A traffic policeman standing on a road sounds a whistle emitting the main frequency of 2.00 kHz. What could be the apparent frequency heard by a scooter driver approaching the policeman at a speed of `36.0km h^(-1) ?` Speed of sound in air `=340 m s^(-1).`

Explanation

To solve the problem of finding the apparent frequency heard by a scooter driver approaching a traffic policeman who is sounding a whistle, we can follow these steps:

### Step-by-Step Solution:

1. **Identify Given Data:**
- Main frequency of the whistle (f₀) = 2.00 kHz = 2000 Hz
- Speed of the scooter driver (vₒ) = 36.0 km/h
- Speed of sound in air (v) = 340 m/s

2. **Convert Speed of the Scooter Driver:**
- Convert the speed from km/h to m/s:
\[ vₒ = 36.0 \text{ km/h} \times \frac{1000 \text{ m}}{1 \text{ km}} \times \frac{1 \text{ h}}{3600 \text{ s}} = 10 \text{ m/s} \]

3. **Apply the Doppler Effect Formula:**
- The formula for the apparent frequency (f) when the source is stationary and the observer is moving towards the source is:
\[ f = f₀ \times \frac{v + vₒ}{v} \]
- Here, \(f₀\) is the original frequency, \(v\) is the speed of sound, and \(vₒ\) is the speed of the observer (scooter driver).

4. **Substitute the Values into the Formula:**
- Substitute \(f₀ = 2000 \text{ Hz}\), \(v = 340 \text{ m/s}\), and \(vₒ = 10 \text{ m/s}\):
\[ f = 2000 \text{ Hz} \times \frac{340 \text{ m/s} + 10 \text{ m/s}}{340 \text{ m/s}} \]
\[ f = 2000 \text{ Hz} \times \frac{350 \text{ m/s}}{340 \text{ m/s}} \]

5. **Calculate the Apparent Frequency:**
- Now calculate the fraction:
\[ f = 2000 \text{ Hz} \times 1.0294 \approx 2058.82 \text{ Hz} \]
- Converting back to kHz:
\[ f \approx 2.06 \text{ kHz} \]

6. **Final Answer:**
- The apparent frequency heard by the scooter driver is approximately **2.06 kHz**.