When we began this chapter, armed only withthe basic notions of integers and counting, we had little idea of the power of theprocesses of abstraction and generalization. Using the set of algebraic “laws,”or properties of numbers, Eq. (22.1),and the definitions of inverse operations (22.2),we have been able here, ourselves, to manufacture not only numbers but usefulthings like tables of logarithms, powers, and trigonometric functions (for theseare what the imaginary powers of real numbers are), all merely by extractingten successive square roots of ten!
本章开始时，我们所拥有的，只是整数和计数的基本表示法，对于次方的抽象化和普遍化过程，我们基本没有什么概念。利用代数的“规律”的集合、或数字的属性、方程（22.1）、及反向操作的定义（22.2），我们在这里，不仅制造出了数字，而且也造出了有用的事物，比如对数表、次方表、和三角函数（因为，这些就是实数的虚数次方）所有这一切，都是通过从10的十个连续的平方根中，提取出来的。

1. There is a definite arithmetic procedure, but the easiest way tofind the square root of any number N is to choose some a fairly close, find N/a , average a′=1/2[a+(N/a)] , and use this average a′ for the next choice for a . The convergence is very rapid—the number of significant figuresdoubles each time.
1、要找到任何数N的平方根，有一个确定的算术过程，但是，最简单的方法则是，选择一个相对较近的数a，然后求出N/a，平均值a′=1/2[a+(N/a)]，使用这个平均值a′，作为下一个a。{重复以上过程}收敛很快--有意义的数字个数，每次都会加倍。

Chapter23.Resonance第23章 共振
23–1Complex numbers and harmonic motion23-1 复数与谐波运动
In the present chapter we shall continueour discussion of the harmonic oscillator and, in particular, the forcedharmonic oscillator, using a new technique in the analysis. In the precedingchapter we introduced the idea of complex numbers, which have real andimaginary parts and which can be represented on a diagram in which the ordinaterepresents the imaginary part and the abscissa represents the real part. If ais a complex number, we may write it as a=ar+iai, where the subscript r means the real part of a , and the subscript i means the imaginary part of a . Referring to Fig. 23–1, wesee that we may also write a complex number a=x+iy in the form x+iy=reiθ , where r2=x2+y2=(x+iy)(x−iy)=aa∗ . (The complex conjugate of a , written a∗ , is obtained by reversing the sign of i in a .) So we shall represent a complex number in either of two forms, areal plus an imaginary part, or a magnitude r and a phase angle θ , so-called. Given r and θ , x and y are clearly rcosθ and rsinθ and, in reverse, given a complex number x+iy , r=(x2+y2)1/2 and tanθ=y/x, the ratio of the imaginary to the real part.
在本章中，我们将继续讨论谐波振荡，特别是，强迫的谐波震荡，在分析中，将使用一个新的技术。在前面一章中，我们介绍了复数的概念，它有实部和虚部，它也可以在一个图中，被表示，其中纵坐标代表虚部，横坐标代表实部。如果a是一个复数，我们可以把它写为a=ar+iai，这里，下标r，意味着a的实部，下标 i，意味着a的虚部。参考图23-1，我们看到，我们可以把复数a=x+iy，写作x+iy=reiθ这种形式，这里r2=x2+y2=(x+iy)(x−iy)=aa∗。（a的共轭复数，写作a∗，可以通过把a中的i的符号取反，得到它。）所以，我们将用两种形式，表示复数：一个实部，加上一个虚部，或者，长度r加上一个相位角θ。给定r和θ，则 x和y，明显就是rcosθ 和rsinθ ，相反, 给定一个复数x+iy ,则 r=(x2+y2)1/2 ，tanθ=y/x ,即虚部与实部的比。

Fig. 23–1.A complex number may berepresented by a point in the “complex plane.” 图23-1 一个复数，可以用“复数平面”中的一个点来表示。

Let us take another example. Suppose wewant to represent a force which is a cosine wave that is out of phase with adelayed phase Δ . This, of course, would be the real part of F0ei(ωt−Δ), but exponentials being what they are, we may write ei(ωt−Δ)=eiωte−iΔ. Thus we see that the algebra of exponentials is much easier thanthat of sines and cosines; this is the reason we choose to use complex numbers.We shall often write
让我们再举一例。假设我们想表示一个力，它是余弦波，相位落后了 Δ。这当然是F0ei(ωt−Δ) 的实部，但是，指数正是它，我们可以写ei(ωt−Δ)=eiωte−iΔ。这样，我们看到，指数的代数，要比正弦和余弦更容易，这就是我们选复数的原因。我们将经常写：
(23.1)
We write a little caret (^ ) over the F to remind ourselves that this quantity is a complex number: here thenumber is
在F上面，我们写一个脱字符(^ )，用来提醒我们，这个量，是一个复数：这里，此数就是：

Now let us solve an equation, using complex numbers, to see whether wecan work out a problem for some real case. For example, let us try to solve
现在，让我们解一个方程，使用复数，看看我们是否能够解决一些真实的问题。例如，让我们尝试解方程：
(23.2)
where F is the force which drives the oscillator and x is the displacement. Now, absurd though it may seem, let us supposethat x and F are actually complex numbers, for a mathematical purpose only. That isto say, x has a real part and an imaginary part times i , and F has a real part and an imaginary part times i . Now if we had a solution of (23.2)with complex numbers, and substituted the complex numbers in the equation, wewould get
这里F是一个力，它驱动着震荡器，x是位移。现在，虽然有些荒唐，但让我们假设x和F，是实际的复数，这只是为了一个数学的目的。也就是说，x有一个实部、和一个虚部乘以i，F有一个实部、和一个虚部乘以i。现在，如果我们有（23.2）的一个解，它有复数，把它带入方程，我们得到：

or

Now, since if two complex numbers are equal, their real parts must beequal and their imaginary parts must be equal, we deduce that thereal part of x satisfies the equation with the real part of the force. We mustemphasize, however, that this separation into a real part and an imaginary partis not valid in general, but is valid only for equations which are linear,that is, for equations in which x appears in every term only in the first power or the zeroth power. Forinstance, if there were in the equation a term λx2 , then when we substitute xr+ixi , we would get λ(xr+ixi)2 , but when separated into real and imaginary parts this would yield λ(x2r−x2i)as the real part and 2iλxrxi as the imaginary part. So we see that the real part of the equationwould not involve just λx2r , but also −λx2i . In this case we get a different equation than the one we wanted tosolve, with xi , the completely artificial thing we introduced in our analysis, mixedin.
现在，由于如果两个复数相等，那么，它们的实部就应相等，它们的虚部也应相等，我们推出x的实部，应该用力的实部，来满足方程。然而，我们必须强调，这种把数分成实部和虚部，一般并不有效，而是只对有些方程有效，这些方程，是线性的，也就是说，在这些方程中，x在每一个项出现时，只是一次方或零次方。例如，如果方程中，有一项 λx2，当我们把xr+ixi带入时，我们就会得到λ(xr+ixi)2，但是，当我们把它分成实部和虚部时，这就会产生实部λ(x2r−x2i)，和虚部2iλxrxi 。于是我们看到，方程的实部，不仅包含λx2r，而是也包含−λx2i。在这种情况下，我们得到的方程，与我们想要求解的，并不相同，它带有xi，一个我们在分析中引入的、完全人造的东西，混入了。

Let us now try our new method for theproblem of the forced oscillator, that we already know how to solve. We want tosolve Eq. (23.2)as before, but we say that we are going to try to solve
强制震荡的这个问题，我们已经知道如何解了，现在，让我们尝试用我们的新方法，来解它。我们想与以前一样，解方程（23.2），但是，我们说，我们要解的方程是：
(23.3)
where F^eiωt is a complex number. Of course x will also be complex, but remember the rule: take the real part tofind out what is really going on. 这里，F^eiωt是一个复数。当然，x也将是一个复数，但是记住规则：我们只取实部，来找出，真正发生的是什么。So we try to solve (23.3)for the forced solution; we shall discuss other solutions later. The forcedsolution has the same frequency as the applied force, and has some amplitude ofoscillation and some phase, and so it can be represented also by some complexnumber x^ whose magnitude represents the swing of x and whose phase represents the time delay in the same way as for theforce. Now a wonderful feature of an exponential function x=x^eiωtis that dx/dt=iωx .所以，我们尝试解方程（20.3），求强制的解；后面，我们将讨论其他的解。强制的解，与所应用的力，有着同样的频率，还有一些震幅，和一些相位，所以，它也可以用某个复数x^来代表，对于该复数，其大小，代表着x的摆幅，其相位，代表着时间的延迟，与对力的方式一样{？}。现在，指数函数x=x^eiωt ，有一个奇妙的特性，即 dx/dt=iωx。When we differentiate an exponential function, we bring down theexponent as a simple multiplier. The second derivative does the same thing, itbrings down another iω , and so it is very simple to write immediately, by inspection, whatthe equation is for x : every time we see a differentiation, we simply multiply by iω. (Differentiation is now as easy as multiplication! This idea ofusing exponentials in linear differential equations is almost as great as theinvention of logarithms, in which multiplication is replaced by addition. Heredifferentiation is replaced by multiplication.) Thus our equation becomes 当我们微分一个指数函数时，我们把指数，带了下来，作为了一个简单的乘数。第二个导数，做了同样的事情，它会带下来另外一个iω，所以，对于x，这个方程究竟是什么，通过检查，可把它直接写出，非常简单：我们每次看到一个微分，我们只是简单地乘以iω。（现在，微分与乘法，同样简单！在线性微分方程中，使用指数这一想法，几乎与对数的发明，一样伟大，在对数中，乘法被加法替代。在这里，微分被乘法替代。）这样，我们的方程就变成：
(23.4)

(We have cancelled the common factor eiωt .) See how simple it is! Differential equations are immediately converted,by sight, into mere algebraic equations; we virtually have the solution bysight, that
（我们消去了因子eiωt。）看啊，它多简单！微分方程，看上去，立即被转化成了单纯的代数方程；我们几乎能看到解，即

since (iω)2=−ω2 . This may be slightly simplified by substituting k/m=ω20, which gives 由于(iω)2=−ω2。把k/m=ω20带入，可以稍微简化一点，即给出：
(23.5)

23–2The forced oscillator with damping23-2 有阻尼的强制震荡
That, then, is how we analyze oscillatorymotion with the more elegant mathematical technique. But the elegance of thetechnique is not at all exhibited in such a problem that can be solved easilyby other methods. It is only exhibited when one applies it to more difficultproblems. Let us therefore solve another, more difficult problem, whichfurthermore adds a relatively realistic feature to the previous one.Equation (23.5)tells us that if the frequency ω were exactly equal to ω0 , we would have an infinite response. Actually, of course, no suchinfinite response occurs because some other things, like friction, which wehave so far ignored, limits the response. Let us therefore add to Eq. (23.2)a friction term.
我们就是这样，用一个精致的数学技术，来分析震荡运动的。但是，这个问题，用其他方法，也很容易解决，所以，在这样一个问题中，技术的精致性，并未得到充分展示。只有当我们把它，应用到更复杂的问题时，它才能被充分展示。因此，让我们来解另外一个更复杂的问题，它给前一问题，另外增加了相对现实的性质。方程（23.5）告诉我们，如果频率 ω准确地等于 ω0，那么，我们将有一个无限的反应。当然，实际上，这种无限的反应，并不会出现，这是因为一些其他的事情，例如摩擦力，我们一直是忽略的，它对反应，会有所限制。因此，现在让我们给方程（23.2），增加一个摩擦项。

Ordinarily such a problem is very difficultbecause of the character and complexity of the frictional term. There are,however, many circumstances in which the frictional force is proportional tothe speed with which the object moves. An example of such friction is thefriction for slow motion of an object in oil or a thick liquid. There is noforce when it is just standing still, but the faster it moves the faster theoil has to go past the object, and the greater is the resistance. So we shallassume that there is, in addition to the terms in (23.2),another term, a resistance force proportional to the velocity: Ff=−cdx/dt. It will be convenient, in our mathematical analysis, to write theconstant c as m times γ to simplify the equation a little. This is just the same trick we usewith k when we replace it by mω20 , just to simplify the algebra. Thus our equation will be
通常，因为摩擦项的特性和复杂性，所以，这种问题，会比较困难。然而，在很多情形中，摩擦力，正比于对象移动的速度。这种摩擦的一个例子，就是在油、或粘稠的液体中，对象在慢速运动。当对象不动时，没有力，但是，它移动得越快，油就越快地从对象边上移过，于是，阻力就越大。所以，我们将假设，除了（23.2）中的项外，还有另外有一项，一个正比于矢速的阻力：Ff = −cdx/dt。在我们的数学分析中，把常数c，写成m乘以γ ，会方便些，这能把方程，简化一点。当我们用mω20，替换k时，使用的是同样的技巧，只是为了简化代数。这样，我们的方程将是：
(23.6)
or, writing c=mγ and k= mω20 and dividing out the mass m ,
或者，用c=mγ 和k= mω20来写，然后，约去质量m：
(23.6a)

Thus again x^ is given by F^ times a certain factor. There is no technical name for this factor, noparticular letter for it, but we may call it R for discussion purposes:
这样，通过 F^乘以某个因子，x^再次被给予。对于这个因子，没有任何技术名称，也没有任何特别的字母，但是，为了讨论的目的，我们可称其为R：

and 及
(23.9)
(Although the lettersγand ω0 are in very common use, this R has no particular name.) This factor R can either be written as p+iq , or as a certain magnitude ρ times eiθ . If it is written as a certain magnitude times eiθ，let us see what it means.Now F^=F0eiΔ , and the actual force F is the real part of F0eiΔeiωt, that is, F0cos(ωt+Δ) . Next, Eq. (23.9)tells us that x^ is equal to F^R . So, writing R=ρeiθ as another name for R , we get
（虽然字母γ和ω0，非常通用，但这个R，并无任何特别的名字。）这个R既可被写作 p+iq，或也可写作某大小ρ，乘以eiθ。如果它被写作某大小ρ，乘以eiθ，让我们看看，其意义为何。现在，F^=F0eiΔ，实际的力 F，就是F0eiΔeiωt的实部，也就是说，F0cos(ωt+Δ)。接下来，方程（23.9）告诉我们，x^ 等于 F^R。于是，把R的另外一个名字，写作 R=ρeiθ，我们得到：

Finally, going even further back, we see that the physical x, which is the real part of the complex x^eiωt, is equal to the real part of ρF0ei(θ+Δ)eiωt. But ρ and F0 are real, and the real part of ei(θ+Δ+ωt)is simply cos(ωt+Δ+θ) . Thus
最后，更往回走一点，我们看到，物理的x，是复数x^eiωt的实部，它等于ρF0ei(θ+Δ)eiωt的实部。但是，ρ和 F0是实数，所以，ei(θ+Δ+ωt)的实部，简单就是cos(ωt+Δ+θ)。这样：
(23.10)
This tells us that the amplitude of the response is the magnitude ofthe force F multiplied by a certain magnification factor, ρ ; this gives us the “amount” of oscillation. It also tells us,however, that x is not oscillating in phase with the force, which has the phase Δ, but is shifted by an extra amount θ . Therefore ρ and θ represent the size of the response and the phase shift of theresponse.
这告诉我们，共振的幅度，就是力F的大小，乘以某个放大因子ρ；这给我们提供了共振的“量”。然而，它也告诉我们，x与力的震荡，并不是同相位的，力有相位Δ，但被漂移了额外的量θ。因此，ρ代表了共振的尺寸，θ代表了共振的相位漂移。

Now let us work out what ρ is. If we have a complex number, the square of the magnitude is equalto the number times its complex conjugate; thus
现在，让我们求ρ。如果我们有一个复数，其大小的平方，就等于该数，乘以其共轭复数；这样
(23.11)
In addition, the phase angle θ is easy to find, for if we write
另外，相位角，很容易发现，因为，如果我们写：

we see that
我们就看到：
(23.12)
It is minus because tan(−θ)=−tanθ . A negative value for θ results for all ω , and this corresponds to the displacement x lagging the force F .
它是负的，因为tan(−θ)=−tanθ。对于θ来说，一个负的值，是对所有ω的结果{？}，这相当于，位移x，滞后于力 F。

Fig. 23–2.Plot of ρ2 versus ω . 图23-2 ρ2对ω的曲线图。

Fig. 23–3.Plot of θ versus ω . 图23-3 θ对ω的曲线图。
Figure 23–2 showshow ρ2 varies as a function of frequency (ρ2 is physically more interesting than ρ , because ρ2 is proportional to the square of the amplitude, or more or less to theenergy that is developed in the oscillator by the force). We see that ifγ is very small, then 1/(ω20−ω2)2 is the most important term, and the response tries to go up towardinfinity when ω equals ω0 . Now the “infinity” is not actually infinite because if ω=ω0, then 1/γ2ω2 is still there. The phase shift varies as shown in Fig. 23–3.
图23-2，显示了，ρ2作为频率的函数，是如何变化的（ρ2物理上，比ρ更有趣，因为ρ2正比于振幅的平方，或者，正比于震荡中所开发出的能量，震荡是由力带来的）。我们看到，如果γ非常小，那么1/(ω20−ω2)2 ，就是最重要的项了，当ω等于ω0，反应趋于无限大。现在，“无限”并非无限，因为，如果ω=ω0 ,那么1/γ2ω2，仍在那里。相位漂移的变化，见图23-3。

In certain circumstances we get a slightlydifferent formula than (23.8),also called a “resonance” formula, and one might think that it represents adifferent phenomenon, but it does not. The reason is that if γ is very small the most interesting part of the curve is near ω=ω0, and we may replace (23.8)by an approximate formula which is very accurate if γ is small and ω is near ω0 . Since ω20−ω2=(ω0−ω)(ω0+ω), if ω is near ω0 this is nearly the same as 2ω0 (ω0−ω)and γω is nearly the same as γω0 . Using these in (23.8),we see that ω20−ω2+iγω≈2ω0 (ω0−ω+iγ/2) , so that
在某些情形下，我们会得到一个公式，它与（23.8），略有不同，它也被称为“共振”公式，有人可能会想，它代表着一个不同的现象，但并非如此。原因是这样，如果γ很小，那么，曲线的最有趣的部分，就在ω=ω0附近，我们可以用一个近似公式，来替换（23.8），如果γ很小，且ω接近ω0，那么，该近似公式，将非常精确。由于ω20−ω2=(ω0−ω)(ω0+ω), 如果ω接近于ω0，那么，这就与2ω0 (ω0−ω)，几乎一样，而γω，就与γω0几乎一样。在(23.8)中使用这些, 我们看到 ω20−ω2+iγω≈2ω0 (ω0−ω+iγ/2) , 于是
(23.13)
It is easy to find the corresponding formula for ρ2 . It is
对于ρ2，找到相应的公式，也很容易。它就是：