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Oscillations
Michael Fowler
3/24/07
Introduction
In this lecture, we will be looking at a wide variety of oscillatory phenomena.
After a brief recap
of undamped simple harmonic motion, we go on to look at a heavily damped oscillator.
We do
that before considering the lightly damped oscillator because the mathematics is a little more
straightforward—for the heavily damped case, we don’t need to use complex numbers.
But they
arise very naturally in the lightly damped case, and are great for understanding the
driven
oscillator and resonance phenomena, as will become apparent in later sections.
Brief Review of Undamped Simple Harmonic Motion
Our basic model simple harmonic oscillator is a mass
m
moving back and forth along a line on a
smooth horizontal surface, connected to an inline horizontal spring, having spring constant
k
, the
other end of the string being attached to a wall. The spring exerts a restoring force equal to –
kx
on the mass when it is a distance
x
from the equilibrium point.
By “equilibrium point” we mean
the point corresponding to the spring resting at its natural length, and therefore exerting no force
on the mass. The inclass realization of this model was an aircar, with a light spring above the
track (actually, we used
two
light springs, going in opposite directions—we found if we just one
it tended to sag on to the track when it was slack, but two in opposite directions could be kept
taut.
The two springs together act like a single spring having spring constant the sum of the
two).
Newton’s Law gives:
2
2
, or
.
dx
Fm
a m
k
x
dt
=
=−
Solving this differential equation gives the position of the mass (the aircar) relative to the rest
position as a function of time:
0
()
cos
(
)
.
xt
A
t
ω
ϕ
=
+
Here
A
is the maximum displacement, and is called the
amplitude
of the motion.
0
t
+
is
called the
phase
.
is called the
phase constant
: it depends on where in the cycle you start, that
is, where is the oscillator at time zero.
The velocity and acceleration are given by differentiating
x
(
t
) once and twice:
00
s
in
(
)
dx
vt
A
t
dt
ωϕ
=
+
and
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2
2
00
2
()
cos
(
)
.
dxt
at
A
t
dt
ω
ϕ
==
−
+
We see immediately that this
x
(
t
) does indeed satisfy Newton’s Law provided
0
is given by
0
/.
km
=
Exercise
: Verify that, apart from a possible overall constant, this expression for
0
could have
been figured out using dimensions.
Energy
The spring stores
potential energy
: if you push one end of the spring from some positive
extension
x
to
x
+
dx
(with the other end of the spring fixed, of course) the force –
kx
opposes the
motion, so you must push with force +
kx
, and therefore do work
kxdx.
To find the
total
potential energy stored by the spring when the end is
x
0
away from the equilibrium point (natural
length) we must find the total work required to stretch the spring from its natural length to an
extension
x
0
.
This means adding up all the little bits of work
kxdx
needed to get the spring from
no extension at all to an extension of
x
0
.
In other words, we need to do an integral to find the
potential energy
U
(
x
0
):
0
2
1
2
0
x
U x
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 Fall '07
 MichaelFowler
 Simple Harmonic Motion, Underdamped Oscillator

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