@DoctorFastest
It looked like his string was a spring of some kind. But it works with a normal string, too; it doesn't have to be able to stretch. The only catch is that the string has to be heavy enough that the air resistance is negligible compared to the momentum, and that the string be flexible rather than stiff. In other words, air resistance makes it so this won't work for a piece of thread, and stiffness makes it so that it won't work with a thick cable. But for most strings, a good approximation.
@kamiltoktas4897
That line at 4:36 is flat as a ruler. Wow.
@suvratsharma8817
someone please tell me at about 1:11 when he tells about the shifting of the graph .......i dont get that how you get to know where the graph will shift or even if it will shift in any direction......im confused , somebody help ....please
@optimusimperat
brillian lecture
@RbtV92
alright thanks Mr. Lewin, now I have more questions than answers, just how I like it.
@shahramkhazaie
Dear friend, the term "d2y" that you have multiplied to both sides of your equation has nothing to do with dt2 (dt squared), i.e. d2y is not "dy squared", that is the second derivitive of y with respect of a variable that is not precised in your calculations ...
@RDash
I'm jealous of his students since they're being taught by the God of Chalk. we are mear mortal peasants in his presence
@hamsterpoop
You are just manipulating arbitrary symbols to reach an answer that you already know. Your wave equation contains no information about what "v" depends on. Without that, you can't use it to model any phenomena.
@GGRSC
awesome
@roygbiv146
thanks
@ashdeash
I am in love with u genius : )
@lovuami
v= dx/dt v2 = dx2/dt2 1/v2 = dt2/dx2 1/(dt2*v2)=1/dx2 d2y/(dt2@v2)=d2y/dx2 there is no need for any string how strange, kindly tell i am new to this
@chrisjohnston2043