klare (alk736) – homework 30 – Turner – (58220)
1
This printout should have 13 questions.
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beFore answering.
001
(part 1 oF 3) 10.0 points
Consider the oscillation oF a massspring
system where
x
=
A
cos(
ω
t
+
φ
)
.
At the time
t
= 0, the mass
m
is at
x
=0
(the equilibrium point) and it is moving with
a positive velocity
v
0
.
k
m
v
0
x
=0
x
±ind the phase angle
φ
.
Consider
x
as
the projection oF a counterclockwise uniForm
circular motion.
1.
φ
=
3
2
π
correct
2.
φ
=0
3.
φ
=
π
4.
φ
=
1
2
π
5.
φ
=
5
4
π
6.
φ
=
3
4
π
7.
φ
=2
π
8.
φ
=
1
4
π
9.
φ
=
7
4
π
Explanation:
A
B
C
D
E
F
G
H
!
"
x
The SHM can be represented by the
x

projection oF a uniForm circular motion:
x
=
A
cos
ω
(
t
+
φ
)
.
At
t
= 0,
x
= 0. ±rom inspection, it should
be either
C
or
G
; at
C
,
v<
0, while at
G
,
v>
0.
002
(part 2 oF 3) 10.0 points
Let the mass be 6
.
45 kg, spring constant
697 N
/
m and the initial velocity 2
.
31 m
/
s.
±ind the amplitude
A
.
Correct answer: 0
.
222216 m.
Explanation:
Let :
m
=6
.
45 kg
,
k
= 697 N
/
m
,
and
v
0
=2
.
31 m
/
s
.
v
=
dx
dt
=

ω
A
sin(
ω
t
+
φ
)
,
so the velocity amplitude or the maximum
speed is
v
max
=
ω
A
;
i.e.
,
v
0
=
ω
A
A
=
v
0
ω
=
v
0
±
m
k
= (2
.
31 m
/
s)
²
(6
.
45 kg)
(697 N
/
m)
=
0
.
222216 m
.
003
(part 3 oF 3) 10.0 points
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View Full Documentklare (alk736) – homework 30 – Turner – (58220)
2
Find the total energy of oscillation at
t
=
T
8
;
i.e.
, at oneeighth of the period.
Consider
what happens to the total energy during os
cillatory motion.
1.
E
=
1
2
√
2
mv
2
0
2.
E
=
mv
2
0
3.
E
=2
mv
2
0
4.
E
=
3
4
mv
2
0
5.
E
=
1
2
mv
2
0
correct
6.
E
=
3
2
mv
2
0
7.
E
=
1
4
mv
2
0
8.
E
=
5
2
mv
2
0
Explanation:
Since the spring force is a conservative force
the total energy is conserved. One may equate
the energy at
t
= 0 where
x
= 0, so
E
=
K
+
U
=
K
max
=
1
2
mv
2
0
.
004
(part 1 of 3) 10.0 points
A block of mass 0
.
2 kg is attached to a
spring of spring constant 26 N
/
m on a fric
tionless track. The block moves in simple har
monic motion with amplitude 0
.
17 m. While
passing through the equilibrium point from
left to right, the block is struck by a bullet,
which stops inside the block.
The velocity of the bullet immediately be
fore it strikes the block is 70 m
/
s and the mass
of the bullet is 4
.
36 g.
26 N
/
m
0
.
2 kg
4
.
36 g
70 m
/
s
Find the speed of the block immediately
before the collision.
Correct answer: 1
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