A train A is travelling at a speed of 108 km ` hr^(-1) ` The train approaches another train B standing on the platform.The engine of the train B below its horn. The frequency of the horn as observed by the driver in train A is 504 Hz. The frequency of the horn of train B is (speed of sound = 330 `ms ^(-1)`

Options

A. 504 Hz
B. 462 Hz
C. 550 Hz
D. 407 Hz

Answer

To solve the problem, we will use the Doppler effect formula for sound. The formula for the apparent frequency \( f' \) as heard by an observer moving towards a stationary source is given by:

\[ f' = f \left( \frac{v + v_o}{v} \right) \]

Where:
- \( f' \) is the observed frequency (504 Hz)
- \( f \) is the source frequency (which we need to find)
- \( v \) is the speed of sound (330 m/s)
- \( v_o \) is the speed of the observer (train A) moving towards the source (train B)

### Step 1: Convert the speed of train A from km/h to m/s
The speed of train A is given as 108 km/h. We convert this to m/s using the conversion factor:

\[ \text{Speed in m/s} = \text{Speed in km/h} \times \frac{5}{18} \]

Calculating:

\[ v_o = 108 \times \frac{5}{18} = 30 \text{ m/s} \]

### Step 2: Substitute the values into the Doppler effect formula
Now we substitute \( f' = 504 \text{ Hz} \), \( v = 330 \text{ m/s} \), and \( v_o = 30 \text{ m/s} \) into the formula:

\[ 504 = f \left( \frac{330 + 30}{330} \right) \]

### Step 3: Simplify the equation
This simplifies to:

\[ 504 = f \left( \frac{360}{330} \right) \]

### Step 4: Solve for the original frequency \( f \)
We can rearrange the equation to solve for \( f \):

\[ f = 504 \times \frac{330}{360} \]

### Step 5: Calculate the value of \( f \)
Calculating \( \frac{330}{360} \):

\[ \frac{330}{360} = \frac{11}{12} \]

Now substituting back:

\[ f = 504 \times \frac{11}{12} \]

Calculating:

\[ f = 504 \div 12 \times 11 = 42 \times 11 = 462 \text{ Hz} \]

### Final Answer
The frequency of the horn of train B is \( 462 \text{ Hz} \).

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