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14.4.
Model:
The airtrack glider attached to a spring is in simple harmonic motion.
Visualize:
The position of the glider can be represented as
x
(
t
)
=
A
cos
ω
t
.
Solve:
The glider is pulled to the right and released from rest at
t
=
0 s
. It then oscillates with a period
T
=
2.0 s
and a maximum speed
v
max
=
40 cm s
=
.
(a)
v
max
=
A
and
=
2
π
T
=
2
2.0 s
=
⇒
A
=
v
max
=
0.40 m s
=
0.127 m
=
12.7 cm
(b)
The glider’s position at
t
=
0.25 s is
x
0.25 s
=
0.127 m
( )
cos
( )
0.25 s
( )
[ ]
=
0.090 m
=
9.0 cm
14.5.
Model:
The oscillation is the result of simple harmonic motion.
Visualize:
Please refer to Figure Ex14.5.
Solve:
(a)
The amplitude
A
=
10 cm.
(b)
The time to complete one cycle is the period, hence
T
=
2.0 s
and
f
=
1
T
=
1
2.0 s
=
0.50 Hz
(c)
The position of an object undergoing simple harmonic motion is
x t
( ) =
A
cos
t
+
φ
0
(
)
.
At
t
=
0 s,
x
0
=
5 cm,
thus
5 cm
=
10 cm
(
)
cos
0 s
(
) +
0
[
]
⇒
cos
0
=
5 cm
10 cm
=
1
2
⇒
0
=
cos
−
1
1
2
=
3
rad or 60
°
14.9.
Solve:
The position of the object is given by the equation
x t
( ) =
A
cos
t
+
0
(
) =
A
cos 2
ft
+
0
(
)
We can find the phase constant
0
from the initial condition:
0 cm
=
4.0 cm
(
)
cos
0
⇒
cos
0
=
0
⇒
0
=
cos
−
1
0
( ) = ±
1
2
rad
Since the object is moving to the right, the object is in the lower half of the circular motion diagram. Hence,
0
=
−
1
2
rad.
The final result, with
f
=
4.0 Hz, is
x t
( ) =
4.0 cm
(
)
cos 8.0
(
)
t
−
1
2
rad
[
]
14.13.
Model:
The airtrack glider attached to a spring is in simple harmonic motion.
Solve:
Experimentally, the period is
T
=
12.0 s
( )
10 oscillations
( )
=
1.20 s.
Using the formula for the period,
T
=
2
m
k
⇒
k
=
2
T
2
m
=
2
1.20 s
2
0.200 kg
(
) =
14.15.
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 Spring '08
 Sharpe
 Simple Harmonic Motion

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