= Solution
For $c>0$ and $l>0$, the zero-endpoint <Dirichlet boundary conditions> select the <Fourier sine basis> functions $\sin(n\pi x/l)$. <Separation of variables> gives temporal factors with angular <frequencies> $\omega_n=n\pi c/l$. Zero initial displacement removes the temporal cosine terms, giving
$$
y(x,t)=\sum_{n=1}^{\infty}a_n\sin(n\pi x/l)\sin(n\pi ct/l).
$$
Differentiating at $t=0$ and using <Fourier orthogonality> gives
$$
\boxed{a_n=\frac2{n\pi c}\int_0^lv(x)\sin(n\pi x/l)\,dx.}
$$
The evaluation property of the given initial velocity means $v(x)=\delta(x-l/2)$ in the sense of a <Dirac delta>, an idealized <impulsively struck fixed-end string>. Thus $a_n=2\sin(n\pi/2)/(n\pi c)$: every even mode vanishes, while the lowest two excited modes have $n=1,3$. Their ordinary <frequencies> and amplitude ratio are
$$
\boxed{f_1=\frac c{2l},\quad f_3=\frac{3c}{2l},\qquad |a_1|:|a_3|=3:1.}
$$
The signed coefficient ratio is $a_3/a_1=-1/3$. The impulse solution is interpreted as a <distributional weak solution>, rather than one with a smooth initial velocity.
\Image[/past-exam-of-the-mathematics-course-of-the-year-1-of-the-natural-sciences-course-of-the-university-of-cambridge/2008/paper-2-midpoint-impulse-modes.png]
{title=The two lowest string modes excited by a midpoint velocity impulse, with their signed relative amplitudes}
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