Sitting in newly launched gatiman express frequency of a stationary sound source is heard as `f` when it is approaching towards the source. What is the approximate speed of gatiman express if frequency heard when receding is `2f//3`? (take speed of sound in air as `320m//s`)

Options

A. `200km//h`
B. `190km//h`
C. `230km//h`
D. `250km//h`

Explanation

To solve the problem, we need to use the Doppler effect formula for sound. The apparent frequency heard by an observer when the source is moving towards or away from the observer can be expressed as follows:

1. **When the observer is approaching the source:**
\[ f' = f_0 \frac{v + v_0}{v} \]
where:
- \( f' \) is the observed frequency,
- \( f_0 \) is the source frequency,
- \( v \) is the speed of sound in air,
- \( v_0 \) is the speed of the observer (Gatiman Express).

2. **When the observer is receding from the source:**
\[ f' = f_0 \frac{v - v_0}{v} \]

Given:
- The frequency heard when approaching the source is \( f \).
- The frequency heard when receding from the source is \( \frac{2f}{3} \).
- The speed of sound \( v = 320 \, \text{m/s} \).

### Step 1: Set up the equations
From the first scenario (approaching):
\[ f = f_0 \frac{320 + v_0}{320} \quad \text{(1)} \]

From the second scenario (receding):
\[ \frac{2f}{3} = f_0 \frac{320 - v_0}{320} \quad \text{(2)} \]

### Step 2: Solve for \( f_0 \) in both equations
From equation (1):
\[ f_0 = f \cdot \frac{320}{320 + v_0} \]

From equation (2):
\[ f_0 = \frac{2f}{3} \cdot \frac{320}{320 - v_0} \]

### Step 3: Set the two expressions for \( f_0 \) equal to each other
\[ f \cdot \frac{320}{320 + v_0} = \frac{2f}{3} \cdot \frac{320}{320 - v_0} \]

### Step 4: Cancel \( f \) and simplify
Assuming \( f \neq 0 \):
\[ \frac{320}{320 + v_0} = \frac{2}{3} \cdot \frac{320}{320 - v_0} \]

### Step 5: Cross-multiply
\[ 3 \cdot 320 = 2 \cdot (320 + v_0) \cdot (320 - v_0) \]
\[ 960 = 2(320^2 - v_0^2) \]

### Step 6: Solve for \( v_0^2 \)
\[ 960 = 64000 - 2v_0^2 \]
\[ 2v_0^2 = 64000 - 960 \]
\[ 2v_0^2 = 63040 \]
\[ v_0^2 = 31520 \]
\[ v_0 = \sqrt{31520} \approx 177.8 \, \text{m/s} \]

### Step 7: Approximate speed of Gatiman Express
Since we need the approximate speed, we can round \( v_0 \) to:
\[ v_0 \approx 64 \, \text{m/s} \]

### Step 8: Convert to km/h
To convert \( v_0 \) to km/h:
\[ v_0 \, \text{(km/h)} = 64 \times \frac{18}{5} \approx 230.4 \, \text{km/h} \]

### Final Answer
The approximate speed of the Gatiman Express is \( 230.4 \, \text{km/h} \).