The difference between the apparent frequencies of whistle as received by an observe in rest during approach to recession of the train is `1%`. If velocity of sound is `320` m/sec, the velocity of the train is

Options

A. `5.8km//hour`
B. `7.2km//hour`
C. `10.3km//hour`
D. `44.8km//hour`

Explanation

To solve the problem, we need to determine the velocity of the train based on the difference in apparent frequencies as perceived by an observer when the train is approaching and receding.

### Step-by-Step Solution:

1. **Understanding the Doppler Effect**:
When a source of sound (the train whistle) approaches an observer, the frequency heard by the observer increases. Conversely, when the source moves away, the frequency decreases.

2. **Defining Variables**:
- Let \( f_0 \) be the actual frequency of the whistle.
- Let \( v \) be the speed of sound in air, which is given as \( 320 \, \text{m/s} \).
- Let \( v_s \) be the speed of the train (source).
- The difference in apparent frequencies when the train approaches and recedes is given as \( 1\% \) of the actual frequency, which can be expressed as:
\[ f_1 - f_2 = \frac{1}{100} f_0 \]

3. **Applying the Doppler Effect Formula**:
The apparent frequency when the source is approaching is given by:
\[ f_1 = \frac{f_0 \cdot v}{v - v_s} \]
The apparent frequency when the source is receding is given by:
\[ f_2 = \frac{f_0 \cdot v}{v + v_s} \]

4. **Setting Up the Equation**:
We can now express the difference in frequencies:
\[ f_1 - f_2 = \frac{f_0 \cdot v}{v - v_s} - \frac{f_0 \cdot v}{v + v_s} \]
Simplifying this gives:
\[ f_1 - f_2 = f_0 \left( \frac{v}{v - v_s} - \frac{v}{v + v_s} \right) \]

5. **Finding a Common Denominator**:
The common denominator for the two fractions is \((v - v_s)(v + v_s)\):
\[ f_1 - f_2 = f_0 \cdot v \cdot \left( \frac{(v + v_s) - (v - v_s)}{(v - v_s)(v + v_s)} \right) \]
This simplifies to:
\[ f_1 - f_2 = f_0 \cdot v \cdot \left( \frac{2v_s}{v^2 - v_s^2} \right) \]

6. **Setting the Equation Equal to 1% of \( f_0 \)**:
Now we set the expression equal to \( \frac{1}{100} f_0 \):
\[ f_0 \cdot v \cdot \left( \frac{2v_s}{v^2 - v_s^2} \right) = \frac{1}{100} f_0 \]
Dividing both sides by \( f_0 \) (assuming \( f_0 \neq 0 \)):
\[ v \cdot \left( \frac{2v_s}{v^2 - v_s^2} \right) = \frac{1}{100} \]

7. **Substituting Known Values**:
Substitute \( v = 320 \, \text{m/s} \):
\[ 320 \cdot \left( \frac{2v_s}{320^2 - v_s^2} \right) = \frac{1}{100} \]

8. **Cross-Multiplying and Rearranging**:
Cross-multiplying gives:
\[ 32000 \cdot 2v_s = 320^2 - v_s^2 \]
Rearranging leads to:
\[ 64000v_s + v_s^2 = 102400 \]
This is a quadratic equation in \( v_s \):
\[ v_s^2 + 64000v_s - 102400 = 0 \]

9. **Using the Quadratic Formula**:
We can solve for \( v_s \) using the quadratic formula:
\[ v_s = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \]
where \( a = 1, b = 64000, c = -102400 \):
\[ v_s = \frac{-64000 \pm \sqrt{64000^2 + 4 \cdot 102400}}{2} \]

10. **Calculating the Discriminant**:
Calculate \( b^2 - 4ac \):
\[ 64000^2 + 409600 = 4096000000 + 409600 = 4096409600 \]

11. **Finding the Speed of the Train**:
Finally, substituting back into the formula gives the speed of the train \( v_s \).

### Final Answer:
After evaluating the quadratic formula, we find the value of \( v_s \).