= Solution
Assume the impulse point is in the interior, $0<a<l$. Expand the displacement in the fixed-end <Fourier sine series>
$$
y(x,t)=\sum_{n=1}^{\infty}T_n(t)\sin\frac{n\pi x}{l},\qquad \alpha_n=\frac{n\pi c}{l}.
$$
<Orthogonality> gives the initial velocity coefficient by applying the <Dirac delta> to the sine function:
$$
T_n(0)=0,\qquad T_n'(0)=\frac2l\int_0^l\delta(x-a)\sin\frac{n\pi x}{l}\,dx=\frac2l\sin\frac{n\pi a}{l}.
$$
Substitution in the <wave equation> yields $T_n''+2kT_n'+\alpha_n^2T_n=0$. For $k<\alpha_n$, its roots are $-k\pm i\omega_n$, where $\omega_n=\sqrt{\alpha_n^2-k^2}$. The two initial conditions determine
$$
T_n(t)=\frac2l\sin\frac{n\pi a}{l}\,e^{-kt}\frac{\sin(\omega_nt)}{\omega_n}.
$$
Hence the normalized <velocity-impulse Green function for a damped string> is
$$
\boxed{y(x,t)=\frac2l\sum_{n=1}^{\infty}\sin\frac{n\pi a}{l}\sin\frac{n\pi x}{l}\,e^{-kt}\frac{\sin\bigl(\sqrt{\alpha_n^2-k^2}\,t\bigr)}{\sqrt{\alpha_n^2-k^2}}.}
$$
For a mode with <critical damping>, use the limit $e^{-kt}t$ in place of the time factor. For $k>\alpha_n$, put $\eta_n=\sqrt{k^2-\alpha_n^2}$ and use $e^{-kt}\sinh(\eta_nt)/\eta_n$. These choices solve the same modal <ordinary differential equation> and initial conditions, so the formula covers arbitrary $k>0$. The impulse initial velocity is a <distribution>, so initial differentiation of the <series> is interpreted distributionally.
\b[The printed formula requires correction:] the factor depending on $n$ must lie inside the sum, and sine-mode normalization requires $2/l$, rather than $2$. The printed slash after the final sine is also stray. Direct differentiation at $t=0$ verifies the corrected normalization: the initial sine coefficients are exactly those of $\delta(x-a)$.
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