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+### Statement
+
+$12.1.29.$ [Insert the problem statement]
+
+### Solution
+
+$12.1.29$ The frequency of a sinusoidal wave incident on a moving metal wall perpendicular
+to its surface changes by $\Delta$ during reflection. The initial frequency of
+the wave is $\nu_0$. Determine the wall speed.
+
+The change in frequency of the electromagnetic wave is due to the relativistic Doppler effect. The frequency of the reflected wave is:
+
+\begin{equation}
+ \nu = \nu_0 \frac{1+v/c}{1-v/c}
+\end{equation}
+
+We can obtain this result using the Lorentz transformation for energy and momentum:
+
+\begin{equation}
+ h \nu = \frac{h \nu_0 - (-v) h \nu_0 / c}{\sqrt{1-v^2/c^2}} \rightarrow \nu = \nu_0 \sqrt{\frac{1+v/c}{1-v/c}}
+\end{equation}
+
+After reflection in the moving frame, the wave will have the same frequency; coming back to the Earth frame:
+
+\begin{equation}
+ h \nu_1 = \frac{h \nu + v h \nu / c}{\sqrt{1-v^2/c^2}} \rightarrow \nu_1 = \nu_0 \frac{1+v/c}{1-v/c}
+\end{equation}
+
+So, the difference is:
+
+\begin{equation}
+ \Delta = \nu_1 - \nu_0 = \nu_0 \left(\frac{1+v/c}{1-v/c} - 1\right) = \nu_0 \frac{2 v/c}{1-v/c}
+\end{equation}
+
+And finally:
+
+\begin{equation}
+ v = \frac{c \Delta }{2 \nu_0 + \Delta}
+\end{equation}
+
+#### Answer
+
+[Insert a concise answer or boxed result]