# String attached to a massless ring in both ends

1. Jun 13, 2017

### Javier Martin

First, sorry if theres grammar mistakes,english is not my native language.
1. The problem statement, all variables and given/known data

Find the normal modes of a string of lenght L with a massles ring ,free to move on the y axis ,attached to each end.

2. Relevant equations
General wave solution: u(x,t)=A·e(kx-wt)i+B·e(-kx-wt)i
Newton second law: F=ma
k=w/v
3. The attempt at a solution
First I use Newton second law on each ring to entablish the contourn conditions, for x=0 and x=L

Fy=m·ay=T·sen(θ)=0 , wich is 0 for being a massless ring
(θ is the angle the string forms at x=0 with the x axis)
for small values of θ we can take tanθ instead of sinθ, and since the definition of derivate is dy/dx=tanθ replacing above for y =u(x,t) and evaluated in x=0 we obtain
∂u/∂x (on x=0)=0 and the same for x=l

Now If we suppose a wave solution
u(x,t)=A·e(kx-wt)i+B·e(-kx-wt)i
and aply the CC I obtain
kn=n·π/l and u(x,t)=A·e-iwt·2cos(n·π·x/L)

My doubt is if this is a correct answer, since I obtain the same k for a string with both ends fixed. Also for n=1
u(0,t)=A and u(L,t)=-A wich doesnt sound right to me for λ1=2·L/π

2. Jun 13, 2017

### BvU

Hello Javier,
Why not ? it satisfies the boundary conditions

Yes, only there you have a sine instead of a cosine. Google 'normal modes open pipe' and check a few pictures.

3. Jun 13, 2017

### Javier Martin

Got it now, thanks a lot