For $f(t)=\sin\omega t$,
$$
\widehat f(p)=\frac{\omega}{p^2+\omega^2},
\qquad
\widehat y(x,p)=\frac{\omega e^{-px/c}}{p^2+\omega^2}.
$$
The <Bromwich inversion formula> gives
$$
y(x,t)=\frac{1}{2\pi i}\int_{\gamma-i\infty}^{\gamma+i\infty}
\frac{\omega e^{p(t-x/c)}}{p^2+\omega^2}\,dp.
$$
When $t>x/c$, close the contour to the left. The residues at $p=\pm i\omega$ sum to $\sin(\omega(t-x/c))$. When $t<x/c$, close it to the right, where there are no poles, and the integral vanishes. Hence
$$
\boxed{y(x,t)=
\begin{cases}
0,&t<x/c,\\
\sin\!\bigl(\omega(t-x/c)\bigr),&t\ge x/c.
\end{cases}}
$$
Equivalently, $y(x,t)=\Theta(t-x/c)\sin(\omega(t-x/c))$ in terms of the <Heaviside step function>. The boundary oscillation travels to the right at <wave speed> $c$ without reflection or dispersion; a point at $x$ begins moving after the propagation delay $x/c$.

Solution

Past exam of the mathematics course of the University of Cambridge / 2020 / ii / Paper 2 / 13E / ii / Solution

For a one-dimensional <wave equation> whose <principal symbol> is $\xi_t^2-c(y)^2\xi_y^2$, a travel-time coordinate is
$$
s(y)=\int^y\frac{dr}{c(r)}.
$$
Its two families of <characteristic curves> are $s(y)-t=\text{constant}$ and $s(y)+t=\text{constant}$ wherever the <wave speed> $c$ is positive.

Travel-time coordinate for a one-dimensional variable-speed wave equation

{title2=$c$}

The wave speed $c$ is the propagation speed in a wave equation such as $u_{tt}=c^2\nabla^2u$.


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