= Solution
The <characteristic impedance of a stretched string> is $Z=T/c=\rho c$, since its <wave speed> obeys $c^2=T/\rho$. Put the endpoint at $x=0$ and use displacement <complex amplitudes> $A$ incident and $B$ reflected:
$$
\Psi=(Ae^{ikx}+Be^{-ikx})e^{-i\omega t},\qquad k=\omega/c.
$$
At the massless end, the string's transverse <force> $-T\Psi_x$ balances the damper's <force> $-\alpha\Psi_t$. Thus $T\Psi_x=-\alpha\Psi_t$, or $Z(A-B)=\alpha(A+B)$. This <viscous termination of a stretched string> has
$$
r=\frac BA=\frac{Z-\alpha}{Z+\alpha},\qquad \mathcal R=|r|^2.
$$
Half the incident mean <power> is reflected when $r=\pm1/\sqrt2$, giving \b[two positive solutions]:
$$
\boxed{\alpha=(3-2\sqrt2)\rho c\quad\text{or}\quad\alpha=(3+2\sqrt2)\rho c}.
$$
They are approximately $0.1716\rho c$ and $5.828\rho c$, in $\mathrm{kg\,s^{-1}}$. The smaller coefficient produces no reflected displacement-phase reversal; the larger produces a $\pi$ reversal. Both dissipate the remaining half of the incident <power>. Choosing the impedance-matched value $\alpha=\rho c$ would instead absorb all of it.
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