Biomedical Engineering Reference
In-Depth Information
We then stretch the system so extending the two springs, and I will call the instantaneous
positions of the two masses
x
1
and
x
2
. The extensions of the springs from their equilibrium
positions are
ξ
1
=
x
1
−
R
1,e
ξ
2
=
x
2
−
R
2,e
Consider the left-hand spring; it exerts a restoring force on particle 1 of
k
1
ξ
1
. Now consider
the right-hand spring. This spring is stretched by an amount (ξ
2
- ξ
1
), and so it exerts a
force of
k
2
(ξ
2
−
−
ξ
1
); this force acts to the left on particle 2 and to the right on particle 1.
Application of Newton's second law gives
m
1
d
2
ξ
1
d
t
2
k
2
(ξ
2
−
ξ
1
)
−
k
1
ξ
1
=
d
2
ξ
2
d
t
2
−
k
2
(ξ
2
−
ξ
1
)
=
m
2
(4.1)
There aremany different solutions to these simultaneous differential equations, but it proves
possible to find two particularly simple ones called
normal modes of vibration
. These
have the property that both particles execute simple harmonic motion at the same angular
frequency. Not only that, every possible vibrational motion of the two particles can be
described as linear combinations of the normal modes.
Having said that it proves possible to find such solutions where both particles vibrate
with the same angular frequency ω, let me assume that there exist such solutions to the
equations of motion such that
ξ
1
(
t
)
=
A
sin(ω
t
+
ϕ
1
)
ξ
2
(
t
)
=
B
sin(ω
t
+
ϕ
2
)
A
,
B
, ϕ
1
and ϕ
2
are constants that have to be determined from the boundary conditions.
Differentiating these two equations with respect to time gives
d
2
ξ
1
(
t
)
d
t
2
=−
ω
2
A
sin(ω
t
+
ϕ
1
)
d
2
ξ
2
(
t
)
d
t
2
=−
ω
2
B
sin(ω
t
+
ϕ
2
)
and substituting these expressions into the equations of motion gives
(
k
1
+
k
2
)
k
2
m
1
ξ
2
=−
ω
2
ξ
1
−
ξ
1
+
m
1
k
2
m
2
ξ
1
−
k
2
m
2
ξ
2
=−
ω
2
ξ
2
(4.2)
These two equations are simultaneously valid only when ω has one of two possible values
called the normal mode angular frequencies. In either case, both particles oscillate with the
same angular frequency.
