## 24Transients

### 24–1The energy of an oscillator

Although this chapter is entitled “transients,” certain
parts of it are, in a way, part of the last chapter on forced
oscillation. One of the features of a forced oscillation which we have
not yet discussed is the *energy* in the oscillation. Let us now
consider that energy.

In a mechanical oscillator, how much kinetic energy is there? It is
proportional to the square of the velocity. Now we come to an
important point. Consider an arbitrary quantity $A$, which may be the
velocity or something else that we want to discuss. When we write $A =
\hat{A}e^{i\omega t}$, a complex number, the true and honest $A$, in
the physical world, is only the *real part*; therefore if, for
some reason, we want to use the *square* of $A$, it is not right
to square the complex number and then take the real part, because the
real part of the square of a complex number is not just the square of
the real part, but also involves the *imaginary* part. So when we
wish to find the energy we have to get away from the complex notation
for a while to see what the inner workings are.

Now the true physical $A$ is the real part of $A_0e^{i(\omega t+\Delta)}$, that is, $A = A_0 \cos\,(\omega t + \Delta)$, where $\hat{A}$, the complex number, is written as $A_0e^{i\Delta}$. Now the square of this real physical quantity is $A^2 = A_0^2 \cos^2\,(\omega t + \Delta)$. The square of the quantity, then, goes up and down from a maximum to zero, like the square of the cosine. The square of the cosine has a maximum of $1$ and a minimum of $0$, and its average value is $1/2$.

In many circumstances we are not interested in the energy at any
specific moment during the oscillation; for a large number of
applications we merely want the average of $A^2$, the *mean* of
the square of $A$ over a period of time large compared with the period
of oscillation. In those circumstances, the average of the cosine
squared may be used, so we have the following theorem: if $A$ is
represented by a complex number, then the mean of $A^2$ is equal
to $\tfrac{1}{2}A_0^2$. Now $A_0^2$ is the square of the magnitude of the
complex $\hat{A}$. (This can be written in many ways—some people
like to write $\abs{\hat{A}}^2$; others write, $\hat{A}\hat{A}\cconj$,
$\hat{A}$ times its complex conjugate.) We shall use this theorem
several times.

Now let us consider the energy in a forced oscillator. The equation
for the forced oscillator is
\begin{equation}
\label{Eq:I:24:1}
m\,d^2x/dt^2+\gamma m\,dx/dt+m\omega_0^2x=F(t).
\end{equation}
In our problem, of course, $F(t)$ is a cosine function of $t$. Now let
us analyze the situation: how much work is done by the outside
force $F$? The work done by the force per second, i.e., the power, is the
force times the velocity. (We know that the differential work in a
time $dt$ is $F\,dx$, and the power is $F\,dx/dt$.) Thus
\begin{equation}
\label{Eq:I:24:2}
P=F\,\ddt{x}{t}=m\biggl[\biggl(\ddt{x}{t}\biggr)\biggl(
\frac{d^2x}{dt^2}\biggr)+\omega_0^2x\biggl(\ddt{x}{t}\biggr)
\biggr]+\gamma m\biggl(\ddt{x}{t}\biggr)^2.
\end{equation}
\begin{align}
P=F\,&\ddt{x}{t}\notag\\[1.25ex]
=m&\biggl[\biggl(\ddt{x}{t}\biggr)\biggl(
\frac{d^2x}{dt^2}\biggr)+\omega_0^2x\biggl(\ddt{x}{t}\biggr)\biggr]\notag\\[.5ex]
\label{Eq:I:24:2}
&+\gamma m\biggl(\ddt{x}{t}\biggr)^2.
\end{align}
But the first two terms on the right can also be written as
$d/dt[\tfrac{1}{2}m(dx/dt)^2 + \tfrac{1}{2}m\omega_0^2x^2]$, as is
immediately verified by differentiating. That is to say, the term in
brackets is a pure derivative of two terms that are easy to
understand—one is the kinetic energy of motion, and the other is the
potential energy of the spring. Let us call this quantity the
*stored energy*, that is, the energy stored in the
oscillation. Suppose that we want the average power over many cycles
when the oscillator is being forced and has been running for a long
time. In the long run, the stored energy does not change—its
derivative gives zero average effect. In other words, if we average
the power in the long run, *all the energy ultimately ends up in
the resistive term $\gamma m(dx/dt)^2$*. There is *some* energy
stored in the oscillation, but that does not change with time, if we
average over many cycles. Therefore the mean power $\avg{P}$ is
\begin{equation}
\label{Eq:I:24:3}
\avg{P} = \avg{\gamma m(dx/dt)^2}.
\end{equation}

Using our method of writing complex numbers, and our theorem that $\avg{A^2} = \tfrac{1}{2}A_0^2$, we may find this mean power. Thus if $x= \hat{x}e^{i\omega t}$, then $dx/dt = i\omega\hat{x}e^{i\omega t}$. Therefore, in these circumstances, the average power could be written as \begin{equation} \label{Eq:I:24:4} \avg{P} = \tfrac{1}{2}\gamma m\omega^2x_0^2. \end{equation}

In the notation for electrical circuits, $dx/dt$ is replaced by the current $I$ ($I$ is $dq/dt$, where $q$ corresponds to $x$), and $m\gamma$ corresponds to the resistance $R$. Thus the rate of the energy loss—the power used up by the forcing function—is the resistance in the circuit times the average square of the current: \begin{equation} \label{Eq:I:24:5} \avg{P} = R\avg{I^2} = R\cdot\tfrac{1}{2}I_0^2. \end{equation} This energy, of course, goes into heating the resistor; it is sometimes called the heating loss or the Joule heating.

Another interesting feature to discuss is how much energy is
*stored*. That is not the same as the power, because although
power was at first used to store up some energy, after that the system
keeps on absorbing power, insofar as there are any heating (resistive)
losses. At any moment there is a certain amount of stored energy, so
we would like to calculate the mean stored energy $\avg{E}$ also. We
have already calculated what the average of $(dx/dt)^2$ is, so we find
\begin{equation}
\begin{aligned}
\avg{E} &= \tfrac{1}{2}m
\avg{(dx/dt)^2} + \tfrac{1}{2}m\omega_0^2
\avg{x^2}\\[1ex]
&=\tfrac{1}{2}m(\omega^2+\omega_0^2)\tfrac{1}{2}x_0^2.
\end{aligned}
\label{Eq:I:24:6}
\end{equation}
Now, when an oscillator is very efficient, and if $\omega$ is near
$\omega_0$, so that $\abs{\hat{x}}$ is large, the stored energy is
very high—we can get a large stored energy from a relatively small
force. The force does a great deal of work in getting the oscillation
going, but then to keep it steady, all it has to do is to fight the
friction. The oscillator can have a great deal of energy if the
friction is very low, and even though it is oscillating strongly, not
much energy is being lost. The efficiency of an oscillator can be
measured by how much energy is stored, compared with how much work the
force does per oscillation.

How does the stored energy compare with the amount of work that is
done in one cycle? This is called the $Q$ of the system, and $Q$ is
defined as $2\pi$ times the mean stored energy, divided by the work
done per cycle. (If we were to say the work done per *radian*
instead of per cycle, then the $2\pi$ disappears.)
\begin{equation}
\label{Eq:I:24:7}
Q=2\pi\,\frac{\tfrac{1}{2}m(\omega^2+\omega_0^2)\cdot
\avg{x^2}}{\gamma m\omega^2\avg{x^2}\cdot
2\pi/\omega}=\frac{\omega^2+\omega_0^2}{2\gamma\omega}.
\end{equation}
$Q$ is not a very useful number unless it is very large. When it is
relatively large, it gives a measure of how good the oscillator
is. People have tried to define $Q$ in the simplest and most useful
way; various definitions differ a bit from one another, but if $Q$ is
very large, all definitions are in agreement. The most generally
accepted definition is Eq. (24.7), which depends
on $\omega$. For a good oscillator, close to resonance, we can
simplify (24.7) a little by setting $\omega=\omega_0$, and we
then have $Q= \omega_0/\gamma$, which is the definition of $Q$ that we
used before.

What is $Q$ for an electrical circuit? To find out, we merely have to translate $L$ for $m$, $R$ for $m\gamma$, and $1/C$ for $m\omega_0^2$ (see Table 23–1). The $Q$ at resonance is $L\omega/R$, where $\omega$ is the resonance frequency. If we consider a circuit with a high $Q$, that means that the amount of energy stored in the oscillation is very large compared with the amount of work done per cycle by the machinery that drives the oscillations.

### 24–2Damped oscillations

We now turn to our main topic of discussion: transients. By a
*transient* is meant a solution of the differential equation when
there is no force present, but when the system is not simply at
rest. (Of course, if it is standing still at the origin with no force
acting, that is a nice problem—it stays there!) Suppose the
oscillation starts another way: say it was driven by a force for a
while, and then we turn off the force. What happens then? Let us first
get a rough idea of what will happen for a very high $Q$ system. So
long as a force is acting, the stored energy stays the same, and there
is a certain amount of work done to maintain it. Now suppose we turn
off the force, and no more work is being done; then the losses which
are eating up the energy of the supply are no longer eating up its
energy—there *is* no more driver. The losses will have to
consume, so to speak, the energy that is stored. Let us suppose that
$Q/2\pi = 1000$. Then the work done per cycle is $1/1000$ of the
stored energy. Is it not reasonable, since it is oscillating with no
driving force, that in one cycle the system will still lose a
thousandth of its energy $E$, which ordinarily would have been
supplied from the outside, and that it will continue oscillating,
always losing $1/1000$ of its energy per cycle? So, as a guess, for a
relatively high $Q$ system, we would suppose that the following
equation might be roughly right (we will later do it exactly, and it
will turn out that it *was* right!):
\begin{equation}
\label{Eq:I:24:8}
dE/dt=-\omega E/Q.
\end{equation}
This is rough because it is true only for large $Q$. In each radian
the system loses a fraction $1/Q$ of the stored energy $E$. Thus in a
given amount of time $dt$ the energy will change by an
amount $\omega\,dt/Q$, since the number of radians associated with the
time $dt$ is $\omega\,dt$. What is the frequency? Let us suppose that the
system moves so nicely, with hardly any force, that if we let go it
will oscillate at essentially the same frequency all by itself. So we
will guess that $\omega$ is the resonant frequency $\omega_0$. Then we
deduce from Eq. (24.8) that the stored energy will vary
as
\begin{equation}
\label{Eq:I:24:9}
E=E_0e^{-\omega_0t/Q}=E_0e^{-\gamma t}.
\end{equation}
This would be the measure of the *energy* at any moment. What
would the formula be, roughly, for the amplitude of the oscillation as
a function of the time? The same? No! The amount of energy in a
spring, say, goes as the *square* of the displacement; the
kinetic energy goes as the *square* of the velocity; so the total
energy goes as the *square* of the displacement. Thus the
displacement, the amplitude of oscillation, will decrease half as fast
because of the square. In other words, we guess that the solution for
the damped transient motion will be an oscillation of frequency close
to the resonance frequency $\omega_0$, in which the amplitude of the
sine-wave motion will diminish as $e^{-\gamma t/2}$:
\begin{equation}
\label{Eq:I:24:10}
x=A_0e^{-\gamma t/2}\cos\omega_0 t.
\end{equation}
This equation and Fig. 24–1 give us an idea of what we
should expect; now let us try to analyze the motion *precisely*
by solving the differential equation of the motion itself.

So, starting with Eq. (24.1), with no outside force, how
do we solve it? Being physicists, we do not have to worry about the
*method* as much as we do about what the solution
*is*. Armed with our previous experience, let us try as a
solution an exponential curve, $x = Ae^{i\alpha t}$. (Why do we try
this? It is the easiest thing to differentiate!) We put this
into (24.1) (with $F(t) = 0$), using the rule that each time
we differentiate $x$ with respect to time, we multiply by $i\alpha$. So
it is really quite simple to substitute. Thus our equation looks like
this:
\begin{equation}
\label{Eq:I:24:11}
(-\alpha^2+i\gamma\alpha+\omega_0^2)Ae^{i\alpha t}=0.
\end{equation}
The net result must be zero for *all times*, which is impossible
unless (a) $A = 0$, which is no solution at all—it stands still, or
(b)
\begin{equation}
\label{Eq:I:24:12}
-\alpha^2+i\alpha\gamma+\omega_0^2=0.
\end{equation}
If we can solve this and find an $\alpha$, then we will have a
solution in which $A$ need not be zero!
\begin{equation}
\label{Eq:I:24:13}
\alpha=i\gamma/2\pm\sqrt{\omega_0^2-\gamma^2/4}.
\end{equation}

For a while we shall assume that $\gamma$ is fairly small compared
with $\omega_0$, so that $\omega_0^2-\gamma^2/4$ is definitely
positive, and there is nothing the matter with taking the square
root. The only bothersome thing is that we get *two* solutions!
Thus
\begin{equation}
\label{Eq:I:24:14}
\alpha_1 =i\gamma/2+\sqrt{\omega_0^2-\gamma^2/4}=
i\gamma/2+\omega_\gamma
\end{equation}
and
\begin{equation}
\label{Eq:I:24:15}
\alpha_2 =i\gamma/2-\sqrt{\omega_0^2-\gamma^2/4}=i\gamma/2-\omega_\gamma.
\end{equation}
Let us consider the first one, supposing that we had not noticed that
the square root has two possible values. Then we know that a solution
for $x$ is $x_1= Ae^{i\alpha_1t}$, where $A$ is any constant
whatever. Now, in substituting $\alpha_1$, because it is going to come
so many times and it takes so long to write, we shall call
$\sqrt{\omega_0^2-\gamma^2/4}=\omega_\gamma$. Thus
$i\alpha_1=-\gamma/2 + i\omega_\gamma$, and we get $x =
Ae^{(-\gamma/2+i\omega_\gamma)t}$, or what is the same, because of the
wonderful properties of an exponential,
\begin{equation}
\label{Eq:I:24:16}
x_1=Ae^{-\gamma t/2}e^{i\omega_\gamma t}.
\end{equation}
First, we recognize this as an oscillation, an oscillation at a
frequency $\omega_\gamma$, which is not *exactly* the
frequency $\omega_0$, but is rather close to $\omega_0$ if it is a good
system. Second, the amplitude of the oscillation is decreasing
exponentially! If we take, for instance, the real part
of (24.16), we get
\begin{equation}
\label{Eq:I:24:17}
x_1=Ae^{-\gamma t/2}\cos\omega_\gamma t.
\end{equation}
This is very much like our guessed-at solution (24.10),
except that the frequency really is $\omega_\gamma$. This is the only
error, so it is the same thing—we have the right idea. But
everything is *not* all right! What is not all right is that
*there is another solution*.

The other solution is $\alpha_2$, and we see that the difference is
only that the sign of $\omega_\gamma$ is reversed:
\begin{equation}
\label{Eq:I:24:18}
x_2=Be^{-\gamma t/2}e^{-i\omega_\gamma t}.
\end{equation}
What does this mean? We shall soon prove that if $x_1$ and $x_2$ are
each a possible solution of Eq. (24.1) with $F = 0$, then
$x_1 + x_2$ is also a solution of the same equation! So the general
solution $x$ is of the mathematical form
\begin{equation}
\label{Eq:I:24:19}
x=e^{-\gamma t/2}(Ae^{i\omega_\gamma t}+Be^{-i\omega_\gamma t}).
\end{equation}
Now we may wonder why we bother to give this other solution, since we
were so happy with the first one all by itself. What is the extra one
for, because of course we know we should only take the real part?
*We* know that we must take the real part, but how did the
*mathematics* know that we only wanted the real part? When we had
a nonzero driving force $F(t)$, we put in an *artificial* force
to go with it, and the *imaginary* part of the equation, so to
speak, was driven in a definite way. But when we put $F(t) \equiv 0$,
our convention that $x$ should be only the real part of whatever we
write down is purely our own, and the mathematical equations do not
know it yet. The physical world *has* a real solution, but the
answer that we were so happy with before is not real, it is
*complex*. The *equation* does not know that we are
arbitrarily going to take the real part, so it has to present us, so
to speak, with a complex conjugate type of solution, so that by
putting them together we can *make a truly real* solution; that
is what $\alpha_2$ is doing for us. In order for $x$ to be real,
$Be^{-i\omega_\gamma t}$ will have to be the complex conjugate
of $Ae^{i\omega_\gamma t}$ that the imaginary parts disappear. So it turns
out that $B$ is the complex conjugate of $A$, and our real solution is
\begin{equation}
\label{Eq:I:24:20}
x=e^{-\gamma t/2}(Ae^{i\omega_\gamma t}+A\cconj e^{-i\omega_\gamma
t}).
\end{equation}
So our real solution is an oscillation with a *phase shift* and a
damping—just as advertised.

### 24–3Electrical transients

Now let us see if the above really works. We construct the electrical circuit shown in Fig. 24–2, in which we apply to an oscilloscope the voltage across the inductance $L$ after we suddenly turn on a voltage by closing the switch $S$. It is an oscillatory circuit, and it generates a transient of some kind. It corresponds to a circumstance in which we suddenly apply a force and the system starts to oscillate. It is the electrical analog of a damped mechanical oscillator, and we watch the oscillation on an oscilloscope, where we should see the curves that we were trying to analyze. (The horizontal motion of the oscilloscope is driven at a uniform speed, while the vertical motion is the voltage across the inductor. The rest of the circuit is only a technical detail. We would like to repeat the experiment many, many times, since the persistence of vision is not good enough to see only one trace on the screen. So we do the experiment again and again by closing the switch $60$ times a second; each time we close the switch, we also start the oscilloscope horizontal sweep, and it draws the curve over and over.) In Figs. 24–3 to 24–6 we see examples of damped oscillations, actually photographed on an oscilloscope screen. Figure 24–3 shows a damped oscillation in a circuit which has a high $Q$, a small $\gamma$. It does not die out very fast; it oscillates many times on the way down.

But let us see what happens as we decrease $Q$, so that the oscillation dies out more rapidly. We can decrease $Q$ by increasing the resistance $R$ in the circuit. When we increase the resistance in the circuit, it dies out faster (Fig. 24–4). Then if we increase the resistance in the circuit still more, it dies out faster still (Fig. 24–5). But when we put in more than a certain amount, we cannot see any oscillation at all! The question is, is this because our eyes are not good enough? If we increase the resistance still more, we get a curve like that of Fig. 24–6, which does not appear to have any oscillations, except perhaps one. Now, how can we explain that by mathematics?

The resistance is, of course, proportional to the $\gamma$ term in the
mechanical device. Specifically, $\gamma$ is $R/L$. Now if we increase
the $\gamma$ in the solutions (24.14) and (24.15)
that we were so happy with before, chaos sets in when $\gamma/2$ exceeds
$\omega_0$; we must write it a different way, as
\begin{equation*}
i\gamma/2+i\sqrt{\gamma^2/4-\omega_0^2}\quad
\text{and}\quad
i\gamma/2-i\sqrt{\gamma^2/4-\omega_0^2}.
\end{equation*}
Those are now the two solutions and, following the same line of
mathematical reasoning as previously, we again find two solutions:
$e^{i\alpha_1 t}$ and $e^{i\alpha_2 t}$. If we now substitute for
$\alpha_1$, we get
\begin{equation*}
x=Ae^{-(\gamma/2+\sqrt{\gamma^2/4-\omega_0^2})t},
\end{equation*}
a nice exponential decay with no oscillations. Likewise, the other
solution is
\begin{equation*}
x=Be^{-(\gamma/2-\sqrt{\gamma^2/4-\omega_0^2})t}.
\end{equation*}
Note that the square root cannot exceed $\gamma/2$, because even if
$\omega_0=0$, one term just equals the other. But $\omega_0^2$ is
taken away from $\gamma^2/4$, so the square root is less than
$\gamma/2$, and the term in parentheses is, therefore, always a
positive number. Thank goodness! Why? Because if it were negative, we
would find $e$ raised to a *positive* factor times $t$, which
would mean it was exploding! In putting more and more resistance into
the circuit, we know it is not going to explode—quite the
contrary. So now we have two solutions, each one by itself a dying
exponential, but one having a much faster “dying rate” than the
other. The general solution is of course a combination of the two; the
coefficients in the combination depending upon how the motion
starts—what the initial conditions of the problem are. In the
particular way this circuit happens to be starting, the $A$ is
negative and the $B$ is positive, so we get the difference of two
exponential curves.

Now let us discuss how we can find the two coefficients $A$ and $B$ (or $A$ and $A\cconj$), if we know how the motion was started.

Suppose that at $t = 0$ we know that $x = x_0$, and $dx/dt = v_0$. If we put $t= 0$, $x = x_0$, and $dx/dt = v_0$ into the expressions \begin{align*} x&=e^{-\gamma t/2}(Ae^{i\omega_\gamma t}+ A\cconj e^{-i\omega_\gamma t}),\\[1ex] dx/dt&=e^{-\gamma t/2}\bigl[ (-\gamma/2+i\omega_\gamma)Ae^{i\omega_\gamma t}+ (-\gamma/2-i\omega_\gamma)A\cconj e^{-i\omega_\gamma t}\,\bigr], \end{align*} \begin{align*} x=e^{-\gamma t/2}(&Ae^{i\omega_\gamma t}+ A\cconj e^{-i\omega_\gamma t}),\\[1ex] dx/dt=e^{-\gamma t/2}\bigl[ &(-\gamma/2+i\omega_\gamma)Ae^{i\omega_\gamma t}+\\ &(-\gamma/2-i\omega_\gamma)A\cconj e^{-i\omega_\gamma t}\,\bigr], \end{align*} we find, since $e^0 =$ $e^{i0} =$ $1$, \begin{align*} x_0&=A+A\cconj=2A_R,\\[1ex] v_0&=-(\gamma/2)(A+A\cconj)+i\omega_\gamma(A-A\cconj)\\[.5ex] &=-\gamma x_0/2+i\omega_\gamma(2iA_I), \end{align*} where $A = A_R + iA_I$, and $A\cconj = A_R - iA_I$. Thus we find \begin{equation} A_R =x_0/2\notag \end{equation} and \begin{equation} \label{Eq:I:24:21} A_I =-(v_0+\gamma x_0/2)/2\omega_\gamma. \end{equation} This completely determines $A$ and $A\cconj$, and therefore the complete curve of the transient solution, in terms of how it begins. Incidentally, we can write the solution another way if we note that \begin{equation*} e^{i\theta}+e^{-i\theta}=2\cos\theta \quad\text{and}\quad e^{i\theta}-e^{-i\theta}=2i\sin\theta. \end{equation*} We may then write the complete solution as \begin{equation} \label{Eq:I:24:22} x=e^{-\gamma t/2}\biggl[ x_0\cos\omega_\gamma t+ \frac{v_0+\gamma x_0/2}{\omega_\gamma}\sin\omega_\gamma t \biggr], \end{equation} where $\omega_\gamma=+\sqrt{\omega_0^2-\gamma^2/4}$. This is the mathematical expression for the way an oscillation dies out. We shall not make direct use of it, but there are a number of points we should like to emphasize that are true in more general cases.

First of all the behavior of such a system with no external force is expressed by a sum, or superposition, of pure exponentials in time (which we wrote as $e^{i\alpha t}$). This is a good solution to try in such circumstances. The values of $\alpha$ may be complex in general, the imaginary parts representing damping. Finally the intimate mathematical relation of the sinusoidal and exponential function discussed in Chapter 22 often appears physically as a change from oscillatory to exponential behavior when some physical parameter (in this case resistance, $\gamma$) exceeds some critical value.