I already graduated with an engineering degree. I don't use a lot of this part of the degree at work but I have always wanted to go back and master the stuff we covered in college. Unfortunately, a lot of school in the US is only about passing a tests. Your videos are helping me go back and actually learn! Thank you so much!!

@DrTrefor

3 жыл бұрын

You are most welcome!

@shayaanrk

Жыл бұрын

Don't worry son, all schools around the world are only about passing the tests. This is all part of the matrix.

@kevinperera6129

9 ай бұрын

This has been my problem too. Thank you

I'm a mechanical engineering student. This really helped. Thank you so much! Love from Italy.

This has genuinely blown my mind because it has linked comparatively simple concepts i learned in A-Level physics with ODEs to describe the world clearly using math's and i understood why. I was always told adding friction into mechanics makes it so much harder, and I'm sure in other places it is, but here its more simple than i expected.

See, I was tracking with you when you started explaining the friction force, like "Okay, I can remember that" but then you explained it in terms of the car or walking, and how it goes against you, and I was like "Okay now THAT makes sense!" I just gotta say, I wish more teachers explained stuff like you do. You make it relatable.

Thanks boss, got a Intro to Vibrations Test on Monday and you make this so much more understandable. Very grateful, and I hope you get a lot of blessings!

Your videos are indubitably helpful. Keep up the good work, Dr. Trefor!

Professor T. Bazett, thank you for a strong Introduction to Mechanical Vibrations in Differential Equations. Mechanical Vibrations is a huge part of Mechanical Engineering, which deals with tools/equipment that vibrate upon impact. Many problems in Mechanical Vibrations are Underdamped.

Wonderful explanations. Clear, Concise, and Clean. Thank you for your time, especially in making sure the "Why?" part is conveyed which is often the hardest to wrap my head around. But I totally get it now!

@wjiuvdjsvhsnsm

4 ай бұрын

You said Exactly what I want to say

I've been struggling with this concept in my physics class all term and you just cleared everything up in like 10 minutes! Thank you so much, your videos are the most helpful!!

Love your videos, ironically I was already subscribed to you but my Analytical Methods professor recommended your video series for our class to watch as a mandatory thing before class! 😁 Your content is making it to universities 🤙

Your engaging bright attitude makes it easier for me to learn this stuff thank you

Thank you so much, you can never imagine how you helped in understanding this part

when everyone's just teaching to help us get marks,teachers like u help us find answers to all our curiosities thankyou so much sir!

Best teacher in the world!

Hi Trefor. Just wanted to tell you that your Videos are great. For me as a 14 year old math and physics nerd from germany you are gold!

@DrTrefor

3 жыл бұрын

Thank you so much! Love to hear from my slightly younger audience, keep at it for sure!

@erikawimmer7908

3 жыл бұрын

@@DrTrefor I will!

@erikawimmer7908

3 жыл бұрын

@@DrTrefor I will!

@pinklady7184

3 жыл бұрын

I wish my young nieces and nephews were as interested in studies as you are.

@erikawimmer7908

3 жыл бұрын

@@pinklady7184 Thank you! Are you a mathmatician or something like that? I am interested in what you are doing. I want to get a Ph.D. in physics later. Oh and dont worry about your nephews. I dont know how old they are but if they are teenagers then the best advice I can give you is to just let them do the stuff they like and dont force them to anything. You go to a hell lot of stress if you are in the middle of puperty. They will find the goals they want to reach in life and I think you will be proud of them whatever their's is!

This video just explained a lot of stuff on solving Second order differential equations and others doubts.. Thanks!

Thank you Sir for this awesome video. The explanation with graph really helps with the understanding.

8:33 OHHHHHH, you're a genius!! That's such a good image!

You're a great teacher, thank you so much

Professor my respect to you ! Amazing video !

this man is literally carry my differential equations class

For anybody who gets into vibration analysis or reliability engineering, in mechanical system you can typically only vary mass or stiffness. Although it can be expensive to modify equipment in the two aforementioned fashions, it can prevent fractures. Changing mass, stiffness or both shifts the frequency response away from the current resonant frequency which is what kills shafts, bearing, supports, pipe fittings.

@dog360

9 ай бұрын

hey could i message you for help brainstorming for a math investigation coursework im doing for school?

Thanks so much for all the content recently, it's been a massive help this semester! I have been meaning to ask, do you have any plans to create more pure-math oriented videos in the future? I noticed on your personal website that your background is in topology. It would be amazing to have some videos which give an intro to point-set topology (or maybe functional analysis which also seems really interesting), considering the clarity with which you are able to explain things!

@DrTrefor

3 жыл бұрын

You are most welcome! I actually am a topologist and so have definitely been meaning to do a series on that. I want to do quite a bit more, but so much to do so little time ha!

Very nice explanation and it is very informative:) Thank you so much sir.

great video man thanks

perfect explanation, thanks :)

( AWESOMENESS !!!!!!!!!! + RESPECT + PASSION + KNOWLEDGE ) * (MATHEMATICS ) = Dr .Trefor Bazett !

Thank you!

Solid video

Thank you so much sir 🔥🔥🔥

First time I ever hear that friction is a function of velocity. I mean it does make a difference if you're moving or not (static vs kinetic friction coefficient), however, once you start moving it becomes a function of the normal force (independent of velocity). Now, an hydraulic/pneumatic dampener does depends on velocity. Anyway, great video as always!!

@angelmendez-rivera351

3 жыл бұрын

This is friction on a spring, and here, there is no normal force, because there is no gravity being considered.

@AMR-555

3 жыл бұрын

@@angelmendez-rivera351 that's definitely not it

@rohitchaudhary3619

3 жыл бұрын

Its viscous force of damper , act like friction force here

@AMR-555

3 жыл бұрын

@@rohitchaudhary3619 I would agree with that, that's the example we always see in Mechanical Vibrations. Usually there's a "damper" parallel to the spring in the schematics.

@carultch

Жыл бұрын

@@AMR-555 Friction is proportional to velocity assuming viscosity governs the fluid mechanics of drag, rather than stagnation pressure. Usually at low velocities and high viscosities (called laminar flow), you get drag forces that are proportional to velocity. Eventually, there is a speed at which there is a transition from laminar flow to turbulent flow, where it is much more accurate to model drag as proportional to the square of speed, rather than to speed itself. Non-fluid friction is independent of velocity altogether, and only depends on normal force, the identity of the surfaces, and whether they are initially at rest, or initially moving. This kind of problem is significantly more difficult to solve, if you have to work with a kind of friction other than viscous damping that is proportional to speed.

you're amazing thank you

Thanks doctor

Do you have a video where friction is modeled as a constant?

Awesome 🤩👌

10:00 At t≈7/3≈2.33 second! [Setting the derivative equal to zero]

Sir, will you recommend laplace transform to solve these differential equations? Actually I used it for a while and got some weird solutions, like extra sine terms and cosine terms... Will it affect my solutions?

@carultch

Жыл бұрын

You can use either the Laplace transform, or the traditional method of assuming the solution has the form of e^(r*t), and solving for the characteristic equation that determines the value of r. It depends on the form of the differential equation given, and whether you are given initial conditions or not. If you are just solving the simple case of a free vibration, and initial conditions aren't really in the picture, then I recommend the more traditional method of assuming the solution has the form of e^(r*t), and solving the characteristic equation for possibly-complex values of r. If instead, you are given an exotic forcing function other than a simple sine or cosine, then I recommend the Laplace transform method. Try a few examples both ways. See if you get consistent results, and which method seems to be easier in which circumstances.

ive never been able to figure out why you need both a sign and a cosine term in the underdamped solution. surely just having the one and moving it around will let you set whatever initial conditions you want

@carultch

Жыл бұрын

You can either have a single amplitude and a phase constant, or you can have a sine and cosine term. They both will give you the same solution, since they are both multiplied by the same exponential decay envelope. Through a trig identity, you can prove that they are equal. Initial value problems with underdamping are significantly easier to solve, if you keep them as an arbitrary linear combination of sine and cosine terms, and then solve for the coefficients based on the initial conditions. If you switch it to a single trig function with an amplitude and phase constant, it is a harder problem to solve. For the situation with no damping, you can proceed either way to find the coefficients. You can use conservation of energy to find the amplitude, and inverse trig to find the phase constant. Or you can use a linear combination of sine and cosine, solve for their amplitudes from initial conditions, and then use trig identities if necessary to translate to a single trig function with an amplitude and phase constant.

❤️ thank you

He is a genius absolutely no doubt about that At 4.25 when the 4 canceled would you not get square root of -mk over m ?

@aleksandreakhvlediani8034

2 жыл бұрын

Yes, but m in the denominator m = sqrt(m)*sqrt(m), one sqrt part will cancel with numerator and you are left with sqrt(-k/m)

Why , if friction is a constant times the normal force, can we say the damping is proportional to the velocity?

Hi, Dr.Trefor thank you for the video, but I would like to ask you that what e^rt means at 2:08~a high school student from Asia.

@Dr. Trefor Bazett, When you arrived at the second order differential equation of ma + cv + kx= 0, why did you ignore the negative sign in front of the c and the k? Because if you do not ignore it, you get a sqrt(c^2 + 4mk). Then from this, how would you explain the motion of the oscillation from the discriminant?

@carultch

Жыл бұрын

The negative signs in front of c and k, are only there when you have them on the opposite side of the equation as m*a. The negative signs switch to positive, when you move all terms to one side. When you solve for the values of r, you get: r = (-c +/- sqrt(c^2 - 4*m*k))/(2*m) When this c^2 - 4*m*k is negative, you get two complex conjugate numbers for r. This corresponds to an exponential decay function at the real value of r as its decay constant, that envelopes a linear combination of sine and cosine waves. Knowing that c^2 - 4*m*k is negative, you "ignore" the negative sign, carry out the root, and then know that it belongs with an imaginary unit when constructing the rest of the solution.

What is the similarity of hooke's law and boyle's law?

Sir I have a question from number system. Could you please make me understand this question kindly which is given below Prove that every positive integer different from 1 can be expressed as a product of a non negative power of 2 and an odd number

@carultch

Жыл бұрын

What's there to prove? Look up the concept of a prime factorization.

" it's gonna be a block that goes... Aneeeeeeer😂😂

1:44 Shouldn't the fiction and the spring force have inverse signs here? Don't they have inverse directions?

plz cover physics 2 topics and materials topics too

is that possible underdamped more faster than critically damped?

I found this video but don't see your Mech Vibration's Playlist on your KZread site. Do you have one?

@DrTrefor

Жыл бұрын

This video is part of my differential equations playlist.

How to deal with a case when the mass of that block also decrease as a function of time? Like m(t) = m(o)- rt . I don't know how to solve it.plzz reply

@DrTrefor

3 жыл бұрын

That could totally be done and just gets a new ODE. It is no longer constant coefficients, but it is still linear which is good.

Sir please recommend me the standard book of linear algebra and mathatical analysis with visualization....

@DrTrefor

3 жыл бұрын

THis is a decent free linear algebra one: open.umn.edu/opentextbooks/textbooks/5

@chg691

3 жыл бұрын

@@DrTrefor thanks sir

I don't get way in the Overdamped case (2th case) r1 and r2 are both always negative. sqrt(c²-4mk) can not be greater than c? Because if can -c + sqrt(c²-4mk) > 0 and so r1 is positive.

@carultch

Жыл бұрын

If they were positive, you'd have exponential growth, rather than exponential decay. This could only happen if time were to go backwards, and the dampers added mechanical energy to the system, instead of subtracting it. -c + sqrt(c^2 - 4*m*k) cannot be greater than zero, as long as m and k are both positive real numbers, which they'd have to be for this situation to be realistic. In the limit as c gets large, the -4*m*k term approaches zero. We end up with -c + sqrt(c^2), which simplifies to zero. In the limit as c approaches zero, we end up with sqrt(-4*m*k), which when divided by (2*m), gives us sqrt(k/m), as we have for the frequency of a free vibration without damping. One way you could make a situation where "k" is negative, is with the falling chain. You have a chain with a uniform mass per unit length, and part of it is hanging off the edge of a table, while the rest of it is sitting on the table in a straight line. After you release the other end, the part hanging over the edge pulls the rest off the table, with an acceleration that is proportional to the fraction that is hanging. In the frictionless case, the motion is modeled with the function cosh(t). This is essentially how you make a negative spring constant, because the force causing the motion is proportion to position and in the same direction as the position.

well there is actually 4 cases , the under damped can be separated into two cases,damped (real part is 0) and underdamped

At 1:58, the force of friction is defined as -cx', but when it's included in the larger equation it's +cx'. Is the sign arbitrary?

@musabothman4362

2 жыл бұрын

the velocity could be - or + ,depending on whether the body is moving upward or downward . and the -ve sign in the equ put so that the friction force is always opposite to the direction of moving

@ohmc7gd7610

2 ай бұрын

​@@musabothman4362can you explain what is c and it's units

Where does that t come from in the critically damped? You didn't explain how you got there

@DrTrefor

3 жыл бұрын

Ah it was in the previous video in the playlist for “repeated roots”.

🔥🔥🔥

Mechanical vibrations? More like "Might good differential revelations!"

I always thought frictional forces were nonlinear, something like First derivative squared

@DrTrefor

2 жыл бұрын

It depends on the speed and the medium. Linear for slower speeds and things like movement in fluids. Fast speeds in air quadratic.

@gustavoespinoza7940

2 жыл бұрын

Oh my god I’m fangirling so hard rn You noticed me senpai

@carultch

Жыл бұрын

@@gustavoespinoza7940 It has to do with the relative ratio of "inertial forces" to viscous forces that govern the flow. We quantify it with a unitless number called Reynolds' number, which is rho*v*x/mu. rho is the density, mu is the viscosity, v is the speed, and x is the characteristic geometry dimension. When Reynolds' number is low, viscosity governs, and the flow is laminar. The drag forces are proportional to speed. When Reynolds' number is high, inertial forces govern, and the flow is turbulent. The drag forces are proportional to speed squared. For mass/spring/damper systems, the problem is significantly more difficult to solve if friction is anything other than a viscous damper. That's most likely what is used in practice in an application such as a car's suspension system.

Put in a source of oscillation and some Green's function and you're 'almost' doing Quantum Field Theory ;-) Great job, man!!!

I think mazda 2 dj has characteristic of critically damped

Do you get money from KZread

@DrTrefor

3 жыл бұрын

A little from ad sense:)

@yaiba8992

3 жыл бұрын

@@DrTrefor oh

#ArsenalAman

coool

sir i give you want to urdu please thanks

first one to comment 🤣😂

@DrTrefor

3 жыл бұрын

Nice one!!

Why would you not show the plot for all 3 cases on a single plot? You're so focused on the mathemaics, you've forgotten the qualitative value of understanding.