3.6
11
2.5
9.8
2
6
l
Ts
g
ππ
==
≈
D
s
3.7
For springs tied in parallel:
() ( )
12
1
2
s
Fx
k
xkx
k kx
=−
−
+
()
1
2
kk
m
ω
+
=
For springs tied in series:
The upward force
is
m
eq
k
x
.
Therefore, the downward force on spring
is
2
k
eq
k
x
.
The upward force on the spring
is
2
k
1
kx
′
where
x
′
is the
displacement
of P, the point at which the springs are tied.
Since the spring
is in equilibrium,
2
k
1
eq
k
′ =
x
.
Meanwhile,
The upward force at P is
1
′
.
The downward force at P is
( )
2
kxx
′
−
.
Therefore,
k
x x
′′
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This note was uploaded on 01/06/2012 for the course PHYSICS 4360C taught by Professor Johnanderson during the Fall '11 term at UNF.
 Fall '11
 JohnAnderson
 Force, Work

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