A whistling train approaches a junction. An observer standing at juction observes the frequency to be `2.2 KHz` and `1.8 KHz` of the approaching and the receding train respectively. Find the speed of the train (speed of sound `300m//s`)

Explanation

To solve the problem of finding the speed of the train using the observed frequencies of the whistling train, we can use the Doppler effect equations for sound. Here’s a step-by-step solution:

### Step 1: Understand the Doppler Effect
The Doppler effect describes how the frequency of sound changes for an observer moving relative to the source of the sound. When the source approaches the observer, the frequency increases, and when it recedes, the frequency decreases.

### Step 2: Write the Doppler Effect Equations
For a source moving towards a stationary observer, the apparent frequency \( f_1 \) is given by:
\[ f_1 = \frac{v}{v - v_s} f_0 \]
For a source moving away from a stationary observer, the apparent frequency \( f_2 \) is given by:
\[ f_2 = \frac{v}{v + v_s} f_0 \]
Where:
- \( f_0 \) is the actual frequency of the source.
- \( v \) is the speed of sound (300 m/s).
- \( v_s \) is the speed of the source (train).

### Step 3: Set Up the Equations
From the problem, we have:
- \( f_1 = 2.2 \, \text{kHz} = 2200 \, \text{Hz} \) (approaching)
- \( f_2 = 1.8 \, \text{kHz} = 1800 \, \text{Hz} \) (receding)

Using the equations:
1. For the approaching train:
\[ 2200 = \frac{300}{300 - v_s} f_0 \quad \text{(1)} \]
2. For the receding train:
\[ 1800 = \frac{300}{300 + v_s} f_0 \quad \text{(2)} \]

### Step 4: Divide the Two Equations
Dividing equation (1) by equation (2):
\[ \frac{2200}{1800} = \frac{300/(300 - v_s)}{300/(300 + v_s)} \]
This simplifies to:
\[ \frac{2200}{1800} = \frac{300 + v_s}{300 - v_s} \]
Calculating the left side:
\[ \frac{2200}{1800} = \frac{11}{9} \]
Thus, we have:
\[ \frac{11}{9} = \frac{300 + v_s}{300 - v_s} \]

### Step 5: Cross Multiply and Solve for \( v_s \)
Cross multiplying gives:
\[ 11(300 - v_s) = 9(300 + v_s) \]
Expanding both sides:
\[ 3300 - 11v_s = 2700 + 9v_s \]
Rearranging gives:
\[ 3300 - 2700 = 11v_s + 9v_s \]
\[ 600 = 20v_s \]
Thus, solving for \( v_s \):
\[ v_s = \frac{600}{20} = 30 \, \text{m/s} \]

### Final Answer
The speed of the train is \( 30 \, \text{m/s} \).

---