Past exam of the mathematical methods II course of the University of Oxford / 2017 resit / 11

For two equal-amplitude waves with the same frequency and wavenumber, choose their phases so thatThe spatial amplitude and temporal oscillation separate. Positions with $\cos kx=0$ are fixed nodes of a standing wave, while positions with $|\cos kx|=1$ are antinodes. Different initial phases merely shift this pattern in space and time.

A travelling wave has a phase profile that moves at its phase velocity; a standing wave has stationary nodes and each point oscillates with its fixed local amplitude. The two counterpropagating waves carry equal opposite time-averaged energy flux, so their superposition has zero net time-averaged energy transport. Instantaneous energy can still move within the standing pattern.

For a uniform string, let $c=\sqrt{T/\mu }$. A separated mode obeys $Y^{\prime \prime }+k^{2}Y=0$, $\omega =ck$. Fixed ends require $Y(0)=Y(L)=0$, soThe first three normal modes have one, two and three half-waves respectively. Their interior nodes of a standing wave are at $x=jL/n$ for $j=1,\dots ,n-1$.

A free end has zero transverse force, hence $T{Y}^{\prime }(L)=0$, rather than zero displacement. With $Y(0)=0$, write $Y=A\sin kx$. The remaining condition is $\cos kL=0$, givingThe first three shapes have one, three and five quarter-waves. Each has a node of a standing wave at the fixed end and an antinode at the free end; the mode frequencies are in the ratio $1:3:5$.

Free ends give $Y^{\prime }(0)=Y^{\prime }(L)=0$, so the oscillatory normal modes areThere is also the zero mode $k_0=0$, with uniform displacement $y_0(x,t)=a+bt$: a free string can translate without a restoring force. It is not a positive-frequency oscillation. The first three oscillatory shapes are $\cos (\pi x/L)$, $\cos (2\pi x/L)$ and $\cos (3\pi x/L)$, with an antinode at each end. The figure includes the uniform mode too, so that the zero-frequency limit is explicit.

At common tension, the lighter segment has three times the wave speed, so its wavenumber at the same frequency is $k_2=k_1/3$. The fixed left end and free right end selectLet $q=k_{1}L$. Since $k_2(4L-L)=q$, continuity of displacement and of transverse force at the junction givesEqual tension makes the second condition continuity of slope. A nontrivial pair of amplitudes requires the coefficient determinant to vanish:For positive frequency, $k_2/k_1=1/3$, soNeither $\sin q=0$ nor $\cos q=0$ gives a nonzero mode, as direct substitution into the two matching equations confirms. The positive roots are $q=\pi /3+n\pi$ or $q=2\pi /3+n\pi$ for $n\ge 0$. The first modes in the figure show the changing wavelength across the junction while retaining continuous displacement and slope. The zero-frequency limit must be solved separately; the fixed and free boundary conditions allow only the trivial static solution.