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!!


when the string is pulled 0.06 the shape of string will be enter image description here

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 $$D_n=\frac{2}{L}\int_{0}^{0.2L}y_1\sin(\frac{n\pi x}{L})dx+\frac{2}{L}\int_{0.2L}^{L}y_2\sin(\frac{n\pi x}{L})dx=\sum_{n=1}^{\infty}\:\big[\frac{3}{4 \pi^2 n^2}\sin(\frac{\pi n}{5})\big]$$


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