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Second, the amplitude of the oscillation is decreasing exponentiallyl If we take,
for instance, the real part of (24.16), we get
gì = Ae~ 1/2 eosu,f. (24.17)
--- Trang 431 ---
This is very mụuch like our guessed-at solution (24.10), except that the frequency
really is «¡„. This is the only error, so i% is the same thing——we have the right
idea. But everything is no all rightl What is not all right is that £here ¡s another
solution.
The other solution is œ¿, and we see that the diference is only that the sign
OŸ ¿J„ 1s reversed:
#ạ = Be~7/2e~1sat, (24.18)
What does this mean? We shall soon prove that if z¡ and #a are each a possible
solution of Eq. (24.1) with ?' = 0, then z¡ + #a is also a solution of the same
cquation! So the general solution #+ is of the mathematical form
œ=e 2U2( Aelsst + Be~ +), (24.19)
Now we may wonder why we bother to give this other solution, since we were
so happy with the frst 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 rmafhematics know that we only wanted the
real part? When we had a nonzero driving force #{f), we put in an artjicial
force to go with it, and the #naginarw part of the equation, so to speak, was
driven in a delnite way. But when we put #{£) = 0, our convention that #
should be only the real part of whatever we write down is purely our own, and
the mathematical equations do not know it yet. 'Phe physical world høs a real
solution, but the answer that we were so happy with before is not real, it 1s
cormmplez. 'The equation does not know that we are arbitrarily going to take the
real part, so 1t has to present us, so to speak, with a complex conjugate type of
solution, so that by putting them together we can maœke a truhụ real solution;
that is what œ¿ is doïng for us. In order for z to be real, Be~“»† will have to be
the complex conjugate of Ae*“+f that the imaginary parts disappear. So it turns
out that Ö is the complex conjugate of A, and our real solution is
a=e TU2( Aesst+ A*esst), (24.20)
So our real solution is an oscillation with a phase shft and a damping—just as
advertised.
24-3 Electrical 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
--- Trang 432 ---
Fig. 24-2. An electrical circuit for demonstrating transients.
inductance Ù after we suddenly turn on a voltage by closing the switch Š. lt 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.
Tt 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
1s only a technical detail. We would like to repeat the experiment many, many
tỉimes, since the persistence of vision is not good enough to see onÌy 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
sereen. Pigure 24-3 shows a damped oscillation in a circuit which has a high @, a
small y. It does not die out very fast; it oscillates many times on the way down.
But let us see what happens as we decrease @, so that the oscillation dies out
more rapidly. We can decrease @ by increasing the resistance # in the circuit.
When we increase the resistance in the circuit, i9 dies out faster (Eig. 24-4).
'Then ïf 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 alll "The question is, is this because our eyes are not good enough?
TÍ 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 1s, of course, proportional to the + term in the mechanical
device. Specifically, + is //E. NÑow iŸ we increase the + in the solutions (24.14)
and (24.15) that we were so happy with before, chaos sets in when +/2 exceeds
œg; we must write i% a different way, as
y~/2+iVW^2/4—,u$ and y/2— iv^22/4- uậ.
--- Trang 433 ---
Figure 24-3
Figure 24-4
Figure 24-5
Figure 24-6
--- Trang 434 ---
Those are now the two solutions and, following the same line of mathematical
reasoning as previously, we again fñnd two solutions: e'*“f and e?%2†, Tf we now
substitute for œ, we get
z— Ae—(/3†V3/4— s8):
a nice exponential decay with no oscillations. Likewise, the other solution is
+ — Be-(%/2~V^2/4-ui)t.
Note that the square root cannot exceed +/2, because even IŸ «o = 0, one term
jusi equals the other. But ø is taken away from +2/4, so the square root is less
than +/2, and the term in parentheses is, therefore, always a positive number.
Thank goodnessl Why? Because ïf it were negative, we would find e raised to a
postfiue factor tỉìmes ‡, which would mean it was explodingl In putting more and
more resistance into the cireuit, we know it is not going to explode—qulite the
contrary. So now we have ©wo solutions, each one by itself a dying exponential,
but one having a much faster “dying rate” than the other. 'Phe 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 circeuit happens to be starting, the A is negative and the
B 1s positive, so we get the diference of bwo exponential curves.
Now let us discuss how we can fnd the two coefficients A and Ö (or Aand 4Š),
1ƒ we know how the motion was started.
Suppose that at # = 0 we know that ø = zo, and đz/đ# = 0ọ. TÝ we put £= 0,
% = #ọ, and d+/đt = 0ọ into the expressions
+ — e~1⁄2(Aefsat + A*e a9,
da /dt = e~1⁄2[(—+x/2 +4) Aezf + (—+/2T— iax„) A*e a1,
we fnd, sinee e? = e9 =1,
#o= A+ A4” =2An,