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

For opposite propagation directions on a uniform string, write the equal-frequency transverse waves asThe cosine addition identity givesThis is a standing wave: the spatial pattern is fixed while its amplitude oscillates in time. Nodes occur at the zeros of the first cosine and are separated by $\pi /k$, half a wavelength; antinodes have displacement amplitude $2A$. The two equal opposite travelling waves carry equal opposite average energy flux, so the net average transported power is zero.

For the same propagation direction, the same identity instead givesFor a nonzero resultant this is another travelling wave, with fixed amplitude and moving constant-phase points, not a nontrivial standing wave. If the prefactor vanishes, the waves cancel identically; the zero disturbance is a trivial exception rather than a standing oscillation. The argument assumes propagating waves with $k,\omega >0$ on the same uniform string.

The wave equation on a string gives $c_j=\sqrt{T/{\rho }_j}$ and $\omega =c_j{k}_j$, henceThe tension is common across the join, and the frequency is unchanged by reflection and transmission at the stationary boundary.

Use the complex time convention $e^{i\omega t}$. The physical displacement is the real part ofThe first term is incident from the left, the second reflected to the left, and the right-side term transmitted to the right. Continuity of displacement gives $A+B=C$. At a massless join, continuity of the transverse force $T{y}_x$ gives $k_1(A-B)=k_{2}C$. Solving these two equations yieldsThese are displacement-amplitude coefficients for reflection at a density step on a string, not power fractions. Since the mean incident power is $T\omega k_1|A{|}^2/2$, the reflected and transmitted power fractions are $|r{|}^2$ and $(k_2/k_1)|t{|}^2$. They sum to one, as expected for the lossless join.

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.