A police woman blows a whistle of frequency 500 Hz. A car speeds past her with a velocity of `72 kmh^(-1).` Find the charge in frequency heard by the driver of the car just as he passes the police woman. Velocity of sound = `350 ms^(-1)`.

Answer

To solve the problem of finding the change in frequency heard by the driver of the car as he passes the police woman, we will use the Doppler effect formula. The steps are as follows:

### Step 1: Convert the speed of the car from km/h to m/s.
The speed of the car is given as 72 km/h. We need to convert this to meters per second (m/s) using the conversion factor \(1 \text{ km/h} = \frac{1}{3.6} \text{ m/s}\).

\[ \text{Speed of car} = 72 \text{ km/h} \times \frac{1 \text{ m/s}}{3.6 \text{ km/h}} = 20 \text{ m/s} \]

### Step 2: Identify the parameters for the Doppler effect.
- Frequency of the whistle (source frequency, \(f\)) = 500 Hz
- Velocity of sound (\(v\)) = 350 m/s
- Velocity of the observer (car, \(v_0\)) = 20 m/s (approaching)
- Velocity of the source (police woman, \(v_s\)) = 0 m/s (stationary)

### Step 3: Calculate the frequency heard by the driver as the car approaches the police woman (\(f_1\)).
Using the Doppler effect formula for a moving observer and a stationary source:

\[ f_1 = f \frac{v + v_0}{v - v_s} \]

Substituting the values:

\[ f_1 = 500 \text{ Hz} \times \frac{350 + 20}{350 - 0} = 500 \text{ Hz} \times \frac{370}{350} \]

Calculating \(f_1\):

\[ f_1 = 500 \text{ Hz} \times 1.0571 \approx 528.57 \text{ Hz} \]

### Step 4: Calculate the frequency heard by the driver as the car moves away from the police woman (\(f_2\)).
Using the Doppler effect formula for a moving observer and a stationary source:

\[ f_2 = f \frac{v - v_0}{v - v_s} \]

Substituting the values:

\[ f_2 = 500 \text{ Hz} \times \frac{350 - 20}{350 - 0} = 500 \text{ Hz} \times \frac{330}{350} \]

Calculating \(f_2\):

\[ f_2 = 500 \text{ Hz} \times 0.9429 \approx 471.43 \text{ Hz} \]

### Step 5: Calculate the change in frequency (\(\Delta f\)).
The change in frequency is given by:

\[ \Delta f = f_1 - f_2 \]

Substituting the values:

\[ \Delta f = 528.57 \text{ Hz} - 471.43 \text{ Hz} \approx 57.14 \text{ Hz} \]

### Final Answer:
The change in frequency heard by the driver of the car is approximately **57.14 Hz**.

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