*Michael Fowler 3/24/07*

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.

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 in-class 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).

_{}

Solving this differential equation gives the position of the mass (the aircar) relative to the rest position as a function of time:

_{}

Here *A* is the maximum displacement, and is called the
**amplitude** of the motion. _{} 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:

_{}

and

_{}

We see immediately that this *x*(*t*) does indeed
satisfy _{} is given by

_{}

*Exercise*: Verify that,
apart from a possible overall constant, this expression for _{}_{ }could have
been figured out using dimensions.

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}):

_{}

So the potential energy plotted as a function of distance from equilibrium is parabolic:

The
oscillator has *total* energy equal to kinetic energy + potential energy,

_{}

when the mass is at position *x*. Putting in the values of *x*(*t*), *v*(*t*) from the equations above, it is easy to check
that *E* is independent of time and equal to _{}, *A* being the
amplitude of the motion, the maximum displacement. Of course, when the oscillator is at *A*, it is momentarily at rest, so has no
kinetic energy.

Suppose now the motion is damped, with a drag force proportional to velocity. The equation of motion becomes:

_{}

Although
this equation looks more difficult, it really isn’t! The important point is that the terms are
just *derivatives* of *x* with respect to time, multiplied by *constants*.
It would be a lot more difficult if we had a drag force proportional to the
square of the velocity, or if the force exerted by the spring were not a
constant times *x* (this means we can’t stretch the string too far!). Anyway, it is easy to find exponential
functions that are solutions to this equation.
Let us guess a solution:

_{}

Inserting this in the equation, using

_{}

we find that it *is* a solution provided that _{} satisfies:

_{}

from which

_{}

Staring at this expression for _{}, we notice that for _{} to be real, we need to
have

_{}

What can that mean?
Remember *b* is the damping parameter—we’re finding that our
proposed exponential solution only works for *large* damping! Let’s analyze
the large damping case now, then after that we’ll go on to see how to extend
the solution to small damping.

It’s worth looking at the case of *very* large damping,
where the two exponential solutions turn out to decay at very different
rates. For *b*^{2} *much*
greater than 4*mk*, we can write

_{}

and then expand the square root using

_{}

valid for small *x*, to find that approximately—for
large *b—*the two possible values of _{} are:

_{}

That is to say, there are two possible highly damped decay modes,

_{}

Note that since the damping *b* is large, _{} is *large*,
meaning *fast* decay, and _{} is *small*,
meaning *slow* decay.

*Question*: what,
physically, is going on in these two different highly damped exponential
decays? Can you construct a plausible
scenario of a mass on a spring, all in molasses, to see why two very different
rates of change of speed are possible?

*Hint*: look again at
the equation of motion of this damped oscillator. Notice that in each of these highly damped
decays, one term doesn’t play any part—but the irrelevant term is a *different*
term for the two decays!

*Answer 1*: for _{}, evidently the *mass* doesn’t play a role. This decay is what you get if you pull the
mass to one side, let go, then, after it gets moving, it will very slowly
settle towards the equilibrium point.
Its rate of approach is determined by balancing the spring’s force
against the speed-dependent damping force, to give the speed. The rate of change of speed—the
acceleration—is so tiny that the *inertial* term—the mass—is negligible.

*Answer 2*: for _{}, the *spring* is negligible. And, this is very *fast*
motion (*b*/*m* >>
*k*/*b*, since we said *b*^{2} >> 4*mk*.) The way to get this motion is to pull the
mass to one side, then give it a very strong kick towards the equilibrium
point. If you give it just the right
(high) speed, all the momentum you imparted will be spent overcoming the
damping force as the mass moves to the center—the force of the spring will be
negligible.

The damped oscillator equation

_{}

is
a linear equation. This means that if *x*_{1}(*t*) is a
solution, and *x*_{2}(*t*) is another solution, that is,

_{}

then just adding the two equations we get:

_{}

It is also clear that *multiplying a solution by a
constant produces another solution*: if *x*(*t*) satisfies the
equation, so does 3*x*(*t*).

This means, then, that given two solutions *x*_{1}(*t*)
and *x*_{2}(*t*), and two arbitrary constants *A*_{1}
and *A*_{2}, the function

*A*_{1}*x*_{1}(*t*)
+ *A*_{2}*x*_{2}(*t*)

is also a solution of the differential equation.

In fact, all possible motions of the highly damped
oscillator have this form. The way to understand this is to realize that the
oscillator’s motion is *completely determined* if we specify at an initial
instant of time *both* the position *and* the velocity of the oscillator. The equation of motion gives the acceleration
as a function of position and velocity, so, at least in principle, we can work out step by step how the mass
must move; technically, we are integrating the equation of motion, either
mathematically, or numerically such as by using a spreadsheet. So, by suitably
adjusting the two arbitrary constants *A*_{1} and *A*_{2},
we can match our sum of solutions to any given initial position and velocity.

To summarize, for the highly damped oscillator any solution is of the form:

_{}

*Exercises on highly damped oscillations*

1. If the oscillator
is pulled aside a distance *x*_{0}, and released from rest at *t*
= 0, what are *A*_{1}, *A*_{2}? Describe the subsequent motion, especially
the very beginning: what is the initial acceleration? (*Hint*:
think carefully about how important the damping term is immediately after
release from rest—you should be able to *guess*
the initial acceleration.)

2. If the oscillator is initially at the equilibrium
position *x*_{0} = 0, but is given a kick to a velocity *v*_{0},
find *A*_{1} and *A*_{2} and describe the subsequent
motion.

The equation for the highly damped oscillator is a linear differential equation, that is, an equation of the form (in more usual notation):

_{}

where *c*_{0}, *c*_{1} and *c*_{2}
are constants, that is, independent of *x*.

For such a linear differential equation, if *f*_{1}(*x*)
and *f*_{2}(*x*) are solutions, so is *A*_{1}*f*_{1}(*x*)
+*A*_{2}*f*_{2}(*x*) for any constants *A*_{1},
*A*_{2}. This is called the ** Principle of Superposition**,
and is proved in general

Even more important, this Principle of Superposition is
valid, using analogous arguments, for linear differential equations in *more
than one variable*, such as the wave equations we shall be considering
shortly. In that case, it gives insight
into how waves can pass through each other and emerge unchanged.

We can go through exactly the same mathematical steps in solving the equation of motion as we did for the heavily damped case: we look for solutions of the form

_{}

_{}

and as before we find there are solutions with

_{}

But the difference is that for *light *damping, by
which we mean *b*^{2} < 4*mk*, the expression inside the square root is
negative! We are going to have to work
with the square root of a negative number.
We do this formally by writing:

_{}

with *i*^{2} = -1 as usual. This gives the
two possible exponential solutions:

_{}

and a general solution

_{}

Of
course, the position of the mass *x*(*t*) has to be a real
number! We must choose *A*_{1}
and *A*_{2} to make sure this is so. If we choose

_{}

where *A* and _{} are real, and
remembering

_{}

we find

_{}

This is the most general real solution of the lightly damped
oscillator—the two arbitrary constants are the amplitude *A* and the phase
_{}. So for small *b*,
we get a cosine oscillation multiplied by a gradually decreasing function, *e*^{-bt}^{/2m}.

This is often written in terms of a **decay time** _{} defined by

_{}

The amplitude of oscillation *A * therefore decays in time as _{}, and the *energy* of the oscillator (proportional to *A*^{2})
decays as _{} This means that in
time_{} the energy is down by a factor 1/*e*, with *e* =
2.71828…

The solution is sometimes written

_{}

where

_{}

Notice that for small damping, the oscillation frequency
doesn’t change much from the undamped value: the change is proportional to the *square*
of the damping.

The
*Q* factor is a measure of the “quality” of an oscillator (such as a
bell): how long will it keep ringing once you hit it? Essentially, it is a measure of how many
oscillations take place during the time the energy decays by the factor of 1/*e*.

*Q
*is defined by:

_{}

so,
strictly speaking, it measures how many radians the oscillator goes around in
time _{}. For a typical bell, _{} would be a few
seconds, if the note is middle C, 256 Hz, that’s _{} so *Q* would be
of order a few thousand.

*Exercise*:
estimate *Q* for the following oscillator (and don’t forget the energy is
proportional to the *square* of the amplitude):

The
yellow curves in the graph above are the pair of functions +*e*^{-bt}^{/2m},
-*
e*^{-bt}^{/2m},
often referred to as the *envelope* of the oscillation curve, as they
“envelope” it from above and below.

There is just one case we haven’t really discussed, and it’s
called “critical damping”: what happens when *b*^{2} – 4*mk*
is exactly zero? At first glance, that
sounds easy to answer: there’s just the one solution

_{}

But that’s not good enough—it tells us that if we begin at *t*
= 0 with the mass at *x*_{0}, it must have velocity *dx*/*dt*
equal to -*x*_{0}*b*/2*m*.
But, in fact, we can put the mass at *x*_{0}
and kick it to any initial velocity we want!
So what happened to the other solution?

We can get a clue by examining the two exponentially falling solutions for the overdamped case as we approach critical damping:

_{}

As we approach critical damping, the small quantity

_{}

_{}

approaches zero. The general solution to the equation has the form

_{}

This is a valid solution for any real *A*_{1}, *A*_{2}.
To find the solution we’re missing, the
trick is to take _{} In the limit of small _{}, we can take _{} and we discover the
solution

_{}

As usual, we can always multiply a solution of a linear differential equation by a constant and still have a solution, so we write our new solution as

_{}

The general solution to the critically damped oscillator then has the form:

_{}

*Exercise*: check that this *is* a solution for
the critical damping case, and verify that solutions of the form *t* times
an exponential *don’t* work for the other (*non*critical damping)
cases.

A
shock absorber is basically a damped spring oscillator, the damping is from a
piston moving in a cylinder filled with oil.
Obviously, if the oil is very thin, there won’t be much damping, a
pothole will cause your car to bounce up and down a few times, and shake you
up. On the other hand, if the oil is *really*
thick, or the piston too tight, the shock absorber will be too stiff—it won’t
absorb the shock, and you will! So we
need to tune the damping so that the car responds smoothly to a bump in the
road, but doesn’t continue to bounce after the bump.

Clearly,
the “Damped Oscillator” graph in the *Q*-factor section above corresponds
to too little damping for comfort from a shock absorber point of view, such an
oscillator is said to be **underdamped**.
The opposite case**, overdamping**, looks like this:

The
dividing line between overdamping and underdamping is called **critical
damping**. Keeping everything constant
except the damping force from the graph above, critical damping looks like:

This
corresponds to _{} in the equation for *x*(*t*)
above, so it is a purely exponential curve. Notice that the oscillator moves
more quickly to zero than in the overdamped (stiff oil) case.

You might think that critical damping is the best solution for a shock absorber, but actually a little less damping might give a better ride: there would be a slight amount of bouncing, but a quicker response, like this:

You can find out how your shock absorbers behave by pressing down one corner of the car and then letting go. If the car clearly bounces around, the damping is too little, and you need new shocks.

We are now ready to examine a very important case: the driven damped oscillator. By this, we mean a damped oscillator as analyzed above, but with a periodic external force driving it. If the driving force has the same period as the oscillator, the amplitude can increase, perhaps to disastrous proportions, as in the famous case of the Tacoma Narrows Bridge.

The equation of motion for the driven damped oscillator is:

_{}

We shall be using_{}for the frequency of the driving force, and_{} for the natural frequency of the oscillator if the damping
term is ignored, _{}

The solution to this differential equation is not unique: as with any second order differential equation, there are two constants of integration, which are determined by specifying the initial position and velocity.

However, as we shall prove below using complex numbers, the
equation *does* have a *unique* steady state solution with *x* oscillating at the same frequency as
the external drive. How can that be
fitted to arbitrary initial conditions?
The key is that we can add to the steady state solution any solution of
the *undriven* equation _{} and we’ll
clearly still have a solution of the full damped driven equation. We know what those undriven solutions look
like: they all die away as time goes on.
So, we can add such a solution to fit the specified initial conditions,
and after a while the system will lose memory of those conditions and settle
into the steady driven solution. The initial deviations from the steady
solution needed to satisfy initial conditions are termed *transients*.

Here’s a pair of examples: the same driven damped oscillator, started with zero velocity, once from the origin and once from 0.5:

Notice that after about 70 seconds, the two curves are the same, both in amplitude and phase.

We begin by writing:

external driving
force = _{}

with *F*_{0} real, so the *real *driving force is just the real part of this,_{}.

So now we’re trying to solve the equation

_{}

We’ll try the complex function, _{}, with *A* a real
number, *x*(*t*) cycling at the same rate as the driving force. We can always take the amplitude *A* to be real: that is *not* a restriction, since we’ve added the
adjustable phase factor _{}. Physically, this factor
allows the solution to lag the driver in phase, as indeed we shall find to be
the case. If we succeed in finding an *x*(*t*)
that satisfies the equation, the real parts of the two sides of the equation
must be equal:

*If *_{}* is a
solution to the equation with the complex driving force, _{} its *real

It’s very easy to check that _{} *is* a solution to the equation, with the right *A* and _{}! Just plug it in and
see what happens. The differentiations
are simple, giving

_{}

To
nail down *A* and _{}, we begin by cancelling out the common factor _{} , then shifting the _{}to the other side, to find

_{}

Now *A* is a real
number, and the right hand side of this equation looks alarmingly complex, so
what’s going on?

Let’s begin to untangle this by diagramming that complex number in the denominator,

_{}.

It has real part _{} and imaginary part _{}.

Its phase is the angle_{}: that is,

_{}.

Putting it in the equation in this _{} notation gives

_{}

_{ }

Now,
remembering that *F*_{0} and *r* are real, we see that *A* *will*
be real (as it must be) *if* _{} is real: so _{}and

_{}

_{ }

where
we’ve written _{}.

So we’ve already solved the differential equation: the
amplitude *A* is proportional to the
strength of the driving force, and that ratio is determined by the parameters
of the undriven oscillator, and the driving oscillation frequency.

The important thing to note about the amplitude *A* is that if the damping *b* is
small, *A* gets *very large* when
the frequency of the driver approaches the natural frequency of the
oscillator! This is called *resonance*,
and is what happened to the

The *phase lag* of
the oscillations behind the driver, _{}, is completely determined by the frequency together with the
physical constants of the undriven oscillator: the mass, spring constant, and
damping strength. So, when the driving
force * _{}*generates the motion

Note that at *low*
frequencies, _{} the oscillator lags
behind by a small angle, but at resonance _{} and for driving
frequencies above _{}, _{}

** To summarize**: we’ve just
established that

with the complex driving force _{}.

So, *equating the real
parts of the two sides of the equation*, since *m*, *b*, *k* are all real,

_{}

*is a solution of the equation with the real
driving force _{}.*

We could have found this out without complex numbers, by
using a trial solution _{}. However, it’s not
that easy—the left hand side becomes a mix of sines and cosines, and one needs
to use trig identities to sort it all out.
With a little practice, the complex method is easier and is certainly
more direct.

Now the total energy of the oscillator is

_{}

Putting in

_{}

gives

_{}

Note that this is *not*
constant through the cycle unless the oscillator is at resonance, _{}

_{ }

We can see from the above that at the resonant frequency, _{}, and from the section above

_{}

so the energy in the oscillator at the resonant frequency is

_{}

recalling that _{}

So *Q*, the quality
factor, the measure of how long an oscillator keeps ringing, also measures the
strength of response of the oscillator to an external driver at the resonant
frequency.

But what happens on going *away* from the resonant frequency?
Let’s assume that *Q* is large,
and the driving force is kept constant.
It won’t take much change in_{}from_{} for the denominator _{} in the expression for *E* to double in size. In fact, for large *Q*, it’s a good approximation to replace _{} by _{}over that variation, and it is then straightforward to check
that the energy in the oscillator drops to one-half its resonant value for _{}

*Exercise*: prove this.

The bottom line is that for increasing *Q*, the response at the resonant frequency gets larger, but this
large response takes place over a narrower and narrower range in driving
frequencies.

An important practical question is: how much *work* is the driver doing to keep this
thing going?

It’s simplest to work with the real solution. Suppose the oscillator moves through _{} in a time _{}, the driving force does work _{}, so

_{}

The important thing is the *average* rate of working of the driving force, the *mean power input*, found by averaging
over a complete cycle:

From _{}, _{}, averaging the power input (the bar above means average over
a complete cycle) and denoting average power by *P*,

_{}

since over one cycle the average _{} and _{} (Remembering _{}at all times, and sine is just cosine moved over, so they
must have the same average over a complete cycle.)

This can be expressed entirely in terms of the driving force and frequency. Since

_{}

_{}

*Exercise
1*: Prove that for a lightly damped oscillator, at resonance the
oscillator extracts the most work from the driving force.

*Exercise 2*: Prove that
any solution of the damped oscillator equation (with *F* = 0) can be added
to the driven oscillator solution, and gives another solution to the driven
oscillator. How do you pick the “right solution”?