# Archived Standing Waves Proof for Open-Open Tubes

1. Sep 17, 2014

### alingy1

My teacher assigned me to prove open-closed and closed-closed standing waves patterns using math.
With closed-closed, it was fairly easy:

\begin{align} D(x=0,t)&=0\\ D(x=L,t)&=0=2A\sin(kx) \end{align}
Isolate $$L$$ to find that $$\lambda=2L/m$$.
Similarly for closed-open.
\begin{align} D(x=0,t)&=0\\ D(x=L,t)&=\pm 1=2A\sin(kx)\\ \lambda&=4L/m. \end{align}
I wanted to prove open-open too. But I am stuck.
I know that:
\begin{align} D(x=0,t)&=2A\cos(\omega t) \\ D(x=L,t)&=2A\cos(\omega t). \end{align}

Where do I go next?

2. Feb 5, 2016

### haruspex

I'm not quite sure what the D() function represents. Judging from the factor 2 on the right hand side it represents the range of 'y', e.g. range of movement of air. That being so, eqn 4 should read $A-(-A)=2A\sin(kL)$ (where, presumably, k and λ are the same thing).
Equations 6 and 7 seem to have gone off in a different direction, using the travelling wave equations at x=0. Applying equation 4 at x=0 and x=L would be the logical continuation.