Burgers equation with sinusoidal bump initial data

Indeed, the characteristic curves $x = \phi(x_0) t + x_0$ along which $u = \phi(x_0)$ is constant are shown below:

Until characteristics intersect, the solution of the PDE is given by the method of characteristics, i.e. by solving $u = \phi(x-ut)$ numerically (I don't know any closed form expression). The breaking time $t_b$ where the solution becomes multi-valued can be computed as described in this post: $$ t_b = \frac{-1}{\inf \phi'} = \frac{1}{\pi} \approx 0.32 . $$ The speed of shock $\dot x_s$ satisfies the Rankine-Hugoniot condition $$ \dot x_s = \frac{1}{2}\big(\tilde u(x_s,t) + 0\big), \qquad x_s(1/\pi) = 1 $$ where $\tilde u(x_s,t)$ solves $\tilde u = \phi(x_s-\tilde ut)$. Hence, it is not easy to compute the shock trajectory $x_s(t)$ analytically in the present case. However, it is possible to derive some analytical expressions if the sinusoidal bump is replaced by a polynomial bump, e.g. the parabola $x\mapsto 4x(1-x)$ or the triangular function $x\mapsto 1-|2x-1|$ displayed below

This is more a comment than an answer, but not possible to edit with the figure in the comments section.

It is easier to see when the solution $u(x,t)$ becomes multivalued on the graph $u(x)$ at various $t$ than on the graph $x(t)$ at various $u$ :

Of course, the solution is on the form of implicit equation : $$u=\sin\left(\pi(x-t\,u) \right)$$ It is very easy to plot the above curves, without having $u(x,t)$ explicitly. The trick is explained in : plot solution that becomes multivalued

Plot the two branches : $\quad x(u)=\frac{1}{\pi}\sin^{-1}(u)+t\,u\quad$ and $\quad x(u)=\frac{1}{\pi}(\pi-\sin^{-1}(u))+t\,u$ .