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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? 'Phis is called the @ of the system, and @ is defined as 27 times the
mean stored energy, divided by the work done per cycle. (If we were to say the
work done per rœđan instead of per cycle, then the 2z disappears.)
1 2 2 2 2 2
sm((Z + 08) - (œ
Q=2n2— sa o ý SẺ ` ĐỘ GÔ — Tổ, (24.7)
“+mú2(#2) - 2/0 2%
Q} ïs not a very useful number unless it 1s very large. When it is relatively large,
1E gives a measure of how good the oscillator is. People have tried to deñne Q
in the simplest and most useful way; various defnitions difer a bit from one
another, but if Q is very large, all deÑnitions are in agreement. 'Phe most generally
accepted defnition is Bq. (24.7), which depends on œ¿. Eor a good oscillator, close
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to resonance, we can simplify (24.7) a little by setting œ = œo, and we then have
Q = ¿0/+, which is the definition of Q that we used before.
'What is @Q for an electrical circuit? To ñnd out, we merely have to translate
L form, R for m+, and 1/C for mới (see Table 23-1). The Q at resonance is
T/R, where œ is the resonance frequency. IÝ we consider a circuit with a hiph
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.
24-2 Damped oscillations
W©e now turn to our main topic of discussion: transients. By a transient 1s
meant a solution of the diferential equation when there is no force present, but
when the system is not simply at rest. (Of course, 1Ÿ it is standing still at the
origin with no force acting, that is a nice problem—it stays therel) Suppose the
oscillation starts another way: say it was driven by a force for a while, and then
we turn of the force. What happens then? Let us first get a rough idea of what
will happen for a very high Q system. 5o long as a force is acting, the stored
energy stays the same, and there is a certain amount of work done to maintain
1t. NÑow suppose we turn of 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 is
energy——there 7s no more driver. 'Phe losses will have to consume, so to speak,
the energy that is stored. Let us suppose that Q/2z = 1000. Then the work
done per cycle is 1/1000 of the stored energy. Is it not reasonable, sinee iE is
oscillating with no driving force, that in one cycle the system will still lose a
thousandth of its energy #⁄, which ordinarily would have been supplied from the
outside, and that it will continue oscillating, always losing 1/1000 of its energy
per cycle? 5o, as a guess, for a relatively high @ 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 Ͽs rightl):
dE/dt = —=uE/Q. (24.8)
Thịs is rough because iE is true only for large Q. In each radian the system loses a
fraction 1/Q of the stored energy #2. Thus in a given amount oŸ tỉme đý the energy
will change by an amount œ đ£/@Q, since the number of radians associated with
the time để is œ d. What is the frequency? Let us suppose that the system moves
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so nicely, with hardly any force, that if we let go i9 will oscillate at essentially the
same frequency all by itself. 5o we will guess that œ is the resonant Írequency œ.
Then we deduce rom Eaq. (24.8) that the stored energy will vary as
EB= Eụe 9/9 = Eue-Ðt, (24.9)
This would be the measure of the energu at any moment. What would the
formula be, roughly, for the amplitude of the oscillation as a function of the
time? The same? Nol "The amount of energy in a spring, say, goes as the sguare
of the displacement; the kinetic energy goes as the sợuare of the velocity; so
the total energy goes as the sguare 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 frequeney close to the resonance frequency œọ, in which the
amplitude of the sine-wave motion will diminish as e~?⁄2:
œ = Aoe"?2 cosugt. (24.10)
This equation and Fig. 24-1 give us an idea of what we should expect; now let
us try to analyze the motion øreciselu by solving the diferential equation of the
motion itself.
` ` ⁄ e-1/2
° e~71/2 cos wo t
_—— —— £
Fig. 24-1. A damped cosine oscillation.
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 rmethod as mụuch as we do about
what the solution 2s. Armed with our previous experience, let us try as a solution
an exponential curve, z = Ac”*f, (Why do we try this? It is the easiest thỉng to
diferentiate!) We put this into (24.1) (with Ƒ{) = 0), using the rule that each
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time we diferentiate + with respect to time, we multiply by ¿ơ. So it is really
quite simple to substitute. 'Thus our equation looks like this:
(Ta? +i>œ+u)Ac'*t =0. (24.11)
The net result must be zero for øÏl t#mes, which is impossible unless (a) A = 0,
which is no solution at all—it stands still, or (b)
=2 +ia+ + ư8 =0. (24.12)
Tf we can solve this and find an ơ, then we will have a solution in which 4 need
not be zerol
œ =11/2#+ Vu — 32/4. (24.13)
For a while we shall assume that + is fairly small compared with œọ, so that
uẩ — 22/4 is definitely positive, and there is nothing the matter with taking the
square root. The only bothersome thing is that we get #œo solutionsl Thus
œi =i2/2+ Vai — +2/4=1+/2+u+x (24.14)
da =i1/2~— vưi — +2/4= i+/2— ư+x. (24.15)
Let us consider the ñrst one, supposing that we had not noticed that the square
root has two possible values. Then we know that a solution for # is zị = Ac'e1t,
where A is any constant whatever. NÑow, in substituting ơ+, because it is goïng to
come so many tỉmes and it takes so long to write, we shall call /„§ — +2/4 = œx.
Thus iœ¡ = —+/2 + 7+, and we get ø = Ae(7/2†1“+)!, or what is the same,
because of the wonderful properties of an exponential,
đị = Ae 11/2/1241, (24.16)
First, we recognize this as an oscillation, an oscillation at a frequency ¿„, which
1s not ezacflu the frequency œọ, but 1s rather close to œọ 1Ý ït is a good system.