P19 聽寫字幕，若有錯誤敬請斧正

OK! Im coming back to the No.1 example but not the easy example of the Second Order Equation with oscillating forcing term cosωt.

We have to know the answer to this problem and a little messy but the messy is not, the messy is straightforward.

So let me begin by looking for the rectangular form, I called is the rectangular form that separates the cosine with this amplitude and sine with this amplitude and the 2 separate pieces.

So if I’m looking for that solution, M and N, the numbers I want to find,how do I proceed?

It a case of indeterminate coefficient M and N and the way determine them is substitute this into the equation and match the cosine term and then you will find M and N.

And we find M and N, we need to equations for two quantities M and N.

Imagine this, substitute in there, I’ll get some some cosigns, so the cosign on one side will match the cosign on the other side.?

And also from the derived I‘ll get some sines and they should match zero because I have no sinωt on the right hand side.

So two equations matching the sine and matching the cosine and I thought those.

Two equations we want know, I just write the answer down.

M involves C minus A ω square, M is coming from the cosine and we get cosine from that term and that term, divided by some number D.

And N is just Bω divided by the same D.

And now I write down D that (C-Aω^2)^2 plus (Bω)^2?

This is what I come out when I from the two equations for M and N.

I just solve those equations this D here is the two by two determinative.

If we think about the linear algebra behind two equations and that one of this.

So the answer now is in terms of A C B and D which is a mixture of A B and C, that is a solution.

Only I always want to show you a different form of the solution and then this case is better form, because the most important physical quality is the magnitude how large dose y get?

What is the amplitude of this?

This is a sinusoid and we remember that every sinusoids can be written in a polar form say is y of t is some amplitude G, the gain time a cosine of omega t with a shift with a lag with an angle alpha two numbers now, that is the gain, and this is the phase shift alpha.

And that is attractive form, because it has only one term, the two numbers G and alpha can put into single term where we can see the magnitude of the oscillating and what does that come out to be??

I won’t go through all the steps, I just write down what G turn out to be.

G turn out to be come from there , one over the square D, G is the square root of M square and N Square, that is the gain.

Let me write the word gain again.

And alpha is always tangent alpha is the N over the M, which is the Bω/(C-Aω^2)

I like that polar form, and I feel I should just do an example but I didn’t do any of the algebra in this video, but you know what the algebra came from, it came from substituting the form we expect for the solution and of course.

That form we expect is the form we get provided omega the driving frequency is different from Omega n.And well know, I guess we're all right, even if Omega is Omega n, because we have a damping term.?

Yeah, so that's the answer, so an example why not, an example y double prime plus y prime plus two y equals cosine of t.?

Okay, that's a simple example. I took omega to be 1 you see in there is Omega and then A is 1, B is one, C is 2. We can evaluate everything. In fact I think M and N are one half, D will be 2.

So what do I know, do I know the rectangular form?

Yes, rectangular form is 1/2(cost+sint)?

That's the rectangular form. Two simple things, but I have to add them in my mind I don't necessarily see how the cosine add to the sine.?

But the sinusoid like identity the polar form gives it to me, so what's in polar form.?

So G, the gain is going to be one over the square root of two.?

Yeah, at the highest point, the cosine and the sine are the same, they're both one over the square root of two.?

Cosine of t minus Pi over 4 is the angle, the phase lag, when I add the cosine and the sign I get a sinusoid that's sitting over Pi over 4, 45 degrees, so that those are the two forms, so in a nice example we certainly got a nice answer.?

Okay, so that is the worked out, more or less worked out. In principle, worked out is the solution to what I think of as the most important application when the forcing term is a cosine.?

So that gives oscillating motion, it gives a phase shift and it gives these formulas.?

The only thing I would add is that is I need to comment on better notation, so I have used in these formulas A, B and C.?

But those have meaning as mass, damping constant, spring constant, B and K.?

It's combinations of those that come in, so let me just take this moment to say better notation or I should say engineering notation instead of A, B,?C, which are mass, damping, spring constant, well that's already better to use letters that have a mean, but the small but very important point is that two combinations of A, B, C—m, b, k are especially good.

One is the natural frequency that we see already square root C over A, or square root k over m. So that gives us one important combination of A and C.?

The other one is the damping ratio and its called Zeta, and that damping ratio is B over the square root of 4AC, you'll say where did that come from or I can use these letters B over the square root of 4 mk.?

That damping ratio is the right dimension less quantity, the dimensions of this ratio just number that those two quantities have the same dimension, and we can see that because in the quadratic formula comes remember that in a quadratic formula comes square root of B squared minus 4 AC.?

You know that these must have the same units, otherwise subtraction would be crime.

So they have the same ratio and therefore the same units and therefore the ratio is dimensionless.?

Write that word dimensionless.?

So conclusion I cloud rewrite the answer in terms of these quantities, Omega and Zeta beam.?

I won't do that here that can wait for another time, but you know to say, since we solved found the solution to the most important applications with cosine Omega t there since we found the solution.?

A appropriate to comment that we could write the answer in terms of Omega n the natural frequency and the Zeta the damping ratio.?

Thank you!