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PROBLEM 19.31 (Continued)
(b) For infinite:
τ
= 20
n
n
π
τω
ω
=
=
2
2(600) 0.5 9.81
⎛⎞
PROBLEM 19.32
The force-deflection equation for a nonlinear spring fixed at one end is 1/2
1.5
F
x= where F is the force,
expressed in newtons, applied at the other end, and x is the deflection expressed in meters. (a) Determine the
deflection x0 if a 4-oz block is suspended from the spring and is at rest. (b) Assuming that the slope of the
force-deflection curve at the point corresponding to this loading can be used as an equivalent spring constant,
determine the frequency of vibration of the block if it is given a very small downward displacement from its
equilibrium position and released.
SOLUTION
(a) 0
Deflection .
x
4oz 0.25lb
W
FW
==
=
2
0
0.25
1.5
x⎛⎞
=⎜⎟
⎝⎠
Equivalent spring constant.
0
0
1/ 2 1/ 2
1.5 1.5
22
4.5 lb/ft
4.5 lb/ft
x
x
e
dF x
dx
dF
dx
k
−−
⎛⎞
⎝⎠
⎛⎞
=
⎜⎟
⎝⎠
=
(b) Natural frequency.
e
k
m
n
f
=
n
n
Copyright © McGraw-Hill Education. Permission required for reproduction or display.
PROBLEM 19.33*
Expanding the integrand in Equation (19.19) of Section 19.4 into a series of even powers of sin
φ
and
integrating, show that the period of a simple pendulum of length l may be approximated by the formula
2
1
4
21sin
2
m
l
g
θ
τπ
⎛⎞
=+
⎜⎟
⎝⎠
where m
θ
is the amplitude of the oscillations.
SOLUTION
Using the Binomial Theorem, we write
()
1/ 2
2
22
2
22
11sin sin
2
1sin sin
1
1sin sin
22
m
m
m
θ
θφ
φ
θφ
−
⎡⎤
⎛⎞
=−
⎢⎥
⎜⎟
⎝⎠
⎣⎦
−
=
+ +⋅⋅⋅⋅
Neglecting terms of order higher than 2 and setting 21
2
sin (1 cos 2 ),
φφ
=− we have
()
/2 2
0
/2 22
0
11
41sin1cos2
222
11
41sinsincos2
4242
m
n
mm
ld
g
ld
g
π
π
θ
τ
φφ
θθ
φ
φ
⎧⎫
⎡⎤
=+ −
⎨⎬
⎢⎥
⎣⎦
⎩⎭
⎧
⎫
=+−
⎨
⎬
⎩⎭
∫
∫
/2
22
0
2
11
4 sin sin sin 2
4282
1
4sin0
24 22
mm
m
l
g
l
g
π
θθ
φφ φ
θ
ππ
⎡
⎤
⎛⎞
=+ −
⎢
⎥
⎜⎟
⎝⎠
⎣
⎦
⎡⎤
⎛⎞
=+ +
⎢⎥
⎜⎟
⎝⎠
⎣⎦
2
1
21sin
42
m
n
l
g
θ
τπ
⎛⎞
=+
⎜⎟
⎝⎠
Copyright © McGraw-Hill Education. Permission required for reproduction or display.
PROBLEM 19.34*
Using the formula given in Problem 19.33, determine the amplitude m
θ
for which the period of a simple
pendulum is 1
2percent longer than the period of the same pendulum for small oscillations.
SOLUTION
For small oscillations, 0
() 2
n
l
g
τπ
=
We want 0
1.005( )
1.005 2
nn
l
g
ττ
π
=
=
Using the formula of Problem 19.33, we write
2
0
0
2
1
()1 sin
42
1.005( )
sin 4[1.005 1] 0.02
2
sin 0.02
2
8.130
2
m
nn
n
m
m
m
θ
ττ
τ
θ
θ
θ
⎛⎞
=+
⎜⎟
⎝⎠
=
=−=
=
=° 16.26
m
θ
=°
Copyright © McGraw-Hill Education. Permission required for reproduction or display.
PROBLEM 19.35*
Using the data of Table 19.1, determine the period of a simple pendulum of length 750 mm
l= (a) for small
oscillations, (b) for oscillations of amplitude 60 ,
m
θ
=
° (c) for oscillations of amplitude 90 .
m
θ
=°
SOLUTION
(a) 2
n
l
g
τπ
=(Equation 19.18 for small oscillations):
2
0.750 m
29.81 m/s
1.737 s
n
τπ
=
= 1.737 s
n
τ
=
(b) For large oscillations (Eq. 19.20),
22
2(1.737 s)
n
Kl
g
K
τπ
π
π
⎛⎞
⎛⎞
=⎜⎟
⎜⎟
⎜⎟
⎝⎠
⎝⎠
=
For 60 ,
m
θ
=°
1.686K
=
(Table 19.1)
2(1.686)(1.737 s)
(60 )
1.864 s
n
τπ
°=
= 1.864 s
n
τ
=
(c) For 90 ,
m
θ
=°
1.854K
=
2(1.854)(1.737 s) 2.05 s
n
τπ
==
Copyright © McGraw-Hill Education. Permission required for reproduction or display.
PROBLEM 19.36*
Using the data of Table 19.1, determine the length in inches of a simple pendulum which oscillates with a
period of 2 s and an amplitude of 90°.
SOLUTION
For large oscillations (Eq. 19.20),
22
n
Kl
g
τπ
π
⎛⎞
⎛⎞
=⎜⎟
⎜⎟
⎜⎟
⎝⎠
⎝⎠
for 90
m
θ
=°
1.854K= (Table 19.1)
2
22
2
(2 s) (2)(1.854)(2) 32.2 ft/s
(2 s) (32.2 ft/s )
[(4)(1.854)]
2.342 ft
l
l
=
=
= 28.1 in.l=
Copyright © McGraw-Hill Education. Permission required for reproduction or display.
PROBLEM 19.37
The uniform rod shown has mass 6 kg and is attached to a spring
of constant 700 N/m.k
=
If end B of the rod is depressed 10 mm
and released, determine (a) the period of vibration, (b) the
maximum velocity of end B.
SOLUTION
700 N/mk
Wmg
=
=
where st
st
2
()
(0.5 )
6(0.1 m) 0.6
1(6)(0.8 m)
12
0.32
Fkx
k
ma mr
I
δ
θδ
α
θθ
αθ
θ
=+
=+
== =
=
=
(a) Equation of motion.
: (0.1 m) (0.5 m) (0.1 m)
C
M I mad W F I ma
αα
Σ=+ − =+
st
(0.1) (0.5 )(0.5 m) 0.32 0.6 (0.1)Wk
θδ θ θ
−+ =+
But in equilibrium, we have st
(0.1 m) (0.5 m) 0Wk
δ
−
=
Thus, 2
2
(0.5) [0.32 0.06]
(700 N/m)(0.5) 0.38
(460.53) 0
k
θ
θ
θθ
θθ
−=+
−=
+=
Copyright © McGraw-Hill Education. Permission required for reproduction or display.
PROBLEM 19.37 (Continued)
Natural frequency and period.
2460.53
21.46 rad/s
22
21.46 rad/s
n
n
n
ω
ω
ππ
τω
=
=
== 0.293 s
τ
=
(b) At end B. 0.010 m
m
x=
(10 mm)(21.46 rad/s)
214.6 mm/s
mmn
vx
ω
=
=
= 0.215 m/s
m
v=
Copyrig
h
ht
© McGra
w
w
-Hill Educ
a
PROBL
E
A belt is
p
to two sp
r
1.5 in. d
o
observed
t
to preven
t
flywheel,
(
mm
a
tion. Permis
s
E
M 19.38
p
laced around
r
ings, each of
o
wn and rel
e
t
o be 0.5 s. K
n
t
slipping, d
e
(
b) the centro
i
n
s
ion require
d
the rim of a
5
constant
k
=
e
ased, the pe
r
n
owing that t
h
e
termine (a) t
i
dal radius of
g
d
for reprodu
5
00-lb flywhe
85 lb/in. If e
n
r
iod of vibr
a
h
e initial tensi
o
he maximum
g
yration of th
e
ction or disp
el and attach
e
n
d C of the b
e
a
tion of the
f
o
n in the belt
i
angular vel
o
e
flywheel.
m
lay.
e
d as shown
e
lt is pulled
f
lywheel is
i
s sufficient
o
city of the
PROBLEM 19.38 (Continued)
(b) Centroidal radius of gyration.
2
2
2
n
kr
I
ω
=
