Recent content by ContagiousKnowledge

$$ R(r) = A\frac{e^{-ikr}}{r} $$

@Abhishek11235 $$ \frac{\partial^2 (rR)}{\partial r^2} = rRk^{2} $$ $$ Ak^{2}cos(\omega t - kr - \phi) = r[\frac{A}{r}cos(\omega t - kr - \phi)]k^{2} $$ $$ R(r) = \frac{A}{r}cos(\omega t - kr - \phi) $$

I'm still trying to figure out what you're asking me to do. What am I supposed to plug in for R(r)?

@Dr Transport I apologize if I'm slow, but I'm not what you're asking me to do. What do you mean by

I see why that is confusing. I was responding to @Dr Transport who said: I think Dr Transport meant that the solution to the wave equation takes the form $$ \frac{1}{r} \cos(\omega t - kr - \phi) $$ As I said, I have tried plugging $$ \frac{A}{r} \cos(\omega t - kr - \phi) = \psi $$ into the...

@Dr Transport @BvU @phyzguy @Abhishek11235 $$ 0 = \frac{A}{r}cos(\omega t - kr' - \phi) $$ $$ R(0) = \frac{A}{r}cos(\omega t + \phi) $$ Plugging $$ \frac{A}{r} \cos(\omega t - kr - \phi) $$ into the wave equation, we can verify that it is a solution and that $$ \omega = kv $$ Presumably, I can...

That makes sense. $$ R(0) = A + B $$ Which just tells us that R(0) is some constant. But you suggested that my solution was incorrect. COuld you elaborate on this? I agree that my solution doesn't seem to be getting me closer to an answer, but I don't know what I should do differently.

sinc(x) = sin(x)/x. Why isn't sinc(0) udnefined if we're dividing by zero? Or are you saying that sinc(x) approaches one as x approaches zero? Could you elaborate on why the solution is wrong?

Okay, I see that, as @phyzguy indicated, since the power of the wave is spread over a larger surface area as one moves farther from r=0, so the displacement will tend to decrease as one moves farther from r=0. Thus ψ(0,t) is finite and greater than zero provided the displacement is greater than...

Since the spherical wave equation is linear, the general solution is a summation of all normal modes. To find the particular solution for a given value of i, we can try using the method of separation of variables. $$ ψ(r,t)=R(r)T(t)ψ(r,t)=R(r)T(t) $$ Plug this separable solution into the...

@TSny $$ Acos(\phi)cos(\omega t) = Gcos(\omega t) $$ $$ G = Acos(\phi) $$ $$ -Asin(\omega t)sin(\phi) = H sin(\omega t) $$ $$ H = -Asin(\phi) $$ $$ \sqrt{G^{2} + H^{2}} = \sqrt{A^{2}cos^{2}(\phi) + A^{2}sin^{2}(\phi)} = A $$ Okay, I should have seen that; thank you for your help. Now, as for ϕ...

@TSny Again, I've been rearranging trig functions for a long time, but it hasn't been very productive. $$ Acos(\omega t + \phi) = A[cos(\omega t)cos(\phi) - sin(\omega t)sin(\phi)] = Gcos(\omega t) + Hsin(\omega t) $$ $$ A = \frac{Gcos(\omega t) + Hsin(\omega t)}{cos(\omega t)cos(\phi) -...

I've been playing with this algebra for a while and I'm afraid I don't see how to rewrite the expression in that form. $$ Gcos( \omega t - \phi) + Hsin( \omega t - \phi) = G[cos( \omega t)cos( \phi ) + sin( \omega t)sin( \phi )] + H[sin( \omega t)cos( \phi ) - cos( \omega t )sin( \phi )] $$...

No, I did not intend to use cosine for both terms. I am accustomed to using -ϕ to indicate a wave moving in the positive direction and +ϕ for a wave moving in the negative direction. As for deriving expressions for the amplitude and phase angle for the normal modes in terms of G and H, could...

Let's try inputting a solution of the following form into the two-dimensional wave equation: $$ \psi(x, y, t) = X(x)Y(y)T(t) $$ Solving using the method of separation of variables yields $$ \frac {v^2} {X(x)} \frac {\partial^2 X(x)} {\partial x^2} + \frac {v^2} {Y(y)} \frac {\partial^2 Y(y)}...