I have the following exercise:
Verify that Ex=f(z-ct)+g(z+ct) is a solution of the one dimensional wave equation.
But I don't get what I should do, I mean, isn't Ex the formula of the one dimensional wave equation itself? U.U
How can I verify it?
Thank you.
6 answers
review section 5.2 of this: https://web.stanford.edu/class/math220a/handouts/waveequation1.pdf
Bobpursley coudl I get an email from you? I need to ask you something.
No emails. Our web safety issues don't allow that.
:( I could really use some assistance better than the provided from a input field of HTML xD, and my questions usually have symbols that require more precision in order for me to understand them.
Is there nothing we can do?
Is there nothing we can do?
Not now, we are working on a more flexible input field. Or a white board.
The equation is:
d^2y/dx^2 = (1/c^2 )[d^2y/dt^2]
here:
d^2x/dz^2 = (1/c^2 )[d^2x/dt^2] with the strange letters for displacements
I will just do the half that is moving right (z+ct)= constant or f(z-ct)
write as f(w) where w = z-ct
df/dz = df/dw * dw/dz = df/dw
d^2f/dz^2 = d^2f/dw^2 dw/dz = d^2f/dw^2
and then
df/dt = df/dw * dw/dt = df/dw (-c) ah hah !
d^2f/dt^2 =d^2f/dw^2 (-c)(-c) = c^2 d^2f/dw^2
so in the end
d^2f/dt^2 /c^2 = d^2f/dw^2 = d^2f/dz^2
which is what we had to show.
note that due to the (-c)(-c) = c*c
it will come out the same for x+ct as for x-ct
d^2y/dx^2 = (1/c^2 )[d^2y/dt^2]
here:
d^2x/dz^2 = (1/c^2 )[d^2x/dt^2] with the strange letters for displacements
I will just do the half that is moving right (z+ct)= constant or f(z-ct)
write as f(w) where w = z-ct
df/dz = df/dw * dw/dz = df/dw
d^2f/dz^2 = d^2f/dw^2 dw/dz = d^2f/dw^2
and then
df/dt = df/dw * dw/dt = df/dw (-c) ah hah !
d^2f/dt^2 =d^2f/dw^2 (-c)(-c) = c^2 d^2f/dw^2
so in the end
d^2f/dt^2 /c^2 = d^2f/dw^2 = d^2f/dz^2
which is what we had to show.
note that due to the (-c)(-c) = c*c
it will come out the same for x+ct as for x-ct
1 answer