Past exam of the mathematical methods II course of the University of Oxford / 2019 resit / 11 / c / Solution

Keep the $e^{i\omega t}$ convention of solution. The point mass has the common displacement $y(0,t)=C{e}^{i\omega t}$, so $A+B=C$ still holds. Its net transverse force is the right-side string force minus the left-side one:After rearranging,Eliminate $C=A+B$ to obtain the reflection at a string junction carrying a point mass: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 $e^{-i\omega t}$ 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 areThis provides an independent conservation check on both the numerator signs and the transmission prefactor.