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 1


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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