A policeman blows a whistle with a frequency of 500 Hz.A car approaches him with a velocity of `15 ms^(-1)`.Calculate the change in frequency as heard by the driver of the car as he passes the policeman. Speed of sound in air is `300 ms^(-1)`. a) 25 Hz b) 50 Hz c) 100 Hz d) 150 Hz

Options

A. 25 Hz
B. 50 Hz
C. 100 Hz
D. 150 Hz

Explanation

To solve the problem, we will follow these steps:

### Step 1: Identify the given values
- Frequency of the whistle (f) = 500 Hz
- Velocity of the car (v0) = 15 m/s
- Speed of sound in air (v) = 300 m/s

### Step 2: Calculate the frequency heard by the driver as the car approaches the policeman
When the car approaches the source of sound (the policeman), the formula for the observed frequency (f') is given by:
\[ f' = \frac{v + v_0}{v - v_s} f \]
Where:
- \( v \) = speed of sound in air
- \( v_0 \) = speed of the observer (the car)
- \( v_s \) = speed of the source (the policeman, which is at rest, so \( v_s = 0 \))

Substituting the values:
\[ f' = \frac{300 + 15}{300 - 0} \times 500 \]
\[ f' = \frac{315}{300} \times 500 \]
\[ f' = 1.05 \times 500 \]
\[ f' = 525 \, \text{Hz} \]

### Step 3: Calculate the frequency heard by the driver as the car moves away from the policeman
When the car moves away from the source of sound, the formula for the observed frequency (f'') is given by:
\[ f'' = \frac{v - v_0}{v - v_s} f \]

Substituting the values:
\[ f'' = \frac{300 - 15}{300 - 0} \times 500 \]
\[ f'' = \frac{285}{300} \times 500 \]
\[ f'' = 0.95 \times 500 \]
\[ f'' = 475 \, \text{Hz} \]

### Step 4: Calculate the change in frequency
The change in frequency (Δf) is given by:
\[ \Delta f = f' - f'' \]
Substituting the values:
\[ \Delta f = 525 - 475 \]
\[ \Delta f = 50 \, \text{Hz} \]

### Final Answer
The change in frequency as heard by the driver of the car as he passes the policeman is **50 Hz**.