A train approaching a platform at a speed of `54 km h^-1` sounds a whistle. An observer on the platform finds its frequency to be 1620 Hz. The train passes the platform keeping the whistle on and without slowing down. What frequency will the observer hear after the train has crossed the platfrom ? The speed of sound in air `= 332 ms^-1`

Explanation

To solve the problem step by step, we will use the Doppler effect formula for sound waves.

### Step 1: Convert the speed of the train to meters per second
The speed of the train is given as \(54 \, \text{km/h}\). To convert this to meters per second, we use the conversion factor \(1 \, \text{km/h} = \frac{1}{3.6} \, \text{m/s}\).

\[ \text{Speed of train} = 54 \, \text{km/h} \times \frac{1}{3.6} = 15 \, \text{m/s} \]

### Step 2: Identify the known values
- Apparent frequency when the train is approaching (\(f\)) = 1620 Hz
- Speed of sound in air (\(v\)) = 332 m/s
- Speed of the train (\(v_s\)) = 15 m/s (as it is approaching the observer)

### Step 3: Use the Doppler effect formula to find the original frequency (\(f_0\))
The Doppler effect formula for the frequency observed when the source is approaching is given by:

\[ f = f_0 \frac{v + v_b}{v - v_s} \]

Where:
- \(f\) = observed frequency (1620 Hz)
- \(f_0\) = original frequency
- \(v_b\) = speed of the observer (0 m/s, since the observer is stationary)
- \(v_s\) = speed of the source (15 m/s)

Substituting the known values into the formula:

\[ 1620 = f_0 \frac{332 + 0}{332 - 15} \]

\[ 1620 = f_0 \frac{332}{317} \]

### Step 4: Solve for \(f_0\)
Rearranging the equation to solve for \(f_0\):

\[ f_0 = 1620 \times \frac{317}{332} \]

Calculating \(f_0\):

\[ f_0 = 1620 \times 0.9545 \approx 1547 \, \text{Hz} \]

### Step 5: Use the Doppler effect formula to find the frequency after the train has crossed the platform
Now, when the train has passed the observer, it is moving away from the observer. The formula for the frequency observed when the source is receding is:

\[ f' = f_0 \frac{v + v_b}{v - v_s} \]

In this case:
- \(v_b = 0\) (observer is stationary)
- \(v_s = -15\) (source is moving away, so we take it as negative)

Substituting the values into the formula:

\[ f' = 1547 \frac{332 + 0}{332 - (-15)} \]

\[ f' = 1547 \frac{332}{332 + 15} \]

\[ f' = 1547 \frac{332}{347} \]

### Step 6: Calculate \(f'\)
Calculating \(f'\):

\[ f' = 1547 \times 0.9565 \approx 1480 \, \text{Hz} \]

### Final Answer
The frequency that the observer will hear after the train has crossed the platform is approximately **1480 Hz**.

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