Q46E - University Physics with Modern Physics

Question

A railroad train is traveling at 25m/s in still air. The frequency of the note emitted by the locomotive whistle is 400Hz. What is the wavelength of the sound waves (a) in front of the locomotive and (b) behind the locomotive? What is the frequency of the sound heard by a stationary listener (c) in front of the locomotive and (d) behind the locomotive?

Step-by-Step Solution

Answer

a) $\lambda =0.798m$

b) $\lambda =0.922m$

c) $f=431Hz$

d) $f=373Hz$

Step 1: Doppler’s effect

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.
• v is the speed of sound.
• ${\mathit{v}}_{\mathbf{L}}$ 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.

Step 2: Calculate wavelength in front of the locomotive

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.

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

Step 3: Calculate wavelength behind the locomotive

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.

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

Step 4: Calculate frequency when train is moving towards the listener

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

Step 5: Calculate frequency when train is moving away from the listener

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