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

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.