a) In phase

( )( )1

9 N

0.15 m 4 kg 5 sk

−=

+160.0 N/m= !

(b) Out of phase 0.15 mmx = −

( )( )1

9 N40.0 N/m

0.15 m 4 kg 5 sk

−= =

− +!

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Chapter 19, Solution 100.

Eq. (19.33) 2

21

m

mf

n

Pkxωω

=−

2 mst n

P k

k mδ ω= =

2

2

3

1

stst

f

n

δ δωω

≥−

2

2

11

3f

n

ωω

− ≤

2

2

2

3f

n

ωω

>

Also 2

2

3

1

stst

f

n

δ δωω

< −−

2 2

2 2

1 41

3 3f f

n n

ω ωω ω

− < − <

2

2

2 4 2 4

3 3 3 3f

fn

k k

m m

ωω

ω< < < < !

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Chapter 19, Solution 101.

2 2 sinB m fM mb kl P l tθ θ ωΣ = = − +&&

2 2 sinm fmb kl P l tθ θ ω+ =&&

2

2 42.4641 rad/s, sinn m fkl

tmb

ω θ θ ω= = =

2

2 2 2

0.14 in. 6.7620.0175 rad

8 in. 1803.2

m

mn f f

P l

mbθω ω ω

±= = ± = =− −

Lower frequency: ( )26.762 0.0175 1803.2 , 37.64 rad/sf fω ω= − =

Upper frequency: ( )26.762 0.0175 1803.2 , 46.79 rad/sf fω ω= − − =

37.6 rad/s 46.8 rad/sfω< < !

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Chapter 19, Solution 102.

Referring to the figure and solution for Problem 19.101

2

232

1401 1 28.309412

4032.2

nkl

b m b bω

= = =

Range 2

2 2

0.1412

m

mn f

P l

mbb

θω ω

= ± =

−

( ) 22 2

3.00533 3.00533

801.422 225225n bb ω= =

−−

Lower frequency 23.00533 9.3499 2.625b b= −

22.625 3.00533 9.3499 0b b+ − =

1.3998 ftb =

Upper frequency 23.00533 9.3499 2.625b b= − +

22.625 3.00533 9.3499 0b b− − =

2.5446 ftb =

16.80 in. 30.5 in.b< < !

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Chapter 19, Solution 103.

sinx m f C C B BF P t m x m xωΣ = = +&& &&

C BM mΣ = − g l Bmθ = l Bx&&

where B Cx x lθ= +

Eliminate and :C Bx x&&

( ) sinθ θ ω+ + = −&&C B m fml m m g P t

( )

( )2 sin

1

θ ωω

−

+ ∴ = −

+

m

C Bf

f C

C B

P

m m gt

m l

m m g

0.02786136sin 3 , and Bt x gθ θ∴ = − = −&&

( )2

0.273320.27332sin 3 , sin 3

3B Bx t x t

−= + =&&

( )30.4 mm, out of phase 180Bmx = ° !

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Chapter 19, Solution 104.

( )B e Bmx k x δ= − −&&

sinB e B e m fmx k x k tδ ω+ =&&

Thus, from Equation (19.31) and ( )19.33′

( ) 2

21

δωω

=−

mm B

f

n

x

22 29.81 m/s

245.25 s0.040 m

en

ST

k g

mω

δ−= = = =

( ) 2

0.5

1245.25

ω=

−m B

f

x

2

1020

1245.42

ω>

− f

In phase 2

20 1 10245.25

fω + − >

2 245.25

2fω <

11.0736 rad/sfω <

11.07 rad/sfω < !

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Out of phase 2

20 1 10245.25

fω − − >

( )2 3245.25

2fω >

19.1801 rad/sfω >

19.18 rad/sfω > !

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Chapter 19, Solution 105.

From Eq. ( )19.33′ 2

1

mm

f

n

xδ

ωω

= −

(a) 2 117 N/m

2 kgnk

mω = =

2 258.5 snω −=

0.1 m

0.17463 m25

158.5

mx = = −

174.6 mmmx = !

(b) Since 2 2,f nω ω< x and δ are in phase and net spring deflection is

x δ− and ( )F k x δ= −

( )( )117 N/m 0.17463 m 0.10 mmF = −

8.73 NmF = !

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Chapter 19, Solution 106.

2

21

mm

f

n

xδ

ωω

=

−

2 2

2

130 lb/ft261.63 s

16 lb

32.2 ft/s

nk

mω −= = =

In phase

( ) 2

2

11

1m m m m

f

n

F k x kδ δωω

= − = − −

30 lbmF <

So 2

2

1 30 lb1 0.46154

6 in.130 lb/ft1

12 in.ftf

n

ωω

− < =

−

or 2

21 0.68241f

n

ωω

− >

2

2 1 0.68241 0.31579f

n

ωω

< − =

( )( )2 20.31579 0.31579 216.63f nω ω< =

2 268.409 sfω −<

So 19.0896 sfω −<

9.09 rad/sfω < !

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Out of phase

( ) 6 in.130 lb/ft 130 65 30

12 in./ftm m m m mF k x x xδ = − = − = + >

There is no value of mx for which

30 lb when and are out of phasemF x δ< !

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Chapter 19, Solution 107.

( )52 sin 2m fl kl k t l

gθ θ δ ω θ

∴ = − + −

&&

7.5 0.2

30 8 sin10 0.1333sin1032.2 12

t tθ θ + = =

&&

( )0.1333

0.0198659 radians23.29 30mθ = =

− +

(a) ( )( )2 2max 1.5 2.98 ft/sm flθ θ ω= =&& !

(b) ( ) ( )( )0.2max 3 4 2 1.5 0.01988 0.516 lb

12R

= − =

!

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Chapter 19, Solution 108.

2; sin2 2A m fl kl

M ml k tθ δ ω θ Σ = −

&&

22 sin , sin

4 2m

f m fkl kl

ml t tδθ θ ω θ θ ω+ = =&&

So ( )

( )2 23600 N/m

450 s4 4 2 kgnk

mω −= = =

( )( )( )( )

( )2 2 2

3600 N/m 0.003 m

2 2 kg 0.4 m2450 225 s

m

mn f

k

ml

δ

θω ω −= =

− −

0.03 rad 1.719= = °

(a) 0.450 rad/sf mω θ = !

(b) 2 22.70 m/sf mlω θ = !

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Chapter 19, Solution 109.

F maΣ =

( ) 2

c cmx k x

δδ δ′ ′= − = −&&

sin2 2

mc c f

kx x k k t

m

δ δ ω+ = − = −&&

From Equation (19.31 and 19.33′) ( )2

2

2

1

m

c mf

n

x

δ

ωω

−=

−

Thus ( )( )2

0.010m

2

181

654

cm

x

−=

−( )

2

2 9.81 m/s

0.015 mn

ST c

gωδ

= =

( ) 2 20.009909 m 654 s

c nm

x ω −= − =

( ) sinc c fmX x tω=

( ) ( ) ( ) ( )( )22 1 0.009909 m 18 sc c c f cm m m

X a x aω −= = − =&&

( ) 23.21 m/s

cm

a = �

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Chapter 19, Solution 110.

2

n

g

lω =

From Eq. (19.33'):

2

1

1

m

m

n

x

δ ωω

=

−

We want 1 or 1 or 1 1m m m

m m m

x

x x

δ δδ

< > − > >

Thus we have

2

1 1 1

n

ωω

− > − >

2

2 0

n

ωω

− > − >

2

0 2

n

ωω

> >

Since 2:

n

g

lω = 2

g

lω > �

2

0

n

ωω

<

(Impossible)

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Chapter 19, Solution 111.

From Eq. (19.33'):

2

1

1

m

m

n

x

δ ωω

=

−

We want 1 1 1

2 or or2 2 2

m m m

m m m

x

x x

δ δδ

> < − < <

Thus we must have

2

1 11

2 2n

ωω

− < − <

2

3 1

2 2n

ωω

− < − < −

2

1 3

2 2n

ωω

< <

Since 2

n

g

lω =

3

2 2

g g

l lω< < �

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Chapter 19, Solution 112.

From Problem 19.115

2

2

1

f

nm

f

n

mr

Mx

ωω

ωω

=

−

215000 N/mk =

2 2215000 N/m

1.075 s200 kg

n

k

mω −= = =

( ) 60.03 kg0.2 m 30 10 m

200 kg

mrM

− = = ×

26

2

30 10 m1.5 mm: 0.0015

1

fm

nf

n

x

ωωω

ω

− ×= ≥

−

2 2

50 50f f

n n

ω ωω ω

− ≥

So

2

50

51

f

n

ωω

≤

0.99015f

n

ωω

≤

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( )1

20.99015 1075 32.464 rad/sfω ≤ =

( )( ) 132.464 rad/s 60 s/min

2 rad/revπ

=

310.1 rev/min=

310 rpmfω ≤ �

26

2

30 10 m1.5 mm: 0.0015

1

fm

nf

n

x

ωωω

ω

− ×= − − ≤

−

( )1

21.01015 1075 33.120 rad/s 316.27 rev/minfω ≥ = =

316 rpmfω ≥ �

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Chapter 19, Solution 113.

18 kgm =

Four springs each of constant 40 kN/m

We note that the motor is constrained to move vertically.

( )4 40 kN/m 160 kN/mk = =

3160 10 N/m

94.281 rad/s 900 rpm18 kgn

k

mω ×= = = =

For 1200 rpm 125.664 rad/sω = = we have

31.5 mm 1.5 10 mmx −= = ×

Eq. (19.33): 2

1

m

m

n

Pkxωω

=

−

Thus: 2

1mm

n

Px

k

ωω

= −

( )2

3 3125.6641.5 10 m 1 1.1648 10 m

94.281− −

= × − = − ×

(out of phase)

( )( )3160 kN/m 1.1648 10 m 186.4 NmP −= × =

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We have found: 186.4 NmP =

For an unbalanced rotor of mass 4 kg,Rm = rotating at 1200 rpm 125.664 rad/s,ω = = with the mass center at a distance r from the axis of rotation, we have,

2m RP m rω=

( )( )

32 2

186.4 N2.95 10 m

4 kg 125.664 rad/sm

R

Pr

m ω−= = = ×

2.95 mmr = !

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Chapter 19, Solution 114.

From Problem 19.115

2

2

1

f

nm

f

n

mr

Mx

ωω

ωω

=

−

Since 2 ,nM kω =

2

2

1

f

m

f

n

mr

kx

ω

ωω

=

−

Before plate is added 2

2 232.2 ft/s644 s

0.6ft

12

ns

gωδ

−= = =

( )2 22

360 lb644 s 7200 lb/ft

32.2 ft/snk Mω −= = =

22

0.9lb

16 0.0017469 lb s /ft32.2 ft/s

m = = ⋅

2

9 20.0017469 lb s /ft 7.5ft 151.64 10 ft s

7200 lb/ft 12

mr

k−⋅ = = × ⋅

After plate is added

nω changes since M changes

2n

k

Mω′ =

′

Since amplitude of vibration is to be less than 32.16 10 in.−× for motor speeds above 300 rpm

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( )

29 2

13

2

2151.64 10 ft s 300

60 s2.16 10ft

12 2300

601

π

π

ω

−−−

× ⋅ ⋅ × = − = ⋅

− ′

m

n

x

Solve for

( )

63

2

149.663 10: 0.18 10

986.961

n

n

ω

ω

−− ×′ − × =

−′

( )2538.89n

k

Mω′ = =

′

720013.361

538.89M ′ = =

( )( )13.361 32.2 430.22 lbW M g′ ′= = =

430.2 lb 360 lb 70.22 lbW∆ = − =

plate 70.2 lbW = !

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Chapter 19, Solution 115.

Rotor

Motor

sinm fF ma p t kx MxωΣ = − = &&

sinm fMx kx p tω+ =&&

sinmf

k px x t

m Mω+ =&&

2n

k

mω =

From Eq. (19.33)

2

1

m

m

f

n

pkxωω

=

−

But 2

2fmn

mrpk M

k k

ωω= =

2fm

n

p mr

k M

ωω =

Thus ( )( )

( )

2

2 Q.E.D1

f

n

mf

n

mM

x

rωω

ωω

=−

!

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Chapter 19, Solution 116.

(a) One collar: ( ) ( )21 1

15 mmm n

kx

mω= =

(b) Two collars: ( ) ( ) ( )2 2

2 2 1

118 mm

2 2m n n

kx

mω ω= = =

2 1

2

n n

ω ωω ω

=

(c) Three collars:

( ) ( ) ( )2 2

3 3 1

3 1

1unknown, , 3

3 3m n n

n n

kx

m

ω ωω ωω ω

= = = =

We also note that the amplitude m

δ of the displacement of the base remains constant.

Referring to Sec. 19.7, Fig. 19.9, we note that, since ( ) ( ) ( ) ( )2 1

2 1

and ,m m

n n

x x

ω ωω ω

> > we must have

( )1

1

n

ωω

< and ( )1

0.mx > However, ( )

2n

ωω

may be either 1 or > 1,< with ( )2m

x being

correspondingly either 0 or 0.> <

1. Assuming ( )2

0:mx >

For one collar:

( )2 21

1 1

15 mm

1 1

m m

m

n n

x

δ δ

ω ωω ω

= + =

− −

j (1)

For two collars:

( )2 22

2 1

18 mm

1 1 2

m m

m

n n

x

δ δ

ω ωω ω

= + =

− −

j (2)

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Dividing (2) by (1), member by member:

2

2

1

2

1

1

11

1.2 ; we find 7

1 2

n

n

n

ωω ω

ωωω

−

= =

−

Substituting into (1): ( ) 1 9015 mm 1 mm

7 7m

δ = − =

For three collars:

( )23

1

90 mm

907mm,

1 41 3

1 3 7

m

m

n

x

δ

ωω

= = =

− −

( )3

22.5 mmmx = �

2. Assuming ( )2

0:mx <

For two collars, we have: 2

1

18 mm

1 2

m

n

δ

ωω

− =

−

(3)

Dividing (3) by (1), member by member:

2

1

2

1

1

1.2

1 2

n

n

ωω

ωω

− − =

−

2 2

1 1

1.2 2.4 1

n n

ω ωω ω

− + = −

2

1

2.2 1.1

3.4 1.7n

ωω

= =

Substitute into (1): ( ) 1.1 915 mm 1 mm

1.7 1.7m

δ = − =

For three collars:

( )23

1

9

9 mm1.7,

1.1 1.61 3

1 3 1.7

m

m

n

x

δ

ωω

= = =

− − −

( )3

5.63 mmmx = − �

( )out of phase

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Points corresponding to the two solutions are indicated below:

COSMOS: Complete Online Solutions Manual Organization System

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© 2007 The McGraw-Hill Companies.

Chapter 19, Solution 117.

(a) One collar: ( ) ( )21 1

9 mmm n

kx

mω= =

(b) Two collars: ( ) ( ) ( )2 2

2 2 1

2 1

13 mm , 2

2 2m n n

n n

kx

m

ω ωω ωω ω

= = = =

(c) Three collars: ( ) ( )2

2

3 3

1 3 1

1unknown , 3

3 3m n

n n n

kx

m

ω ω ωωω ω ω

= = = =

We also note that the amplitude m

δ of the displacement of the base remains constant.

Referring to Sec. 19.7, Fig. 19.9, we note that, since ( ) ( )2 1

2 1

and ,m m

n n

x x

ω ωω ω

< >

we must

have

2

1

n

ωω

>

and ( )

20.

mx < However,

1n

ωω

may be either 1 or > 1,< with ( )1m

x being

correspondingly either 0 or 0.> <

1. Assuming ( )1

0:mx >

For one collar:

( )2 21

1 1

9 mm

1 1

m m

m

n n

x

δ δ

ω ωω ω

= + =

− −

j (1)

For two collars:

( )2 22

2 1

3 mm

1 1 2

m m

m

n n

x

δ δ

ω ωω ω

= − =

− −

j (2)

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Dividing (1) by (2), member by member:

2

2

1

2

1

1

1 24

3 ; we find, 5

1

n

n

n

ωω ω

ωωω

−

− = =

−

From Eq. (1) 1.8 mmm

δ =

For three collars:

( )23

1

1.8 mm1.286 mm

41 3

1 35

m

m

n

x

δ

ωω

= = = − − −

( )3

1.286 mmmx = �

(out of phase)

2. Assuming ( )1

0:mx <

For one collar, we have now:

2

1

9 mm

1

δ

ωω

− =

−

m

n

(3)

Dividing (3) by (2), member by member:

2

1

2

1

1 2

3

1

n

n

ωω

ωω

−

=

−

2 2

1 1

3 3 1 2 ,n n

ω ωω ω

− = −

2

1

2

n

ωω

=

Substituting into (3): ( )( )9 mm 1 2 9 mmm

δ = − − =

For three collars:

( ) ( )23

1

9 mm1.800 mm

1 3 2

1 3

m

m

n

x

δ

ωω

= = = −−

−

( )3

1.800 mmmx = �

(out of phase)

COSMOS: Complete Online Solutions Manual Organization System

Vector Mechanics for Engineers: Statics and Dynamics, 8/e, Ferdinand P. Beer, E. Russell Johnston, Jr.,

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Points corresponding to the two solutions are indicated below:

COSMOS: Complete Online Solutions Manual Organization System

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© 2007 The McGraw-Hill Companies.

Chapter 19, Solution 118.

2

2

1

m

m m f

f

n

P

kx P mrω

ωω

= =

−

( )20.8

lb s /ft, 0.5 ft, 20,000 lb/ft16 32.2

m r k= ⋅ = =

2 2 220,000

1610 rad/s400

32.2

n nω ω= =

2

sin , sinm f f m fx x t x x tω ω ω= = −&&

( )22

2

2

2

acc. amplitude

1

f

f m

f

n

mr

kx

ω

ωωω

∴ = =

−

( ) ( )( )

22

22

2

2 2

0.00077640.4 ft/s

1 1 20,0001610 1610

ωω

ω ω± = =

− −

f

f

f f

mr

k

( )22 2) 0.0007764 4.9689 8000 0f fω ω+ + − =

2

1332.55 36.5 rad/sf fω ω= =

( )22 2) 0.0007764 4.9689 8000 0f fω ω− − + =

There are no real roots, 2

acc. 0.4 ft/s∴ >

Answer: 36.5 rad/sfω <

349 rpmfω < �

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Chapter 19, Solution 119.

2

2 22

22

2 , ,

1

f

m f m n

f

n

mrkkP mr x

M

ω

ω ωωω

= = =

−

With 2

2

22 36 lb , 40 rad/s

12

fm f

f

mrkx

M

k

ωω π

ω= = ± =

−

Solving for r 2

2

2 12

2

fm

f

Mkx

kr

m

ω

ω

−

= ±

Data : 2 36 lbmkx =

33400 lb/in. 40.8 10 lb/ftk = = ×

600 lb

32.2 ft/sM =

3 22

3.5 oz16 oz/lb

6.79348 10 lb s /ft32.2 ft/s

m −

= = × ⋅

40 rad/sfω π=

( )( )

( )

( )( )

212

3

23 2 1

600 lb40 s

32.2 ft/s36 lb 1

2 40.8 10 lb/ft

0.43725 ft2 6.79348 10 lb s /ft 40 s

π

π

−

− −

− × = ± =

× ⋅r

5.25 in.r = !

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Chapter 19, Solution 120.

(1) From Equation (19.33) 2

1

m

m

f

n

Pkxωω

=

−

Force transmitted, ( ) 2

1

m

T mmf

n

PkP kx kωω

= = −

Thus

Transmissibility ( )

2

1

1

T m

m f

n

P

P ωω

= =

−

!

(2) From Equation ( )19.33′

Displacement transmitted 2

1

mm

f

n

xδωω

=

−

Transmissibility 2

1

1

m

m f

n

x

δ ωω

= =

−

!

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Chapter 19, Solution 121.

G describes a circle about the axis AB of radius .r e+

Thus ( ) 2n fa r e ω= +

Deflection of the shaft is.

Thus F kr=

nF ma=

( ) 2fkr m r e ω= +

22nn

k km

mω

ω= =

( ) 22 fn

kkr r e ω

ω= +

2

2

2

21

f

n

f

n

e

r

ωωωω

=−

(a) Resonance occurs when , i.e., f n rω ω= → ∞

580000 N/m

146.57 rad/s 1399.6 rpm27 kgn

k

mω = = = =

1400 rpmn fω ω= = !

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(b) ( )

26

62

1200150 10 m

1399.6416.28 10 m

12001

1399.6

r

−

−

× = = ×

−

0.416 mmr = !

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Chapter 19, Solution 122.

Total spring constant ( )2 350 lb/ft 700 lb/ftk = =

(a) 2 2

2

700 lb/ft45.08 s

500 lb

32.2 ft/s

nk

mω −= = =

15 ft, 2 in. 0.16667 ftmλ δ= = =

sin ,mx

y x vtδλ

= =

2

sin , and ,m fv

y tv

λ πδ τ ωλ τ

= = =

so 2 2

0.16667sin where15f f

v vy t

π πω ωλ

= = =

From Equation ( )19.33′ , 2

21

mm

f

n

xδ

ωω

=

−

Resonance: 1245.08 s ,

15f nvπω ω −= = =

16.029 ft/s 10.93 mphv = = !

(b) ( )

1

52802 30

3600 18.431s15f

πω −

= =

( )0.16667

0.0255 ft18.431

145.08

mx = = −−

!

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Chapter 19, Solution 123.

In steady state vibration, block A does not move and therefore remains in its original equilibrium position

(a) Block A 0FΣ =

sinm fkx P tω= − (1)

Block B ,BF m xΣ = &&

0Bm x kx+ =&&

2sin , /m n n BX x t k mω ω= =

From (1) sin sinm n m fkx t P tω ω= −

2 rad/s,n f m mkx Pω ω= = = −

12 sB

k

M−=

22

44 lb4 s 5.4658 lb/ft

32.2 ft/sk −

⇒ = =

5.47 lb/ftk = !

(b) ,5 lb

0.91477 ft5.4658 lb/ftm m mkx P x

−= − = = −

0.915 ftmx = − !

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Chapter 19, Solution 124.

2

2

sin

1

mf

f

n

x tδ ω

ωω

= −

sinm fy tδ ω=

relative motionz =

2

2

sin

1

mm f

f

n

z x y tδ δ ω

ωω

= − = − −

2

2

2 2

2 2

11

1 1

fm

nm m

f f

n n

z

ωδ

ωδω ωω ω

= − = − −

(a)

2 2

2

2 2

2

75025150

1.041724750

1 1150

f

m n

m f

n

z

ωω

δ ωω

= = = = − −

!

Error 4.17%=

(b)

2

2

2

2

1

1

f

m n

m f

n

z

ωω

δ ωω

= =−

( )2

2

2 21 2 150 106.07 Hz

2 2f

f nn

f fωω

= = = =

106.1 Hznf = !

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Chapter 19, Solution 125.

From Problem 19.124

2

222

2 2

2 2

or

1 1

f

m fnm m m n

f f

n n

z z

ωδ ωωδ ω

ω ωω ω

= =− −

Now 2 22

2

, so

1

mm m f m n

f

n

aa zδ ω ω

ωω

= =−

2 2480

0.074381760

f

n

ωω

= =

Then 2 1

1.08031 0.07438

m n

m

z

a

ω = =−

Error 8.03%= !

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Chapter 19, Solution 126.

Since cc c> we use equation (19.42), where

1 20, 0λ λ< <

1 21 2

t tx c e c eλ λ= + (1)

1 21 1 2 2

t tdxv c e c e

dtλ λλ λ= = + (2)

(a) 00 0t , x x v= = =

From (1) and (2) 0 1 2x c c= +

1 1 2 20 c cλ λ= +

Solving for 1c and 2c

2 11 0 2 0

2 1 2 1

c x c xx

λ λλ λ λ

−= =− −

Substituting for 1c and 2c in (1)

1 222 1

2 1

t txx e eλ λλ λ

λ λ = − −

For 0x = When ,t ≠ ∞ we must have

( )2 12 1 21 2

1

0 tt te e e λ λλ λ λλ λλ

−− = = (3)

Recall that

1 20, 0.λ λ< < Choosing 1 2 and λ λ so that 1 2 0,λ λ< < we have

22 1

1

0 1 and 0λ λ λλ

< < − >

Thus a positive solution for 0t > for Equation (3) cannot exist since it would require that e raised to a positive power be less than 1, which is impossible. Thus x is never 0.

The x t− curve for this case is shown

!

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(b) 00, 0,t x v v= = = Equations (1) and (2), yield

1 2 0 1 1 2 20 c c v c cλ λ= + = +

Solving for 1 2 and ,c c 0 21 2

2 1 2 1

,v v

c cλ λ λ λ

= − =− −

Substituting into (1)

2 10

2 1

t tvx e eλ λ

λ λ = − −

For 0,x = t α≠

2 1t te eλ λ=

For 1 2, ;cc c λ λ> ≠ Thus no solution can exist for t, and x is never 0.

The x t− curve for this motion is as shown

!

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Chapter 19, Solution 127.

Substitute the initial conditions, 0 00, ,t x x v v= = = in Equations (1) and (2) of Problem 19.126

0 1 2 0 1 1 2 2x c c v c cλ λ= + = +

Solving for 1 2 and ,c c ( ) ( )0 2 0 0 1 0

1 22 1 2 1

,v x v x

c cλ λ

λ λ λ λ− −

= − =− −

And substituting in (1) ( ) ( )2 10 1 0 0 2 0

2 1

1 t tx v x e v x eλ λλ λλ λ

= − − − −

For 0,x t= ≠ ∞ ( ) ( )2 10 1 0 0 2 0

t tv x e v x eλ λλ λ− = −

( ) ( )( )

2 1 0 2 0

0 1 0

t v xe

v xλ λ λ

λ− −

=−

( )0 2 0

2 1 0 1 0

1ln

v xt

v x

λλ λ λ

−=− −

This defines one value of t only for 0,x = which will exist if the natural log is positive,

i.e., if 0 2 0

0 1 0

1.v x

v x

λλ

− >−

Assuming 1 2 0λ λ< <

This occurs if 0 1 0v xλ<

!

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Chapter 19, Solution 128.

For light damping, cc c<

Equation (19.46) ( )20 0sin

ct

mx x e tω φ − = +

At given maximum displacement, ,n nt t x x= =

( ) 20 0sin 1,

nc

tm

n nt x x eω φ − + = =

At next maximum displacement, 1 1,n nt t x x+ += =

( ) 120 1 1 0sin 1,

nc

tm

n nt x x eω φ+

−

+ ++ = =

But 1 2D n D nt tω ω+ − = π

12

n nD

t tω+

π− =

Ratio of successive displacements:

( )1

1

22

20 2

1 20

nn n

D

n

ct ccm t t mn m

ctn m

x x ee e

xx e

πω+

+

− +− −

−+= = =

Thus 1

ln n

n D

x c

x mω+

π= (1)

From Equations (19.45) and (19.41) 2

1D nc

c

cω ω

= −

2

12

cD

c c

m cω = −

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Chapter 19, Solution 129.

As in Problem 19.128, for maximum displacements nx and n kx + at ( )0 and , sin 1n n k nt t tω φ+ + =

and ( )sin 1.n n ktω φ+ + =

( ) ( )( )2 20 0

c cn n km m

t tn n kx x e x x e +− −

+= =

Ratio of maximum displacements

( )

( )( )2

2

2

0

0

cnm c

n n kmc

n km

tt tn

tn k

x x ee

x x e

−−

+−

+

−

+= =

But ( ) 22D n k D n n n k

D

t t k t t kπω ω π

ω+ +− = − =

Thus 2

; ln2

n n

n k D n k D

x c k x ck

x m x m

π πω ω+ +

= + =

(2)

But from Problem 19.128 Equation (1)

1

log decrement ln n

n D

x c

x m

πω+

= =

Comparing with Equation (2)

1log decrement ln Q.E.D.n

n k

x

k x += !

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Chapter 19, Solution 130.

Fig 19.11

Eq. (19.46)

( ) ( )2

0sin

c

mt

Dx x e tω φ−

= +

(a) Maxima (positive or negative) when 0x =&

( ) ( ) ( ) ( )2 2

0 0sin cos

2

c c

m m

t t

D D D

cx x e t x e t

m

ω φ ω ω φ− −− = + + +

&

Thus zero velocities occur at times when

( ) 20, or tan D

D

mx t

c

ωω φ= + =& (1)

The time to the first zero velocity, 1,t is

1

1

2tan

D

D

m

ct

ω φ

ω

− − = (2)

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The time to the next zero velocity where the displacement is negative is

1

1

2tan

D

D

m

ct

ω φ π

ω

− − + ′ = (3)

Subtracting (2) from (3)

1 1Q.E.D

2 2

D D

D

t tπ π τ τ

ω π⋅′ − = = =

(b) Zero displacements occur when

( )sin 0 or at intervals oftθω φ+ =

, 2 ,Dt nω φ π π π+ =

Thus, ( ) ( ) ( ) ( )1 0

1 0

2and

D D

t

t

π φ π φω ω= − −

′ =

Time between ( ) ( )1 10 0

20 Q.E.D

2 2

D D

s

D

t tπ π πτ τω π

−′ ′= − = = =

Plot of Equation (1)

(c) The first maxima occurs at 1, ( )1Dtω φ+

The first zero occurs at ( )( )1 0Dtω φ π+ =

From the above plot ( )( ) ( )1 102

D Dt t

πω φ ω φ+ − + >

or ( ) ( )1 1 1 10 0Q.E.D

2 4

D

D

t t t tπ τω

− > − >

Similar proofs can be made for subsequent max and min

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Chapter 19, Solution 131.

(a) From Problem 19.128, 2

1

2

ln

1

cn

n

c

c

cx

xc

c

π

+

=

−

For 1

1.25 in. and 0.75 in.n nx x += =

2

21.25

ln 0.51080.75

1

c

c

c

c

c

c

π = =

−

2 22

1 10.5108

c

c

c

π + =

2

10.006566

151.3 1c

c

c

= = +

0.0810

c

c

c

= �

(b) 2 2c

kc m km

m= =

( )2

36 lb2 175 lb/ft 25.022 lb s/ft

32.2 ft/s

= = ⋅

( )( )0.0810 25.022 2.037c = =

2.04 lb s/ftc = ⋅ �

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Chapter 19, Solution 132.

c8 rad/s, 2 32 N s/m, 0.01875n nc

k cc m

m cω ω= = = = ⋅ =

( )2

0.1511 7.99859363 rad/s, sint

d n dc

cx c e t

cω ω ω−

= − = =

At 0,t = 1 10.4 m/s, 0.050008791 mdx c cω= = =&

First peak at 1 2

0.1963840654 d

tπ

ω

= =

( )0.15 0.196381 1 0.04855714 m−∴ = =x c e

(a) Log decrement ( )( )2

2 0.01875,

1 0.01875

π=

− Log decrement 0.1178= !

(b) 1

3

1.2657x

x∴ =

3 0.0384 m 38.4 mmx = = !

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Chapter 19, Solution 133.

(a) A critically damped system regains its equilibrium position in the

shortest time.

1320 2ck

c c mm

= = =

Then

2 2

2

13202 2

7790 lb/ft1800 lb

32.2 ft/s

cc

km

= = = !

(b) For a critically damped system: Equation (19.43):

( )1 2ntx c c t e ω−= +

We take 0t = at maximum deflection 0x

Thus ( ) ( ) 00 0, 0x x x= =&

Using the initial conditions

( ) ( ) 00 1 1 00 0 , so x x c e c x= = + =

( )0 2ntx x c t e ω−= +

And ( )0 2 2n nt t

nx x c t e c eω ωω − −= − + +&

( ) 0 2 2 00 0 , so n nx x c c xω ω= = − + =&

Thus ( )0 1 ntnx x t e ωω −= +

For ( )0 1, 1 , with

3 3nt

n nx k

x t em

ωω ω−= = + =

We obtain 1

2

7790 lb/ft11.807 s

1800 lb

32.2 ft/s

nω −= =

And, since 2.289ntω =

2.2890.19387

11.807t = =

0.1939 st = !

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Chapter 19, Solution 134.

From the given data

2150 rad/s, 2 58.7878 N s/mn c n

kc m

mω ω= = = = ⋅

( )( )( )

22.4 kg 9.81 m/sStretch at equilibrium 0.0654 m

2 180 N/m= =

( )( )( )

23.3 kg 9.81 m/sInitial stretch 0.089925 m

2 180 N/m= =

With 2

7.5, 150 1 12.1474 rad/s

58.7878c dc c ω < = − =

From Figure 19.11 1 1 2

0.25862 s2 2 12.1474D

πτ = =

( )1.5625 0.25862Minimum 0.024525x e−= −

0.001637 m= −

( )( )Minimum force in one spring 180 N/m 0.0654 0.01637= −

8.825 N= 8.82 N!

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Chapter 19, Solution 135.

From the given data

2150 rad/s

n

k

mω = =

2 58.7878 N s/mc nc mω= = ⋅

( )( )( )

22.4 kg 9.81 m/sStretch at equilibrium 0.0654 m

2 180 N/m= =

( )( )( )

23.3 kg 9.81 m/sInitial stretch 0.089925 m

2 180 N/m= =

1 2

1 2:

t t

cc c x c e c e

λ λ> = +

So ( ) ( )22.4 kg 60 N s/m 360 0λ λ+ ⋅ + =

( ) ( )( )( )( )

260 60 4 360 2.4

2 2.4λ

− ± −=

1 210, 15λ λ= − = −

1 2

1 2

1 1 2 2

, 0.024524 m solve for ,

0

t x c c

c c

x c cλ λ= = = +

= = + &

1 2

0.073575; 0.04905c c= = −

1 20.1 0.1

1 1 2 2x c e c e

λ λλ λ= +&

0.1065 m/sx = −&

106.5 mm/sv =

�

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Chapter 19, Solution 136.

For small angles:

sin , cos 1θ θ θ! !

0.15 , 0.45 , 0.15n B Gy y yδ θ δ θ δ θ= = =

(a) ( )0 0 effM MΣ = Σ

( )0.15 0.15 17.658 0.45 0.15s D tF f I maα− + − = + (1)

( )( ) ( )( )0.15s A ST STA AF k y kδ δ θ δ= + = +

( )0.45D BF c y cδ θ= = ""

( )221 1 0.6 0.03 m12 12

I ml m= = =

Kinematics , 0.15 0.15taα θ α θ= = ="" ""

From (1)

[ ] ( ) ( ) ( )( )20.03 m 0.0225 m 0.45 0.15 0.15 2.6487 0ST Ac kθ θ θ δ+ + + + − ="" " (2)

In the equilibrium position 0 0MΣ =

( ) ( ) ( )( )0.15 0.15 17.658 0ST Ak δ − =

( ) ( )0.15 2.6487ST Ak δ =

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Equation (2) becomes

0.0525 0.2025 0.225 0m c kθ θ θ+ + ="" "

With 1.8, 2.2, 73m c k= = =

Have 0.0945 0.446 1.643 0θ θ θ+ + ="" " !

(b) Substituting teλ into the differential equation, we obtain

20.0945 0.4455 1.6425 0λ λ+ + =

Roots 1, 2 2.3571 3.4387λ = − ± i

Since the roots are complex conjugates (light damping) the solution of the differential equation is: Equation (19.46)

( )2.35710 sin 3.4387te tθ θ φ−= + (3)

Initial conditions: ( )0 0.023 mByδ =

( ) 0.023 m0 0.05111 rad0.45 m

θ⇒ = =

And ( )0 0θ ="

From (3): ( ) 00 0.05111 sinθ θ φ= =

( ) 0 00 0 2.3571 sin 3.4387 cosθ θ φ θ φ= = − +"

3.4387tan 1.458872.3571

φ = =

0.96989 radφ =

And 00.05111 0.061965 rad

sinθ

φ= =

Thus ( )2.35710.061965 sin 3.4387 0.96989te tθ −= +

Hence ( ) 52.5 2.4183 10 radθ −= − ×

( ) ( )62.5 24.2 10 radθ −= × !

Above horizontal

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Chapter 19, Solution 137.

4

0.612

crθ θ =

& & 2

21 1 10 4

2 2 32.2 12mr θ θ =

&& &&

+ 2 2 2

4 7.5 1 10 4 1 6 156 6

12 12 2 12 3 12AΜ g gθ θ θ

Σ = − − = +

& &&

(a) 0.1143 0.0667 3.75 0θ θ θ+ + =&& & !

3.75

5.7278 rad/s0.1143nω = =

( )2 0.1143 1.3094c nc ω= =

(b) 0.0667

/ 0.05091.3094cc c = = !

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Chapter 19, Solution 138.

(a) No Damping:

( )2 250 lb/in. 500 lb/in. = 6000 lb/ftk = =

( ) ( )2

6000 lb/ft31.08 rad/s 4.947 Hz

2000 lb 32.2 ft/sn

k

mω = = = =

297 rpmf = !

(b) Damped Motion:

9 lb s/in. = 108 lb s/ftc = ⋅ ⋅

( )2

200 lb2 2 31.08 rad/s 386.09 lb s/ft

32.2 ft/sc nc mω

= = = ⋅

108 lb s/ft

0.27973386.09 lb s/ftc

c

c

⋅= =⋅

From Eq. (19.53):

22

1

1 2

m

m

n c n

x

cc

δ ω ωω ω

= − +

For maximum amplitude we set equal to zero the derivative with respect to ωω n

of the square of the

denominator.

2 2

2 1 2 8 0ω ω ωω ω ω

− − + = n n c n

c

c

Rearranging we obtain

2 2

4 1 2 0ω ωω ω

− + = n n c

c

c

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2 2

1 2ωω

= − n c

c

c

Thus ( )2

21 2 0.27973 0.8435

ωω

= − = n

0.91842n

ωω

= ( ) ( )( )0.91842 0.91842 31.08 rad/snω ω= =

28.544 rad/s 4.543 Hzω = = 273 rpmω = !

(c) Amplitude:

From Eq. (19.53): 2 22

2ω ωω ω

= − +

m

m

n c n

Pkx

c

c

For part (a) with 125 lbmP = and nω ω=

[ ] ( )( ) 22

125 lb6000 lb/ft

0.03724 ft1 1 2 0.27973 1

mx = = − +

0.447 in.mx = !

For part (b) with 125 lbmP = and 0.91842ωω

=n

( ) ( )( )2 22

125 lb6000 lb/ft

0.03879 ft

1 0.91842 2 0.27973 0.91842mx = =

− +

0.465 in.mx = !

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Chapter 19, Solution 139.

(a) No Damping: Same as Problem 19.138

6000 lb/ft,k = 31.08 rad/s,nω = 297 rpmf = !

(b) Damped Motion:

12 lb s/in. = 144 lb s/ftc = ⋅ ⋅

2 386.09 lb s/ftc nc mω= = ⋅ (Same as in Problem 19.138)

144 lb s/ft

0.37297386.09 lb s/ftc

c

c

⋅= =⋅

For maximum amplitude we use the same equation used in Problem 19.138

( )2 2

21 2 1 2 0.37297 0.72179

ωω

= − = − = n c

c

c

0.84958ωω

=n

( ) ( )( )0.84958 0.84958 31.08 rad/snω ω= =

26.405 rad/s 4.2025 Hzω = = 252 rpmω = !!!!

(c) Amplitude:

From Eq. (19.53): 2 22

1 2ω ωω ω

= − +

m

m

n c n

Pkx

c

c

For part (a) with 125 lbmP = and nω ω=

[ ] ( )( ) 22

125 lb6000 lb/ft

0.02793 ft1 1 2 0.37297 1

mx = = − +

0.335 in.mx = !

For part (b) with 125 lb=mP and 0.84958ωω

= n

( ) ( )( )2 22

125 lb6000 lb/ft

0.0301 ft

1 0.84958 2 0.84958 0.37297mx = =

− +

0.361 in.mx = !

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Chapter 19, Solution 140.

From Eq. ( )19.53 ′

Magnification Factor

( ) ( )( )22

2

1

1 2

m

f f

n c n

mx

P

k cc

ω ωω ω

= − +

Find value of c

c

c

for which there is no maximum for m

m

x

P

k

as f

n

ωω

increases

( )( ) ( )

( ) ( )( )

2

2

2

2

2 22

2 2

2 1 1 4

0

1 2

c

c

m f

m n

ff f

nn n

xcd

Pc

k

dcc

ωω

ωω ωωω ω

− − − + = =

− +

2

2 2

2 2 4 0f

nc

c

c

ωω

− + + =

2 2

21 2

f

nc

c

c

ωω

= −

For 2

2

1

2c

c

c

≥ there is no maximum for ( )m

mx

P

k

and the magnification factor will decrease as n

fωω

increases

1

2c

c

c

≥ 0.707

c

c

c

≥ �

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Chapter 19, Solution 141.

From Eq. (19.53′)

( ) ( )( )2 22

1Magnification factor

1 2c

mf f

n n

mxPk c

cω ωω ω

= = − +

Find value of n

fωω

for which m

mxPk

is a maximum

( )( ) ( )

( ) ( )( )

2

2

2

2

2 22 22

2 1 1 4

0

1 2

m fm n c

ff fnn c n

x cd P ck

d cc

ωω

ωω ωωω ω

− − + = = −

− +

2 2

2 2 4 0f

n c

c

c

ωω

− + + =

For small c

c

c 1 f

f nn

ωω ω

ω≈ ≈

For

[ ] ( ) 22

11,

1 1 2 1m

c

f m

n

xP

ck c

ωω

= = − +

( )1

2m

m cx cP ck

= !

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Chapter 19, Solution 142.

From Eq. (19.52)

( ) ( )2 22

mm

f f

Px

k m cω ω=

− +

22 22

900 rad/s 8882.6 s60fπω − = =

36 lb480 lb/in. 5760 lb/ft

0.075 in.st

Wk

δ= = = =

( ) ( )2 22

0.646.2516 lb ft 8882.6 s1232.2 ft/s

m fP m rω − ′= =

5.7470 lb=

(a) 0c =

( )( )22

5.747 lb

36 lb5760 lb/ft 8882.6 s32.2 ft/s

mx−

= −

5.747 lb

4170.9 lb/ft=

−

0.0013779 ft= −

0.01653 in.mx = !

(out of phase)

(b) ( ) 2

36 lb2 2 5760 lb/ft 160.496 lb s/ft

32.2 ft/scc km

= = = ⋅

( )0.055 160.496 lb s/ft 8.8273 lb s/ftc = ⋅ = ⋅

0.0013513 ftmx =

0.01622 in.mx = !

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Chapter 19, Solution 143.

From Eq. (19.53′)

( ) ( )( )22 2

1 2

m

f f

n c n

m

Pkx

cc

ω ωω ω

=

− +

(1)

22 29.81 m/s

1635 s0.06 mn

st

gωδ

−= = =

( )2

2 2300 987 s30fπω − = =

2987

0.603651635

f

n

ωω

= =

( )( )( )2 20.11 kg 0.075 m 987 s 8.1424 Nm fP m rω − ′= = =

( )( )2 245 N/m 1635 s 73.575 kN/m,nk mω −= = =

0.11067 mmmP

k=

(a) Then, from Equation (1)

( ) ( )( )3

3

22

0.11067 10 m0.25 10 m

1 0.60365 4 0.60365ccc

−− ×× =

− +

or 2

0.19597 0.15707 2.4147c

c

c

= +

Then 0.1269c

c

c= !

(b) ( )( )1222 2 45 kg 1635 s 3639.2 N s/mc nc mω −= = = ⋅

( )( )0.1269 3639.2 N s/m 462 N s/mc = ⋅ = ⋅ !

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Chapter 19, Solution 144.

From solution of Prob. 19.113, we have

160 kN/mk = 18 kgm =

94.281 rad/snω = 900 rpm

From Eq. (19.41): ( )( )2 2 18 94.28 3394 N s/mc nc mω= = = ⋅

For 1200 rpm 125.664 rad/sω = =

31.5 mm 1.5 10 mmx −= = × 350 N s/mc = ⋅

From Eq. (19.53): 2 22

1 2ω ωω ω

= − +

m

m

n c n

Pkx

c

c

3

2 221.5 10 m

125.664 350 125.6641 2

94.281 3394 94.281

mPk−× =

− +

( ) ( )

3

2 21.5 10 m

0.823760.7765 0.2749

m mP Pk k−× = =+

31.2356 10mP

k−= × ( )( )3 31.2356 10 160 10 197.7 NmP −= × × =

For rotor of mass 4 kg and rad/sm ω′ = = 125.664

( )( )

22 2

197.7 N,

4 kg 125.664 rad/sm

mP

P m r rm

ωω

′= = =′

33.13 10 mr −= × 3.13 mmr = !

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Chapter 19, Solution 145

( )( )2

2 3 11085 10 kg 0.150 m s

3m fP mrπω − − = = ×

3

2 7.5 10 N41.666

180 kgnω ×= =

Then ( )2 2 3 2sin , 9 10 m/sf m f f mx x t xω ω φ ω −= − − − = ×&&

So

( ) ( )2

2

2 220.009 m/s

1 2

m

f f

n c n

fPk

cc

ω

ω ωω ω

= − +

( )

( )

4 2

2

22

2

2

10.009

4

mf f

n

f

nc

P

kc

c

ω ωω

ωω

− − =

0.487=c

c

c!

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Chapter 19, Solution 146.

From Eq. (19.52)

( ) ( )2 22

mm

f f

Px

k m cω ω=

− +

( )( ) 2

2 1200 2

60f

πω

=

2 215791 sfω −=

( )4 7.5 kips/ft 30 kips/ftk = =

( )2 22

30 lb15791 s 14712

32.2 ft/sm fP m e e eω − ′= = =

( )( ) ( ) ( )2 2

0.10 14712

12 20030,000 15791 490 1579132.2

e= − +

0.16027e=

0.051995 ft 0.62395 in.e∴ = =

0.624 in.e = !

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Chapter 19, Solution 147.

From Equation (19.48), the motion of the machine is, ( )sinm fx x tω φ= −

The force transmitted to the foundation is,

( )Springs sins m fF kx kx tω φ= = −

( )Dashpot cosD m f fF cx cx tω ω φ= = −&

( ) ( )sin cosT m f f fF x k c tω φ ω ω φ = − + −

or recalling the identity,

( )2 2sin cos sinA y B y A B y φ+ = + +

2 2sin

B

A Bψ =

+

2 2cos

A

A Bψ =

+

( ) ( )22 sinT m f fF x k c tω ω φ ψ = + − +

Thus the amplitude of TF is

( )22m m fF x k cω= + (1)

From Eq. (19.53):

( ) ( )2 22

1 2

m

f f

n c n

m

Pkx

cc

ω ωω ω

= − +

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Substituting for mx in Equation (1)

( )

( ) ( ) 2

2

22

1

1 2c

f

f f

n n

m

m

cP k

F

cc

ω

ω ωω ω

+=

− +

(2)

2n

k

mω =

And Equation (19.41) 2 2

nc n

cc m m

ωω= =

2 2f f f

c nn

c c c

k cm

ω ω ωωω

= =

Substituting in (2)

( )( )

( ) ( )( )

2

2 22

1 2 Q.E.D.

1 2

f

c n

m

f f

n c n

mcP c

F

cc

ωω

ω ωω ω

+ = − +

!

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Chapter 19, Solution 148.

( ) 12 2 0.8 Hz 1.6 sf nfω π π π −= = =

( )2 2 14 175 N/m7.6923 s , 2.7735 s

91 kgn nk

mω ω− −= = = =

2 21.6

3.28462.7735

f

n

ω πω

= =

( )( )1

365 N s/m0.72309

2 2 91 kg 2.7735 sc n

c c

c mω −⋅= = =

2

0.52286c

c

c

=

(a) ( ) ( )( )

[ ] ( )( )2

89 N 1 4 0.52286 3.2846 89 1 6.8695

5.2194 6.86951 3.2846 4 0.52286 3.2846mF

+ + = =+ − +

70.9689 N=

71.0 NmF = !

(b) 89

0, 38.9565.2194

mc F= = =

39.0 NmF = !

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Chapter 19, Solution 149.

Energy is dissipated by the dashpot.

From Equation (19.48) the deflection of the system is ( )sinm fx x tω φ= −

The force on the dashpot, DF cx= &

( )cosD m f fF cx tω ω φ= −

The work done in a complete cycle with 2

ff

cπ

ω=

( )2

0 i.e Force Distancef

dE F dxπ

ω= ×∫

( )cosm f fdx x t dtω ω φ= −

( )2

02 2 2cosfm f fE cx t dt

πω ω ω φ= −∫

( )( )2

1 2coscos

2

f

D

tt

ω φω φ

− − − =

( )22 2

0

1 2cos

2f f

m f

tE cx dt

πω ω φ

ω− −

= ∫

( )2

2 2

0

2sin

2

ffm f

f

tcxE t

πωω φω

ω

− = −

( )( )2 2

2 2sin 2 sin

2m f

f f

cxE

ω π π φ φω ω

= − − −

2 Q.E.D.m fE cxπ ω= !

0

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Chapter 19, Solution 150.

(a)

( )2

2:

st

dx d d xF ma W k x c m

dt dt dt

δδ δ Σ = − + − − − =

Recalling that ,st

W kδ= we write

2

2

d x dx dm c kx k c

dt dtdt

δδ+ + = + (1)

Motion of wheel is a sine curve, sinm f tδ δ ω= the interval of time needed to travel a distance L at a

speed v, is ,

Lt

v= which is the period of the road surface.

Thus 2 2 2

f Lf v

v

L

π π πωτ

= = =

And 2

sin cosm

m f fLv

dt t

dt

δ δ πδ δ ω ω= =

Thus Equation (1) is,

( )2

sin cosf f f m

d x dxm c kx k t c t

dt dtω ω ω δ+ + = + �

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(b) From the identity

( )2 2sin cos sinA y B y A B y ψ+ = + +

2 2

sinB

A B

φ =+

2 2

cos

A

A B

φ =+

We can write the differential equation

( ) ( )2

22

2sinm f f

d x dxm c kx k c t

dtdtδ ω ω ψ+ + = + +

1tan

fc

k

ωψ −=

The solution to this equation is (analogous to Equations 19.47 and 19.48, with

( )22

m m fP k cδ ω= +

( )sin ,m fx x tω ϕ ψ= − + (where analogous to Equations (19.52)) �

( )

( ) ( )

22

2 22

f

m

f f

k c

x

k m c

δ ω

ω ω

+=

− +�

2

tanf

f

c

k m

ωφ

ω=

−�

tanfc

k

ωψ = �

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Chapter 19, Solution 151.

Since the origins of coordinates are chosen from the equilibrium position, we may omit the initial spring compressions and the effect of gravity

For load A

( ) ( ); sin 2A m f B A B A AF ma P t k x x c x x mxωΣ = + − + − =! ! !! (1)

For load B

( ) ( ); 2 2B B A B A B B BF ma k x x c x x kx cx mxΣ = − − − − − − =! ! ! !! (2)

Rearranging Equations (1) and (2), we find:

( ) ( )2 sinA A B A B m fmx c x x k x x P tω+ − + − =!! ! ! !

3 3 2 0B B A B Amx cx cx kx kx+ − + − =!! ! ! !

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Chapter 19, Solution 152.

For a mechanical system oscillations take place if c

c c< (lightly damped)

But from Equation (19.41),

2 2c

kc m km

m= =

Therefore

2c km< (1)

From table 19.2:

c R→

m L→

1

kC

→ (2)

Substituting in (1) the analogous electrical values in (2), we find that oscillations will take place if,

( )12R L

C

<

2L

RC

< �

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Chapter 19, Solution 153.

Electrical system Mechanical system

The mechanical analogue of closing a switch S is the sudden application of a constant force of magnitude P to the mass. (a) Final value of the current corresponds to the final velocity of the mass, and since the capacitance is zero, the spring constant is also zero

F maΣ =

2

2dx d xP C mdt dt

− = (1)

Final velocity occurs when 2

2 0d xdt

=

finalfinal final

0dx dxP C vdt dt

− = =

finalPvC

=

From table 19.2: , , v L P E C R→ → →

Thus finalELR

= !

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(b) Rearranging Equation (1), we have 2

2d x dxm C P

dtdt+ =

Substitute 2

; t tdx P d xAe A edt C dt

λ λλ− −= + = −

t t Pm A e C Ae PC

λ λλ − − − + + =

0 cm Cm

λ λ− + = =

Thus c tmdx pAe

dt c

− = +

At 0 t = 0 0 dx P PA Adt C C

= = + = −

1c tmdx pv e

dt c

−

= = −

From table 19.2: , , , v c p E c R m L→ → → →

1R tLEL e

R

−

= −

For 11 , 1E RL tR e L

= − =

LtR

= !

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Chapter 19, Solution 154.

We note that both the spring and the dashpot affect the motion of point A.

Thus one loop in the electrical circuit should consist of a capacitor

1k

c

⇒

and a resistance ( ).c R⇒

The other loop consists of ( )sin sin ,m f m fP t E tω ω→ an inductor

( ) m L→ and the resistor ( )c R→

Since the resistor is common to both loops, the circuit is

�

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Chapter 19, Solution 155.

Loop 1 (Point A) 1 2 1 1

1 2

1 1 , , k k c R

c c→ → →

Loop 2 (Mass m) 2 2 2

2

1 , , k m L c R

c→ → →

With 2

2

1k

c→ common to both loops

�

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Chapter 19, Solution 156.

(a) Mechanical system

Point A

0FΣ = ( ) 0A m A

dc x x kxdt

− + = �

Mass m

( )2

2 sin

mm A m f

d d xF ma c x x P t m

dt dtωΣ = − − = −

( )2

2sin

mm A m f

d x dm c x x P t

dtdtω+ − = �

(b) Electrical Analogue

From table 19.2

m L→

c R→

1k

C→

x q→

P E→

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Substituting into the results from part (a), the analogous electrical characteristics,

( ) 10

A m n

dR q q qdt C

− + =

�

( )2

2sin

mm A m f

d q dL R q q E t

dtdtω+ − = �

Note: These equations can also be obtained by summing the voltage drops around the loops in the circuit

of Problem 19.155.

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Chapter 19, Solution 157.

(a) Mechanical system:

0FΣ =

Point A ( )1 1 2 0AA A m

dxk x c k x xdt

+ + − =

( )1 1 2 2 0AA m

dxc k k x k xdt

+ + − = !

Mass m

( )2

2 2 2 m mA m

dx d xF ma k x x c mdt dt

Σ = − − =

( )2

2 22 0m mm A

d x dxm c k x xdtdt

+ + − = !

(b) Electrical analogue Substituting into the results from part (a) using the analogous electrical characteristics from table 19.2 (see left),

11 2 2

1 1 1 0AA m

dqR q qdt C C C

+ + − =

!

( )2

222

1 0m mm A

d q dqL R q qdt Cdt

+ + − = !

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Chapter 19, Solution 158.

For simple harmonic motion and 1 40 in. 3.333 ft:= =

232.2 ft/s 3.1082 rad/s3.333 ftn

gl

ω = = =

Then, with

( )sinm ntθ θ ω φ= +

And with

( ) ( )0 5 0.08727 rad, and 0 0:θ θ= ° = =!

For

( ) ( ) ( ) 50 0 cos 0 , with 0 0.08727 rad2 180m n mπ πθ θ ω φ φ θ θ= = + ⇒ = = = =!

Then

( )5 sin 3.1082 rad/s180 2

tπ πθ = +

( ) ( )0.08727 rad sin 3.1082 rad/s2

t π = +

(a) At 1.6 s,t =

( )( )5 sin 3.1082 rad/s 1.6 s 0.022496 rad 1.288180 2

π πθ = + = = °

1.288θ = °! (b)

( )cosm n ntθ θ ω ω φ= +!

( ) ( )( )5 3.1082 rad/s cos 3.1082 rad/s 1.6 s180 2

π π = +

0.262074 rad/s=

( )( )3.3333 ft 0.262074 rad/s 0.874 ft/sv lθ= = =!

( )1.6 s 0.874 ft/sv = !

continued

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( )2 sinm n ntθ θ ω ω φ= − +!!

( ) ( )( )25 3.1082 rad/s cos 3.1082 rad/s 1.6 s180 2

π π = − +

20.21733 rad/s= −

Now

( ) ( )2 2n la a a= +

( )( )2

22 23.3333 ft 0.26207 rad/s 0.22894 m/snva ll

θ= = = =!

( )( )22 23.333 ft 0.21733 rad/s 0.72443 m/sla lθ= = − = −!!

20.75974 m/sa = 20.759 ft/sa = !

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Chapter 19, Solution 159.

(a) First, calculate the spring constant

1 224 24

P P P

kδ δ δ= + = + =

Hence, 1 1 1

12 kN/m24 24k

= + ⇒

Then ( )

2 20.40558 s

12000 kN/m

50 kg

n

k

m

π πτ = = =

0.405 sn

τ = �

1

2.4656 Hzn

n

fτ

= = 2.47 Hznf = �

(b)

2 15.492 rad/sn n

fω π= =

( ) ( )0.060 m sin 15.492 rad/sx t φ = +

( )( ) ( )0.060 m 15.492 rad/s cos 15.492 rad/sx t φ = + &

( )( ) ( )20.060 m 15.492 rad/s sin 15.492 rad/sx t φ = − + &&

Then max

0.92951 m/sv = �

2

max14.400 m/sa = �

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Chapter 19, Solution 160.

( )eff: 0

2A A B

lM M lN x mg

Σ = Σ − + =

1

2B

xN mg

l

= +

( )eff

1: 0

2y y A

xF F N mg mg

l

Σ = Σ + + − =

1

2A

xN mg

l

= −

Thus 1

2A k A k

xF N mg

lµ µ = = −

1

2B k B k

xF N mg

lµ µ = = +

( )effx

F FΣ = Σ

A BF F mx− = &&

1 1

2 2k k

x xmg mg mx

l lµ µ − − + =

&&

20

fmx gx

l

µ+ =&&

2 2k

n

g

l

µω =

1 2

2 2

n k

n

gf

l

ω µπ π

= = �

COSMOS: Complete Online Solutions Manual Organization System

Vector Mechanics for Engineers: Statics and Dynamics, 8/e, Ferdinand P. Beer, E. Russell Johnston, Jr.,

Elliot R. Eisenberg, William E. Clausen, David Mazurek, Phillip J. Cornwell

© 2007 The McGraw-Hill Companies.

Chapter 19, Solution 161.

(a)

( )eff:

o oM MΣ = Σ

( ) ( )0.6 , 0.6C st B stC B

F k l F kδ θ θ δ = − = +

0.6 0.6C B

F F Iθ− = &&

( ) ( )0.6 0.6 0.4 0.6st stC B

k l k Iδ θ θ δ θ − − + = && (1)

At equilibrium ( ) ( ) ( )0 , , B st C stB C

F k F kθ δ δ= = =

( ) ( )0 0.6 0.6A st stC B

M k kδ δΣ = = − (2)

Substitute (2) into (1)

0.72 0I kθ θ+ =&&

( )( )22 2270 kg 0.5 m 67.5 kg mI mh= = = ⋅

( )( )20.72 0.72 m 13000 N/m 9360 N mk = = = ⋅

Thus 2 2

2

0.72 9360 N m138.67 s

67.5 kg mn

k

Iω −⋅= = =

⋅

2

2 20.5335 s

138.67 sn

n

π πτω −

= = =

0.534 sn

τ = �

COSMOS: Complete Online Solutions Manual Organization System

Vector Mechanics for Engineers: Statics and Dynamics, 8/e, Ferdinand P. Beer, E. Russell Johnston, Jr.,

Elliot R. Eisenberg, William E. Clausen, David Mazurek, Phillip J. Cornwell

© 2007 The McGraw-Hill Companies.

(b) ( ) ( )sin , cosm n m n n

t tθ θ ω φ θ θ ω ω φ= + = +&

max m nθ θ ω=&

2 1138.67 s 11.776 s

nω − −= =

0.025 m0.04167 rad

0.6 m

C

m

x

r

θ = = =

( )( )1max0.04167 rad 11.776 s 0.49067 rad/sθ −= =&

0.491 rad/sω = �

COSMOS: Complete Online Solutions Manual Organization System

Vector Mechanics for Engineers: Statics and Dynamics, 8/e, Ferdinand P. Beer, E. Russell Johnston, Jr.,

Elliot R. Eisenberg, William E. Clausen, David Mazurek, Phillip J. Cornwell

© 2007 The McGraw-Hill Companies.

Chapter 19, Solution 162.

α θ= &&

21

2D

WI r

g=

( )effC C

M MΣ = Σ

( ) ( ) 2sin sin rat

wwr wr I

gβ θ β θ α− − − = +

( ) ( )sin sin 2 sin cosr Wrβ θ β θ θ β − − + = −

sin ta rθ θ θ≈ = &&

( )2 222 cos 0

2

w Wr r wr

g gθ β θ

+ + =

&&

( )2 cos 4 cos

4 /2

2

n

wg gw

W W w rw r

β β= = ++

(1)

0

0

2 20

4

4

w g

Wr

w

π πβ τ= = =

+

0

2 42

cos 4

4 4

n

g

W Wr r

w w

π πτ τβ

= = =

+ +

21 1

cos2 4

β = =

75.5β = °�

COSMOS: Complete Online Solutions Manual Organization System

Vector Mechanics for Engineers: Statics and Dynamics, 8/e, Ferdinand P. Beer, E. Russell Johnston, Jr.,

Elliot R. Eisenberg, William E. Clausen, David Mazurek, Phillip J. Cornwell

© 2007 The McGraw-Hill Companies.

Chapter 19, Solution 163.

( )0 0 effM MΣ = Σ

( )2 2

22 2 2

AC AC BD AC

L L Lm g k I I mθ α α − = + +

21,

2BD AC BD AC

m m m I I I mL= = = = =

2

21 12 0

6 4 2 2

L LmL k mgθ θ

+ + − =

&&

( )2

2 2610

024 2 2 5

n

kL mgkL mgLmL

mLθ θ ω

−+ − = ⇒ =

&&

2

24 lb40 in. 3.3333 ft, 24 lb, 50 lb/ft,

32.2 ft/sL mg k m= = = = =

( )( )

( )2 2

2

6 50 lb/ft 3.3333 ft 24 lb68.908 s

24 lb5 3.3333 ft

32.2 ft/s

nω − − = =

18.301s

nω −=

1.321Hz2

n

nf

ωπ

= = �

COSMOS: Complete Online Solutions Manual Organization System

Vector Mechanics for Engineers: Statics and Dynamics, 8/e, Ferdinand P. Beer, E. Russell Johnston, Jr.,

Elliot R. Eisenberg, William E. Clausen, David Mazurek, Phillip J. Cornwell

© 2007 The McGraw-Hill Companies.

Chapter 19, Solution 164.

4 in. 0.33333 ftr = =

15 in. 1.25 ftl = =

Position 1

( )2 2

1 disk rod

1 1

2 2m A m

T J Jθ θ= +& &

2

1 disk

10

2D

V J m r= =

( ) 2

rod

1

3A AB

J m l= ⋅

Position 2

20T =

( ) ( )2

2

11 cos

2 2m AB m

lV k r m gθ θ= + −

2

21 cos 2sin

2 2

m m

m

θ θθ− = ≈

2 2 2

2

1

2 2 2

AB

m m

m g lV kr θ θ = +

1 1 2 2T V T V+ = +

COSMOS: Complete Online Solutions Manual Organization System

Vector Mechanics for Engineers: Statics and Dynamics, 8/e, Ferdinand P. Beer, E. Russell Johnston, Jr.,

Elliot R. Eisenberg, William E. Clausen, David Mazurek, Phillip J. Cornwell

© 2007 The McGraw-Hill Companies.

2 2 2 2 2 21 1 1 1 10 0

2 2 3 2 2 2D AB m m AB m

lm r m l kr m gθ θ θ + + = + +

&

m n mθ ω θ=&

2 2 2 2 2 21 1

2 3 2D AB n m AB m

lm r m l kr m gω θ θ + = +

2

2

2 2

1

2

1 1

2 3

AB

n

D AB

kr m gl

m r m l

ω+

=+

22

2

2 22 200

11

22

1 11 1 1

2 32 3

ABAB

n

ABAB

g kr W lkr W l

W r W lW r W lg

ω

++ = = ++

( ) ( )( ) ( )( )

( )( ) ( )( )

22

2 2

132.2 ft/s 25 lb/ft 0.33333 ft 6 lb 1.25 ft

2

1 110 lb 0.33333 ft 6 lb 1.25 ft

2 3

+ =+

257.109 s

−=

7.5571 rad/sn

ω =

20.8314 s

n

n

πτω

= =

0.831 sn

τ = �

COSMOS: Complete Online Solutions Manual Organization System

Vector Mechanics for Engineers: Statics and Dynamics, 8/e, Ferdinand P. Beer, E. Russell Johnston, Jr.,

Elliot R. Eisenberg, William E. Clausen, David Mazurek, Phillip J. Cornwell

© 2007 The McGraw-Hill Companies.

Chapter 19, Solution 165.

Kinematics 2 2A C A C

r rθ θ θ θ= =

2A C

θ θ=& &

Let ( ) 2A m m C m

θ θ θ θ= =

( )2m C

mθ θ=& &

Position 1

( ) ( )2 2

2 22 2

1

1 1 1 1 1 12 2 2

2 2 2 2 2 2 2 2A m C m AB m CD m AB m CD m

l lT I I I I m mθ θ θ θ θ θ = + + + + +

& & & & & &

( )( )2 214 2 8

2AI m r mr= =

( )( )2 21 1

2 2CI m r mr= =

2 21 1

12 12AB CDI ml I ml= =

2 2 2 2

2 2 2

1

18 4

2 2 12 3 4m

r l l lT m r l θ

= + + + + +

&

COSMOS: Complete Online Solutions Manual Organization System

Vector Mechanics for Engineers: Statics and Dynamics, 8/e, Ferdinand P. Beer, E. Russell Johnston, Jr.,

Elliot R. Eisenberg, William E. Clausen, David Mazurek, Phillip J. Cornwell

© 2007 The McGraw-Hill Companies.

2 2 2

1 1

1 510 0

2 3m

T m r l Vθ = + =

&

Position 2

( ) ( )2 20 1 cos 1 cos 2

2 2m m

l mglT V mg θ θ= = − − + −

For small angles 2

21 cos 2sin

2 2

m m

m

θ θθ− = ≈

2 21 cos2 2sin 2

m m mθ θ θ− = ≈

2

2 2

2

1 32

2 2 2 2 2

m

m m

l mglV mg mgl

θ θ θ= − + =

1 1 2 2

m n mT V T V θ ω θ+ = + =&

2 2 2 2 21 5 1 310 0 0

2 3 2 2m n m

m r l mglθ ω θ + + = +

3

2 2

2 2 2 25

3

9

10 60 10n

gl gl

r l r lω = =

+ +

2 22 60 10

29

n

n

r l

gl

πτ πω

+= = �

COSMOS: Complete Online Solutions Manual Organization System

Vector Mechanics for Engineers: Statics and Dynamics, 8/e, Ferdinand P. Beer, E. Russell Johnston, Jr.,

Elliot R. Eisenberg, William E. Clausen, David Mazurek, Phillip J. Cornwell

© 2007 The McGraw-Hill Companies.

Chapter 19, Solution 166.

(a) Position 1

( )2 2 2

1

1 1

2 2m m

T m l mlθ θ= =& &

10V =

Position 2

( ) ( )2

2 2

10 sin 1 cos

2m m

T V k a Wlθ θ= = − −

For small angles sinm m

θ θ≈

2

21 cos 2sin

2 2

m m

m

θ θθ− = ≈

2 2

2 2 2

2

1 1

2 2 2 2

m m

mV ka Wl ka Wl

θ θ θ = − = −

1 1 2 2T V T V+ = +

2 2 2 21 10 0

2 2m m

ml ka Wlθ θ + = + − &

COSMOS: Complete Online Solutions Manual Organization System

Vector Mechanics for Engineers: Statics and Dynamics, 8/e, Ferdinand P. Beer, E. Russell Johnston, Jr.,

Elliot R. Eisenberg, William E. Clausen, David Mazurek, Phillip J. Cornwell

© 2007 The McGraw-Hill Companies.

m n m

Wm

gθ ω θ= =&

2 2 2 2

n m

Wl ka Wl

gω θ = −

( )2 2

2

21n

Wg

ka Wl g ka

l Wllω

−= = −

2

1 11

2 2n n

g kaf

l Wlω

π π

= = −

�

(b) 0nf =

2

1 0ka

Wl− >

Wla

k> �

COSMOS: Complete Online Solutions Manual Organization System

Vector Mechanics for Engineers: Statics and Dynamics, 8/e, Ferdinand P. Beer, E. Russell Johnston, Jr.,

Elliot R. Eisenberg, William E. Clausen, David Mazurek, Phillip J. Cornwell

© 2007 The McGraw-Hill Companies.

Chapter 19, Solution 167.

2

2, 1

1

m

fm m m

nf

n

P

kx P x k

ωωω

ω

= = − −

1

2

1000 lb/ft22.43 s

64 lb

32.2 ft/s

n

k

mω −= = =

( )2

1

1

0.75 in. 10 s1 1000 lb/ft 50.077 lb

12 in./ft 22.43 smP

−

−

= − =

50.1 lbmP = �

COSMOS: Complete Online Solutions Manual Organization System

Vector Mechanics for Engineers: Statics and Dynamics, 8/e, Ferdinand P. Beer, E. Russell Johnston, Jr.,

Elliot R. Eisenberg, William E. Clausen, David Mazurek, Phillip J. Cornwell

© 2007 The McGraw-Hill Companies.

Chapter 19, Solution 168.

( ) F ma k x mxδΣ = − = &&

kx x

mδ + =

&&

From Equation (19.31 and 19.33 )′

2

21

mm

f

n

x

δωω

=

−

1 2 2 212 2 s s

2f f ffω π π π ω π− − = = = =

( )( )

2 270 N/m

194.44 s0.36 kg

n

k

mω −= = =

0.2 mm

δ =

2

0.2 m0.21070 m

1194.44

mx

π= =

−

(In Phase)

Check to see whether cord goes slack

Static deflection 0.8 lb

0.16 ft5 lb/ft

ST

W

kδ = = =

COSMOS: Complete Online Solutions Manual Organization System

Vector Mechanics for Engineers: Statics and Dynamics, 8/e, Ferdinand P. Beer, E. Russell Johnston, Jr.,

Elliot R. Eisenberg, William E. Clausen, David Mazurek, Phillip J. Cornwell

© 2007 The McGraw-Hill Companies.

Since x and δ are in phase the maximum

Deflection of the cord is 0.21070 0.20 0.01070 mmm mx δ− = − =

Which is less than the static deflection of 50.5 mm

211 mmmx = �

COSMOS: Complete Online Solutions Manual Organization System

Vector Mechanics for Engineers: Statics and Dynamics, 8/e, Ferdinand P. Beer, E. Russell Johnston, Jr., Elliot R. Eisenberg, William E. Clausen, David Mazurek, Phillip J. Cornwell © 2007 The McGraw-Hill Companies.

Chapter 19, Solution 169.

In equilibrium the force in the spring is mg

For small angles

sin cos 1θ θ θ≈ ≈

2Blyδ θ=

Cy lδ θ=

(a)Newton�s Law ( )effA AM MΣ = Σ

2 2 2 2t

mgl l l lk mg cl l I maθ θ α − + − = +

!

Kinematics 2 2tl laα θ α θ= = =!! !!

2 22 0

2 2l lI m cl kθ θ θ

+ + + =

!! !

221

2 3lI m ml + =

3 3 04

c km m

θ θ θ + + =

!! ! !

(b) Substituting teλθ = into the differential equation obtained in (a), we obtain the characteristic equation,

2 3 3 04

c km m

λ λ + + =

continued

COSMOS: Complete Online Solutions Manual Organization System

Vector Mechanics for Engineers: Statics and Dynamics, 8/e, Ferdinand P. Beer, E. Russell Johnston, Jr., Elliot R. Eisenberg, William E. Clausen, David Mazurek, Phillip J. Cornwell © 2007 The McGraw-Hill Companies.

And obtain the roots

23 3 3

2

c c km m mλ

− − =

∓

The critical damping coefficient, ,cc is the value of c, for which the radicand is zero.

Thus 23 3cc k

m m =

3c

kmc = !