= Solution
For the <reflection of a string wave from a freely moving endpoint mass>, let $c=\sqrt{T/\mu}$, $k=\omega/c>0$ and use the complex <time-harmonic wave> convention
$$
y(x,t)=\operatorname{Re}\left\{A_i\left(e^{ikx}+R e^{-ikx}\right)e^{-i\omega t}\right\},
\qquad x\le0.
$$
The string pulls its endpoint toward negative $x$, so its transverse <force> on the attached <mass> is $-Ty_x(0,t)$. <Newton's second law> gives the <boundary condition> $my_{tt}(0,t)=-Ty_x(0,t)$. Substitution yields
$$
m\omega^2(1+R)=iTk(1-R),\qquad
\boxed{R=\frac{iTk-m\omega^2}{iTk+m\omega^2}
=\frac{Z+i m\omega}{Z-i m\omega},\quad Z=\frac Tc=\sqrt{T\mu}.}
$$
Here $Z$ is the <characteristic impedance of a stretched string>. Since the incident and reflected <waves> share the same string, their mean <power> ratio is the squared modulus of this displacement <reflection coefficient>:
$$
\boxed{\mathcal R=|R|^2=1.}
$$
There is \b[no remaining fraction of the mean incident energy]: the freely moving <mass> provides no <dissipation> and no onward transmission channel. It exchanges <kinetic energy> with the string during each cycle but has zero mean absorbed <power>. Indeed the <energy flux of a transverse string wave> satisfies
$$
-Ty_xy_t\big|_{x=0}
=my_{tt}y_t
=\frac{d}{dt}\left(\frac12my_t(0,t)^2\right),
$$
whose cycle average is zero. As checks, $m\to0$ gives $R=+1$, the <free-end reflection of a transverse string wave>, while $m\to\infty$ at fixed $\omega>0$ gives $R=-1$, the fixed-end limit.
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