21–2The harmonic oscillator 21-2 谐波震荡

Fig. 21–1.A mass on a spring: a simpleexample of a harmonic oscillator. 图21-1 弹簧上的一个质量：一个简单的谐波震荡例子。
Perhaps the simplest mechanical systemwhose motion follows a linear differential equation with constant coefficientsis a mass on a spring: first the spring stretches to balance the gravity; onceit is balanced, we then discuss the vertical displacement of the mass from itsequilibrium position (Fig. 21–1). Weshall call this upward displacement x , and we shall also suppose that the spring is perfectly linear, inwhich case the force pulling back when the spring is stretched is precisely proportionalto the amount of stretch. That is, the force is −kx (with a minus sign to remind us that it pulls back). Thus the masstimes the acceleration must equal −kx :
md2x/dt2=−kx. (21.2)
For simplicity, suppose it happens (or we change our unit of time measurement)that the ratio k/m=1 . We shall first study the equation
d2x/dt2=−x. (21.3)
Later we shall come back to Eq. (21.2)with the k and m explicitly present.
有些力学系统，其运动，具有常数系数的线性微分方程，或许其中最简单的，就是弹簧上的质量：首先，弹簧伸展，以平衡重力，一旦平衡后，我们将讨论质量距其平衡位置的垂直位移（图21-1）。我们将把它，称为向上的位移x，我们也将假设，弹簧是完全线性的，在这种情况下，当弹簧伸展时，其往回拉的力，就是准确地正比于伸展的量。也就是说，力就是-kx（负号用来提醒我们，力是往回拉的）。这样，质量乘以加速度，就应等于-kx：
md2x/dt2=−kx. (21.2)
为简单起见，假设（或者，我们改变了我们的时间测量的单位），比率k/m=1。我们将首先研究方程：
d2x/dt2=−x. (21.3)
稍后，我们将返回方程（21.2），届时，k和m的意义，将完全清楚。

Now to go further with the original problem,we restore the time units to real seconds. What is the solution then? First ofall, we might think that we can get the constants k and m in by multiplying cost by something. So let us try the equation x=Acost ; then we find dx/dt=−Asint , and d2x/dt2= −Acost= −x . Thus we discover to our horror that we did not succeed in solvingEq. (21.2),but we got Eq. (21.3)again! That fact illustrates one of the most important properties of lineardifferential equations: if we multiply a solution of the equation by anyconstant, it is again a solution. The mathematical reason for this isclear. If x is a solution, and we multiply both sides of the equation, say by A, we see that all derivatives are also multiplied by A , and therefore Ax is just as good a solution of the original equation as x was. The physics of it is the following. If we have a weight on aspring, and pull it down twice as far, the force is twice as much, theresulting acceleration is twice as great, the velocity it acquires in a giventime is twice as great, the distance covered in a given time is twice as great;but it has to cover twice as great a distance in order to get back tothe origin because it is pulled down twice as far. So it takes the same timeto get back to the origin, irrespective of the initial displacement. In otherwords, with a linear equation, the motion has the same time pattern, nomatter how “strong” it is.
现在，我们继续最初的问题，我们把时间的单位，恢复为真正的秒。那么，解答是什么呢？首先，我们可以想，我们可以通过让cost，乘以某物，来得到常数k和m。让我们先试一下方程x=Acost ；因此，我们得到 dx/dt=−Asint , 和 d2x/dt2=−Acost= −x 。我们非常吃惊地发现，我们并没有成功地解开方程（21.2），而是又得到了方程（21.3）。这一事实，展示了线性方程最重要的特性之一，即对于方程的解答，如果我们用任何常数，来乘之，它就又是一个解答。此事的数学原因，非常清楚。如果x是一个解答，我们用某物乘以方程的两边，比如说A，我们看到，所有的导数，也被乘以A，因此，Ax就正如x一样，都是原方程的解答。此事的物理原因如下。如果我们有一个重量，在弹簧上，且我们把它拉下来的距离，{与上一次比}，是两倍远，那么，力就也是两倍，最终的加速度，也是两倍；在给定时间内，所走距离，也是两倍，在给定时间内，它所要求的矢速，也是两倍，在给定时间内，所走距离也是两倍；但是，因为它被拉开了两倍远的距离，所以，为了回到原点，它要走的距离，必须是两倍。所以，无论最初的位移是多少，它回到原点所需的时间，是一样的。换句话说，对于一个线性方程，无论运动有多“强”，运动都有着同样的时间模式。

That was the wrong thing to do—it onlytaught us that we can multiply the solution by anything, and it satisfies thesame equation, but not a different equation. After a little cut and try to getto an equation with a different constant multiplying x , we find that we must alter the scale of time. In other words,Eq. (21.2)has a solution of the form
x=cosω0t. (21.4)
(It is important to realize that in the present case, ω0is not an angular velocity of a spinning body, but we run out ofletters if we are not allowed to use the same letter for more than one thing.)The reason we put a subscript “0 ” on ω is that we are going to have more omegas before long; let us rememberthat ω0 refers to the natural motion of this oscillator. Now we try Eq. (21.4)and this time we are more successful, because dx/dt=−ω0sinω0tand d2x/dt2= −ω20cosω0t= −ω20x . So at last we have solved the equation that we really wanted tosolve. The equation d2x/dt2=−ω20xis the same as Eq. (21.2)if ω20=k/m .
这样做事情，是错的—它只教会我们，我们可以用任何事物，来乘以这个解答，这样做，满足了同样的方程，但是并没有满足不同的方程。为了得到一个方程，它是用不同的常数，乘以x，我们做了一些探索和尝试，我们发现，应该改变时间的尺度。换句话说，方程（21.2）要有一个如下形式的解答：
x=cosω0t. (21.4)
（有一点，很重要，即在当前案例中，ω0不是一个旋转物体的角速度，而是，对于不止一个事物，如果不允许我们使用同样字母的话，我们将会用光字母。）之所以要在ω上，放一个下标“0”，乃是因为，不久之后，我们将会有更多的ω；让我们记住，ω0指向的：是这个振动的自然运动。现在，我们试解方程（21.4），这次，我们更成功，因为dx/dt=−ω0sinω0t 和d2x/dt2= −ω20cosω0t= −ω20x 。所以，至少我们所解的方程，就是我们真正想解的。如果ω20=k/m，那么，d2x/dt2=−ω20x就与方程（21.2），是一样的。

The next thing we must investigate is thephysical significance of ω0 . We know that the cosine function repeats itself when the angle itrefers to is 2π . So x=cosω0t will repeat its motion, it will go through a complete cycle, when the“angle” changes by 2π . The quantity ω0t is often called the phase of the motion. In order to change ω0tby 2π , the time must change by an amount t0 , called the period of one complete oscillation; of course t0must be such that ω0t0=2π . That is, ω0t0 must account for one cycle of the angle, and then everything willrepeat itself—if we increase t by t0 , we add 2π to the phase. Thus
(21.5)
Thus if we had a heavier mass, it would take longer to oscillate backand forth on a spring. That is because it has more inertia, and so, while theforces are the same, it takes longer to get the mass moving. Or, if the springis stronger, it will move more quickly, and that is right: the period is lessif the spring is stronger.
下面，我们要研究的，就是ω0的物理意义。我们知道，对于余弦函数，当它所指角度，是 2π时，它就会重复自己。所以，对于x=cosω0t，当“角度”变化了2π时，它就会经历一个完整的周期，然后重复其运动。量 ω0t，通常被称为运动的相位。为了通过2π，来改变ω0t，时间必须改变一个量 t0，它被称为一个震荡的完整周期；当然， t0应满足ω0t0=2π。也就是说，ω0t0应该覆盖角度的一个循环，然后，每个事情，都将重复其自己--如果我们让t，增加了 t0，就等于我们让相位，增加了2π。这样
(21.5)
这样。如果我们有一个较重的质量，那么，它在弹簧上，来回震荡，需时更长。这是因为，其惯性更大，于是，当力一样时，要让质量移动，就需时更长。或者，如果弹簧更强，它移动地就会更快，这是对的：如果弹簧更强，周期就会更小。

Note that the period of oscillation of amass on a spring does not depend in any way on how it has been started,how far down we pull it. The period is determined, but the amplitude ofthe oscillation is not determined by the equation of motion (21.2).The amplitude is determined, in fact, by how we let go of it, by what wecall the initial conditions or starting conditions.
注意，对于质量上的弹簧，其震荡周期，并不以任何方式，依赖于它是如何开始的，我们把它下拉地有多远。周期是确定的，但震荡的振幅，并不是通过运动方程（21.2）而确定的。事实上，振幅，是通过我们如何让它走、通过我们称为的初始条件或开始条件，而确定的。

Actually, we have not quite found the mostgeneral possible solution of Eq. (21.2).There are other solutions. It should be clear why: because all of the casescovered by x=acosω0t start with an initial displacement and no initial velocity. But it ispossible, for instance, for the mass to start at x=0 , and we may then give it an impulsive kick, so that it has some speedat t=0 . Such a motion is not represented by a cosine—it is represented by asine. To put it another way, if x=cosω0t is a solution, then is it not obvious that if we were to happen towalk into the room at some time (which we would call “t=0 ”) and saw the mass as it was passing x=0 , it would keep on going just the same? Therefore, x=cosω0tcannot be the most general solution; it must be possible to shift thebeginning of time, so to speak. As an example, we could write the solution thisway: x=acosω0 (t−t1) , where t1 is some constant. This also corresponds to shifting the origin of timeto some new instant. Furthermore, we may expand
cos(ω0t+Δ)=cosω0tcosΔ−sinω0tsinΔ,
and write
x=Acosω0t+Bsinω0t,
where A=acosΔ and B=−asinΔ . Any one of these forms is a possible way to write the complete,general solution of (21.2):that is, every solution of the differential equation d2x/dt2=−ω20xthat exists in the world can be written as
(a) x =acosω0 (t−t1),
or
(b) x=acos(ω0t+Δ)
or
(c) x=Acosω0t+Bsinω0t. (21.6)
实际上，方程（21.2）的可能的最普遍的解答，我们尚未发现。还有其他解答。应该搞清楚为什么：因为，由x=acosω0t所覆盖的情况，在开始时，都有初始位移，而没有初始速度。但可能会有这种情况，开始时，质量在x=0，我们一时冲动，踢了它一脚，于是，在t=0，它有一定速度。这种运动，并未被一个余弦来代表，而是被一个正弦来代表。换一种说法，如果x=cosω0t是一个解答，那么，下面的情况，是否很明显呢：即如果我们在某个时间（我们可称其为“t=0”），碰巧走到这个房间里，刚好看到，质量正在通过x=0，它会继续这样走吗？因此，x=cosω0t不可能是最普遍的解答；这么说吧，改变时间的开始，应该是可能的。作为一个例子，我们可以这样来写解答：x=acosω0 (t−t1)，这里t1是某个常数。把时间的原点，改到某个新的瞬间，这个解答，也可与这种改动相应。另外我们可以扩展：
cos(ω0t+Δ)=cosω0tcosΔ−sinω0tsinΔ,
而写：
x=Acosω0t+Bsinω0t,
这里A=acosΔ，B=−asinΔ 。这些形式中的任何一个，都可能是方程（21.2）的完整的、普遍的解答；也就是说，对于微分方程d2x/dt2=−ω20x，这个世界上存在的、关于它的每个解答，都可写成：
(a) x =acosω0 (t−t1),
或
(b) x=acos(ω0t+Δ)
或
(c) x=Acosω0t+Bsinω0t. (21.6)

Some of the quantities in (21.6)have names: ω0 is called the angular frequency; it is the number of radians bywhich the phase changes in a second. That is determined by the differentialequation. The other constants are not determined by the equation, but by howthe motion is started. Of these constants, a measures the maximum displacement attained by the mass, and is calledthe amplitude of oscillation. The constant Δ is sometimes called the phase of the oscillation, but that is aconfusion, because other people call ω0t+Δ the phase, and say the phase changes with time. We might say that Δ is a phase shift from some defined zero. Let us put itdifferently. Different Δ ’s correspond to motions in different phases. That is true, butwhether we want to call Δ the phase, or not, is another question.
方程（21.6）中的一些量，是有名字的：ω0被称角频率；它是在一秒内，相位改变的弧度的个数。这是由微分方程决定的。其他的常数，并不是由方程决定的，而是由运动如何开始决定的。在这些常数中，a所测量的，是质量达到的最大位移，被称为振幅。常数 Δ，有时被称为震荡的相位，但这里有个混淆，因为，其他人把ω0t+Δ称为相位，且说，相位随时间而变。我们可以说，Δ是距所定义的零位的相位偏差。让我们换一种说法。不同的Δ，相应于在不同相位中的运动。这是真的，但是，我们是否要把Δ称为相位，则是另外一个问题。

21–3Harmonic motion and circular motion[A1] 21-3 谐波震荡与圆运动

Fig. 21–2.A particle moving in a circularpath at constant speed. 图21-2 一个粒子，以恒速，绕圆运动。
The fact that cosines are involved in thesolution of Eq. (21.2)suggests that there might be some relationship to circles. This is artificial,of course, because there is no circle actually involved in the linear motion—itjust goes up and down. We may point out that we have, in fact, already solvedthat differential equation when we were studying the mechanics of circularmotion. If a particle moves in a circle with a constant speed v , the radius vector from the center of the circle to the particleturns through an angle whose size is proportional to the time. If we call thisangle θ=vt/R (Fig. 21–2) thendθ/dt= ω0= v/R . We know that there is an acceleration a= v2/R= ω20R toward the center. Now we also know that the position x , at a given moment, is the radius of the circle times cosθ , and that y is the radius times sinθ :
x=Rcosθ, y=Rsinθ.
方程（21.2）的解，牵扯到余弦函数，这一事实表明，可能与圆，有些关系。当然，这是人工的，因为，实际上并没有圆，牵扯到线性运动中—它只是在上下地走。我们可以指出，当我们研究圆运动的力学时，事实上，我们已经解了微分方程。如果一个粒子，以恒速v，绕圆运动，那么，从圆的中心到粒子的半径矢量，其角度在转，该角度的大小，正比于时间。如果我们把这个角度，称为θ=vt/R (图 21–2)，那么，dθ/dt= ω0= v/R。我们知道，有一个加速度a= v2/R= ω20R，它是向心的。现在，我们也知道了，在给定时刻，位置x就是，圆的半径乘以cosθ，y就是，半径乘以sinθ：
x=Rcosθ, y=Rsinθ.

Now what about the acceleration? What is the x -component of acceleration, d2x/dt2? We have already worked that out geometrically; it is the magnitude ofthe acceleration times the cosine of the projection angle, with a minus signbecause it is toward the center.
ax=−acosθ=−ω20Rcosθ=−ω20x. (21.7)
现在加速度又如何呢？加速度d2x/dt2的x分量是什么？我们已经从几何上得出了；它就是：加速度的大小，乘以投影角的余弦，因为方向向内，所以加个负号：
ax=−acosθ=−ω20Rcosθ=−ω20x. (21.7)

In other words, when a particle is moving in a circle, the horizontalcomponent of its motion has an acceleration which is proportional to thehorizontal displacement from the center. Of course we also have the solutionfor motion in a circle: x=Rcosω0t . Equation (21.7)does not depend upon the radius of the circle, so for a circle of any radius,one finds the same equation for a given ω0. Thus, for several reasons, we expect that the displacement of a masson a spring will turn out to be proportional to cosω0t, and will, in fact, be exactly the same motion as we would see if welooked at the x -component of the position of an object rotating in a circle withangular velocity ω0. As a check on this, one can devise an experiment to show that theup-and-down motion of a mass on a spring is the same as that of a point goingaround in a circle. In Fig. 21–3 anarc light projected on a screen casts shadows of a crank pin on a shaft and ofa vertically oscillating mass, side by side. If we let go of the mass at theright time from the right place, and if the shaft speed is carefully adjustedso that the frequencies match, each should follow the other exactly. One canalso check the numerical solution we obtained earlier with the cosine function,and see whether that agrees very well.
换句话说，当一个粒子，做圆运动，其运动的水平分量，就有一个加速度，它正比于距圆心的水平位移，当然，对于圆运动，我们也有解答：x=Rcosω0t。方程（20.7），并不依赖于圆的半径，于是，对于任何半径的圆，给定ω0，，人们就会发现同样的公式。这样，由于几个原因，我们期待，弹簧上质量的位移，最终将是，正比于cosω0t ，事实上，如果一个对象，以角速度ω0，绕圆运动，那么，对象位置的x分量的变化，将与弹簧质量运动，是一样的。为了检查这一点，人们可以设计实验，以显示，弹簧上质量的上下运动，与一个点绕圆运动，是一样的。在图21-3中，被投影到平面上弧形光，形成两个影子，并排而立，一个是：曲柄销在一个转动轴上的影子，另一个是：垂直震荡着的质量的影子。如果我们让质量，在正确的时间，从正确的位置出发，如果传动轴的速度，经过仔细调整，以让其频率匹配，那么，两个影子，将会惟妙惟肖。前面，我们曾用余弦函数，得到过数值解，人们也可用它，来检查一致性，是否很好。

Fig. 21–3.Demonstration of the equivalencebetween simple harmonic motion and uniform circular motion. 图21.3 简单的谐波运动，与均匀的圆运动，之间等效的类比。

Here we may point out that because uniformmotion in a circle is so closely related mathematically to oscillatory up-and-downmotion, we can analyze oscillatory motion in a simpler way if we imagine it to bea projection of something going in a circle. In other words, although thedistance y means nothing in the oscillator problem, we may still artificiallysupplement Eq. (21.2)with another equation using y , and put the two together. If we do this, we will be able to analyze ourone-dimensional oscillator with circular motions, which is a lot easierthan having to solve a differential equation. The trick in doing this is to usecomplex numbers, a procedure we shall introduce in the next chapter.
这里，我们可以指出，由于一个圆上的均匀运动，与一个上下震荡运动，数学上非常接近，所以，对于震荡运动，我们可以类似地分析，即把它想象成，一个某物绕圆运动的投影。换句话说，虽然距离 y，在震荡运动中，毫无意义，我们仍然可以用另外一个使用y的方程，人工地替换方程（21.2），然后，把这两个，放在一起。如果我们这样做，我们就可以用圆运动，来分析我们的一维震荡，这比解微分方程，要容易地多。这样做的诀窍，是利用复数，其步骤，下章介绍。

That is the end of our solution, but thereis one physically interesting thing to check, and that is the conservation ofenergy. Since there are no frictional losses, energy ought to be conserved. Letus use the formula
x=acos(ω0t+Δ);
then
v=−ω0asin(ω0t+Δ).
Now let us find out what the kinetic energy T is, and what the potential energy U is. The potential energy at any moment is kx2/2, where x is the displacement and k is the constant of the spring. If we substitute for x , using our expression above, we get
U=kx2/2=ka2cos2(ω0t+Δ) /2.
Of course the potential energy is not constant; the potential neverbecomes negative, naturally—there is always some energy in the spring, but the amountof energy fluctuates with x . The kinetic energy, on the other hand, is mv2/2 , and by substituting for v we get
T=mv2/2=mω20a2sin2(ω0t+Δ) /2.
Now the kinetic energy is zero when x is at the maximum, because then there is no velocity; on the otherhand, it is maximal when x is passing through zero, because then it is moving fastest. Thisvariation of the kinetic energy is just the opposite of that of the potentialenergy. But the total energy ought to be a constant. If we note that k=mω20, we see that
T+U=mω20a2[cos2 (ω0t+Δ)+sin2 (ω0t+Δ)]/2=mω20a2/2.
The energy is dependent on the square of the amplitude; if we havetwice the amplitude, we get an oscillation which has four times the energy. Theaverage potential energy is half the maximum and, therefore, half thetotal, and the average kinetic energy is likewise half the total energy.
我们的解答，到此为止，但是，还有一个物理上感兴趣的事情，要检查，它就是能量守恒。由于没有摩擦损失，所以，能量应该是守恒的。让我们用公式：
x=acos(ω0t+Δ);
然后：
v=−ω0asin(ω0t+Δ).
现在，让我们找出，动能T是什么，势能U又是什么。任何时刻的势能，都是kx2/2，这里x是位移，k是弹簧常数。如果我们用上面的表达式，替换x，我们就得到：
U=kx2/2=ka2cos2(ω0t+Δ) /2 。
当然，势能不是常数；自然，势能永不为负，弹簧中总有一些能量，但是，能量的量，随着x而波动。另一方面，动能则是mv2/2，把v替换掉，我们得到：
T=mv2/2=mω20a2sin2(ω0t+Δ) /2 。
现在，当x最大时，动能为零，因为，这时没有矢速；另一方面，当x通过零时，动能最大，因为这时速度最快。动能的这种变化，与势能的变化，正好相反。但是，总的能量，应该是一个常数。如果我们注意到k=mω20，我们看到：
T+U=mω20a2[cos2 (ω0t+Δ)+sin2 (ω0t+Δ)]/2=mω20a2/2 。
能量依赖于振幅的平方；如果振幅乘以2，那么，震荡的能量，就要乘以4。平均势能，是最大势能的一半，从而，是总能量的一半，同理，平均动能，也是总能量的一半。

21–5Forced oscillations 21-5强制震荡
Next we shall discuss the forcedharmonic oscillator, i.e., one in which there is an external driving forceacting. The equation then is the following:
下面，我们将讨论强制的和谐震荡，在这种震荡中，有一个外部的驱动力。因此，方程就是：
md2x/dt2=−kx+F(t). (21.8)
We would like to find out what happens in these circumstances. Theexternal driving force can have various kinds of functional dependence on thetime; the first one that we shall analyze is very simple—we shall suppose thatthe force is oscillating:
我们希望找出，在这种情形下，会发生什么。外部的驱动力，对于时间的依赖，有各种不同的形式；我们将分析的第一种，非常简单--我们将假设，力是震荡的：
F(t)=F0cosωt. (21.9)

Notice, however, that this ω is not necessarily ω0: we have ω under our control; the forcing may be done at different frequencies.So we try to solve Eq. (21.8)with the special force (21.9).What is the solution of (21.8)?One special solution, (we shall discuss the more general cases later) is
然而，要注意，这个ω，并不必然就是ω0：ω在我们的掌控之中；可以用不同的频率，来做此强制。所以，我们用特殊的力（21.9），来解方程（21.8）。（21.8）的解是什么呢？一个特殊的解如下：（稍后，我们将讨论更一般的解）
x=Ccosωt, (21.10)
where the constant is to be determined. In other words, we mightsuppose that if we kept pushing back and forth, the mass would follow back andforth in step with the force. We can try it anyway. So we put (21.10)and (21.9)into (21.8),and get
这里，常数待定。换句话说，我们可以假定，如果我们不断地来回推，那么，质量将会跟着力的步伐，来回走。我们可以用任何方式，尝试此事。于是，我们把（21.10）和（21.19），代入（21.8），就得到：
−mω2Ccosωt=−mω20Ccosωt+F0cosωt. (21.11)
We have also put in k=mω20 , so that we will understand the equation better at the end. Nowbecause the cosine appears everywhere, we can divide it out, and that showsthat (21.10)is, in fact, a solution, provided we pick C just right. The answer is that C must be
我们把k=mω20 ，也带入了，这样，最终，我们对方程，将会理解的更好。现在，因为余弦似乎到处都有，我们可以把它除掉，这就指出，假设我们选对C的话，事实上（21.10）就是一个解。答案就是，C应该是：
C=F0/m(ω20−ω2). (21.12)

That is, m oscillates at the same frequency as the force, but with an amplitudewhich depends on the frequency of the force, and also upon the frequency of thenatural motion of the oscillator. It means, first, that if ω is very small compared with ω0 , then the displacement and the force are in the same direction. Onthe other hand, if we shake it back and forth very fast, then (21.12)tells us that C is negative if ω is above the natural frequency ω0 of the harmonic oscillator. (We will call ω0 the natural frequency of the harmonic oscillator, and ω the applied frequency.) At very high frequency the denominator maybecome very large, and there is then not much amplitude.
这就是说，m震荡的频率，与力的频率一样，但是，其振幅，依赖于力的频率，和自然振荡运动的频率。这就意味着，首先，如果与ω0相比，ω非常小，那么，位移和力，就在同一方向。另一方面，如果我们非常快地来回的摇动它，那么，（21.12）就告诉我们，如果ω大于自然的谐波震荡的频率ω0，那么，C就是负的。（我们将把ω0，称为自然的谐波振荡的频率，而把ω，称为应用频率）。频率很高时，分母就会变得很大，因此，振幅就不会很大。