A locomotive approaching a crossing at a speed of 20 `ms^(-1)` sounds a whistle of frequency 640 Hz when 1 km from the crossing. There is no wind and the speed of sound in air is 330 `ms^(-1)`. What frequency is heard by an observer `sqrt(3)` km on the straight road from the crossing at right angle :-

Options

A. 600 Hz
B. 630 Hz
C. 660 Hz
D. 720 Hz

Answer

To solve the problem step by step, we will use the principles of the Doppler effect for sound.

### Step 1: Identify the Given Data
- Speed of the locomotive (source), \( V_s = 20 \, \text{ms}^{-1} \)
- Frequency of the whistle (source frequency), \( f_0 = 640 \, \text{Hz} \)
- Speed of sound in air, \( V = 330 \, \text{ms}^{-1} \)
- Distance of the locomotive from the crossing, \( d = 1 \, \text{km} = 1000 \, \text{m} \)
- Distance of the observer from the crossing, \( d_o = \sqrt{3} \, \text{km} = 1000\sqrt{3} \, \text{m} \)

### Step 2: Calculate the Angle and the Component of Velocity
The observer is positioned at a right angle to the direction of the train. We can visualize the situation as a right triangle where:
- The distance from the locomotive to the crossing is one leg (1 km).
- The distance from the crossing to the observer is the other leg (\(\sqrt{3}\) km).

Using the right triangle, we can find the angle \( \theta \) using:
\[ \cos \theta = \frac{\text{adjacent}}{\text{hypotenuse}} = \frac{1000}{1000\sqrt{3}} = \frac{1}{\sqrt{3}} \]

### Step 3: Calculate the Effective Velocity of the Source
The effective velocity of the source towards the observer is given by:
\[ V_{s,\text{effective}} = V_s \cdot \cos \theta = 20 \cdot \frac{1}{\sqrt{3}} = \frac{20}{\sqrt{3}} \, \text{ms}^{-1} \]

### Step 4: Apply the Doppler Effect Formula
The frequency heard by the observer (\( f' \)) can be calculated using the Doppler effect formula:
\[ f' = f_0 \cdot \frac{V + V_o}{V - V_s} \]
Where:
- \( V_o = 0 \) (the observer is stationary)
- \( V_s = V_{s,\text{effective}} = \frac{20}{\sqrt{3}} \)

Substituting the values:
\[ f' = 640 \cdot \frac{330 + 0}{330 - \frac{20}{\sqrt{3}}} \]

### Step 5: Simplify the Denominator
Calculate the denominator:
\[ 330 - \frac{20}{\sqrt{3}} = 330 - 11.547 = 318.453 \]

### Step 6: Calculate the Final Frequency
Now substituting back into the equation:
\[ f' = 640 \cdot \frac{330}{318.453} \]

Calculating this gives:
\[ f' \approx 640 \cdot 1.036 = 662.4 \, \text{Hz} \]

### Step 7: Final Result
The frequency heard by the observer is approximately \( 662.4 \, \text{Hz} \).