Name: 物理學：第 16.1 章 有阻尼的簡諧運動（20 箇中的 1 個） 基本方程：無阻尼 (Physics: Ch 16.1 Simple Harmonic Motion with Damping (1 of 20) Basic Equation: No Damping)
Uploaded: 2024-10-20T05:28:16.000Z
Duration: 8 min 25 s

關係子句

In this playlist we're going to take a closer look at simple harmonic motion and in particular how the equation is derived what it means physically relative to what's happening when there's simple harmonic motion and then we're also going to look at it in terms of the damping because a lot of simple harmonic motion involves damping and we want to see how the equation itself is derived and how to utilize that equation as well and there's several solutions that we need to look at depending upon what kind of damping we're dealing with.

在這個播放列表中，我們將仔細研究簡諧運動，特別是方程是如何推導出來的，它對簡諧運動的物理意義是什麼，然後我們還將從阻尼的角度來研究它，因為很多簡諧運動都涉及到阻尼，我們想看看方程本身是如何推導出來的，以及如何利用這個方程。

But before we look at the details of damping let's take a look again at the general undamped simple harmonic motion equation.

不過，在瞭解阻尼的細節之前，讓我們再來看看一般的無阻尼簡諧運動方程。

So here we graphically show when we have an object hanging from a spring the spring has a certain amount of spring constant K and the object has a certain amount of mass M.

是以，我們在這裡用圖表說明，當一個物體懸掛在彈簧上時，彈簧有一定的彈簧常數 K，物體有一定的品質 M。

Once we allow it to balance so let's say we hang it on the at the equilibrium point then we can give it a push upward or we can pull downward and then it'll begin to oscillate up and down.

一旦我們讓它達到平衡，比方說，我們把它掛在平衡點上，然後我們可以把它向上推，或者向下拉，然後它就會開始上下襬動。

The gravity portion is is then negated by the fact that the spring has been extended somewhat before we start the oscillatory motion.

在我們開始擺動運動之前，彈簧已經伸長了一些，這就抵消了重力部分。

Here at this position this position in this position that the mass is what we call at the equilibrium point.

在這個位置上，這個位置上的品質就是我們所說的平衡點。

That is where there is no net force acting on the object and therefore no acceleration and the position is therefore at zero.

此時，物體上沒有淨力作用，是以沒有加速度，位置是以為零。

That's the equilibrium point and in this particular example let's assume that the motion is upward so velocity in this point is upward.

這就是平衡點，在這個特定的例子中，我們假設運動是向上的，是以該點的速度是向上的。

It will then reach its maximum distance away from the equilibrium point when X equals the amplitude of the motion capital A stands for amplitude.

當 X 等於運動的振幅時，它與平衡點的距離將達到最大值。

At that moment V will be zero and the acceleration will be in the negative direction.

此時 V 將為零，加速度方向為負。

As long as the object is above the equilibrium point acceleration will be negative.

只要物體位於平衡點上方，加速度就是負值。

When it's below the equilibrium point like like it is over here then the acceleration will be in the positive direction.

當它低於平衡點時，就像現在這樣，加速度就會向正方向移動。

Notice it'll be coming back down moving through the equilibrium point but in this case it's moving on downward negative velocity.

注意，它將通過平衡點向下運動，但在這種情況下，它是以向下的負速度運動。

At this moment in time the position is zero the acceleration is zero then it reaches its maximum elongation below the equilibrium point when X equals negative the amplitude.

此時，位置為零，加速度為零，當 X 等於振幅的負值時，達到平衡點以下的最大伸長。

So we have positive amplitude negative amplitude at this moment the velocity is zero the acceleration is upward and then it reaches back to the equilibrium point just like it was over here with an upward velocity X is equal to zero and there's no acceleration that moment because there's no net force acting on the mass.

是以，我們有正振幅和負振幅，此時速度為零，加速度向上，然後回到平衡點，就像剛才一樣，向上的速度 X 等於零，此時沒有加速度，因為沒有淨力作用在品質上。

And so that's how it is continuous up and down and notice that there's some relationship to a sine or a And then we moved a paper past that pencil at a constant speed as this is oscillating up and down you would actually the pen would actually make that sine wave or that cosine wave as the paper is moving and as the object is oscillating up and down.

然後，我們以恆定的速度將紙張從鉛筆上移過，當鉛筆上下襬動時，鉛筆實際上會隨著紙張的移動和物體的上下襬動而產生正弦波或餘弦波。

So there's some relationship between simplomatic motion and the sine or the cosine function.

是以，簡諧運動與正弦或餘弦函數之間存在某種關係。

Now we'll see mathematically why that is the case.

現在我們從數學角度來看看為什麼會這樣。

Starting with Newton's second law where f equals ma we can turn the ma equals f and then we know that the force exerted on the mass by the spring is equal to minus kx. k is the spring constant and X is a distance away from the equilibrium point.

從牛頓第二定律 f 等於 ma 開始，我們可以將 ma 變為 f，然後就知道彈簧對品質施加的力等於減去 kx。k 是彈簧常數，X 是距離平衡點的距離。

Notice the negative sign because if the mass is in a positive position the spring is then pushing in the negative direction that's why the negative is there.

注意負號，因為如果品質處於正位置，彈簧就會向負方向推動，這就是負號出現的原因。

Then we realize that the acceleration is essentially the second derivative of position with respect to time so we can replace a by that and then notice that if we move the minus kx to the left we end up with an equation here equal to zero and also notice if we divide everything by m then we have d square x dt square with other words the second derivative of x with respect to time plus k over m times x equals zero.

然後我們意識到，加速度本質上是位置相對於時間的二階導數，是以我們可以用它來代替 a，然後注意，如果我們將減去 kx 的部分向左移動，我們會得到一個等於零的等式，同時注意，如果我們將所有部分除以 m，我們會得到 d 平方 x dt 平方，換句話說，x 相對於時間的二階導數加上 k 乘以 m 再乘以 x 等於零。

Again this is an undamped case and if we then replace this with x double dot. x double dot simply means the second derivative of x with respect to time and then if we allow Omega to be equal to the square root of k over m which essentially that's the definition of Omega we can now write this as x double dot plus Omega squared x equals zero which is the second order differential equation of an undamped system.

x 雙點指的是 x 相對於時間的二階導數，如果我們允許 Omega 等於 k 對 m 的平方根，這就是 Omega 的定義，我們就可以把它寫成 x 雙點加 Omega 平方 x 等於零，這就是無阻尼系統的二階微分方程。

Now the solution to that can either be the sine or the cosine.

現在的解可以是正弦，也可以是餘弦。

Why is there a potential for a sine or cosine?

為什麼存在正弦或餘弦的可能性？

Well it depends on the initial condition, the initial time.

這取決於初始條件和初始時間。

If time equals zero here notice that the amplitude is zero when time equals zero.

如果時間等於零，這裡注意到時間等於零時振幅為零。

Well that would happen when we have a sine function.

當我們使用正弦函數時，就會出現這種情況。

When time equals zero the sine of zero is zero and x will be at zero so that's what we have over here.

當時間等於零時，零的正弦值為零，X 將為零，這就是我們的結果。

However if time equals zero when the object is up here then we have a cosine function because notice now when time equals zero the cosine of zero is one and x equals a which is what we have over here.

但是，如果物體在上面時時間等於零，那麼我們就有了一個餘弦函數，因為現在注意到，當時間等於零時，零的餘弦值是 1，而 x 等於 a，這就是我們這裡的情況。

So whether or not we use the sine or the cosine to describe the oscillatory motion or the simple harmonic motion simply depends on the initial condition.

是以，我們是用正弦還是餘弦來描述振盪運動或簡諧運動，完全取決於初始條件。

Where's the object when time equals zero?

當時間等於零時，物體在哪裡？

Now of course it could be over here on the way down it could be over here at time equals zero and then we have to modify the equation by putting a negative sign in front or by having a phase angle and we'll show you how to do that in the later video.

當然，它也可能在下降過程中出現在這裡，也可能在時間等於零時出現在這裡，然後我們必須在前面加上負號或相位角來修改等式，我們將在後面的視頻中為您演示如何修改等式。

Now we want to show you also that these two solutions are indeed solutions of this second order differential equation.

現在，我們還要向大家證明，這兩個解確實是這個二階微分方程的解。

Well first of all we can take the first derivative with respect to time and the derivative of the sine is the cosine and the derivative of the angle omega t is omega so end up with a omega cosine of omega t.

首先，我們可以求出關於時間的一階導數，正弦的導數就是餘弦，而角度 omega t 的導數就是 omega，所以最終得出 omega t 的 omega 餘弦。

If we then take the second derivative the derivative of cosine is the negative sign and again we have to multiply times the derivative of the omega square sine of omega t.

如果我們求二階導數，餘弦的導數是負號，我們必須再次乘以歐米茄 t 的歐米茄平方正弦的導數。

Then realizing that a sine omega t, a sine omega t is equal to x we can replace that so we're left with minus omega squared times x.

這樣，我們就可以用減去歐米茄的平方乘以 x 來代替正弦歐米茄 t。

And then if we move to the left side then we end up with x double dot plus omega squared x equals zero which is exactly what we had over there.

然後，如果我們移動到左邊，我們就會得到 x 雙點加歐米茄平方 x 等於零的結果，這正是我們在那邊得到的結果。

So therefore we understand that this is indeed a solution to this second order differential equation.

是以，我們可以理解為，這確實是這個二階微分方程的解。

We can do the same with the cosine function.

我們可以用餘弦函數做同樣的運算。

Again take the first derivative with respect to time.

再次求出與時間有關的一階導數。

The derivative of the cosine is the negative sine.

X double dot as we call it is equal to again the derivative of sine is the negative cosine.

我們所說的 X 雙點等於正弦的導數也就是負餘弦。

The derivative of the sine is the positive cosine.

And then we'll multiply the times the derivative of omega t which is omega.

然後乘以歐米茄 t 的導數，即歐米茄。

We have x double dot equals minus omega squared x.

我們有 x 雙點等於減去歐米茄平方 x。

Now you can see again we get the exact same second order differential equation showing that x equal a cosine of omega t is also a solution of this differential equation.

現在你可以再次看到，我們得到了完全相同的二階微分方程，表明 x 等於歐米茄 t 的餘弦值也是這個微分方程的解。

So that's the basic differential equation describing the oscillator motion of an object.

這就是描述物體振盪運動的基本微分方程。

Remember that k is the spring constant. m is the mass.

請記住，k 是彈簧常數，m 是品質。

And the square root of k over m. m is indeed the angular speed or the angular frequency of the motion.

But that's the basic construct of symplemonic motion as we use the equation f equals ma which is then converted to the second order differential equation in terms of omega which is the angular frequency or angular speed of the oscillator motion.

但這就是交感運動的基本構造，因為我們使用 f 等於 ma 的方程，然後將其轉換為以歐米茄為組織、部門的二階微分方程，歐米茄是振盪器運動的角頻率或角速度。

字幕列表影片播放

物理學：第 16.1 章有阻尼的簡諧運動（20 箇中的 1 個）基本方程：無阻尼 (Physics: Ch 16.1 Simple Harmonic Motion with Damping (1 of 20) Basic Equation: No Damping)

essentially

assume

constant

realize

物理學：第 16.1 章 有阻尼的簡諧運動（20 箇中的 1 個） 基本方程：無阻尼 (Physics: Ch 16.1 Simple Harmonic Motion with Damping (1 of 20) Basic Equation: No Damping)

essentially

assume

constant

realize

物理學：第 16.1 章有阻尼的簡諧運動（20 箇中的 1 個）基本方程：無阻尼 (Physics: Ch 16.1 Simple Harmonic Motion with Damping (1 of 20) Basic Equation: No Damping)