Past exam of the mathematical methods II course of the University of Oxford / 2021 / 7 / iv

Put the fixed end at $x=0$, with the mechanical string occupying $x\le 0$. Let $H$ be the incident peak amplitude, set $u=x/L$ and $\tau =ct/L$, and take $t=0$ when the incident front first reaches the end. The symmetric triangular pulse in the source has initial profile $HT(u)$, whereThe fixed-end reflection of a triangular pulse follows from the Dirichlet boundary condition $y(0,t)=0$ and the superposition principle, giving the reflected displacementThe minus sign is the inversion on reflection. At $\tau =1/4$, the resulting profile is zero outside $[-3/4,0]$ and consists of the straight segments throughThe segment beside the wall is steeper because incident and inverted reflected pulses overlap there.

At $\tau =1/2$, the two triangular pulses coincide with opposite signs:The instantaneous sketch is the straight equilibrium line. This is destructive interference, not the disappearance of the wave energy. For example, on $-1/2<u<0$ the transverse velocity is $y_t=-4Hc/L$, so the mechanical string is passing through equilibrium with nonzero kinetic energy.

At $\tau =3/4$, the reflected pulse dominates the overlap. The total displacement is zero outside $[-3/4,0]$ and consists of straight segments throughThis is the inverted version of the quarter-reflected profile. The entire negative pulse has not yet separated from the incident pulse.

At $\tau =1$, the incident pulse has left the physical half-line, so only the inverted reflected pulse remains. Its straight segments pass throughIt subsequently travels toward decreasing $x$ at the original wave speed $c$, with unchanged width $L$ and peak amplitude magnitude $H$.