()
2
2
222
9.8m/s
2
2
2 9.8m/s
0.080m
0.050m
0.56m/s
kg
vg
y
y
g
y
y
m
§·
=−=−
=
−
¨¸
©¹
=
A
(c) Let
m
be the original mass and
Δ
m
be the additional mass. The new angular frequency
is
′
=+
ω
km m
/(
)
Δ
. This should be half the original angular frequency, or
1
2
km
. We
solve
)
/
+=
Δ
1
2
for
m
. Square both sides of the equation, then take the
reciprocal to obtain
m
+
Δ
m
= 4
m
. This gives
m
=
Δ
m
/3 = (300 g)/3 = 100 g = 0.100 kg.
(d) The equilibrium position is determined by the balancing of the gravitational and
spring forces:
ky
= (
m
+
Δ
m
)
g
. Thus
y
= (
m
+
Δ
m
)
g
/
k
. We will need to find the value of
the spring constant
k
. Use
k
=
m
2
=
m
(2
π
f
)
2
. Then
(
)
2
22
0.100 kg 0.300 kg 9.80 m/s
+
=0
.
2
0
0
m
.
2
0.100 kg 2
2.24 Hz
mm
g
y
mf
ππ
+
Δ
=
×
This is measured from the initial position.
37. (a) The object oscillates about its equilibrium point, where the downward force of
gravity is balanced by the upward force of the spring. If
A
is the elongation of the spring
at equilibrium, then
g
A
=
, where
k
is the spring constant and
m
is the mass of the
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This note was uploaded on 06/03/2011 for the course PHY 2049 taught by Professor Any during the Spring '08 term at University of Florida.
 Spring '08
 Any
 Physics, Force, Gravity, Mass

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