1. ## Pizzicato (plucking string)

String attatched at x=0 and x=L, string is at rest, initially y is
1) y(0,t)=0
2) y(L,t)=0
Initial conditions: dy/dt=0 with t=0

y(x,0)= 2H*x/L 0<=x<=1/2L
2H(1-x/L) 1/2L<=x<=L

Just struggling to see where the boundary condition y(x,0) is generated from?
I guess it has something to do with the conditions before,
any help will be appreciated, many thanks.

2. ## Re: Pizzicato (plucking string)

Originally Posted by breitling
String attatched at x=0 and x=L, string is at rest, initially y is
1) y(0,t)=0
2) y(L,t)=0
Initial conditions: dy/dt=0 with t=0

y(x,0)= 2H*x/L 0<=x<=1/2L
2H(1-x/L) 1/2L<=x<=L

Just struggling to see where the boundary condition y(x,0) is generated from?
I guess it has something to do with the conditions before,
any help will be appreciated, many thanks.
Try plotting y(x,0) as a function of x. Does that plot look like what a pizzicato string looks like just before it's released?

3. ## Re: Pizzicato (plucking string)

It does yes, obviously you have the two ranges for x, but how to get the functions do you substitute into the derivative at t=0. Many thanks.

4. ## Re: Pizzicato (plucking string)

Originally Posted by breitling
It does yes, obviously you have the two ranges for x, but how to get the functions do you substitute into the derivative at t=0. Many thanks.
I'm not sure I know what you're asking. The boundary condition for y(x,0) comes from a model of what the string looks like just before release. I think that answers the question in the OP.

What do you mean by "how to get the functions"? Which functions? What do you mean by "how to get"? Are you asking how to solve the pde?