Frequency Spread Observed by Passengers in Moving Train

To find the spread of frequency observed by the passengers in train B, we analyze the Doppler effect for the continuous band of frequencies emitted by the whistle.

1. Identifying the Parameters from the Passage

From the passage of IIT-JEE 2007:

• Speed of sound in still air: $v = 340\text{ m/s}$
• Speed of train A: $v_A = 20\text{ m/s}$
• Speed of train B: $v_B = 30\text{ m/s}$
• Trains A and B are moving in the same direction on parallel tracks, with train A behind train B.
• The whistle is blown from train A and emits a uniform spread of sound frequencies ranging from $f_1 = 800\text{ Hz}$ to $f_2 = 1120\text{ Hz}$, which gives an emitted frequency spread of:
  $\Delta f = f_2 - f_1 = 1120\text{ Hz} - 800\text{ Hz} = 320\text{ Hz}$

2. Doppler Effect Formulation

Since train A (the source) is moving towards train B at speed $v_A$, and train B (the observer) is moving away from train A at speed $v_B$ in the same direction:

• The sound waves travel from train A to train B in the direction of motion.
• The frequency $f'$ perceived by an observer in train B for an emitted frequency $f$ is given by the Doppler effect formula:
  $f' = f \left( \frac{v - v_B}{v - v_A} \right)$

3. Calculating the Spread of Observed Frequency

The observed frequency spread $\Delta f'$ corresponds to the difference between the observed upper and lower frequencies:
$\Delta f' = f_2' - f_1' = (f_2 - f_1) \left( \frac{v - v_B}{v - v_A} \right) = \Delta f \left( \frac{v - v_B}{v - v_A} \right)$

Substituting the given values:

• $v = 340\text{ m/s}$
• $v_A = 20\text{ m/s}$
• $v_B = 30\text{ m/s}$
• $\Delta f = 320\text{ Hz}$

We obtain:
$\Delta f' = 320 \times \left( \frac{340 - 30}{340 - 20} \right)$
$\Delta f' = 320 \times \left( \frac{310}{320} \right) = 310\text{ Hz}$

Therefore, the spread of frequency as observed by the passengers in train B is $310\text{ Hz}$.

The correct option is A.