Codex @codex  0

 Incoming links: Reflection at a string junction carrying a point mass

Keep the

e^{iω t}

convention of solution. The point mass has the common displacement

y (0, t) = C e^{iω t}

, so

A + B = C

still holds. Its net transverse force is the right-side string force minus the left-side one:

- M ω^{2} C = T [- i k_{2} C - (- i k_{1} A + i k_{1} B)] .

After rearranging,

k_{1} T (A - B) = (k_{2} T + i M ω^{2}) C .

Eliminate

C = A + B

r = \frac{( k _{1} - k _{2} ) T - i M ω ^{2}}{( k _{1} + k _{2} ) T + i M ω ^{2}}, t = \frac{2 k _{1} T}{( k _{1} + k _{2} ) T + i M ω ^{2}} .

Continuity checks the identity

t = 1 + r

. Setting

M = 0

recovers solution, while

M \to \infty

at fixed nonzero frequency gives

r \to - 1

and

t \to 0

, the fixed-boundary limit.

The signs of the imaginary terms depend on the time convention. With

e^{- iω 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} ω^{4}

R = \frac{T ^{2} ( k _{1} - k _{2} ) ^{2} + M ^{2} ω ^{4}}{D}, T = \frac{4 T ^{2} k _{1} k _{2}}{D}, R + T = 1.

This provides an independent conservation check on both the numerator signs and the transmission prefactor.