Blowing a horn of 500 Hz frequency, a car A is moving in the opposite direction to car B. Both the cars are approaching ead1 other at `30"km h"^(-1)`. The person in the car B hears the horn of A which according to B has a frequency `v_A`. Given `30"km h"^(-1) = (25)/(3)"m s"^(-1)`. The frequency `v_A` as measured by B when they are approaching each other is

Options

A. 500 Hz
B. 526 Hz
C. less than 500 Hz
D. none of these

Explanation

To solve the problem, we will use the Doppler effect formula for sound waves. The formula for the apparent frequency (f') perceived by a listener when both the source and the observer are moving towards each other is given by:

\[ f' = \frac{(v + v_L)}{(v - v_S)} \cdot f \]

Where:
- \( f' \) = apparent frequency (frequency perceived by the listener)
- \( f \) = true frequency of the source (500 Hz in this case)
- \( v \) = speed of sound in air (approximately 330 m/s)
- \( v_L \) = speed of the listener (car B) towards the source (car A)
- \( v_S \) = speed of the source (car A) towards the listener (car B)

### Step 1: Convert the speed of the cars to m/s
Both cars are moving at 30 km/h, which is given as \( \frac{25}{3} \) m/s.

### Step 2: Assign values to the variables
- True frequency \( f = 500 \) Hz
- Speed of sound \( v \approx 330 \) m/s
- Speed of listener \( v_L = \frac{25}{3} \) m/s (moving towards the source, hence positive)
- Speed of source \( v_S = \frac{25}{3} \) m/s (moving towards the listener, hence also positive)

### Step 3: Substitute the values into the Doppler effect formula
Now we can substitute these values into the formula:

\[ f' = \frac{(330 + \frac{25}{3})}{(330 - \frac{25}{3})} \cdot 500 \]

### Step 4: Simplify the equation
First, calculate \( 330 + \frac{25}{3} \) and \( 330 - \frac{25}{3} \):

- Convert 330 to a fraction:
\[ 330 = \frac{990}{3} \]
- Now add and subtract:
\[ 330 + \frac{25}{3} = \frac{990 + 25}{3} = \frac{1015}{3} \]
\[ 330 - \frac{25}{3} = \frac{990 - 25}{3} = \frac{965}{3} \]

### Step 5: Substitute back into the formula
Now substitute these back into the formula:

\[ f' = \frac{\frac{1015}{3}}{\frac{965}{3}} \cdot 500 \]

### Step 6: Simplify further
The \( \frac{3}{3} \) cancels out:

\[ f' = \frac{1015}{965} \cdot 500 \]

### Step 7: Calculate the final frequency
Now calculate \( \frac{1015}{965} \):

\[ f' \approx 1.052 \cdot 500 \approx 526 \text{ Hz} \]

### Final Answer
The frequency \( v_A \) as measured by B when they are approaching each other is approximately **526 Hz**.

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