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

Take an ideal flexible mechanical string with constant tension $T>0$ and uniform linear mass density $\mu >0$. Let $x$ measure distance along its straight equilibrium position and let $y(x,t)$ be its small transverse displacement. Neglect damping, bending stiffness and external distributed forces, and assume $|y_x|\ll 1$.

A segment of reference length $\Delta x$ has mass $\mu \Delta x$. The endpoint tension forces have transverse resultantsince $\sin \theta \simeq \tan \theta =y_x$ for small slopes. Newton's second law, divided by $\Delta x$ and taken to the limit, givesHere $t$ is time, the subscripts denote partial derivatives, and $c$ is the wave speed. The transverse acceleration is $y_{tt}$; $y_{xx}$ describes the small-slope curvature responsible for the net transverse force.

Put $u=ct-x$ and assume $f$ has continuous second derivatives. The chain rule yieldsThus the wave equation is satisfied. A point of fixed wave phase $u=u_0$ has $x=ct-u_0$, so for $c>0$ this travelling wave moves in the $\boxed{\text{positive }x\text{ direction at speed }c}$.

For $\Delta t=\Delta x/c$,Equivalently $\boxed{y(x,t+\Delta t)=y(x-c\Delta t,t)}$. Thus every value of the displacement is translated by the same distance $c\Delta t$: the entire travelling wave profile keeps its shape. The wave speed is independent of the profile, as expected for this nondispersive wave equation.

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 $H$.