Thanks for the awesome video, I have a quick question though. I understand everything up to the point around 5 minutes in where you introduced gamma into the equation. In my textbook at this point in the explanation, the author solved for the leading constants A and B at t = 0, and then he got B=0 and A=x_0 where x(0) = x_0 is the starting position of the spring. The final equation he got was x(t)=(x_0)*cos(a*t) where a:=sqrt(k/M). I didn't see anything resembling gamma after this and I'm confused what it's purpose is in the equation. Thanks again for the help and let me know if you need me to clarify anything.
Professor Trefor Bazett, thank you for introducing Undamped Mechanical Vibrations and Hooke Law in Differential Equations. DR. Bazett, please correct the sign error in the characteristic equation. Professor Bazett, you have mr^2 - k equal zero instead of mr^2 +k equal zero.
I’m really confuse When we talking about the Characteristic Equation why there is a sign minus before the « k » ? Same for the i we know there some complex because there no y’ right ? Thanks for your help guys
Thank you Dr. For all your great educational videos , and for sure you have talent to deliver education and information. I have a question please : In your derivation you assumed that acceleration is in the positive x direction and thats why when we apply newton's law projecting on the positive x direction we get the equation m.r(2) + k =0 with two imaginery solutions , but what if we initially assume that the acceleration is in the negative x direction then when we apply newton's law projecting on the positive x direction we will get m.r(2) - k =0 with two real solutions ?? , what is wrong with my logic ?? , i am sure that iam missing something here !.. Thank you in advance
I'm not really sure tbh, but what I would do is see if you can find a list of prerequisites for the typical ME course in this subject, and see how many of them you can knock off on your own on youtube
@ 3.22- you have cleverly manipulated the real & distinct roots r1 and r2 into 'imaginary' ones and gone ahead. But actually the roots are real is'nt it? That is r1= +Sqrt(W_0 t) and r2= - Sqrt(W_0 t) for which the general solutions will be of the type x= c1. e^r1t + c2.e^r2t. The solution is purely real and there are no Cosines/Sines here. In fact the solutions are exponential type. This is the method shown in your earlier video (569- "Constant Coefficient ODEs: Real & Distinct vs Real & Repeated vs Complex Pair". Can you kindly explain please Professor?
where did the pythagorean theorem came from?? I mean why did the 2 undetermined coefficients had a geometric relationship suddenly being perpendicular?? just came out of nowhere
good video sir but how you have written x= Acoswt + B sinwt we got the roots -+iw right so on substitution using the eulars formula we get real part on adding (A+B)coswt,(A-B)sinwt will be imaginary
i’d recommend watching other videos in the playlist, but whenever we have a linear differential equation (DE where derivatives aren’t inside functions and their coefficients are functions of the independent variable) this is always a good guess. the reason is because taking the nth derivative of e^rt yields (r^n)e^rt, so you end up with a polynomial equation in terms of r when you divide out the e^rt which we know how to solve. though you probably either already know this now or have given up learning odes
it is to increase range of A and B values if we only have A = cos(gamma) and B = sin(gamma) their maximum and minimum values would be 1 and -1 respectively(since sin and cos max and min values are 1 and -1) multiplying that by a constant lets it have any value in the range (-infinity,infinity)
You multiply C/C to get cos gamma and sin gamma. Because you have A/C = cos gamma and B/C is sin gamma Rewriting in terms of gamma and keeping the C you have done what commenter before said
At 3:25, Equation has minus sign error. Only by considering plus sign you could get imaginary value of r.
Good catch, thanks! Yes the ODE is correct, then the minus sign appeared out of nowhere and immediately vanished again after. Oops:)
@@DrTrefor : the next video would be uploaded next week?
Wednesday:)
appeared and disappeared like a virtual particle
@@RahulSharma-oc2qd But I didn't understand what you mean by only a positive sign we can get an imaginary value of r?
Calling 'g' shift rather than 'phase' makes huge sense. Well done Prof.
Ah yes breakfast and math 😁
lol the best time to do math imo!
Dinner and math here. lol
Characteristic equation has a minus mistake 03:35
Good catch, thank you!
Watch Bazett spring into action!
lol i see what you did there:D
@@DrTrefor I fixed your name in the post. Apologies, mate.
Thanks for the awesome video, I have a quick question though. I understand everything up to the point around 5 minutes in where you introduced gamma into the equation. In my textbook at this point in the explanation, the author solved for the leading constants A and B at t = 0, and then he got B=0 and A=x_0 where x(0) = x_0 is the starting position of the spring. The final equation he got was x(t)=(x_0)*cos(a*t) where a:=sqrt(k/M). I didn't see anything resembling gamma after this and I'm confused what it's purpose is in the equation. Thanks again for the help and let me know if you need me to clarify anything.
Really great explanations! Can you make a video on the method of variation of parameters?
I do plan to!
Professor Trefor Bazett, thank you for introducing Undamped Mechanical Vibrations and Hooke Law in Differential Equations. DR. Bazett, please correct the sign error in the characteristic equation. Professor Bazett, you have mr^2 - k equal zero instead of mr^2 +k equal zero.
Sir can you do mathematical physics
I’m really confuse
When we talking about the Characteristic Equation why there is a sign minus before the « k » ?
Same for the i we know there some complex because there no y’ right ?
Thanks for your help guys
Thank you so much Dr. Bazett!!!! You have no idea how much your works benefit students from all around the world😭😭🙇🙇🙏🙏🙏
Thank you Dr. For all your great educational videos , and for sure you have talent to deliver education and information.
I have a question please :
In your derivation you assumed that acceleration is in the positive x direction and thats why when we apply newton's law projecting on the positive x direction we get the equation m.r(2) + k =0 with two imaginery solutions , but what if we initially assume that the acceleration is in the negative x direction then when we apply newton's law projecting on the positive x direction we will get m.r(2) - k =0 with two real solutions ?? , what is wrong with my logic ?? , i am sure that iam missing something here !..
Thank you in advance
Is there a course that goes into the theory of Controls without having to take a boatload of mechanical engineering courses?
I'm not really sure tbh, but what I would do is see if you can find a list of prerequisites for the typical ME course in this subject, and see how many of them you can knock off on your own on youtube
Undamped mechanical? More like "Phenomenal lectures that are unparalleled!" 👍
lmao ive seen you on other channels, keep it up
@@thegoofiestgoooberr Oh man, I intend to! In the meantime, feel free to join in the fun.
Thank you for your time .
My pleasure!
This was very helpfull thanks
U r amazing
@ 3.22- you have cleverly manipulated the real & distinct roots r1 and r2 into 'imaginary' ones and gone ahead. But actually the roots are real is'nt it? That is r1= +Sqrt(W_0 t) and r2= - Sqrt(W_0 t) for which the general solutions will be of the type x= c1. e^r1t + c2.e^r2t. The solution is purely real and there are no Cosines/Sines here. In fact the solutions are exponential type.
This is the method shown in your earlier video (569- "Constant Coefficient ODEs: Real & Distinct vs Real & Repeated vs Complex Pair". Can you kindly explain please Professor?
He made a sign mistake when writing the equation. The solutions are supposed to be imaginary
The characteristic equation should have +k, not -k
where did the pythagorean theorem came from?? I mean why did the 2 undetermined coefficients had a geometric relationship suddenly being perpendicular?? just came out of nowhere
ohhhhhh because of the trig function we assumed
Thank you very much sir 🔥🔥🙏
This time I am first ..XD
Nice one, I think so!!
Will being able to solve this problem bring my father back?
He can't promise miracles
Thanks for the explanation! My textbook always confuses me with so much ugly theory when it covers real world applications.
Thanks for this video ! Would love to see more of dyanamics videos
More on there way!
For god the third time, thank you for being out here as a life savior to my finals
good video sir but how you have written x= Acoswt + B sinwt we got the roots -+iw right so on substitution using the eulars formula we get real part on adding (A+B)coswt,(A-B)sinwt will be imaginary
thanks!
Just wondering. How/Why did you guess x=e^rt?
i’d recommend watching other videos in the playlist, but whenever we have a linear differential equation (DE where derivatives aren’t inside functions and their coefficients are functions of the independent variable) this is always a good guess. the reason is because taking the nth derivative of e^rt yields (r^n)e^rt, so you end up with a polynomial equation in terms of r when you divide out the e^rt which we know how to solve. though you probably either already know this now or have given up learning odes
waw... i salute you dr
Why are we introducing C at 4:30?
it is to increase range of A and B values
if we only have A = cos(gamma) and B = sin(gamma) their maximum and minimum values would be 1 and -1 respectively(since sin and cos max and min values are 1 and -1)
multiplying that by a constant lets it have any value in the range (-infinity,infinity)
You multiply C/C to get cos gamma and sin gamma. Because you have A/C = cos gamma and B/C is sin gamma
Rewriting in terms of gamma and keeping the C you have done what commenter before said
you are one the most underrated TH-cam professors! Thank you!
Why can I not just use either cos or sin from the start? Both are oscillating anyway?
I you guess a solution, how do you know it works? We should generally go from the physics to get an equation and only then solve it so we can be sure
@@DrTrefor OK, got you. I was wondering if I missed a mathematical reason for this. Thanks for your reply! :)