Solution manual vector mechanics engineers dynamics 8th beer chapter 19

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Chapter 19, Solution 2.

Eq. 19.15:

vm = xmω n

am = xmω n2

Given data

vm = 0.2 m/s

am = 4 m/s 2

vm = xmω n :

0.2 m/s = xmω m

(1)

am = xmω n2 :

4 m/s 2 = xmω m2

(2)

Divide Equ. (2) by Equ. (1):

ωn =

Eq. (1):

4 m/s 2
= 20 rad/s
0.2 m/s

0.2 m/s = xm ( 20 rad/s )
xm = 0.01 m

Frequency

fn =

ω n 20 rad/s
=
2π
2π

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.

xm = 10 mm !
f n = 3.18 Hz !


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Chapter 19, Solution 3.

x = xm sin ω nt ,

ωn = 6

cycle 2π rad
×
s
cycle

x = xm sin 12π t
x& = 12π xm cos12π t

&&
x = −144π 2 xm sin12π t

12π xm = 4 ft/s
xm =

4
= 0.1061 ft
12π
xm = 1.273 in. !
Max Acc. = 144π 2 ( 0.1061) = 150.8 ft/s 2 !

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 4.

Simple Harmonic Motion

δs =

W
20 lb
=
= 0.2222 in.
k
90 lb/in.

ωn =

(a)

k
=
m

( 90 )(12 )
 20 


 32.2 

= 41.699 rad/s = 2π f

Amplitude = δ s = xm = 0.222 in.
xm = 0.222 in. !
f =

41.699 rad/s
= 6.6366
2π
f = 6.64 Hz !

(b)

vm = ω n xm = ( 41.699 rad/s )( 0.2222 in.) = 9.2655 in./s
vm = 9.27 in./s !

am = ω n2 xm = ( 41.699 rad/s ) ( 0.2222 in.) = 386.36 in./s 2
2

= 32.197 ft/s 2
am = 32.2 ft/s 2 !

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 5.

Simple Harmonic Motion
x = xm sin (ω nt + φ )

(a)

k
=
m

ωn =

9000 lb/ft

( 70 lb ) ( 32.2 lb/s2 )

= 64.343 rad/s

2π
= 0.90765 s
ωn

τn =

τ n = 0.0977 s !
fn =

1

τn

= 10.240 Hz
f n = 10.24 Hz !

t = 0: x0 = 0, x&0 = v0 = 10 ft/s

(b) At

x0 = 0 = xm sin (ω n ( 0 ) + φ ) ⇒ φ = 0
x&0 = v0 = xmω n cos (ω n ( 0 ) + φ ) = xmω n

Substituting
or

10 ft/s = xm 64.343 rad/s
xm = 0.1554 ft = 1.865 in.
xm = 1.865 in. !

am = xmω n2 = ( 0.15542 ft )( 64.343 rad/s )

2

= 643.4 ft/s 2
am = 643 ft/s 2 !

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 6.

In Simple Harmonic Motion
(a)

am = xmω n2

Substituting

50 m/s 2 = ( 0.058 m ) ω n2

or

ω n2 = 862.07 ( rad/s )

2

ω n = 29.361 rad/s
fn =

Now

f in Hz =

Then

So

ωn

29.361 rad/s
=
= 4.6729 Hz
2π
2π

(

1 cycle
1
=
Hz
(1 min )( 60 s/min ) 60

f Hz
4.6729 Hz
=
= 280.37 r/min
1
Hz
60

1
60

)

280 rpm

and

(b)

vm = xmω n = ( 0.058 m )( 29.361 rad/s ) = 1.7029 m/s
vm = 1.703 m/s

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 7.

Simple Harmonic Motion

θ = θ m sin (ω nt + φ )

(a)

ωn =

2π

τn

=

2π
(1.35 s )

= 4.833 rad/s

θ& = θ mω n cos (ω nt + φ )

θ&m = θ mω n
vm = lθ&m = lθ mω n

Thus,

θm =

vm
lω n

ωn =

g
l

(1)

For a simple pendulum

Thus,
l =

g

ω n2

=

9.81 m/s 2

( 4.833 rad/s )

2

= 0.420 m

From (1)

θm =

vm
0.4 m/s
=
lω n
( 0.42 m )( 4.833 rad/s )
= 0.197 rad

or

11.287°

θ m = 11.29° !

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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(b) Now
at = lθ&&

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

θ&& = −θ mω n2 sin (ω nt + φ )
θ&&m = θ mω n2

( at )m
or

( at )m

= lθ mω n2

= ( 0.42 m )( 0.197 rad )( 4.833 rad/s )

2

= 1.9326 m/s 2

( at )m

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.

= 1.933 m/s 2 !


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Chapter 19, Solution 8.

Simple Harmonic Motion:

ωn =

fn =

k
=
m

60 lb/ft
= 6.2161 rad/s
50 lb
32.2 ft/s 2

ω n 6.2161
=
Hz
2π
2π

xm = 2.4 in. = 0.2 ft

(a)

xm = 0.2 ft !
f n = 0.989 Hz !

am = xmω n2 = ( 0.2 ft )( 6.2161 rad/s )

2

= 7.728 ft/s 2
F f = mam = µ s mg

(b)
or

µs =

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.

am
7.728 ft/s 2
=
= 0.240 !
g
32.2 ft/s 2

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Chapter 19, Solution 9.

k = 6 lb/in. = 72 lb/ft,
F = ma :

vm = 55 in./s,

−kx = mx&&

ω2 =

Thus:

Eq. (19.15):

&&
x+

W = 4 lb.

k
x=0
m

k
72

=
= 579.6
m  4 


 32.2 

ω = 24.025 rad/s

vm = xmω

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

xm = 2.2845 in.

xm = 2.28 in. !

(

am = xmω 2 = ( 2.2845 in.) 579.6 rad 2 /s 2

)

am = 1324.1 in./s 2

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.

am = 110.3 ft/s 2 !

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Chapter 19, Solution 10.

π

x = 60 cos (10π t ) + 45sin 10π t − 
3

π
π

= 60 cos (10π t ) + 45 sin10π t cos − cos10π t sin 
3
3

= 22.5sin10π t + 21.02886 cos10π t

(1)

Now
xm sin(10π t + φ ) = xm sin10π t cos φ + xm cos10π t sin φ

(2)

Comparing (1) and (2) gives
22.5 = xm cos φ ,

21.02866 = xm sin φ

τn =

(a)

(b)

2π
2π
=
= 0.2 s !
ω n 10π

xm2 = (22.5)2 + (21.02866) 2
xm = 30.8 mm !

(c)

tan φ =

21.02866
22.5

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.

φ = 0.7516 rad = 43.1° !

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Chapter 19, Solution 11.

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

ω = 600 rpm = 62.832 rad/s
am = xmω 2

Eq. (19.15):

(b)

SI:

xm =

US:

xm =

9.81

( 62.832 )

xm =

g

( 62.832 )2

= 2.4849 × 10−3 m

2

32.2

xm = 2.48 mm

= 0.008156 ft

( 62.832 )2

xm = 0.0979 in.

ω = 1200 rpm = 125.664 rad/s
am = xmω 2

Eq. (19.15):

SI:

xm =

US:

xm =

9.81

(125.664 )

2

32.2

(125.664 )2

xm =

g

(125.664 )2

= 621.2 × 10−6 m

xm = 0.621 mm

= 0.002039 ft

xm = 0.0245 in.

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.


COSMOS: Complete Online Solutions Manual Organization System

Chapter 19, Solution 12.

Simple Harmonic Motion, thus
x = xm sin (ω nt + φ )

ωn =

k
=
m

400 N/m
= 16.903 rad/s
1.4 kg

x(0) = 0 = xm sin(0 + φ ) ⇒ φ = 0

Now
Then

x& (0) = xmω n cos(0 + 0)
2.5 m/s = ( xm )m(16.903 rad/s) ⇒ xm = 0.14790 m

or

x = ( 0.14790 m ) sin (16.903 rad/s ) t 

Then
(a)

At

x = 0.06 m: 0.06 m = (0.14790 m)sin (16.903 rad/s ) t 

 0.06 m 
sin −1 

 0.14790 m  = 0.02471 s
t =
16.903 rad/s

or

t = 0.0247 s !

(b)

Now
x& = xmω n cos (ω nt )
&&
x = − xmω n2 sin (ω nt )

Then, for t = 0.024713 s
x& = ( 0.1479 m )(16.903 rad/s ) cos (16.903 rad/s )( 0.024713 s ) 
= 2.285 m/s

x& = 2.29 m/s !

And
&&

x = − ( 0.1479 m )(16.903 rad/s ) sin (16.903 rad/s )( 0.024713 s ) 
2

= −17.143 m/s 2

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.

&&
x = 17.14 m/s 2 !


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Chapter 19, Solution 13.

Referring to the figure of Problem 19.12
x = xm sin (ω nt + φ )
x& = xmω n cos (ω nt + φ )
&&
x = − xmω n2 sin (ω nt + φ )

Using the data from Problem 19.13:

φ = 0, xm = 0.1479 m, ω n = 16.903 rad/s
x, x&, &&
x are

And

x = ( 0.14790 m ) sin (16.903 rad/s ) t 

So, at t = 0.9 s,
x = ( 0.1479 m ) sin (16.903 rad/s )( 0.9 s ) 
= 0.0703 m

x = 70.3 mm !

x& = ( 0.1479 m )(16.903 rad/s ) cos (16.903 rad/s )( 0.9 s ) 
= −2.19957 m/s

x& = 2.20 m/s !

&&
x = ( 0.1479 m )(16.903 rad/s ) sin (16.903 rad/s )( 0.9 s ) 
2

= −20.083 m/s 2

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.

&&
x = 20.1 m/s 2 !


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Chapter 19, Solution 14.

(a)
x = xm sin(ω nt + φ )

ωn =

k
=
m

9000 lb/ft
( 70 lb )

(32.2 lb/s )
2

= 64.343 rad/s

τn =
With the initial conditions:

2π

ωn

= 0.9765 s
x(0) = 15 in. = 1.25 ft, x& (0) = 0

1.25 ft = xm sin ( 0 + φ )

x& (0) = 0 = xmω n cos(0 + φ ) ⇒ φ =

π
2

xm = 1.25 ft

Then

π

x(t ) = (1.25 ft) sin  64.343t + 
2

π

x(1.5) = (1.25 ft) sin  64.343(1.5 s) +  = −0.80137 ft
2

π

x& (1.5) = (1.25 ft)(64.343) cos  64.343 (1.5 s ) +  = −61.726 ft/s
2

In 1.5 s, the block completes
1.5 s
= 15.361 cycles
0.09765 s/cycle

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.


COSMOS: Complete Online Solutions Manual Organization System

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)

π

&&
x(1.5) = −(1.25 ft)(64.343 rad/s) 2 sin (64.343 rad/s)(1.5 s) + 
2

= 3317.68 ft/s 2

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.

&&
x = 3320 ft/s 2 !


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Chapter 19, Solution 15.

m=

10 lb
= 0.31056 lb ⋅ s 2 /ft
2
32.2 ft/s

With the given properties:

ωn =

k
=
m

50 lb/ft
= 12.6886 rad/s
0.31056 lb ⋅ s 2 /ft

From free fall of the collar
v0 =

2 gh =

2 g(1.5 ft ) =

3g = 9.82853 ft/s

The free-fall time is thus:

t1 =

2y
=
g

2 (1.5 ft )
g

=

3
= 0.30523 s
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:
W
10 lb
δ st =

=
= 0.2 ft
k
50 lb/ft
Then mx + kx = 0, where x is measured positively up from the
position of static deflection. The solution is:
x = xm sin (ω nt + φ ) , with velocity
x = xmω n cos (ω nt + φ )
Now to determine xm and φ , impose the conditions at impact and
count the time from there.
Thus:
At impact:
t = 0, x = δ st = 0.2 ft and v = −9.82853 ft/s (down)
or

0.2 ft = xm sin φ

−9.82853 ft/s = (12.6886 rad/s ) xm cos φ

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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Solving for xm and φ
xm = −0.800 ft

φ = −0.25268 rad

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, t2 is the time such
that
x ( t2 ) = 0 ⇒ 12.6886t2 + φ =
12.6886t2 − 0.25268 =

or

(a)

π
2

π
2

Hence,
t2 = 0.14371 s
Thus, the period of the motion is

τ = 2 ( t1 + t2 ) = 2 ( 0.30523 s + 0.14371 s )
= 0.89788 s

(b)

τ = 0.898 s W

After 0.4 seconds, the velocity is

x ( 0.4 ) = xmω n cos ⎡⎣ω n ( 0.4 − t1 ) + 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

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.

v ( 0.4 ) = 5.91 ft/s W


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Chapter 19, Solution 16.

θ = θ m sin ω nt ,
ωn =

θ& = θ mω n cos ω nt, ω n =

9.81
= 2.8592 rad/s, t = 0:
1.2

θ& =

g
l

0.18

= θ mω n
1.2

∴ θ m = 0.052462 radians

At

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

θ = − 0.047826 radians = − 2.74° !

(a)
(b)

v = 1.2 ( 0.052462 )( 2.8592 ) cos ( 2.8592 )(1.5 )
v = 74.0 mm/s !

a = 1.2 ( 0.052462 )( 2.8592 ) sin ( 2.8592 )(1.5 )
2

a = 469 mm/s 2 !

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.


COSMOS: Complete Online Solutions Manual Organization System

Chapter 19, Solution 17.

(a)
x = xm sin (ω nt + φ )
x0 = xm sin ( 0 + φ ) = 0.75 ft
x&0 = 0 = xmω n cos ( 0 + φ ) , ⇒ φ =

π
2

∴ xm = 0.75 ft

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

π

x& = ( 0.75 ft ) ω n cos  ω nt + 
2

π
π π

x& ( 0 ) = v = 0 = ( 0.75 ft ) ω n cos  ω nt +  , ⇒ ω nt +
=
2
2
2



And
π

a = − g = − ( 0.75 ft ) ω n2 sin  ω nt + 
2

−g = −( 0.75 ft ) ωn2

or

ωn =

32.2 ft/s 2
= 6.5524 rad/s
0.75 ft

Then

ωn =

k
10 lb
2
, ⇒ k = mω n2 =
6.5524 rad/s )
2(
m
32.2 ft/s

= 13.333 lb/ft

k = 13.33 lb/ft !

(b)

ω n = 6.5524 rad/s

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.


COSMOS: Complete Online Solutions Manual Organization System

At t = 1.6 s:

π

x = ( 0.75 ft ) sin ( 6.5524 rad/s )(1.6 s ) +  = −0.36727 ft
2

x = −0.367 ft above equilibrium !

π


v = x& = ( 0.75 ft )( 6.5524 rad/s) cos ( 6.5524 rad/s)(1.6 s) +  = 4.2848 ft/s
2


v = 4.28 ft/s !

π
2

a = &&
x = −( 0.75 ft )( 6.5524 rad/s) sin ( 6.5524 rad/s)(1.6 s) +  = 15.768 ft/s2
2

a = 15.77 ft/s 2 !

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.


COSMOS: Complete Online Solutions Manual Organization System

Chapter 19, Solution 18.

Determine the constant k of a single spring equivalent to the three springs shown.
Springs 1 and 2:

δ = δ1 + δ 2 ,

P1
P
P
= 1 + 1
k′
k1 k2

and

Hence

k′ =

k1k2
k1 + k2

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

P = P1 + P2 ,

So

and

kδ = k ′δ + k3δ

Now

k = k ′ + k3 =
or

(a)

P = kδ , P1 = k ′δ , P2 = k3δ

k =

k1k2
+ k3
k1 + k2

( 3.5 kN/m )( 2.1 kN/m ) + 2.8 kN/m
( 3.5 kN/m ) + ( 2.1 kN/m )

= 4.11 kN/m = 4.11 × 103 N/m

τn =

2π
=
k
m
fn =

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.

2π
4.11 × 103 N/m
13.6 kg
1

τn

=

= 0.361 s

1
= 2.77 Hz
0.3614 s


COSMOS: Complete Online Solutions Manual Organization System

x = xm sin (ω nt + φ )

(b)
And

xm = 44 mm = 0.044 m

ω n = 2π f n =

( ( 2π ) 2.77 Hz ) = 17.384 rad/s

And

x = ( 0.044 m ) sin (17.4 rad/s ) t + φ 
x& = ( 0.044 m )(17.4 rad/s ) cos (17.4 rad/s ) t + φ 
Then

vmax = ( 0.044 m )(17.4 rad/s ) = 0.766 m/s
2

&&
x = − ( 0.044 m )(17.4 rad/s ) sin (17.4 rad/s ) t + φ 
Then

2

amax = ( 0.044 m )(17.4 rad/s ) = 13.3 m/s 2
vmax = 0.765 m/s

amax = 13.31 m/s 2

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.


COSMOS: Complete Online Solutions Manual Organization System

Chapter 19, Solution 19.

(a) First, calculate the spring constant
P = ( 24 kN/m ) δ + (12 kN/m ) δ + (12 kN/m ) δ = ( 48 kN/m ) δ

∴ k = 48 kN/m

Then

ωn =

k
=
m

τn =

40 × 103 N/m
= 30.984 rad/s
50 kg
2π

ωn

fn =

= 0.20279 s
1

τn

= 4.9312 Hz

τ n = 0.203 s !
f n = 4.93 Hz !

(b) Now
x = xm sin (ω nt + φ )

And, since

x0 = 0.060 m
x = ( 0.060 m ) sin ( 30.984 rad/s ) t + φ 

x& = ( 0.060 m )( 30.984 rad/s ) cos ( 30.984 rad/s ) t + φ 

&&
x = − ( 0.060 m )( 30.984 rad/s ) sin ( 30.984 rad/s ) t + φ 
2

Hence

vmax = 1.859 m/s !
amax = 57.6 m/s 2 !

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.


COSMOS: Complete Online Solutions Manual Organization System

Chapter 19, Solution 20.

Equivalent spring constant
k ′ = 2k + 2k = 4k

(Deflection of each spring is the same.)

2π
k

10 lb
32.2 ft/s 2

(τ n )1 = 6.8 s =

⇒

k
 2π 
=

10
 6.8 
32.2

2

k = 0.2651 lb/ft

(τ n )2

=

=

2π
4k
6 lb
32.2 ft/s 2
2π

(

4 ( 0.2651 lb/ft ) 32.2 ft/s 2
6 lb

)

= 2.633 s

τ n = 2.63 s

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.


COSMOS: Complete Online Solutions Manual Organization System

Chapter 19, Solution 21.
Equivalent Springs

k1k2
k1 + k2

Series:

ks =

Parallel:

k p = k1 + k2

τs =

2π

ωs

=

2π
;
ks
m

τp =

2π

ωp

=

2π
kp
m

2

2
τ 
kp
( k + k2 )2
k + k2
5
= 1
= 1
 s =  =
τp 
ks
k1k2
( k1k2 )
2
 
( k1 + k2 )

( 6.25)( k1k2 ) = k12 + 2k1k2 + k22
k1 =

( 4.25) k2 m ( 4.25)2 k22 − 4k22
2
k1
= 2.125 m 3.516
k2
k1
1
= 4 or
k2
4

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.