What is the solution for a string under tension with given boundary conditions?

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Homework Statement



A string of mass M and length L is attached to walls at either end, and is
under tension T. The string is held at rest in the following shape: y=B for L/2<x<3L/4
and y= 0 for the rest of the string. The string is released
at t =0.(a) Find y(x, t). You may express your answer as an infinite
series, so long as you have defined all the symbols in your series

Homework Equations


y(x)=Bsinkx
k=npi/L L =length of string n=mode no
Y(x,y)=\SigmaBsin(kx)cos(wt) (sm over all n)

The Attempt at a Solution


i don't understand how to incorporate the boundary conditions in given equation...i have been trying and all i come up with are two equations sin(3npix/2L) for 3l/4 and sin(kx) for L/2 ...Dont know how to go any further
 
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maylie said:

Homework Statement



A string of mass M and length L is attached to walls at either end, and is
under tension T. The string is held at rest in the following shape: y=B for L/2<x<3L/4
and y= 0 for the rest of the string. The string is released
at t =0.(a) Find y(x, t). You may express your answer as an infinite
series, so long as you have defined all the symbols in your series

Homework Equations


y(x)=Bsinkx
k=npi/L L =length of string n=mode no
Y(x,y)=\SigmaBsin(kx)cos(wt) (sm over all n)

You probably mean$$
y(x,t)=\sum_{n=1}^\infty B_n \sin(\frac{n\pi} L x)\cos(\omega_n t)$$If you put ##t=0## and call your initial position of the string ##f(x)## you have$$
y(x,0)=f(x) = \sum_{n=1}^\infty B_n \sin(\frac{n\pi} L x)$$This is a Fourier series problem. Use that theory to figure out ##B_n##. We will just gloss over the fact that a continuous string can't be put in that initial shape.
 
are you sure we have to use Fourier analysis because we arent taught this yet we had to use the above equations !
 
Yes, I am sure. How did you get that solution in the first place? Did you use separation of variables? Fourier Series is standard material to go with that in problems like this. Maybe your teacher is trying to give you a "preview of coming attractions".
 
no i used brute force . got it! thank you so much :)
 
You're welcome. But you haven't "got it" until you know the formulas for ##B_n## and ##\omega_n##.
 
There are two things I don't understand about this problem. First, when finding the nth root of a number, there should in theory be n solutions. However, the formula produces n+1 roots. Here is how. The first root is simply ##\left(r\right)^{\left(\frac{1}{n}\right)}##. Then you multiply this first root by n additional expressions given by the formula, as you go through k=0,1,...n-1. So you end up with n+1 roots, which cannot be correct. Let me illustrate what I mean. For this...
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