Past exam of the mathematical methods II course of the University of Oxford / 2017 / 10

For a short length $\delta x$, the kinetic energy is $\frac{1}{2}\mu y_t^2\delta x$. For small slope, its increase in length is $[\sqrt{1+y_x^2}-1]\delta x\simeq \frac{1}{2}{y}_x^2\delta x$, so its elastic potential energy is $\frac{1}{2}T{y}_x^2\delta x$. For a travelling wave $y=f(x-vt)$, $y_t=-v{f}^{\prime }$ and $y_x=f^{\prime }$. Using $v^2=T/\mu$ givesWriting a pulse as $f(x)=Ag(x/w)$ yields $E=(T{A}^2/w)\int [g^{\prime }(s){]}^{2}ds$. Thus energy is proportional to amplitude squared for a fixed shape and width. Width must also be retained when comparing pulses on different strings.

At fixed frequency $k_j=\omega /v_j$, so the displacement reflection and transmission coefficients areThe heavy-string speed is $v_h=20\mathrm{m}{\mathrm{s}}^{-1}$ and the light-string speed is $v_l=40\mathrm{m}{\mathrm{s}}^{-1}$. ThereforeA displacement amplitude greater than the incident amplitude does not mean energy is created: the transmitted pulse also has a different width and propagates in a different medium.

After the first encounter, the reflected pulse travels backwards through the heavy string with amplitude $A_i/3$ and width $w$. The transmitted pulse has amplitude $4A_i/3$ and speed $40\mathrm{m}{\mathrm{s}}^{-1}$. Its temporal duration is unchanged at the stationary junction, so its spatial width doubles to $2w$. Both pulses retain the original displacement sign.

Reflection at a fixed end has displacement coefficient $-1$. The light-string pulse therefore returns from the wall with amplitude $-4A_i/3$ and width $2w$. It is inverted and travels left at $40\mathrm{m}{\mathrm{s}}^{-1}$. Taking the first junction encounter as time zero, its centre reaches the wall after $2/40=0.05\mathrm{s}$ and returns to the junction after $0.10\mathrm{s}$; a finite pulse takes its own duration to complete each encounter.

At the return encounter, transmission from light to heavy multiplies the amplitude by $2/3$, giving a backward heavy-string pulse of amplitude $-8A_i/9$. Its speed halves, so its width returns to $w$. Reflection within the light segment multiplies the incoming amplitude by $-1/3$, giving a right-going pulse of amplitude $4A_i/9$ and width $2w$. The earlier $A_i/3$ reflected pulse is already farther along the heavy string.

The sketches show these as separate pulses in the short-pulse regime $w/20\ll 0.10\mathrm{s}$. For longer pulses the corresponding components overlap and must be superposed rather than interpreted as isolated visible humps.

Each light-string round trip combines wall reflection $-1$ with junction reflection $-1/3$, multiplying the remaining pulse amplitude by $1/3$. The initial directly reflected heavy-string pulse has amplitude $A_i/3$. The subsequent outgoing heavy pulses have amplitudesThey all have width $w$ in the same heavy medium. For nonoverlapping pulses their energy fractions therefore sum toThe first transmission leaves energy fraction $8/9$ in the light segment; each return encounter leaves only $1/9$ of the previous stored energy there, so this residual tends to zero.

For reflection from a finite string segment with a fixed far end, if pulses overlap, squared amplitudes cannot simply be added because of interference. The same result follows frequency by frequency: with round-trip phase $q=e^{2i{k}_l{L}_l}$, $L_l=2\mathrm{m}$, the total amplitude coefficient isFor real frequencies $|q|=1$, so $|1/3-q|=|1-q/3|$ and $|R|=1$. Thus the ideal lossless fixed-ended segment returns all incident finite-pulse energy even when individual echoes overlap; their relative phases must be retained.