= Solution
Keep the $e^{i\omega t}$ convention of <11/b/solution>. The point mass has the common displacement $y(0,t)=Ce^{i\omega t}$, so $A+B=C$ still holds. Its net transverse <force> is the right-side string force minus the left-side one:
$$
-M\omega^2C=T\big[-ik_2C-(-ik_1A+ik_1B)\big].
$$
After rearranging,
$$
k_1T(A-B)=(k_2T+iM\omega^2)C.
$$
Eliminate $C=A+B$ to obtain the <reflection at a string junction carrying a point mass>:
$$
\boxed{r=\frac{(k_1-k_2)T-iM\omega^2}{(k_1+k_2)T+iM\omega^2},\qquad
t=\frac{2k_1T}{(k_1+k_2)T+iM\omega^2}.}
$$
Continuity checks the identity $t=1+r$. Setting $M=0$ recovers <11/b/solution>, while $M\to\infty$ at fixed nonzero frequency gives $r\to-1$ and $t\to0$, the fixed-boundary limit.
The signs of the imaginary terms depend on the time convention. With $e^{-i\omega t}$ and conjugate spatial conventions, both coefficients are complex conjugated; the source's displayed signs agree with the convention used here.
There is no damping in the point mass, so its cycle-averaged absorbed <power> is zero. Writing $D=T^2(k_1+k_2)^2+M^2\omega^4$, the power fractions are
$$
\mathcal R=\frac{T^2(k_1-k_2)^2+M^2\omega^4}{D},\qquad
\mathcal T=\frac{4T^2k_1k_2}{D},\qquad\mathcal R+\mathcal T=1.
$$
This provides an independent conservation check on both the numerator signs and the transmission prefactor.
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