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PROBLEM 19.15 (Continued)
Solving for and
m
x
φ
0.26975 m
m
x=−
0.25531 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, 2
tis the time such
that
(b) After 0.4 seconds, the velocity is
(
)
(
)
1
0.4 cos 0.4
mn n
xx t
ω
ωφ
⎡
=
−+
⎣
PROBLEM 19.16
A small bob is attached to a cord of length 1.2 m and is released from rest when
5.
A
θ
=°
Knowing that d = 0.6 m, determine (a) the time required for the bob to
return to Point A, (b) the amplitude .
C
θ
SOLUTION
As the pendulum moves between Points A and B, the length of the pendulum is 1.2 m.
AB
ll==
2
The falling from A to B is one quarter period.
As the pendulum moves between Points B and C, the length of the pendulum is
1.2 m 0.6 m 0.6 m.
BC
ll== − =
The motion from B to C and back to B is one half period
As the pendulum moves from B to A, the length is again 1.2 meters.
(a) Time required to return to A.
S
O
O
LUTION
PROBLE
M
A 25-kg bloc
k
downward fr
o
of the result
i
amplitude of
t
M
19.17
k
is supporte
d
o
m its equilib
r
i
ng motion,
(
t
he motion is
For a st
a
()a
(b)
d
by the sprin
g
r
ium position
(
b) the maxi
m
30 mm.
a
tic load P the
1
2
PP
kk
δ
=
+
k
m
ω
=
=
For
3
m
x
=
g
arrangement
and released,
m
um velocit
y
total elongat
i
2
3
P
P
k
+
3
3.6923 10
25 kg
×
0 mm = 0.03
shown. If the
determine (a
)
y
and acceler
a
i
on of the spri
n
3
N/m 12.1
5
=
m
block is mov
)
the period a
n
a
tion of the
b
n
g is:
5
3rad/s
ed vertically
n
d frequency
b
lock if the
PROBLEM 19.21
A 50-kg block is supported by the spring arrangement shown. The block is moved
vertically downward from its equilibrium position and released. Knowing that the
amplitude of the resulting motion is 60 mm, determine (a) the period and frequency
of the motion, (b) the maximum velocity and maximum acceleration of the block.
SOLUTION
(a) First, calculate the spring constant
(
)
(
)
(
)
(
)
24 kN/m 12 kN/m 12 kN/m 48 kN/mP
δ
δδ δ
=++=
48 kN/mk
∴
=
Then
(b) Now
(
)
sin
mn
xx t
ω
φ
=
+
And, since 00.060 mx=
