- #1
Javier Martin
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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.
General wave solution: u(x,t)=A·e(kx-wt)i+B·e(-kx-wt)i
Newton second law: F=ma
k=w/v
First I use Newton second law on each ring to entablish the contourn conditions, for x=0 and x=LFy=m·ay=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
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 which doesn't sound right to me for λ1=2·L/π
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=LFy=m·ay=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
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 which doesn't sound right to me for λ1=2·L/π