91Ó°ÊÓ

The Doppler’s effect is given by the formula, ${\mathit{f}}_{\mathbf{L}}\mathbf{=}\frac{\mathbf{v}\mathbf{+}{\mathbf{v}}_{\mathbf{L}}}{\mathbf{v}\mathbf{+}{\mathbf{v}}_{\mathbf{s}}}{\mathit{f}}_{\mathbf{s}}$, where

• ${\mathit{f}}_{\mathbf{L}}$is the frequency observed by the listener.
• vis the speed of sound.
• role="math" localid="1664343839235" ${\mathit{v}}_{\mathbf{L}}$is the speed of the listener.
• ${\mathit{v}}_{\mathbf{s}}$is the speed of the source of sound.
• ${\mathit{f}}_{\mathbf{s}}$is the frequency of the source of sound.

In front of the train, the relative velocity of the sound in the frame of the train is less than the speed of sound by the value of the velocity of the train.

$$\begin{gathered} \mathrm{λ}=\frac{v-v_{train}}{f_s} \\ \mathrm{λ}=\frac{344-25}{400} \\ \mathrm{λ}=0.7975m \end{gathered}$$

Behind the locomotive, the relative velocity of the sound in the frame of the train is higher than the speed of sound by the value of the velocity of the train.

$$\begin{gathered} \mathrm{λ}=\frac{v-v_{train}}{f_s} \\ \mathrm{λ}=\frac{344+25}{400} \\ \mathrm{λ}=0.9225m \end{gathered}$$

$$\begin{gathered} f_L=\frac{v+v_L}{v+v_s}{f}_s \\ {f}_L=\frac{344+0}{344-25}×400 \\ {f}_L=431.35Hz \end{gathered}$$

$$\begin{gathered} f_L=\frac{v+v_L}{v+v_s}{f}_s \\ {f}_L=\frac{344+0}{344+25}×400 \\ {f}_L=372.9Hz \end{gathered}$$