Harmonic oscillator: Differential equation

[@prathamnaik8501]

No where literally nowhere anybody explained HO in this way. He gave reasons for every single step which I always wondered why we do like that! Pure beauty.

[@brandonhughes645]

I often wonder if physics at top universities are easier because I find myself constantly coming here for a better explaination. Even the lecture notes online help me with my tutorials and assignments

[@beoptimistic5853]

kzbin.info/www/bejne/joKsk6FobMmCoKc 💐

[@vinyltherapy9410]

our professors have no excuse, this guys first language is undoubtedly not English and elucidates these concepts better than a native speaker!

[@dirac1425]

Vaya, en verdad es una inspiración este hombre, pensar que estudio en la uni, ¡orgullo peruano!

[@biajidejajchowdhury1058]

Very different way of introducing quantum mechanics- I liked it.

[@matrixate]

Talk about clear and concise.

[@김윤성-w3g]

Much better than the Griffiths explaination! Thanks

[@michaellam4082]

dang, didn't realize until now he actually wrote the schrodinger's equation in a much cleaner fashion that's mathematically this much easier to read, and I can see the concepts this much more easily! Not the most important when it comes to solving the problem but in terms of arithmetics things get cleaned up very nicely!

[@clairemuhonja4844]

What are the mechanisms responsible for energy loss of a harmonic ascillator

[@rogermarceloramirezramirez6120]

El buen Barton :")

[@zacharythatcher7328]

Can anyone expand upon the statement around 3:40 that the first derivative in the Taylor series going to zero at a minimum will leave the Taylor approximation of the potential looking like a quadratic? Is this just assuming that cubed and higher terms fall off fast? Why can you assume that?

[@sangraampatwardhan1573]

We can assume that when we are considering a very small interval

[@sangraampatwardhan1573]

Can someone please explain why a^2=h/mw is actually used? It's obtained by equating the units right? So the correct form must be a^2 = r. h/mw right, where r is some number?

[@yyc3491]

your 'r' is actually the unitless 'u' which is extracted out. considering x now equals a·u.

[@ShubhamSingh-lz1mb]

[E]=[h^2/ma^2] = [mw^2x^2]. Equate the two expressions for [E]. You will get h^2/m^2w^2 = x^2.a^2 . As we talking dimensionally, x and a have the same dimensions of length. Therefore they can be treated as equal and the RHS becomes a^4, and therefore a^2 = h/mw.

[@omarfarukh8200]

sir how do you got h cross / m omega

[@beoptimistic5853]

kzbin.info/www/bejne/joKsk6FobMmCoKc 💐

[@ShubhamSingh-lz1mb]

[E]=[h^2/ma^2] = [mw^2x^2]. Equate the two expressions for [E]. You will get h^2/m^2w^2 = x^2.a^2 . As we talking dimensionally, x and a have the same dimensions of length. Therefore they can be treated as equal and the RHS becomes a^4, and therefore a^2 = h/mw.

[@kelvinadimas8851]

@@ShubhamSingh-lz1mb whats the tradeoff from equating [h^2/ma^2] = [mw^2x^2], we know the dimension is energy but certainly h^2 posses a value since its a constant, and we know dimension shouldnt have a definite value (its like a variable)

[@drscott1]

👍🏼

[@shyamtripathi6817]

What about this: kzbin.info/www/bejne/fITPdqSPpJ6AeqM This MIT professor vs that indian professor.