Second-Order Ordinary Differential Equations: Solving the Harmonic Oscillator Four Ways

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Here we introduce the second-order ordinary differential equation (ODEs) for a mass on a spring. In Newton's Second Law, F=ma, the acceleration a is the second derivative of the position x(t), giving a second order differential equation. We solve this equation four ways: 1) by guessing the solution, 2) using Taylor Series, 3) by guessing a different form of the solution, and 4) by writing as a matrix system of equations.
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This video was produced at the University of Washington
%%% CHAPTERS %%%
0:00 Introduction
1:00 Deriving the Spring-Mass Equations from F=ma
6:59 Method 1: Guess the Solution!
11:57 Method 2: Taylor Series Solution
24:18 Method 3: Guess Again!
33:20 Method 4: Write as a Matrix System of Equations

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28 июн 2024

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Комментарии : 51

@peterborisenko5767 Год назад

Thanks for not cutting fire alarm!

@schenzur Год назад

I'm loving this whole series, thanks a lot professor for putting forward a whole course!

@bitterthread6794 Год назад

Keep uploading the content Professor. I am taking Engineering courses and this has been very useful for even providing motivation how all things are fundamentally related .

@hoseinzahedifar1562 Год назад

A perfect topic by an excellent teacher... thank you.

@Pedritox0953 Год назад

Great video! Would be great videos of Lagrangian's dynamics

Год назад

Seems like magic. Thank you very much!

@ErikScott128 Год назад

This is how I wish differential equations was taught when I was in undergrad, and methods 1 and 3 are the approach I'm taking for a video I want to make on rotorcraft flapping. Around time 31:42, you start trying to solve c1 and c2 in terms of the initial values x0 and v0. My response to this is, why bother? You already established that c1 and or c2 must be complex valued in order to make the final equation real. They are arbitrary constants, and i is a constant as well. So, when I demonstrate this, I just say "let a=c1+c2, let b=(c1-c2)i" and then establish a=x0 and b=v0. No need for complex analysis, and the result matches the intuitive guess (assuming you were able to guess the sine component) and the Taylor series result, while still being completely general.

@riteshparmar4687 10 месяцев назад

Thanque very much for this beautiful lecture.

@FA-tq9ip Год назад

Around 17:28 I think you left the minus sign out on the board which I have put in single quotes (although you do say it): => C3 = '-'1/3! * V0 - (from an example I calculated it also seems like there should not be a minus here: => C5 = '-'1/5! * V0)

@FA-tq9ip Год назад

Just saw that you spotted and corrected these later on in the video :)

@Buyson Год назад

What about Laplace Transform? I think we could use it as well.

@BarkanUgurlu Год назад

I came to write this. :)

@TheRsmits Год назад

The Laplace Transform is basically guessing a mixture of exponential solutions (3rd way) and converting analytic problems into algebraic problems. In this case, the guess is also that it is a discrete mixture.

@VTdarkangel Год назад

There is also the characteristic equation. That being said, I suspect that he is going to hit in that with the eigenvalue determinant or if he covers Laplace Transforms.

@fryingpanm6977 Год назад

thank you so much sir

@xiangwenyan3772 Год назад

jajaja, a fire alarm short break

@timgorringe6668 10 месяцев назад

great lecture! is x_o cos(t) - v_o sin(t) soln dimensionally incorrect? (taylor series soln)

@muthukamalan.m6316 Год назад

how do we visualize second order DE equations like we understand by seeing slope field in 1st order DE ?

@curtpiazza1688 3 месяца назад

I learned so much in less than 40 min.! Thanx! ❤ 😂

@waynelast1685 8 месяцев назад

Please tell me how you use this display board technology?

@joseluizpereiradacunha7244 7 месяцев назад

MUITO OBRIGADO

@halilibrahimEgilmez 5 месяцев назад

in time 30:00 if λ is complex like a±ib then the answer should be x=e^at (x_0 cos(⁡bt) + v_0 sin⁡(bt) ) and here a=0 and b=1 then x=x_0 cos⁡t+v_0 sin⁡(t)

@arthurcoward6979 Год назад

So how do you get the notes you're writing to appear correctly? That is, writing right to left when I'm pretty sure you would have to be writing backwards to achieve that since you're facing the viewer. Enjoying the talks, btw!

@carultch 10 месяцев назад

He probably mirrors the video, so it appears normal. If you saw him in person, your side of his writing would appear backwards.

@aris.mavridis Год назад

I'm having a bit of trouble using the second method for any k/m instead of just 1. Would the Xo and Vo terms be multiplied by k/m as well? I tried to find such solutions online but had no luck.

@carultch 10 месяцев назад

In the general solution: x = x0*cos(w*t) + "v0*sin(w*t)" v0 isn't really the coefficient on the sine term, unless w=1 (which would be the case if k/m were 1, since w=sqrt(k/m)). I put this term in quotes, because it really needs to be something else. We want dx/dt to equal v at t=0. Let A and B be defined such that the general solution is: x = A*cos(w*t) + B*sin(w*t) Take the derivative: dx/dt = -A*w*sin(w*t) + B*w*cos(w*t) Evaluate at t = 0: x(0) = A dx/dt at t=0 = B*w So as you can see, v0 = B*w. Thus, when we solve for the coefficient B, this really is equal to v0/w. Thus, the general solution is: x = x0*cos(w*t) + v0/w * sin(w*t)

@aris.mavridis 5 месяцев назад

Hello @@carultch, thank you for the reply. I was asking about the second method followed in the video, that is, the Taylor series solution.

@hsenagrahdeers 25 дней назад

Hi, did you find what you were looking for somewhere/figured it out somehow? I'm attempting the same rn.

@gregtroll Год назад

I feel smarter after watching this. Sometimes your “t”s look like “+”s. I have to pay attention

@seannibecker5500 3 месяца назад

Are you actually writing left to right from your perspective?

@danieljulian4676 6 месяцев назад

Linear operator! Superposition! Great! 27:50 So, I guess we're assuming some linear algebra already. Not the first time in this series that I've seen messages from the future. It's not that free stuff is useless, but anyone watching these is sinking a half-hour at a time into following the presentation. It seems more that these videos are intended as refresher, or as quick review of formal lecture material given in a physical classroom by Steve himself. Anyone else is going to have to forego any pretense of rigor or fill in a lot of blanks with a lot of legwork. This series is not a substitute for a complete course.

@klammer75 Год назад

I was sooooo close to skinning that cat…phew!🧐😉😆

@yoyoyowazupperson 5 месяцев назад

hello. Can someone please explain how x0 got divided it by 2 at 32:35? this makes no sense to me… method III after the fire alarm in particular is very confusing to me

@hilalvenus 4 месяца назад

Because c1=c2.

@salah3beed 5 месяцев назад

28:00 ✨ok that was just the fire alarm✨

@benarcher372 Год назад

Strange. In Method II the answer is x(t) = cos(t)x_zero + sin(t)v_zero. But here we are adding a length to a velocity!? The answer x(t) is a length. Perhaps the second term should have been multiplied by 't'? Am I missing something here?

@Oberbremser 10 месяцев назад

the simplification not to write (k/m) which has dimension 1/second^2 causes the "problem" with units. I had same thought as you to add length and velocity which is not possible due to different units. But considered the (k/m) factor and the units are then correct as in corresponding terms of the taylor serie (1/second^2) (1/second^4) ... appears forcing each term to have dimension of length. i.g: - 1/(3!)*v0*t^3*(1/second^2) (1/second^2) stems from k/m and so on.

@benarcher372 10 месяцев назад

@@Oberbremser Thanks for your comment. I need to think about this. Old brain..

@mintakan003 Год назад

The Taylor series is too tedious for most humans. But if it's general, isn't it something that can be handled by computers? Computer solvers? (And I mean "symbolic" solvers, not "data fitting".) Or is the "suspend variables" method actually used?

@bendavis2234 Год назад

I jumped at the fire alarm at ~27:30

@hilalvenus 4 месяца назад

I was looking for this comment!

@Roxas99Yami Год назад

the harmonic approxillator hehe

@user-bn9kz6hs4l 8 месяцев назад

bro the fire alarm XD

@lioneloddo Год назад

And there are also 2 other ways to establish the differential equation of the oscillator. 1. From the conservation of the energy : d/dt(1/2.m.V**2 + 1/2.K.X**2) = 0 then mV*d/dt(V)+KX*d/dt(X)=0 and then ... 2. From the momentum : d/dt(mV)=kX then ... Many ways to establish the equation, many ways to solve the equation. Why are there so many ways to deal with the oscillator ? Why are there so many points of view to deal with the oscillator ? I really don't know what it means ...