A train blowintg its whistle moves with constant - speed on a straight tack towards obsever and then crosses him . If the ratio and difference between the actual and apparent frequencies be 3:2 in the two cases , then the speed of train is [v is speed of sound ]

Options

A. `(2v)/(3) `
B. `(v)/(5)`
C. `(v)/(3)`
D. `(3v)/(2)`

Answer

To solve the problem, we will use the Doppler effect formulas for sound. The situation involves a train moving towards an observer and then away from the observer. We are given the ratio of the apparent frequencies when the train approaches and when it departs.

### Step-by-Step Solution:

1. **Identify Variables**:
- Let \( f \) be the actual frequency of the whistle.
- Let \( v \) be the speed of sound.
- Let \( v_s \) be the speed of the train.
- The apparent frequency when the train is approaching the observer is \( f_1 \).
- The apparent frequency when the train is moving away from the observer is \( f_2 \).

2. **Doppler Effect Formulas**:
- When the train approaches the observer:
\[ f_1 = f \frac{v}{v - v_s} \]
- When the train is moving away from the observer:
\[ f_2 = f \frac{v}{v + v_s} \]

3. **Set Up the Ratio**:
- We are given that the ratio of the apparent frequencies is:
\[ \frac{f_1}{f_2} = \frac{3}{2} \]
- Substitute the expressions for \( f_1 \) and \( f_2 \):
\[ \frac{f \frac{v}{v - v_s}}{f \frac{v}{v + v_s}} = \frac{3}{2} \]

4. **Simplify the Equation**:
- The \( f \) and \( v \) cancel out:
\[ \frac{v}{v - v_s} \cdot \frac{v + v_s}{v} = \frac{3}{2} \]
- This simplifies to:
\[ \frac{v + v_s}{v - v_s} = \frac{3}{2} \]

5. **Cross Multiply**:
- Cross multiplying gives:
\[ 2(v + v_s) = 3(v - v_s) \]

6. **Expand and Rearrange**:
- Expanding both sides:
\[ 2v + 2v_s = 3v - 3v_s \]
- Rearranging gives:
\[ 2v + 2v_s + 3v_s = 3v \]
\[ 5v_s = 3v - 2v \]
\[ 5v_s = v \]

7. **Solve for Speed of the Train**:
- Thus, the speed of the train is:
\[ v_s = \frac{v}{5} \]

### Final Answer:
The speed of the train is \( \frac{v}{5} \).

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