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So, let's say we have String from $0$ to $L$.

If we define the displacement as $\xi (x,t)$, the equation that governs the system is

$$\big(\dfrac{\partial ^2}{\partial t^2}-c^2\dfrac{\partial ^2}{\partial x^2}\big) \xi(x,t)=0$$

Our professor gave us initial conditions as following: $$\xi (x,t_o) = \sin\big(\dfrac{2 \pi x}{L}\big)$$ and $$\dot{\xi} (x,t_o) = \sin^2\big(\dfrac{2 \pi x}{L}\big)$$

But from what I have studied online: http://farside.ph.utexas.edu/teaching/315/Waves/node24.html

the most general solution for a stationary wave is $$\xi_ν = \xi_o \sin(ν\pi \dfrac{x}{L})\cos(\omega_ν t +\phi_ν)$$

from which I can't relate to my Initial Condition.

Since this is a Homework-Related question and I know it's not right and polite to ask for exact solutions, ANY advice to guide me to the solution is MUCH appreciated

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  • $\begingroup$ You need to show us what you've done to solve the problem. And $\dot{\xi} (x,t_o) = \sin^2\big(\dfrac{2 \pi x}{L}\big)$ makes no sense at all: to what variable is $\xi(x,t)$ derived? That should be a partial derivative ($\partial$) to either $x$ or $t$. $\endgroup$ – Gert Nov 6 '17 at 20:45
  • $\begingroup$ By $\dot{\xi} $ I meant partial derivative to t. I couldn't really find any solution, and I don't really get how the Initial Velocity is inconsistent with the boundary conditions. @jcandy $\endgroup$ – user174411 Nov 6 '17 at 21:25
  • $\begingroup$ @Gert I mean partial derivative to t. And can't solve it since ALL solutions online suggest $$\xi(x,t) = X(x) \cdot T(t) $$ which can't satisfy the initial velocity AND displacement at once. $\endgroup$ – user174411 Nov 6 '17 at 21:27
  • $\begingroup$ @MichaelSeifert My mistake. There is no problem. $\endgroup$ – jcandy Nov 6 '17 at 21:38
  • $\begingroup$ $\xi(x,t) = X(x) \cdot T(t)$, yes, because that's how it's DONE! Have a snifter of this: sciencemadness.org/talk/… (I'm the author). $\endgroup$ – Gert Nov 6 '17 at 22:29
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You make the statement that

[T]he most general solution for a stationary wave is $$\xi_ν = \xi_o \sin(ν\pi \dfrac{x}{L})\cos(\omega_ν t +\phi_ν).$$

But this isn't quite true. What you have here are the so-called modes of the string; the most general solution for a stationary wave on a string is a superposition (possibly infinite) of these fundamental modes: $$ \xi(x,t) = \sum_{\nu=1}^\infty \xi_{o\nu} \sin(ν\pi \dfrac{x}{L})\cos(\omega_ν t +\phi_ν), $$ where the $\xi_{o\nu}$ and $\phi_\nu$ are constants to be determined. In particular, assuming that $\dot{\xi}$ means $\partial_t \xi$, writing the solution in this form implies that $$ \xi(x,0) = \sum_{\nu=1}^\infty \xi_{o\nu} \sin(ν\pi \dfrac{x}{L})\cos(\phi_ν) $$ and $$ \dot{\xi}(x,0) = -\sum_{\nu=1}^\infty \xi_{o\nu} \omega_\nu \sin(ν\pi \dfrac{x}{L})\sin(\phi_ν). $$ In principle, if one knows the functions $\xi(x,0)$ and $\dot{\xi}(x,t)$, the right-hand side of each equation is then a Fourier series for these functions; and one can use the techniques of Fourier analysis to determine the coefficients $\xi_{0\nu}$ and $\phi_\nu$.

In practice, BTW, this form for the Fourier series is going to make your problem hard to solve. An equivalent form for this series is the following: $$ \xi(x,t) = \sum_{\nu=1}^\infty \sin(ν\pi \dfrac{x}{L})\left( A_\nu \cos(\omega_ν t) + B_\nu \sin(\omega_\nu t) \right). $$ I would encourage you to (A) satisfy yourself that these two forms are equivalent, and (B) try applying the above logic to this form to solve your problem. The algebra will be much simpler than trying to solve for $\xi_{0\nu}$ and $\sin \phi_\nu$.

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