Oscillations
AP Physics 1 Topic 7 7:48 English narration · English + 中文 subtitles burned in
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Transcript
Two identical springs, two identical masses.
两根一模一样的弹簧，两个一模一样的物体。
Pull one of them far down.
把其中一个往下拉得很远， 另一个只拉开一点点。
Pull the other only a little.
同时松手。
Let go of both at the same moment.
哪一个先回来——幅度大的，还是幅度小的？
Which one gets back first — the big bounce, or the small one?
它们始终同步。
<slow>They stay in step.</slow> Bounce after bounce, they cross the middle together.
一次又一次，它们同时经过中间。
How far you pull it does not change the time.
你拉多远，并不改变时间。
This is Unit Seven: oscillations. One rule defines them.
这是第七单元：振动。
Two formulas give their timing. Three graphs describe them. And one energy idea ties the whole thing together.
一条规则定义它们，两个公式给出它们的时间， 三张图像描述它们，还有一个能量思想把这一切串在一起。
Let's begin.
让我们开始吧。
Here is the plan.
这是我们的计划。
First we define simple harmonic motion and find the test that decides whether a motion counts.
首先定义简谐运动，并找出判断一种运动算不算简谐运动的标准。
Then period and frequency.
然后是周期和频率。
Then reading the motion off graphs.
接着从图像中读出运动。
Finally, energy — kinetic and potential trading places while the total never changes.
最后是能量——动能和势能互相转换，而总量始终不变。
Oscillations are everywhere.
振动无处不在。
A tuning fork vibrates hundreds of times each second, and your ear hears that vibration as a note; a bigger, heavier fork vibrates more slowly, so it sounds lower.
一支音叉每秒振动几百次，你的耳朵把这种振动听成一个音； 更大、更重的音叉振动得更慢，所以声音更低。
A mass on a spring bounces up and down.
弹簧上的物体上下振动。
Anything nudged away from a stable resting place tends to be pulled back, and to oscillate.
任何东西一旦离开稳定的静止位置，都会被拉回去，从而振动起来。
Watch one oscillation from three sides at once.
同时从三个角度看同一个振动。
On the left, a mass bounces on a spring. In the middle, a point travels steadily around a circle. On the right, a graph draws a smooth curve as time passes.
左边，物体在弹簧上上下振动； 中间，一个点绕着圆周匀速运动；右边，随着时间推移，图像画出一条平滑的曲线。
<slow>These are three pictures of the same motion.</slow> The mass, the point and the curve stay locked together, and that repeating curve is what simple harmonic motion looks like.
这是同一种运动的三幅画面。 物体、圆周上的点和曲线始终同步， 而那条重复的曲线，就是简谐运动的样子。
So what makes a motion simple harmonic — SHM for short?
那么，什么样的运动才是简谐运动（SHM）？
One condition.
只有一个条件。
Push the object away from its resting place and a force pulls it back — the restoring force.
把物体推离静止位置， 就会有一个力把它拉回来——这就是回复力。
For simple harmonic motion that force must be proportional to the displacement: pull it twice as far, and the spring pulls back twice as hard.
对简谐运动来说，这个力必须与位移成正比： 拉远一倍，弹簧回拉的力就大一倍。
The minus sign in the law is doing real work — it says the force always points back toward the middle.
式子里的负号很关键—— 它表示这个力始终指向中间。
This is the test to carry into the exam.
这就是你要带进考场的判据。
On the left the mass hangs below its resting place: the displacement points down, the acceleration points up.
左边，物体停在静止位置下方：位移向下，加速度向上。
On the right it sits above: displacement up, acceleration down.
右边，它在上方：位移向上，加速度向下。
Either way the acceleration is opposite to the displacement and aims at the middle.
无论哪种情况，加速度都与位移方向相反， 并指向中间。
Proportional in size, opposite in direction: that is simple harmonic motion.
大小成正比，方向相反：这就是简谐运动。
A pendulum passes it only while the swing stays small.
单摆只有在摆角很小时才满足它。
Now the timing.
现在来看时间。
The period is the time for one complete cycle.
周期是完成一个完整循环所用的时间。
The frequency is how many cycles happen each second, in hertz, and the two are upside-down versions of each other.
频率是每秒发生多少个循环， 单位是赫兹，两者互为倒数。
There is a third: the angular frequency, omega, which is two pi times the frequency — it is what appears inside the cosine.
还有第三个：角频率 omega，等于二 pi 乘以频率——它出现在余弦函数里面。
For a mass on a spring, more mass means a longer period, a stiffer spring a shorter one.
对弹簧上的物体来说，质量越大周期越长， 弹簧越硬周期越短。
For a small pendulum only the length and gravity matter — the mass of the bob never appears.
对小幅摆动的单摆，只有摆长和重力有影响—— 摆锤的质量根本不出现。
And notice what is missing from both: the amplitude.
注意两个公式里都没有的东西：振幅。
Let's use it.
我们来用一用。
A mass of zero point two five kilograms hangs on a spring of stiffness one hundred newtons per metre. Find the period, then the frequency.
一个零点二五千克的物体挂在劲度为一百牛每米的弹簧上， 求它的周期，然后求频率。
Pause here and try it yourself.
先暂停，自己试一试。
Start from the spring formula.
从弹簧的周期公式出发。
Divide the mass by the stiffness: that is zero point zero zero two five.
用质量除以劲度，得到零点零零二五。
Its square root is zero point zero five.
它的平方根是零点零五。
Multiply by two pi.
乘以二派。
That gives a period of zero point three one seconds.
这样得到周期为零点三一秒。
One divided by that period gives about three point two hertz.
一除以这个周期，约得三点二赫兹。
Here is that motion as a graph: displacement against time, a clean sine curve.
把这个运动画成图像：位移对时间，是一条干净的正弦曲线。
Two numbers describe it completely.
两个数字就能完整描述它。
The first is the amplitude — the greatest distance the object reaches from the middle, marked on the vertical axis.
第一个是振幅——物体离中间最远的距离，标在竖轴上。
The second is the period — the time from one peak to the next, marked along the bottom.
第二个是周期——从一个波峰到下一个波峰的时间，标在下方。
Read those two off a graph and you can describe the whole motion.
从图上读出这两个数，就能描述整个运动。
Now the question the exam keeps asking: where is it fastest, and where does it accelerate most?
现在来看考试反复考的问题：它在哪里最快，在哪里加速度最大？
Watch the two arrows.
看这两个箭头。
At the two ends the object stops for an instant, so the velocity is zero — but it is furthest out, so the acceleration is largest.
在两端，物体会瞬间停下，所以速度为零——但它离中间最远，所以加速度最大。
In the middle the force is zero, so the acceleration is zero — yet the object races through at top speed.
在中间，力为零，所以加速度为零——可物体正以最高速度冲过去。
Here is the trap: speed and acceleration are never large at the same place.
陷阱就在这里：速度和加速度绝不会在同一位置同时最大。
Put displacement, velocity and acceleration on three graphs, one above the other, and the pattern jumps out.
把位移、速度和加速度画成上下三张图，规律立刻显现。
Displacement is a cosine curve here, starting at the maximum.
这里位移是一条余弦曲线，从最大值开始。
Velocity is the same shape shifted a quarter of a cycle in phase, so it peaks where displacement crosses zero.
速度是同样的形状，平移了四分之一个周期， 所以它在位移过零点时达到峰值。
Acceleration is displacement turned upside down.
加速度则是位移上下翻转的结果。
Learn this quarter-cycle pattern once, and graph questions become easy marks.
把这个四分之一周期的规律记住一次，图像题就是轻松得分。
Energy tells the same story in a different language.
能量用另一种语言讲同一个故事。
Watch the two bars.
看这两根能量条。
At the ends of the swing the bob stops, high up: all the energy is potential.
在摆动的两端，摆锤停下来，位置最高： 能量全部是势能。
At the bottom it moves fastest and sits lowest: all the energy is kinetic.
在最低点，它速度最快、位置最低：能量全部是动能。
In between, the two share it.
在中间的位置，两者分享能量。
Kinetic becomes potential, potential becomes kinetic, twice every cycle — and with no friction, the total never changes.
动能变势能，势能变动能，每个周期两次—— 而且在没有摩擦时，总能量始终不变。
Now plot the same energies against displacement, not time.
现在把同样的能量画成对位移的图，而不是对时间。
Potential energy is a valley: zero in the middle, largest at the two ends.
势能是一个低谷： 在中间为零，在两端最大。
Kinetic energy is the mirror image: largest in the middle, zero at the ends.
动能正好相反：在中间最大，在两端为零。
Add them and you get the flat line across the top — the same total everywhere.
把它们相加，就得到顶部那条水平线——每个位置的总量都相同。
That flat line is conservation of energy, drawn.
那条水平线就是画出来的能量守恒。
And the total goes with the amplitude squared: double the amplitude, four times the energy.
而总能量正比于振幅的平方： 振幅翻倍，能量就是四倍。
One more.
再来一道例题。
A mass of zero point five kilograms sits on a spring of stiffness two hundred newtons per metre, with an amplitude of zero point one metres.
一个零点五千克的物体放在劲度为二百牛每米的弹簧上， 振幅是零点一米，求它的最大速度。
Find its greatest speed.
先暂停，自己算一算。
Pause here and work it out. First the total energy: one half, times the stiffness, times the amplitude squared.
先求总能量：二分之一乘以劲度，再乘以振幅的平方，结果是一焦耳。
That is one joule. At the middle all of that is kinetic.
在中间位置，这些能量全部是动能。
So one half times the mass times the speed squared equals one joule — and that gives two metres per second.
于是二分之一乘以质量再乘以速度的平方等于一焦耳—— 解出来是二米每秒。
Four quick checks before we finish.
结束前做四个快速检查。
Double the amplitude — does the period change?
振幅翻倍，周期会变吗？
No.
不会。
The period never depends on how far it swings.
周期从不取决于摆动的幅度。
Where is the velocity greatest?
速度在哪里最大？
In the middle, at the equilibrium position.
在中间，也就是平衡位置。
Where is the acceleration greatest?
加速度在哪里最大？
At the two extremes.
在两个端点。
And does a pendulum's period depend on the mass of the bob?
单摆的周期取决于摆锤的质量吗？
No — only on its length and on gravity.
不取决于——只取决于摆长和重力。
Three habits that save marks.
三个能保住分数的习惯。
First, the period does not depend on the amplitude, so changing how far the object swings leaves the timing untouched.
第一，周期不取决于振幅， 所以改变物体摆动的幅度，并不会改变时间。
Second, speed and acceleration peak in different places: speed in the middle, acceleration at the ends, never together.
第二，速度和加速度在不同的位置达到最大：速度在中间，加速度在两端，绝不会同时。
Third, energy goes with the amplitude squared, so doubling the amplitude multiplies the energy by four, not by two.
第三，能量正比于振幅的平方，所以振幅翻倍，能量变成四倍，而不是两倍。
Get those three right and Unit Seven is yours.
把这三点做对，第七单元就是你的了。