Let: $$y(s; t) =\sum\limits_{n=1}^{+\infty} B_n \cos\left(\dfrac{n\pi ct}{ L}\right) \sin\left(\dfrac{n\pi s}{L}\right),$$ be a solution of the vibrating string problem. Suppose that the string is further constrained at its midpoint, so that $y\left(\dfrac{L}{2}, t\right) = 0$ for all t. What condition does this impose on the coefficients $ B_n$
*Mark A. Pinsky " PDE and boundary-value problems with application." p.150 *
I think with $s=\dfrac{L}{2}$ we have $y(s,t)= \sum\limits_{n=1}^{+\infty} B_n \cos\left(\dfrac{n\pi ct}{ L}\right) $, now i don't known to find conditions for $B_n$
At the midpoint, $$ \sin{\left(\frac{n\pi}{L} \frac{L}{2}\right)} = \sin{\left( \frac{n\pi}{2} \right)} = \begin{cases} (-1)^{(n-1)/2} & n \text{ odd} \\ 0 & n \text{ even} \end{cases}, $$ so you get $$ y(L/2,t) = \sum_{n=1}^{\infty} (-1)^{k} B_{2k+1} \cos{\left( \frac{(2k+1)\pi c t}{L} \right)}. \tag{*} $$ A theorem about Fourier series tells you that this can only be zero for all $t$ if all the $B_{2k+1}$ are $0$. (Think of it like this: you're basically solving the same problem on the interval $[0,L/2]$ by imposing this condition)
Or, look at it a different way: the right-hand side is a periodic function with period $2\pi c/L$ (or something like that). The left-hand side is supposed to be $0$, which is also a periodic function with period $2\pi c/L$. Therefore you can find the Fourier coefficients by multiplying both sides by $\cos{(m\pi c t/L)} $, integrating and using the orthogonality relations: on the left, you always get zero. On the right, you get $$ \sum_{n=1}^{\infty} (-1)^{k} B_{2k+1} \int_0^{2\pi c/L}\cos{\left( \frac{(2k+1)\pi c t}{L} \right)} \cos{\left( \frac{m\pi c t}{L} \right)} \, dt = \frac{\pi c}{L}(-1)^m B_m, $$ if there is a $k$ in the sum with $2k+1=m$. It follows that all the $B_m$ in (*), i.e. those with odd $m$, must be zero.