Find the first four terms in the Fourier series for a solution of the wave equation

Question

The question is to find the first four terms in the Fourier series for $u(x,t), t>0$. It is for a plucked string of length L, has zero initial displacement (I.e. $u(x,0)=0, 0<x<L$) and initial velocity is given by:

$$u_t(x,0)=g(x)= \begin{cases} 0 & 0<x<\frac L4 \\ g_0 & \frac L4<x<\frac {3L}4 \\ 0 & \frac {3L}4<x<L \end{cases} $$

Can I just use the general solution of $u(x,t)$ for $u(x,0)=0$ which is

$$u(x,t)=\sum_{n=1}^\infty B_n \sin \frac{(n\pi ct)}{L} \sin \frac{(n\pi x)}{L}.$$

Then find

$$B_n=\frac 2{n \pi c} \int_0^L \sin \frac{(n\pi x)}{L} g(x) dx. $$

Then sub in values for n=1,..,4 ? It doesn't give me specific $c$ or $L$ values which is why I'm a bit confused and not sure if I'm approaching this right.

Thanks

Answer 1

Here is how you advance

$$ B_n=\frac 2{n \pi c} \int_0^L \sin \frac{(n\pi x)}{L} g(x) dx=\frac{2}{n \pi c} \int_{L/4}^{3L/4} g_0\sin \frac{(n\pi x)}{L} dx $$

$$\implies B_n = {\frac {2\,g_{{0}}L \left( \cos \left( \frac{n\pi}{4} \right) -\cos \left( \frac{3n\pi}{4} \right) \right) }{{n}^{2}{\pi }^{2}c}}.$$

Now, subs $B_n$ back in the highlighted equation and derive the desired number of terms.

Note: The case $n=0$ can be found from the formula by using the limit or just find it directly from the original in tegral.