A train has just completed a U-curve in a trach which is a semi circle. The engine is at the forward end of the semi circular part of the trach while the last carriage is at the rear end of the semi circular track. The driver blows a whistle of frequency 200 Hz. Velocity of sound is `340(m)/(s)`. Then the apparent frequency as observed by a passenger in the middle of the train, when the speed of the train is 30 m/s, is

Options

A. 219 Hz
B. 188 Hz
C. 200 Hz
D. 181 Hz

Explanation

To find the apparent frequency observed by a passenger in the middle of the train, we can use the Doppler effect formula. Here's a step-by-step solution:

### Step 1: Identify the given values
- Frequency of the whistle (f) = 200 Hz
- Velocity of sound (v) = 340 m/s
- Speed of the train (v_s) = 30 m/s

### Step 2: Determine the direction of motion
Since the train is moving in a semicircular path and the observer (passenger) is in the middle of the train, we need to consider the relative motion of the source (the whistle) and the observer.

### Step 3: Calculate the components of velocity
The train is moving with a speed of 30 m/s. When the train is at the semicircular part, the angle between the direction of the train and the line joining the observer and the source is 45 degrees. Therefore, we need to find the component of the velocity along the line joining the source and the observer.

Using trigonometry:
- The component of the velocity of the train along the line joining them is given by:
\[ v_{component} = v_s \cdot \cos(45^\circ) = 30 \cdot \frac{1}{\sqrt{2}} = \frac{30}{\sqrt{2}} \approx 21.21 \text{ m/s} \]

### Step 4: Apply the Doppler effect formula
The apparent frequency (f') can be calculated using the Doppler effect formula:
\[ f' = f \left( \frac{v + v_o}{v - v_s} \right) \]
Where:
- \(v\) = velocity of sound = 340 m/s
- \(v_o\) = velocity of the observer (passenger) = 30 m/s (since the observer is moving with the train)
- \(v_s\) = velocity of the source (the whistle) = 30 m/s (the source is also moving with the train)

Substituting the values:
\[ f' = 200 \left( \frac{340 + 30}{340 - 30} \right) = 200 \left( \frac{370}{310} \right) \]

### Step 5: Calculate the apparent frequency
Now, calculate:
\[ f' = 200 \cdot \frac{370}{310} \approx 200 \cdot 1.1935 \approx 238.7 \text{ Hz} \]

### Final Answer
The apparent frequency as observed by a passenger in the middle of the train is approximately **238.7 Hz**.

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