I think just use dimensional analysis.
The wave travels 2.5 m in 1.0 s so its velocity is 2.5 m/s. Divide this by 6 Hz (s^-1) and get 0.4167 m.
Formula might be \lambda \nu = v .
I need to take the \nabla^2 of x^2+y^2+z^2. This is how far I got
\begin{gather*}
\nabla^2 = \frac{d^2}{dr^2} + \frac{2}{r} \frac{d}{dr} + \frac{1}{r^2}(\frac{1}{sin^2\theta}\frac{d^2}{d\Phi^2} + \frac{1}{sin\theta}\frac{d}{d\theta} sin\theta\frac{d}{d\theta})\\
\nabla^2(r^2sin^2\theta...
For the following:
\begin{gather*}
\hat{\Omega} = \frac{d}{d\theta}sin \theta \frac{d}{d\theta}\\function = e^{i\theta}
\end{gather*}
Use the operator on the function and is it an eigenfunction of \hat{\Omega}?
Thanks. I don't think it is.
There is also another problem with...
The problem I did like this used
\begin{multline*}
\Sigma F_{y}=F_{b}-F_{t}-F_{g}\\F_{g}=ma
\end{multline*}
with gravity and tension in the negative y direction.
The problem says: confirm that the 1st excited state wavefunction of a 1D HO given by the Hermitian equation
H_{1}(y)= 2y y = \frac{x}{\alpha}, \alpha = (\frac{\hbar^2}{mk})^\frac{1}{4}
is a solution of the Schrodinger equation and that the energy is \frac{3}{2}\hbar\omega...
Can some1 help me solve a first energy level Schrodinger (\psi_{1})with a the Hermite polynomial and also show that it equals to \frac{3}{2}\hbar \omega?
I got as far as
\newcommand{\pd}[3]{ \frac{ \partial^{#3}{#1} }{ \partial {#2}^{#3} } }\frac{\hbar^2}{2 m} \ \pd{\Psi}{x}{2} + V \Psi =...