Alternatively, professor, you can consider it a *bonus*, which is rewarding for those who has worked out one of the previous year's final where this situation happened in almost exactly the same manner.

Observe indeed $u$ must be spherical symmetric as is the boundary. Let $v=ru$, then \eqref{6-1}-\eqref{6-2}become, once identity \eqref{6-3} is known,

\begin{align} &v_{tt}-v_{rr} =0, \label{6-1'}\\ &v|_{t=0}=0, &&v_t|_{t=0}= \left\{\begin{aligned} &\sin(r) &&r<\pi,\\ &0 &&r\ge \pi,\end{aligned}\right.\qquad \label{6-2'} \end{align}

which is easily solved with a combined use of even continuation and D'Alembert's:

\begin{equation}v=\left\{\begin{aligned}&0&&r>t+\pi, \\ &\sin r\sin t&&0<r<-t+\pi, \\ &\frac{\cos(r-t)+1}{2}&&|t-\pi|<r<t+\pi,\\&0&&0<r<t-\pi.\end{aligned}\right.\qquad\label{6-4}\end{equation}

So then

\begin{equation}u=\left\{\begin{aligned}&0&&r>t+\pi, \\ &\frac{\sin r\sin t}{r}&&0<r<-t+\pi, \\ &\frac{\cos(r-t)+1}{2r}&&|t-\pi|<r<t+\pi,\\&0&&0<r<t-\pi.\end{aligned}\right.\qquad\label{6-5}\end{equation}