Damn, this is so beautiful. When the something as usual as waves are just infinite combinations of oscillations with fixed ends. When you can prove it with math, it just blows your mind.
Shouldn't the equation for 'k' at 48:05 include 'a'? sin(x) = 0 happens when x = nπ, so sin[(N+1)ka] happens when (N+1)ka = nπ. Solving for k gives k = nπ/[a(N+1)] Also, (assuming 'a' is 1) shouldn't k range from 1 to 2N+1? Since sin(x) hits 0 again at π, but it doesn't start repeating itself until 2π
Why is omega^2*A= A_tt? Aren't newtons laws complete derivativea? Why is it partial? Isn't A constant in the first coupled equations, so that the derivative only applies to the exponencial term? Why does the derivative operator apply now to A, even going from a total derivative to a partial one?
Making the system continuous means A is now a continuous function of x. To describe the amplitude of the system at any place at any time you need to describe it in terms of both x and t, so we are in multivariable calculus with the partial derivatives. That's the wave function!
Damn, this is so beautiful. When the something as usual as waves are just infinite combinations of oscillations with fixed ends. When you can prove it with math, it just blows your mind.
Shouldn't the equation for 'k' at 48:05 include 'a'?
sin(x) = 0 happens when x = nπ, so sin[(N+1)ka] happens when (N+1)ka = nπ. Solving for k gives
k = nπ/[a(N+1)]
Also, (assuming 'a' is 1) shouldn't k range from 1 to 2N+1? Since sin(x) hits 0 again at π, but it doesn't start repeating itself until 2π
k=2*pi/lambda and lambda||min=2*a so n
@@ashutoshpanigrahy7326 I am not able to understand your comment, can you please explain it again? Thanks
a is just a constant so I think there is no problem in answer because 2 sin (0) and sin(0) are same
There should definitely be an a, I think he just forgot it. Otherwise the boundary condition would not hold.
Why is omega^2*A= A_tt? Aren't newtons laws complete derivativea? Why is it partial? Isn't A constant in the first coupled equations, so that the derivative only applies to the exponencial term? Why does the derivative operator apply now to A, even going from a total derivative to a partial one?
I think instead of A there should be wave function.
Making the system continuous means A is now a continuous function of x. To describe the amplitude of the system at any place at any time you need to describe it in terms of both x and t, so we are in multivariable calculus with the partial derivatives. That's the wave function!
good job.i have exam day affter tommoro
❤
I didn't like the approximations, it is too stilted!