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Chapter 14
Page 1
CHAPTER 14 – Oscillations
1.
In one period the particle will travel from one extreme position to the other (a distance of 2
A
) and back
again.
The total distance traveled is
d
= 4
A
= 4(0.15 m) =
0.60 m
.
2. (
a
) We find the spring constant from the elongation caused by the weight:
k
=
mg
/?
x
= (3.7 kg)(9.80 m/s
2
)/(0.028 m) =
1.30
×
10
3
N/m
.
(
b
) Because the fish will oscillate about the equilibrium position, the amplitude will be the distance
the fish was pulled down from equilibrium:
A
= 2.5 cm
.
The frequency of vibration will be
f
= (
k
/
m
)
1/2
/2p = [(1.30
×
10
3
N/m)/(3.7 kg)]
1/2
/2p =
3.0 Hz
.
3.
We find the spring constant from the compression caused by the increased weight:
k
=
mg
/
x
= (80 kg)(9.80 m/s
2
)/(0.0140 m) = 5.60
×
10
4
N/m.
The frequency of vibration will be
f
= (
k
/
m
)
1/2
/2p = [(5.60
×
10
4
N/m)/(1080 kg)]
1/2
/2p =
1.15 Hz
.
4. (
a
) Because the motion starts at the maximum extension, we have
x
=
A
cos (
ω
t
) =
A
cos (2p
t
/
T
) =
(8.8 cm) cos [2p
t
/(0.75 s)]
.
(
b
) At
t
= 1.8 s we get
x
= (8.8 cm) cos [2p(1.8 s)/(0.75 s)] =
– 7.1 cm
.
5. (
a
) We find the effective spring constant from the frequency:
f
1
= (
k
/
m
1
)
1/2
/2p;
10 Hz = [
k
/(0.60
×
10
–3
kg)]
1/2
/2p, which gives
k
=
2.4 N/m
.
(
b
) The new frequency of vibration will be
f
2
= (
k
/
m
2
)
1/2
/2p = [(2.37 N/m)/(0.40
×
10
–3
kg)]
1/2
/2p =
12 Hz
.
6.
The general expression for the displacement is
x
=
A
cos (
t
+
φ
)
, so
x
(
t
= 0) =
A
cos
.
(
a
) –
A
=
A
cos
, which gives cos
= – 1, so
= p (or –p)
.
(
b
) 0 =
A
cos
, which gives cos
= 0, so
= p/2 (or 3p/2)
.
(
c
)
A
=
A
cos
, which gives cos
= + 1, so
= 0
.
(
d
)
!
A
=
A
cos
, which gives cos
=
!
, so
= p/3 (or –p/3)
.
(
e
) –
A
/2 =
A
cos
, which gives cos
= –
!
, so
= 2p/3 (or 4p/3)
.
(
f
)
A
/v2 =
A
cos
, which gives cos
= 1/v2, so
= p/4 (or –p/4)
.
7.
Because the mass is released at the maximum displacement, we have
x
=
x
max
cos (
t
);
v
= –
v
max
sin (
t
);
a
= –
a
max
cos (
t
).
(
a
) We find
t
from
v
= –
!
v
max
= –
v
max
sin (
t
), which gives
t
= 30°.
Thus the distance is
x
=
x
max
cos (
t
) =
x
max
cos 30° =
0.866
x
max
.
(
b
) We find
t
from
a
= –
!
a
max
= –
a
max
cos (
t
), which gives
t
= 60°.
Thus the distance is
x
=
x
max
cos (
t
) =
x
max
cos 60° =
0.500
x
max
.
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View Full Document Chapter 14
Page 2
8. (
a
) We find the effective spring constant from the frequency:
f
1
= (
k
/
m
1
)
1/2
/2p;
2.5 Hz = [
k
/(0.050 kg)]
1/2
/2p, which gives
k
=
12 N/m
.
(
b
) Because the size and shape are the same, the spring constant, which is determined by the
buoyant force, will be the same.
The new frequency of vibration will be
f
2
= (
k
/
m
2
)
1/2
/2p = [(12 N/m)/(0.25 kg)]
1/2
/2p =
1.1 Hz
.
9.
For a general displacement, we have
x
=
A
cos (
ω
t
+
φ
);
v
= –
A
sin (
t
+
).
We find
t
+
from
v
= –
!
A
= –
A
sin (
t
+
), which gives
t
+
= 30°.
Thus the displacement is
x
=
A
cos (
t
+
) =
A
cos (30°) =
0.866
A
.
10. The dependence of the frequency on the mass is
f
= (
k
/
m
)
1/2
/2p.
Because the spring constant does not change, we have
f
2
/
f
= (
m
/
m
2
)
1/2
;
(0.48 Hz)/(0.88 Hz) = [
m
/(
m
+ 1.25 kg)]
1/2
, which gives
m
=
0.53 kg
.
11. In the equilibrium position, the net force is zero.
When the mass is pulled
down a distance
x
, the net restoring force is the sum of the additional forces
from the springs, so we have
F
net
= ?
F
2
+ ?
F
1
= –
k
2
x
–
k
1
x
= – (
k
1
+
k
2
)
x
,
which gives an effective force constant of
k
1
+
k
2
.
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This note was uploaded on 02/05/2010 for the course PHYSICS 37,38,39 taught by Professor Fielding during the Spring '09 term at Los Angeles Valley College.
 Spring '09
 Fielding
 Physics

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