# A partial solution to the wave equation

I have been presented with the one-dimensional wave equation:

$$\frac{\partial^2 y}{\partial x^2}=\frac1{c^2} \frac{\partial^2 y}{\partial t^2}$$

for a particular string of length $$L$$ that is fixed at two ends.

Initially, I had to Formulate the boundary conditions for this problem. I said that they were:

$$y(0, t) = y(L, t) = 0$$

Secondly, I had to separate the variables, etc etc to find the components to make the solution. Here I obtained, using $$y(x, t) = X(x)T(t)$$

$$X(x) = A\sin(px) + B\cos(px)$$ $$T(t) = A'\sin(pct) + B'\sin(pct)$$

Using the formulated boundary conditions on $$X(x)$$, I obtained that $$p_n = \frac {n\pi}{L}$$ and also that $$B = 0$$

Using this, I had to write down a general solution of the wave equation. This, I assume is as follows:

$$y(x,t) = \sum _{n=0}^{\infty }\:\big[C_n\sin(p_nct)+D_ncos(p_nct)\big]\sin(p_nx)$$ where $$C_n =AA'$$ and $$D_n = AB'$$

From here, I had to prove the orthonormality relation for $$\phi_n (x)=\sin(\frac{n\pi x}{L})$$, which was simple. For reference:

$$\int_{0}^{L} \phi_n(x)\phi_m(x)dx=\frac L2 \delta_{nm}$$ (I'm not 100% sure if this is relevant to the last part of the question)

And finally (now the part that i'm stuck with): "Now assume that the string is initially (at $$t=0$$) pulled by $$0.06$$ at $$x=\frac L5$$ and then released. Determine the corresponding partial solution of the wave equation.

I attempted to obtain the solution using $$y(\frac{L}{5}, 0)$$ which were the conditions given, to obtain: $$\sum_{n=1}^{\infty}\:D_n\sin(p_n\frac{L}{5})=\sum_{n=1}^{\infty}\:D_n\sin(\frac{n \pi}{5}) = 0.06$$ from my original solution.

At the initial moment, the velocity is $$0$$ leading to all $$C_n = 0$$.

Then my attempt to determine $$D_n$$ yielded $$\frac{3}{25}$$ (using the method akin to Fourier Series), instead of $$\frac{3}{4 \pi^2 n^2}$$ (the correct result, as per below).

I'm not sure why I'm not getting the correct answer for $$D_n$$

The final answer should be:

$$\sum_{n=1}^{\infty}\:\big[\frac{3}{4 \pi^2 n^2}\sin(\frac{\pi n}{5})\big]\sin(\frac{\pi n x}{L})\cos(\frac{\pi c n t}{L})$$

But I'm just very stuck at the moment (for finding the coefficient). Any hints and help will be appreciated!!

## 1 Answer

when the string is pulled 0.06 the shape of string will be

so you will have two equations which are: $$(y_1=0.3\frac{x}{L})$$and $$(y_2=-0.075\frac{x}{L}+{0.075})$$ so $$D_n=\frac{2}{L}\int_{0}^{0.2L}y_1\sin(\frac{n\pi x}{L})dx+\frac{2}{L}\int_{0.2L}^{L}y_2\sin(\frac{n\pi x}{L})dx=\sum_{n=1}^{\infty}\:\big[\frac{3}{4 \pi^2 n^2}\sin(\frac{\pi n}{5})\big]$$