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PROFESSOR: Welcome to the
applet Amplitude and Phase:

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Second Order II.

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This applet is a sibling of
the applet Amplitude and Phase:

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First Order, which I used
to introduce the MIT Mathlet

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collection.

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Its color coding,
placement conventions,

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and its functionalities
are identical to that one.

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This applet represents
a mechanical system

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driven by the motion of
the far end of the dashpot.

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Let's begin by animating the
system to see how this works.

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You can see the
dashpot is moving

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up and down sinusoidally,
driving a mass in yellow.

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And the mass is also
constrained by a spring

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at the top, attached to
a fixed wall at the top.

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Perhaps we should
animate this again,

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so you can see the
whole thing at work.

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This applet, and its siblings,
show only the steady state

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or periodic solutions
to these equations.

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They don't allow you to
pick initial conditions.

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They don't represent the
effect of transients.

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In the applet, x represents
the position of the mass,

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and you can see it
read off on this scale.

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It's represented in yellow
on the graphing window here.

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And we declare x to be
zero when the spring is

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exerting no force on the mass.

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It's at rest.

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We declare x to be
the output signal

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of the system in the mathlet.

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y denotes the position
of the plunger,

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and it's read off
by this scale here.

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We declare y to be the
input signal of the system.

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Now the force
exerted by a dashpot

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is proportional to
the relative velocity

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of the cylinder and the
piston sliding inside of it.

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The proportionality
constant is called b.

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As a consequence,
the right-hand side

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of the equation controlling
the position of the mass

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is b y dot, b times the
time derivative of y.

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This is a good example in
which the right-hand side

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of a linear equation
in standard form

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is not just the input signal.

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In this case, it's not even a
multiple of the input signal.

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It's a multiple of the
derivative of the input signal.

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The most important
case to study is

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that in which the input
signal is sinusoidal,

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and that's what is
represented here.

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In the applet, we take the
amplitude of the input signal

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to be one.

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If the amplitude of the
input signal is doubled,

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so is its derivative,
and so by superposition

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the amplitude of the output
signal would also be doubled.

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So setting the input's
amplitude equal to one

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isn't really a restriction.

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In this situation, when
the input amplitude is one,

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the output amplitude is the
same as the gain of the system.

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We can see how this
output amplitude depends

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upon the input frequency
by opening the Bode Plot

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window here.

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This opens two windows.

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The top one represents
the amplitude

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of the system response, the
gain, as a function of omega.

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And we can start
with omega small.

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When omega equals zero, the
plunger isn't moving at all,

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and the mass has no
reason to move either.

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So x equals zero, the spring
is relaxed, there's no motion.

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When omega increases, the
system response becomes greater.

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But you'll notice
something interesting.

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In this system,
when omega is small,

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the system response
leads the input signal.

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That is to say, the
phase lag is negative.

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And that's represented here.

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This is a little confusing.

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This is a graph of the negative
of the phase lag or the phase

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gain.

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And that's a positive angle, in
this system, for omega small.

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You can see the effect.

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The mass seems to be
pulling the plunger,

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although that's not
actually what's happening.

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When omega increases,
the amplitude

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of the system response increases
until a critical moment

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when omega takes on
some critical value.

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This is the resonant
frequency of the system.

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And at this point,
the system response

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is identical to
the input signal.

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Shall we see what this looks
like by animating the system?

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Now the plunger seems to
be locked to the piston.

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That's not actually the case,
but the system is simply

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operating in harmony here.

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It's the resonant frequency.

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When omega increases
still further,

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then the system
response falls off.

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The phase lag becomes
positive, the response

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falls behind the input signal.

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And that's all visible from
the Bode plot pictures.

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As I look at this, I notice
something interesting.

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When the system response
reaches a maximum or a minimum,

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that's exactly the same moment
as when the system response

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curve crosses the
input signal curve.

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So in other words, the
amplitude of the output signal

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equals the value
of the input signal

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at the moment when they cross.

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Let's see if that is
always the case when

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I change the frequency here.

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That always seems
to be the case.

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Very interesting.

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And in fact, perhaps we should
change the values of b and k

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to see whether that
continues to be the case.

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Yes it just seems to
drag up and down there.

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And similarly, if I change the
value of the spring constant k,

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again that peak seems to move up
and down along the blue curve.

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Very odd.

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We'll come back to that.

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Here are some
further observations

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that you can see from
experimenting with this applet.

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First of all, let's
watch what happens

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if I change the value of
the damping constant b.

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I'm going to watch the Bode plot
over here, the amplitude Bode

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plot.

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Well, it changes, but one
thing that doesn't change

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is the position of the
resonant peak there.

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In other words, the
resonant frequency

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seems to be independent
of the value of b,

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of the damping constant.

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The second thing you can
observe, as I vary b,

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is when b is large, the
resonant hump is quite broad,

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but as b becomes smaller,
that spike becomes narrower

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and narrower and narrower.

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And the third thing you can
notice from looking at this

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is as b gets to
be small, the flip

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between phase lag of close
to minus pi over 2 to a phase

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lag of close to plus pi over 2
happens more and more abruptly

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as I change the frequency
from something small

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and cross through
that resonant peak.

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Now the system response is
ahead of the input signal.

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But it very quickly
flips to being

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behind it as you cross
the resonant frequency.

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And that transition happens
more and more rapidly

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as b gets to be small.

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Each one of these
three observations

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can be verified by calculation.

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The relationship between
the gain and the phase lag

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can be illustrated very well
using the Nyquist plot, which

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I'll open using this key.

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This is a plot of
the complex plane,

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and on it is drawn a
complex number in yellow.

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And that complex number has
a magnitude and an angle.

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The magnitude is the gain.

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This yellow strut is the same
length as this radius vector.

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And the angle is the
negative of the phase lag.

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It's the phase gain.

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So when the angle's positive
here, it goes through zero here

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and becomes negative
down here, when

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the phase lag becomes positive.

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So you can see why it is
that the resonant peak occurs

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at exactly the same instant
as when the phase lag is zero.

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That says that this
trajectory, this Nyquist plot,

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goes through this point,
which is the point

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1 in the complex plane,
where the angle is zero

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and the magnitude is 1.

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This trajectory is in fact
independent of either, both,

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of the system parameters.

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If I change them,
many things change,

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but the shape of this
trajectory, the Nyquist plot,

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is independent of those
two system parameters.

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And in fact, it's a circle
of radius 1/2 and center 1/2.

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You can verify that by
calculation as well.

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That fact explains
the observation

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we made earlier, that the
amplitude of the system

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response, the gain, seems to be
equal to the value of the input

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signal when that
maximum is achieved.

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In other words, the claim
is that the gain equals

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the cosine of the phase lag.

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The gain is the cosine
of the phase lag.

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This curve is given by the
equation the radius equals

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the cosine of the angle.

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r equals cosine theta
is a polar equation

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for this particular circle.

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And so the fact that this is a
circle is the same as the fact

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that we observed earlier, that
the solution curve crosses

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the input curve just when it
reaches a maximum or a minimum.

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Well, in addition to the spring
system shown in this applet,

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this very same equation
models the response

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of an AM radio receiver.

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The input signal is the
incoming radio signal,

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which contains
electromagnetic waves

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of many different frequencies.

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The output signal
feeds to the amplifier.

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You tune the radio to a
particular angular frequency

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by adjusting k, the spring
constant, which in the circle

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model is the capacitance, so
that the resonant frequency is

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the desired frequency
that you want to tune to.

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Then the output generated
by the other frequencies

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is much less than the output
generated by frequency omega_R.

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How much less depends on how
sharp the resonance spike is.

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The sharper the spike,
the better the tuning.

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You can make the spike
sharper by decreasing

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the value of b, which in the AM
radio model, is the resistance.