Name: 驅動振盪器上 (Driven Oscillator上)
Uploaded: 2024-10-20T15:22:08.000Z
Duration: 10 min

In this last section, we would like to add yet another ingredient to our oscillator.

在最後一部分，我們要為振盪器添加另一種成分。

情態副詞

So far, we have looked at an oscillator with a force that depends on the separation, basically to the equilibrium point, Hooke's law.

到目前為止，我們已經研究了一個振盪器，它的力取決於分離度，基本上取決於平衡點，即胡克定律。

Then we added some damping, which was a force that was actually trying to reduce, was against the velocity, so a force that was proportional to the velocity but with a minus sign.

然後，我們添加了一些阻尼，這實際上是一種試圖減小速度的力，是以是一種與速度成正比但帶有負號的力。

And finally, we want to see how we can actually include energy into the system, so how we can inject energy into the system.

最後，我們想看看如何才能真正將能量納入系統，也就是如何將能量注入系統。

And we are going to do this by including a sinusoidal driving force.

為此，我們將加入正弦驅動力。

So basically what we are doing, we are including energy into the system in an oscillatory manner.

是以，我們現在所做的，基本上是以振盪的方式將能量輸入系統。

So now that we've said this, we can write the equation, Newton's law.

既然說到這裡，我們就可以寫出牛頓定律的方程式了。

And what we see, the three terms on the left, the terms that we saw before for the damped oscillator, so just as a reminder, the acceleration, Hooke's law, and the damping factor, and the right-hand side will be a driving force.

我們看到，左邊的三個項，就是我們之前看到的阻尼振盪器的項，所以提醒一下，加速度、胡克定律和阻尼係數，右邊將是一個驅動力。

So we are starting to accumulate the frequencies and symbols and so on.

是以，我們開始積累頻率和符號等等。

So I'd like to insist on the fact that omega, in this case, is a parameter, if you will.

是以，我堅持認為，Ω 在這裡是一個參數。

It's an external, it's the frequency at which you're actually applying the force.

這是外部因素，是你實際施力的頻率。

And you see that the force is actually time-dependent.

你會發現，力實際上是隨時間變化的。

So it's an oscillatory force, F0 being a constant.

是以，這是一個振盪力，F0 是一個常數。

Now we are going to rewrite this equation again so that we put an x dot dot without a pre-factor.

現在，我們要再次重寫這個等式，這樣我們就可以在不帶前置因子的情況下打上 x 點。

We've done this a number of times already.

And we have to find a solution to this differential equation.

我們必須找到這個微分方程的解。

Well, the solution to this equation is the usual approach that we learn in ordinary differential equation.

那麼，這個方程的解法就是我們在常微分方程中學到的通常方法。

The first is a solution to the equation without the right-hand side, which we actually know already because this is the damped oscillator that we saw in the previous screencast.

第一個是不帶右邊的方程解，實際上我們已經知道了，因為這就是我們在上一期截屏視頻中看到的阻尼振盪器。

And two, we want a solution that reproduces the right-hand side.

其次，我們需要一個能重現右邊的解決方案。

So we call that a complementary function and then a particular function.

是以，我們稱之為補充功能和特殊功能。

So the complementary function, we've discussed it at length in the damped oscillator screencast, where we have the under-damped, over-damped, and critically damped.

是以，我們在阻尼振盪器截屏視頻中詳細討論了互補函數，其中包括欠阻尼、過阻尼和臨界阻尼。

So this is a solution of the equation in the blue box without the right-hand side.

是以，這就是藍色方框內方程的解，沒有右邊。

And the particular solution, we are going to try as a solution a cosine.

我們將嘗試用餘弦來解決這個問題。

And we are going to see that the cosine actually should work.

我們將看到，餘弦實際上應該起作用。

So the particular solution, let's try to see how this particular solution that we suggest here with d and delta being elements that we have to determine, let's see if it's a possible solution to the equation in blue.

是以，讓我們試著看看我們在這裡提出的 d 和 delta 是我們必須確定的元素的特定解，看看它是否是藍色方程的可能解。

So we are going to introduce the solution xp in that equation and see if it's possible.

是以，我們要在方程中引入解 xp，看看是否可行。

So as always, I know that students always remember trigonometry very well.

是以，我知道學生們總是能很好地記住三角函數。

But it's probably a good idea to remember that trigonometry is always your friend.

但最好記住，三角函數永遠是你的朋友。

So some equations that you might find useful.

是以，一些方程式可能會對你有所幫助。

And then what we do, we insert xp as a possible solution on the equation on the top.

然後，我們將 xp 作為一個可能的解插入上面的等式中。

And then we end up with those two solutions, that equation.

最後我們就得到了這兩個解決方案，也就是這個等式。

And so what we see here is something that's interesting, is that we have a term in front of cosine omega t and a term in front of sine omega t.

有趣的是，我們在餘弦歐米茄 t 和正弦歐米茄 t 前面都有一個項。

However, cosine omega t and sine omega t are linearly independent functions.

然而，餘弦 omega t 和正弦 omega t 是線性獨立函數。

Therefore, in order for this equation to be always true for any time, we need the two coefficients in front of cosine omega t and in front of sine omega t to be zero.

是以，為了使該方程在任何時間都始終成立，我們需要餘弦 omega t 前面和正弦 omega t 前面的兩個係數為零。

And in fact, by doing so, we will be able to determine the two unknowns, which are d and delta.

事實上，通過這樣做，我們就能確定兩個未知數，即 d 和 delta。

So let's start with the simplest one, which is this term.

那麼，我們就從最簡單的這個詞開始吧。

In order to ensure that the function is zero, the other term will also need to be zero.

為了確保函數為零，另一項也必須為零。

我們幾分鐘後再討論這個問題。

So we see directly that for this to be zero, if we divide by cosine delta left and right, we see that we find that the tangent of delta should be two omega beta divided by omega zero squared minus omega squared.

是以，我們可以直接看到，要使其為零，如果我們用餘弦 delta 左右相除，我們就會發現 delta 的正切值應該是兩個歐米茄-貝塔除以歐米茄零的平方減去歐米茄的平方。

Just to remind you, omega zero is the natural frequency of the undamped oscillator that we've discussed since the beginning of this chapter.

提醒一下，Ω 0 是我們在本章開始時討論過的無阻尼振盪器的固有頻率。

Omega is the frequency of the driving force that we have introduced.

歐米茄是我們引入的驅動力的頻率。

And delta is really, when you think about it, when you go back to the equation that we used as xp, delta is a de-phasing between the application of the driving force and the response.

仔細想想，回到我們使用的 xp 等式，delta 其實就是驅動力和響應之間的去相位。

So delta is really kind of a delay, if you will, between the application of the force and the maximum of the response.

是以，如果你願意，德爾塔實際上是施力與最大反應之間的一種延遲。

是以，我們有了這個解決方案。

Again, it's a good idea to remember some trigonometry.

同樣，最好能記住一些三角函數。

And that allows us to calculate sine delta and cosine delta.

這樣我們就能計算正弦三角和餘弦三角。

This is pretty elementary from trigonometry.

So we had this equation that has to be equal to zero.

是以，我們有一個必須等於零的等式。

我們已經重點討論了第二部分。

現在，讓我們來看看這第一部分。

And that allows me to find the value of a, actually the value of d as a function of a.

這樣我就能找到 a 的值，實際上就是 d 的值與 a 的函數關係。

And then when we see this, we see that in the denominator, we have a cosine delta and a sine delta.

看到這裡，我們會發現在分母中，有一個餘弦三角函數和一個正弦三角函數。

But we already calculated sine delta and cosine delta in the previous slide.

但在上一張幻燈片中，我們已經計算了正弦三角和餘弦三角。

So when we substitute, we find that the value of d, which is the amplitude of the solution, the particular solution, is going to be given by this equation here.

是以，當我們代入時，會發現 d 的值，也就是解的振幅，即特定解的振幅，將由這裡的方程給出。

So the amplitude is going to depend on a number of things.

It's going to depend on the damping, it's going to depend on the driving frequency, and also on the natural frequency.

這取決於阻尼、驅動頻率和自然頻率。

So when we put everything together, we see that the particular solution of the problem is completely known now.

是以，當我們把所有東西放在一起時，就會發現問題的具體解決方案現在已經完全為人所知了。

The particular solutions are completely known because we just described delta and we described d.

因為我們剛剛描述了 delta 和 d，所以特定的解決方案是完全已知的。

So this is the solution, this exact solution.

這就是解決方案，一個精確的解決方案。

And then as I discussed already, delta will be the phase difference between the maximum of the driving force and the maximum of the resultant motion.

然後，正如我已經討論過的，delta 將是驅動力最大值與運動結果最大值之間的相位差。

Interestingly enough, this phase difference, delta, actually depends on the driving frequency.

有趣的是，這種相位差（delta）實際上取決於驅動頻率。

Now, for example, I'm sorry, for example, just an example on the last slide, on the last line of this slide, we see that if there is no driving frequency, of course, there is no delay, obviously, there is a delay of pi over 2 at the frequency omega equal omega 0 and a dephasing of pi when the frequency is extremely large.

現在，舉個例子，對不起，舉個例子，就在最後一張幻燈片上，在這張幻燈片的最後一行，我們可以看到，如果沒有驅動頻率，當然也就沒有延遲，很明顯，在頻率ω等於ω0時，會出現π大於2的延遲，當頻率非常大時，會出現π的消相現象。

So that's going to give, you know, like the response of the system is actually delayed compared to the application of the force.

是以，與施加的力相比，系統的反應實際上是延遲的。

So what have we done in the past seven minutes and a half?

那麼，在過去的七分半鐘裡，我們都做了些什麼呢？

Well, we found a solution to the damped oscillator with a driving force.

好了，我們找到了帶驅動力的阻尼振盪器的解決方案。

And we saw that the solution is the sum of a particular solution and of the complementary solution.

我們看到，解是特定解與互補解的總和。

So Xc is what we did in the previous screencast and Xp is the particular solution.

是以，Xc 就是我們在前一個截屏中所做的，而 Xp 則是特定的解決方案。

So here is something that's very important to notice.

What I'm going to argue about now is that in steady state, so in the long time scales, what matters is Xp.

我現在要論證的是，在穩態下，所以在長時間尺度上，重要的是 Xp。

In fact, Xc, the complementary, represents a transient effect.

事實上，互補的 Xc 代表了一種瞬時效應。

So transient effect, I affect the die out quickly.

是以，瞬時效應會很快影響熄滅。

So I reproduce the plot there that we saw in the previous screencast into the damped oscillator.

是以，我將之前截屏中看到的曲線圖複製到阻尼振盪器中。

And we see that in each case, the amplitude after a while goes to zero.

我們看到，在每種情況下，振幅都會在一段時間後歸零。

So Xc, after a sufficiently long time, Xc will go to zero.

是以，在足夠長的時間後，Xc 將歸零。

So after you start the motion, if you wait long enough, the solution in blue will actually no longer matter.

是以，在您開始運動後，如果等待的時間足夠長，藍色的溶液實際上就不再重要了。

And so the only thing that's going to survive is Xp.

是以，唯一能倖存下來的就是 Xp。

So this is the situation in the steady state.

And you can ask, what is long enough time?

你可以問，什麼是足夠長的時間？

Well, the time that's long enough will be the one where the exponential, which remember the exponential was e to the power minus beta t, when that exponential is low enough.

那麼，足夠長的時間就是指數的時間，記得指數是 e 的冪次減去貝塔 t，當指數足夠低的時候。

So in other words, when the time, when beta t is large enough, so in other words, when the time is much larger than 1 over beta.

是以，換句話說，當時間，當貝塔 t 足夠大時，是以，換句話說，當時間遠遠大於貝塔的 1 時。

So if you have a very damp situation, so beta is very large, the time it takes to get to steady state is very small.

是以，如果情況非常潮溼，β 值非常大，那麼達到穩定狀態所需的時間就非常短。

So at that time, we are going to suppose that the blue solution is actually not going to matter anymore because we waited long enough so that the exponential got to zero.

是以，我們假設藍色溶液實際上已經不再重要，因為我們等待的時間足夠長，以至於指數歸零。

 And when we do that, we see that depending on the frequency at which we drive the system, we are going to see that Xc is going to go to zero.

當我們這樣做時，我們會發現，根據我們驅動系統的頻率，我們會發現 Xc 將歸零。