MA 36600 LECTURE NOTES: MONDAY, APRIL 6
Applications
SpringMass System:
One Mass, Two Springs.
Now say that we have a mass
m
attached to
two
springs with constants
k
1
and
k
2
, respectively. We assume that these springs are attached to the opposite
ends of a track of fixed length
‘
, with the mass
m
in between. We continue to assume that the mass is under
the influence of an external force
F
(
t
). Denote
x
=
x
(
t
) as the
displacement from equilibrium
on the track
at time
t
. We seek a differential equation for
x
.
Let
L
1
and
L
2
denote the lengths to which the first and second springs are stretched at equilibrium,
respectively. Using Hooke’s Law, we see that the force exerted on the mass from the first spring is

k
1
L
1
(pulling to the left), whereas the force exerted on the mass from the second spring is +
k
2
L
2
(pulling to the
right). Since we are at equilibrium, the sum of the forces must be zero:

k
1
L
1
+
k
2
L
2
= 0
L
1
+
L
2
=
‘
=
⇒
L
1
=
k
2
k
1
+
k
2
‘,
L
2
=
k
1
k
1
+
k
2
‘.
In particular, if the springs are identical, then
k
1
=
k
2
, and so
L
1
=
L
2
=
‘/
2.
Denote
u
=
u
(
t
) as the
position
of the mass at time
t
. Then we have
x
=
u

L
1
=
L
2

(
‘

u
)
=
⇒
u
=
x
+
L
1
=
x
+
k
2
k
1
+
k
2
‘.
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 Spring '09
 MA
 Force, Mass, k2

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