Q44E - University Physics with Modern Physics

Question

In Example $\mathbf{16}\mathbf{.}\mathbf{18}$  (Section $\mathbf{16}\mathbf{.}\mathbf{8}$ ), suppose the police car is moving away from the warehouse at  $\mathbf{20}\mathbf{m}/\mathbf{s}$ . What frequency does the driver of the police car hear reflected from the warehouse?

Step-by-Step Solution

Answer

The driver will hear a frequency of  $267\text{\hspace{0.17em}Hz}$.

Step 1:Given data

Speed of the car going away is  $20 \mathrm{m}/\mathrm{s}$

Step 2: Doppler’s effect

The Doppler’s effect is given by the formula, 

 ${\mathbf{f}}_L\mathbf{=}\frac{\mathbf{v}\mathbf{+}{\mathbf{v}}_L}{\mathbf{v}\mathbf{+}{\mathbf{v}}_S}{\mathbf{f}}_S$

where

• ${\mathbf{f}}_{\mathbf{L}}$  is the frequency of the wave that is heard by the listener.
• $\mathit{v}$  is the wave speed.
• ${\mathit{v}}_{\mathbf{L}}$  is the listener’s speed.
•  ${\mathit{v}}_{\mathbf{s}}$ is the sound source’s speed.
• ${\mathbf{f}}_{\mathbf{s}}$  is the frequency of the source of sound.

Step 3: Consider warehouse as a listener

$\begin{array}{c}{f}_{Warehouse}=\frac{v+v_L}{v+v_S}{f}_S\\ =\frac{344\text{\hspace{0.17em}\hspace{0.17em}m}/\text{s}+0}{344\text{\hspace{0.17em}m}/\text{s}+20\text{\hspace{0.17em}m}/\text{s}}\times 300\text{\hspace{0.17em}Hz}\\ =283.52\text{\hspace{0.17em}Hz}\end{array}$

Step 4: Consider the driver as listener

$\begin{array}{c}{f}_{driver}=\frac{v+v_L}{v+v_S}{f}_S\\ =\frac{344\text{\hspace{0.17em}m}/\text{s}-20\text{\hspace{0.17em}m}/\text{s}}{344\text{\hspace{0.17em}m}/\text{s}+0}\times 283.52\text{\hspace{0.17em}Hz}\\ =267\text{\hspace{0.17em}Hz}\end{array}$

So, the driver hears a frequency of  $267\text{\hspace{0.17em}Hz}$ .