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A string of length $2L$ is fixed at both ends. The displacements on it satisfy the equation ${\partial^2 y\over \partial t^2}=\nu{\partial^2 y\over \partial x^2}$ . Also, $y(x,0)=0$ and for $t<0$, it oscillates in its fundamental mode. At $t=0$, the change in ${\partial y \over \partial t}=c\delta(x-L)$. How do I find $y(x,t)$ for $t>0$?

Thanks.

I know the form of the general solution to the (free) wave equation, I just don't know how to apply boundary conditions. It would be great if someone would kindly elaborate.

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Fixed at both ends means that $y(0,t)=y(L,t)=0$ for all $t$. This will result in solutions that are sine waves with periods that are integer fractions of $2L$. With no dissipation, each mode will oscillate forever. If you find the set of modes, you can find the velocity at each point when the string passes through $x=0$. The initial condition gives you the velocity at each point at $t=0$ and at that time $x=0$. So you need to find a superposition of the modes that have the given velocity at $t=0$.

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  • $\begingroup$ Thanks, Ross. So I know that the general solution is $y(x,t)=\sum_{n=1}^\infty \sin({n\pi x \over 2L})(a_n \cos({n\pi \nu t\over 2L})+b_n \sin({n\pi \nu t\over 2L}))$. But I am still confused about the boundary condition. Isn't $\delta(x-L)=\infty$ at $x=L$? $\endgroup$ – Alex Nov 1 '11 at 21:09
  • $\begingroup$ @Alex: yes. Usually you would just be given ${\partial y \over \partial t}$ at $t=0^+$ and I would take it that way: $y'(L,0^+)=c, y'(x,0^+)=0 \text{for} x \ne L$, but that is not the statement you have. $\endgroup$ – Ross Millikan Nov 1 '11 at 21:54
  • $\begingroup$ Thanks, Ross, so how could I impose the BC to find the wave function? Incidentally if I found the form of the wave for $t<0$ (fundamental mode) and I am given that $y(x,0)=0$. Does this mean that I can find coefficients of the $t<0$ wave using this? Or does the wave no longer hold? Thanks again! :) $\endgroup$ – Alex Nov 1 '11 at 22:05
  • $\begingroup$ @Alex: You are given it oscillates in its fundamental mode for $t<0$ and $y(x,0)=0$, which says $a_n=0$ and $b_n=0 \text{for} n>1$. I don't see how you find $b_0$ for $t<0$. The fixed BC were used to find the sine expansion in your earlier comment. For $t>0$ you Fourier analyze the motion at $t=0^+$ to find the $b_n$. The $a_n$ are all $0$ because $y(x,0)=0$ $\endgroup$ – Ross Millikan Nov 1 '11 at 22:16
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If I understand correctly, this problem is about point force $\delta(t)\delta(x-L)$ acting on the string. This is probably suggested by the word "change" in the question.

Indeed, if we integrate wave equation with r.h.s. $\delta(t)f(x)$ in time over $[-\epsilon, \epsilon]$ we can get in the limit $\epsilon\to0$ that there is a jump in ${\partial y}/{\partial t}$ at $t=0$: $$ \left.\frac{\partial y}{\partial t}\right|_{t=-0}^{t=+0}=f(x). $$

That is we have $$\left.\frac{\partial y}{\partial t}\right|_{t=+0}=c\delta(x)+\left.\frac{\partial y}{\partial t}\right|_{t=-0}$$

By linearity, total solution will be sum of the one existed for $t<0$ and the one generated by the point force impact at $t=0$.

For $t<0$ the string oscillates with its fundamental mode and $y(0,x)=0$, thus for $t<0$ we have $$ y=A \sin{\omega t} \sin(kx) $$ where $\omega=\pi\nu/2L$ and $k = \pi/2L$.

For the solution generated by the force solution can be written as on the linked image page 114 from Morse, Ingard "Theoretical Acoustics", 1968

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