String attached to a massless ring in both ends

[Javier Martin]

First, sorry if there's grammar mistakes,english is not my native language.
1. Homework Statement
Find the normal modes of a string of length L with a massles ring ,free to move on the y-axis ,attached to each end.

Homework 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

The Attempt at a Solution

First I use Newton second law on each ring to entablish the contourn conditions, for x=0 and x=LF_y=m·a_y=T·sen(θ)=0 , which 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
k_n=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 which doesn't sound right to me for λ₁=2·L/π

 

[BvU]

Hello Javier,

which doesn't sound right to me

Why not ? it satisfies the boundary conditions

since I obtain the same k for a string with both ends fixed

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

 

[Javier Martin]

Got it now, thanks a lot