A train in approaching a station with a uniform velocity of `72kmph` and the frequency of the whistle of that train is `480Hz`. The apparent increase in the frequency of that whistle heard by a stationary observer on the platform is (Velocity of sound in air is `340m//s`)

Options

A. `60Hz`
B. `45Hz`
C. `30Hz`
D. `15Hz`

Explanation

To solve the problem, we will use the Doppler effect formula for sound. The formula for the apparent frequency (N1) heard by a stationary observer when the source of sound is moving towards the observer is given by:

\[ N_1 = N \left( \frac{V + V_o}{V - V_s} \right) \]

Where:
- \( N \) = actual frequency of the source (480 Hz)
- \( V \) = speed of sound in air (340 m/s)
- \( V_o \) = speed of the observer (0 m/s, since the observer is stationary)
- \( V_s \) = speed of the source (train) towards the observer

### Step 1: Convert the speed of the train from km/h to m/s
The speed of the train is given as 72 km/h. We need to convert this to meters per second (m/s).

\[ V_s = 72 \, \text{km/h} \times \frac{1000 \, \text{m}}{1 \, \text{km}} \times \frac{1 \, \text{h}}{3600 \, \text{s}} = 20 \, \text{m/s} \]

### Step 2: Substitute the values into the Doppler effect formula
Now we can substitute the known values into the Doppler effect formula.

\[ N_1 = 480 \left( \frac{340 + 0}{340 - 20} \right) \]

### Step 3: Simplify the equation
Calculate the denominator:

\[ 340 - 20 = 320 \]

Now substitute this back into the equation:

\[ N_1 = 480 \left( \frac{340}{320} \right) \]

### Step 4: Calculate the fraction
Now calculate the fraction:

\[ \frac{340}{320} = 1.0625 \]

### Step 5: Calculate the apparent frequency
Now multiply this by 480 Hz:

\[ N_1 = 480 \times 1.0625 = 510 \, \text{Hz} \]

### Step 6: Calculate the increase in frequency
The increase in frequency is given by:

\[ \Delta N = N_1 - N = 510 \, \text{Hz} - 480 \, \text{Hz} = 30 \, \text{Hz} \]

### Final Answer
The apparent increase in the frequency of the whistle heard by the stationary observer is **30 Hz**.

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