24–2Damped oscillations 24-2 阻尼震荡
We now turn to our main topic ofdiscussion: transients. By a transient is meant a solution of thedifferential equation when there is no force present, but when the system isnot simply at rest. (Of course, if it is standing still at the origin with noforce acting, that is a nice problem—it stays there!) 现在，我们转向我们讨论的主要话题：瞬态。通过瞬态，我们的意思是：没有力在场时的一个微分方程的解；但此时，这个系统，也不是简单地处于静止状态（当然，如果它在最初状态，保持静止，没有任何力作用于它，这是一个好的问题--它就一直呆在那儿！）Suppose theoscillation starts another way: say it was driven by a force for a while, andthen we turn off the force. What happens then? Let us first get a rough idea ofwhat will happen for a very high Q system.假设震荡，以另一种方式开始：比如说，它被一个力，驱动一会儿，然后，我们把这个力取消了。会发生什么呢？首先，对于一个高Q系统，在此情况下，会发生什么，让我们先对此，得到一个大致的概念。So long as a force is acting, the stored energy stays the same, andthere is a certain amount of work done to maintain it. Now suppose we turn offthe force, and no more work is being done; then the losses which are eating upthe energy of the supply are no longer eating up its energy—there is nomore driver. 只要一个力在起作用，那么，被存储的能量，就是一样的，力做了一定量的功，以保持它。现在，假设我们把这个力停掉，不继续做功了；本来，损耗消耗的，是供给的能量，现在，损耗不消耗这个能量了—不再继续有驱动了。 The losses willhave to consume, so to speak, the energy that is stored. Let us suppose that Q/2π=1000. Then the work done per cycle is 1/1000 of the stored energy. 这么说吧，损耗将会消耗那些被存储的能量。让我们假定Q/2π=1000。那么，每个循环所做的功，就是被存储的能量的1/1000。Is it not reasonable,since it is oscillating with no driving force, that in one cycle the systemwill still lose a thousandth of its energy E , which ordinarily would have been supplied from the outside, and thatit will continue oscillating, always losing 1/1000 of its energy per cycle? So, as a guess, for a relatively high Qsystem, we would suppose that the following equation might be roughlyright (we will later do it exactly, and it will turn out that it wasright!): 这不合理吗？，由于它在震荡，没有驱动力，那么，在一个循环中，系统将仍会损耗其能量 E的千分之一，而这个损耗，通常是从外部提供的，系统将会继续震荡，每次循环，总损耗其能量的1/1000吗？所以，作为一个猜测，对于一个相对高的Q系统，我们将建议，下面的方程，只是大致正确（稍后，我们将准确地证明它，结果将显示，它是正确的！）
dE/dt=−ωE/Q. (24.8)

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 ωdt/Q , since the number of radians associated with the time dt is ωdt . What is the frequency? Let us suppose that the system moves sonicely, with hardly any force, that if we let go it will oscillate atessentially the same frequency all by itself. So we will guess that ω is the resonant frequency ω0 . Then we deduce from Eq. (24.8)that the stored energy will vary as
这个公式，大致正确，因为它只对大的Q为真。在每个弧度，系统都会失去所存储的能量的1/Q。这样，在一个被给予的时间dt内，能量的变化，将是ωdt/Q，由于，与时间dt相联的弧度数，是 ωdt。频率是什么呢？我们假定。系统运行良好，几乎没有什么力，所以，如果我们让它震荡，那么，它将会完全靠它自己，以同样的频率震荡。所以，我们将猜测，ω就是共振频率ω0。因此，我们从方程（24.8）可推出，所存储的能量，将会以下式变化：
(24.9)

This would be the measure of the energy at any moment. What wouldthe formula be, roughly, for the amplitude of the oscillation as a function ofthe time? The same? No! The amount of energy in a spring, say, goes as the squareof the displacement; the kinetic energy goes as the square of thevelocity; so the total energy goes as the square of the displacement.Thus the displacement, the amplitude of oscillation, will decrease half as fastbecause of the square. In other words, we guess that the solution for thedamped transient motion will be an oscillation of frequency close to theresonance frequency ω0 , in which the amplitude of the sine-wave motion will diminishas e−γt/2 :
这是任何一个瞬间能量的测量公式，如果震荡的振幅，是时间的函数，那么，这个公式，大致上说，又会是什么样子呢？会是同样的吗？比如说，在一个弹簧中能量，是随着位移的平方走；而动能，则是随着矢速的平方走；所以，总的能量，随着位移的平方走。这样，位移、即震荡的振幅，因为平方的关系，{其减少}是能量减少的一半{？}。换句话说，我们猜测，阻尼瞬态运动的解，将会是一个震荡，其频率，接近于共振频率 ω0，在此震荡中，正弦波运动的振幅，将会以e−γt/2衰减：
(24.10)
This equation and Fig. 24–1 giveus an idea of what we should expect; now let us try to analyze the motion preciselyby solving the differential equation of the motion itself. 这个方程和图24-1，给我们提供了，我们所期待的想法；对于此运动本身，有微分方程，现在，让我们通过解此方程，尝试去准确地分析此运动。

Fig. 24–1.A damped cosine oscillation. 图24-1 一个阻尼的余弦震荡。

So, starting with Eq. (24.1),with no outside force, how do we solve it? Being physicists, we do not have toworry 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 exponentialcurve, x=Aeiαt . (Why do we try this? It is the easiest thing to differentiate!) Weput this into (24.1)(with F(t)=0 ), using the rule that each time we differentiate x with respect to time, we multiply by iα . So it is really quite simple to substitute. Thus our equation lookslike this:
于是，从方程（24.1）开始，没有任何外力，我们如何解它呢？作为物理学家，我们非常关心解是什么，至于得到解的方法是什么，则无需如此关心。我们可以用以前的经验，武装起来，让我们尝试一个指数曲线的解，x=Aeiαt （为什么我们要试这个，因为它最容易微分）。把这个代入（24.1）（用F(t)=0），使用如下规则，即每次对x求时间的微分，就是乘以 iα。于是，替换确实非常简单。这样，我们的方程，看上去就是：
(24.11)
The net result must be zero for all times, which is impossibleunless (a) A=0 , which is no solution at all—it stands still, or (b)
净结果应该一直是零，这是不可能的，除非(a) A=0, 这是根本无解，方程还在那里没动，或 (b)：
(24.12)
If we can solve this and find an α , then we will have a solution in which A need not be zero!
如果我们可以解这个，找到一个α，那么，我们就有了一个解，在其中，A无需为零。
(24.13)
For a while we shall assume that γ is fairly small compared with ω0 , so that ω20−γ2/4 is definitely positive, and there is nothing the matter with takingthe square root. The only bothersome thing is that we get two solutions!Thus
这一阵子，我们的将假定，与ω0相比，γ相对非常小，于是，ω20−γ2/4就肯定为正，没有任何事情，可以影响取平方根。唯一的困扰，就是我们得到了两个解！这样：
(24.14)
and 和
(24.15)

Let us consider the first one, supposing that we had not noticed thatthe square root has two possible values. Then we know that a solution for xis x1=Aeiα1t, where A is any constant whatever. Now, in substituting α1 , because it is going to come so many times and it takes so long towrite, we shall call （ω20−γ2/4）1/2=ωγ. Thus iα1=−γ/2+iωγ, and we get x=Ae(−γ/2+iωγ)t , or what is the same, because of the wonderful properties of anexponential, 让我们首先考虑第一个，假设我们没有注意到，平方根有两个可能的值。因此，我们知道，x的解是 x1=Aeiα1t, 这里 A任何常数，不管它了。现在，替换α1，因为它要来很多次，且写它很花时间，我们将调用（ω20−γ2/4）1/2=ωγ 。这样，iα1=−γ/2+iωγ , 且我们得到 x=Ae(−γ/2+iωγ)t ，或者，同样的东西，因为指数的奇妙属性
(24.16)
First, we recognize this as an oscillation, an oscillation at afrequency ωγ, which is not exactly the frequency ω0 , but is rather close to ω0 if it is a good system. Second, the amplitude of the oscillation isdecreasing exponentially! If we take, for instance, the real part of (24.16),we get 首先，我们认出，这是一个震荡，其频率，为 ωγ，并不准确地是ω0，如果系统好的话，就会非常接近ω0。其次，振幅是指数衰减的！如果们取（24.16）的实部，我们得到：
(24.17)
This is very much like our guessed-at solution (24.10),except that the frequency really is ωγ. This is the only error, so it is the same thing—we have the rightidea. But everything is not all right! What is not all right is that thereis another solution.
这与我们猜出来的解（24.10），非常像，除了频率是ωγ。这是唯一的错误，所以，它是同样的事情—{说明}我们的想法正确。但是，并不是所有的事情，都正确了！因为，还有另外一个解。

The other solution is α2, and we see that the difference is only that the sign of ωγ is reversed:
另一个解，是α2，我们看到，唯一区别就是，ωγ的符号反了：
(24.18)
What does this mean? We shall soon prove that if x1 and x2 are each a possible solution of Eq. (24.1)with F=0 , then x1+x2 is also a solution of the same equation! So the general solution xis of the mathematical form 这意味着什么呢？我们很快将会证明，如果x1和x2是当F=0时，方程（24.1）的两个可能的解，那么x1+x2，就也是同一方程的解！所以，普遍解x，就具有如下数学形式：
(24.19)

Now we may wonder why we bother to give this other solution, since wewere 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 knowthat we must take the real part, but how did the mathematics know thatwe only wanted the real part?
现在，我们可能会好奇，为什么我们要麻烦自己，给出另一个解，由于我们与第一个解，相处甚欢，它已经完全够用了。此另一解，目的何在？因为，当然我们知道，我们只应该取实部？我们知道，我们只应该取实部{？重复}，但是数学，又如何能够知道这一点呢？When we had a nonzero driving force F(t) , we put in an artificial force to go with it, and the imaginarypart of the equation, so to speak, was driven in a definite way. But when weput F(t)≡0 , our convention that x should be only the real part of whatever we write down is purely ourown, and the mathematical equations do not know it yet. 当我们有了一个非零的驱动力 F(t)时，我们在方程中，放入一个人为的力，跟着走，且方程的虚数部分，这么说吧，是以确定的方式，被驱动的{？}。但是，当我们设F(t)≡0 时，我们的约定，即无论我们写下的是什么，x只应该是其实部，这一约定，将只是我们自己的，数学方程，并不知道它。The physical world has a real solution, but the answer thatwe were so happy with before is not real, it is complex. The equationdoes not know that we are arbitrarily going to take the real part, so it has topresent us, so to speak, with a complex conjugate type of solution, so that byputting them together we can make a truly real solution; that is what α2is doing for us. 物理世界，有一个真实的解，但是，我们以前相处甚欢的那个答案，并不是实数，而是一个复数。方程并不知道，我们将人为地取实部，所以，这么说吧，它将给我们提供一个复数共轭类型的解，于是，通过把它们放在一起，我们就可以得出真正真实的解；这就是α2为我们做的事情。 In order for x to be real, Be−iωγtwill have to be the complex conjugate of Aeiωγtthat the imaginary parts disappear. So it turns out that B is the complex conjugate of A , and our real solution is 为了让x是实数，Be−iωγt就必须的是 Aeiωγt的复数共轭，这样，虚部就消失了。所以，结果就是，B是A的复数共额，我们真实的解就是：
(24.20)
So our real solution is an oscillation with a phase shift and adamping—just as advertised.
所以，我们的真实解，就是一个震荡，它有一个相位漂移，和一个阻尼--正如提前所宣告。

24–3Electrical transients 24-3 电子瞬态

Fig. 24–2.An electrical circuit for demonstratingtransients. 图24-2 一个电路，目的是演证瞬态。
Now let us see if the above really works.We construct the electrical circuit shown in Fig. 24–2, inwhich we apply to an oscilloscope the voltage across the inductance Lafter we suddenly turn on a voltage by closing the switch S. It is an oscillatory circuit, and it generates a transient of somekind. 现在，让我们看看，上面所讲，是否可行。如图24-2，我们搭了一个电路，示波器与电感L并联，在我们合上开关S之后，L和示波器上，都会突然地被加上电压。这是一个震荡电路，它会产生某种瞬态。It corresponds to a circumstance in which we suddenly apply a forceand the system starts to oscillate. It is the electrical analog of a dampedmechanical oscillator, and we watch the oscillation on an oscilloscope, wherewe should see the curves that we were trying to analyze. 它相当于这种情形：一个系统，我们突然给它加力，然后，它开始震荡。它是阻尼机械振荡的一个电子类比，我们在示波器上，观察震荡，在那里，我们应该看到，我们想要分析的曲线。 (The horizontal motion of the oscilloscope is driven at a uniformspeed, while the vertical motion is the voltage across the inductor. The restof the circuit is only a technical detail. We would like to repeat theexperiment many, many times, since the persistence of vision is not good enoughto see only one trace on the screen. So we do the experiment again and again byclosing the switch 60 times a second; each time we close the switch, we also start theoscilloscope horizontal sweep, and it draws the curve over and over.) In Figs. 24–3to 24–6we see examples of damped oscillations, actually photographed on anoscilloscope screen. Figure 24–3 showsa damped oscillation in a circuit which has a high Q , a small γ . It does not die out very fast; it oscillates many times on the waydown.
（示波器的水平运动，是匀速驱动的，而其垂直运动，是通过电感的电压。电路的其他部分，只不过是技术细节。这个实验，我们要重复很多、很多次，由于视野的持久性，不够好，不足以在屏幕上，只看到一次示踪{？可实验解答}。所以，我们要反复做这个实验，每秒合上开关60次；每次我们合上开关，我们也开始了示波器上的水平扫描，它就会反复地画曲线。）在图24-3至图24-6中，我们看到了阻尼振荡的例子，它们是在一个示波器屏幕上，实际拍的照片。图24-3显示了一个电路中的阻尼震荡，它有一个高的Q值，和一个小的γ。它不会很快消失，在衰减的路上，它会震荡很多次。

Figure 24–3 图24-3

Figure 24–4 图24-4

Figure 24–5 图24-5

Figure 24–6 图24-6

But let us see what happens as we decrease Q, so that the oscillation dies out more rapidly. We can decrease Qby increasing the resistance R in the circuit. When we increase the resistance in the circuit, itdies out faster (Fig. 24–4).Then if we increase the resistance in the circuit still more, it dies outfaster still (Fig. 24–5). Butwhen 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 increasethe 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 canwe explain that by mathematics?
但是，让我们看看，当我们减少Q时，会发生什么，于是，震荡消失地更快了。通过增加电路中的电阻R，我们可以减少Q。当我们增加电路中的电阻时，它衰减的更快了（图24-4）。然后，如果我们继续增加电阻，它就消失地更快（图24-5）。但是，当我们增加电阻，超过一定量时，我们将根本看不到任何震荡！现在的问题是，这是因为我们的眼睛不够好吗？如果我们继续增加电阻，我们就会得到图24-6所示曲线，似乎没有任何震荡，除一个外。现在，我们如何通过数学，来解释呢？

The resistance is, of course, proportionalto the γ term in the mechanical device. Specifically, γ is R/L . Now if we increase the γin the solutions (24.14)and (24.15)that we were so happy with before, chaos sets in when γ/2 exceeds ω0 ; we must write it a different way, as
当然，电阻正比于机械装置中的γ项。具体地说，γ就是R/L 。我们曾与方程（24.14）和（24.15），相处甚欢，现在，如果我们增加方程的解中的γ，那么，当γ/2 超过ω0时，混乱就会发生；我们应该换一种方式来写它：
Those are now the two solutions and, following the same line ofmathematical reasoning as previously, we again find two solutions: eiα1tand e iα2t . If we now substitute for α1 , we get 这些就是现在的两个解，与以前一样，遵循着同样的数学推导的道理，我们又发现了两个解：eiα1t和 e iα2t 。如果现在，我们替换α1，我们得到：

a nice exponential decay with no oscillations. Likewise, the othersolution is 一个良好的指数衰减，不带震荡的。同样，另外一个解是：

Note that the square root cannot exceed γ/2 , because even if ω0=0 , one term just equals the other. But ω20is taken away from γ2/4 , so the square root is less than γ/2 , and the term in parentheses is, therefore, always a positive number.Thank goodness! 注意，平方根不能超过γ/2 ,因为，即便ω0=0，一项也只是刚刚等于另一项。但是ω20，被从γ2/4中拿掉了，所以，平方根要小于γ/2，因此，括号中的项，总是正数。谢天谢地！
Why? Because if it were negative, we wouldfind e raised to a positive factor times t , which would mean it was exploding! In putting more and more resistanceinto the circuit, we know it is not going to explode—quite the contrary. So nowwe have two solutions, each one by itself a dying exponential, but one having amuch faster “dying rate” than the other. The general solution is of course acombination of the two; the coefficients in the combination depending upon how themotion starts—what the initial conditions of the problem are. In the particularway this circuit happens to be starting, the A is negative and the B is positive, so we get the difference of two exponential curves.为什么？因为，如果它是负的，那么，我们将会发现e的 ‘一个正因子乘以 t’的次方，这意味着，它会爆炸！随着电路中放入的电阻，越来越多，我们知道，它不会爆炸，而是恰恰相反。所以，现在我们就有两个解，每个本身，都是指数衰减的，但是，一个的“衰减速率”，要比另一个，快很多。通解，当然是这两个解的合成；在合成中的系数，依赖于运动是如何开始的—即问题的初始条件是什么。在这个电路开始的具体方式中，A是负的，B是正的，于是，我们就得到了两个衰减曲线之间的差别。

Suppose that at t=0 we know that x=x0 , and dx/dt=v0 . If we put t=0 , x=x0 , and dx/dt=v0 into the expressions
假设在t=0 ，我们知道了 x=x0 , 和dx/dt=v0 。如果我们把 t=0 , x=x0 , 和 dx/dt=v0 带入表达式：

we find, since e0= ei0= 1 ,
我们发现，由于e0= ei0= 1，

where A=AR+iAI , and A∗=AR−iAI. Thus we find
这里 A=AR+iAI , and A∗=AR−iAI。这样，我们发现
AR=x0/2
and
和
(24.21)
This completely determines A and A∗ , and therefore the complete curve of the transient solution, in termsof how it begins. Incidentally, we can write the solution another way if wenote that
这就完全决定了A和A∗，因此，也就用瞬态是如何开始，完全地规定了瞬态的解的完整曲线。顺便说一下，我们如果注意到
eiθ + e-iθ=2cosθ and eiθ - e-iθ=2isinθ.
我们就可以把解，用另外一种方式写出来。
We may then write the complete solution as
因此，我们可以把完整的解写为：
(24.22)
where ωγ=+(ω20−γ2/4)1/2 . This is the mathematical expression for the way an oscillation diesout. We shall not make direct use of it, but there are a number of points weshould like to emphasize that are true in more general cases.
这里，ωγ=+(ω20−γ2/4)1/2。一个震荡，就是以这种方式衰减的，这就是其数学表达式。我们不会直接利用它，但是，有几点，我们要强调一下，在更普遍的情况下，它们为真。

First of all the behavior of such a systemwith no external force is expressed by a sum, or superposition, of pureexponentials in time (which we wrote as eiαt ). This is a good solution to try in such circumstances. The values of αmay be complex in general, the imaginary parts representing damping.Finally the intimate mathematical relation of the sinusoidal and exponential functiondiscussed in Chapter 22often appears physically as a change from oscillatory to exponential behaviorwhen some physical parameter (in this case resistance, γ ) exceeds some critical value.
首先，这样一个系统，没有任何外力，其表现，是通过时间中的若干纯粹指数（我们把其写作eiαt）的总和、或叠加，来表示的。在此情形下，用这个解，就是一个好的尝试。α的值，一般可以是复数，虚部代表着阻尼。最后，在第22章，我们讨论了正弦函数，与指数函数之间的、密切的数学关系；当某些物理参数（在这种情况下，就是γ）超过了临界值时，此数学关系，就会作为一种从震荡表现，到指数表现的变化，而经常物理性地出现{？}。

求大佬列一下步骤，随便带数

1 Linear Systemsand Review 第25章线性系统和回顾
25–1Linear differential equations 25-1 线性微分方程
In this chapter we shall discuss certainaspects of oscillating systems that are found somewhat more generally than justin the particular systems we have been discussing. For our particular system,the differential equation that we have been solving is
振动系统的某些方面，被发现，不仅在我们刚刚讨论过的具体系统中，而且，更具普遍性，这一章，我们将讨论它们。对于我们的具体系统，我们已经解过了的微分方程就是：
(25.1)
Now this particular combination of “operations” on the variable xhas the interesting property that if we substitute (x+y)for x , then we get the sum of the same operations on x and y ; or, if we multiply x by a , then we get just a times the same combination. This is easy to prove. Just as a“shorthand” notation, because we get tired of writing down all those lettersin (25.1),we shall use the symbol L(x) instead. When we see this, it means the left-hand side of (25.1),with x substituted in. With this system of writing, L(x+y)would mean the following: 现在，这个关于变量x 的具体的“操作”组合，有一个有趣的特性，就是如果我们用(x+y)，替换x，那么，我们就得到：关于x和y的同样操作的总和；或者，如果我们用a乘以x，那么，我们就得到：a乘以同样的组合。这很容易证明。正如一种“速记”表示法，因为，反复写（25.1）中的字母，我们已经厌烦了，我们将用符号 L(x)，来代替它。当我们看到此符号时，就意味着（25.1）的左边，且被带入的是x。用这种写的方法，L(x+y)就意味着：
(25.2)
(We underline the L so as to remind ourselves that it is not an ordinary function.) Wesometimes call this an operator notation, but it makes no differencewhat we call it, it is just “shorthand.”
（我们给L加下划线，是为了提醒我们，它不是一个普通函数。）我们有时称它为操作符表示法，但是，我们叫它什么，无所谓，它只是一种“速记”。