Free Vibrations of a Single Degree of Freedom (SDOF) System with Viscous Damping

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@keremoznacar34 4 ай бұрын

It's been a long time since you uploaded the video but I have a question. At 10:19 why you didn't put the x dot and x but only x double dot? Thank you so much for the informative content !

@Freeball99 4 ай бұрын

It's a typo! There should be an x-dot in the 2nd term.

@samuelholt5454 8 жыл бұрын

Excellent instruction Thanks for the awesome video

@sayanjitb 4 жыл бұрын

Dear prof. at the time reference 12:49, where you assumed \zeta = 0, and that led to roots r_1 and r_2. But under square root, there was a minus sign so that would have made roots are complex quantity. You missed that imaginary 'i'. Can you please check it! However, your video tutorials are very informative and resourceful. It is very privilege to me to get this on the youtube platform. TIA

@Freeball99 4 жыл бұрын

I didn't assume that ζ=0. Instead, I renamed the radical part ω_d, the damped frequency of vibration. Similar to how we defined ω_n, it is implied that the roots are complex (although I probably should have made this more clear - perhaps adding the i would have helped). For real roots, the system is over-damped and does not oscillate at all. In this case ω_d becomes negative, which is meaningless since frequencies cannot be negative.

@Caffeinegoesinface 8 жыл бұрын

Great video. Thanks

@yihshawn 10 жыл бұрын

C bar doesn't really matter in this case as they are canceled out eventually. I would rather treat it as a habitual action for the sake of physical description. Most of the physics equation has a form of A*e^coswt, so we can gain information of A (typically amplitude). In pure mathematics, e^rt is sufficient to describe any number (including complex number?) so C bar looks useless here. Just my 2 cents, not very good in Maths either

@Freeball99 10 жыл бұрын

Shawn, not sure I understand exactly what you mean, but a few things spring to mind. Hopefully this addresses it. Simple harmonic motion is governed by 2nd order ordinary differential equations. In a previous video kzbin.info/www/bejne/a4LCcmRrnZyMnbM I demonstrated how to derive this using complex notation AND how this can be converted from the complex form to a real form. In the video above, it was never my goal to solve a particular problem for a given set of initial conditions (as I had done in a previous video) , but rather to show how to incorporate damping into the derivation and solution of the characteristic equation. You are correct that Cbar doesn't have an effect here because it cancels, however it is important to note that it is non-zero (for non-trivial solutions). With regards to your math above...e^rt is NOT sufficient to describe complex numbers...for this you need to add an i (where i^2 = -1). So if you just assume solutions of the form: x = Cbar * e^(irt) then this will handle the complex case too. You will find, however, that this assumption leads to the same final result that I presented. I encourage you to use whichever approach is easier to understand, but from my experience, the students taking this introductory vibrations class often battle with complex analysis which is why I chose this method. Hope this helps.

@ashutosh15shukla 10 жыл бұрын

why take c Bar in teh value of x ( x=c' e^rt). i'm weak in math, tell me if that's some mathematical rule, I'd ignore it and learn it like that

@Freeball99 10 жыл бұрын

Ashutosh, this is just a dummy variable - I could have called it anything. In many of mathematical texts, this is written as just 'c', but I didn't want to confuse it with the damping constant, 'c' I decided to use Cbar instead. You can call this whatever you want because it cancels out anyway.

@harrytaylor2479 3 жыл бұрын

First year engineering sent me here with no explanation as to what is going on. Where are these equations coming from? Is there an intermediate step that I have missed?

@Freeball99 3 жыл бұрын

This video is part of a playlist. There are 5 videos preceding this one which should offer more explanation. kzbin.info/aero/PL2ym2L69yzkZJ1fY3SQ1JCyvZIoJYXQGZ

@harrytaylor2479 3 жыл бұрын

@@Freeball99 thanks very much!

@sharonleung4546 9 жыл бұрын

I am not sure...why do you ignore the weight of block in the free body diagram?

@Freeball99 8 жыл бұрын

+Sharon leung there is no assumption that gravity is present. Another way to think of it is that the mass is moving along a horizontal, frictionless plane. It turns out that the inclusion of gravity simply changes the static equilibrium point (which is the point about which the oscillations occur), but does not change the natural frequencies nor modes of vibration of the system.