468
Oscillatory Motion
P15.71
(a)
When the mass is displaced a distance
x
from
equilibrium, spring 1 is stretched a distance
x
1
and
spring 2 is stretched a distance
x
2
.
By Newton’s third law, we expect
kx
11
22
=
.
When this is combined with the requirement that
xx x
=+
12
,
FIG. P15.71
we find
x
k
kk
x
1
2
=
+
L
N
M
O
Q
P
The force on either spring is given by
F
xm
a
1
=
+
L
N
M
O
Q
P
=
where
a
is the acceleration of the mass
m
.
This is in the form
Fkxm
a
eff
==
and
T
m
k
mk
k
eff
+
ππ
bg
(b)
In this case each spring is distorted by the distance
x
which the mass is displaced. Therefore,
the restoring force is
Fk
k
x
=−
+
and
k
eff
so that
T
m
=
+
2
π
.
P15.72
Let
A
represent the length below water at equilibrium and
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This note was uploaded on 12/14/2011 for the course PHY 203 taught by Professor Staff during the Fall '11 term at Indiana State University .
 Fall '11
 Staff
 Physics, Mass

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