= Solution
Each light-string round trip combines wall reflection $-1$ with junction reflection $-1/3$, multiplying the remaining pulse amplitude by $1/3$. The initial directly reflected heavy-string pulse has amplitude $A_i/3$. The subsequent outgoing heavy pulses have amplitudes
$$
A_n=-\frac89\left(\frac13\right)^{n-1}A_i,\qquad n=1,2,\ldots.
$$
They all have width $w$ in the same heavy medium. For nonoverlapping pulses their <energy> fractions therefore sum to
$$
\boxed{\frac{E_{\rm out}}{E_i}
=\frac19+\frac{64}{81}\sum_{n=0}^\infty\left(\frac19\right)^n
=\frac19+\frac89=1.}
$$
The first transmission leaves energy fraction $8/9$ in the light segment; each return encounter leaves only $1/9$ of the previous stored energy there, so this residual tends to zero.
For <reflection from a finite string segment with a fixed far end>, if pulses overlap, squared amplitudes cannot simply be added because of <interference>. The same result follows frequency by frequency: with round-trip phase $q=e^{2ik_lL_l}$, $L_l=2\,\mathrm m$, the total amplitude coefficient is
$$
R(q)=\frac13-\frac{8q/9}{1-q/3}=\frac{1/3-q}{1-q/3}.
$$
For real frequencies $|q|=1$, so $|1/3-q|=|1-q/3|$ and $|R|=1$. Thus the ideal lossless fixed-ended segment returns all incident finite-pulse energy even when individual echoes overlap; their relative phases must be retained.
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