A person is standing on a railway platform.An engine blowing a whistle of frequency 640 Hz approaches him with a speed of 72 km /hr,The frequency of the note heard by the person is (velocity of sound is 340 m/s )

Options

A. 650 Hz
B. 660 Hz
C. 675 Hz
D. 680 Hz

Answer

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

\[ f' = f \times \frac{v + v_o}{v - v_s} \]

Where:
- \( f' \) = apparent frequency
- \( f \) = original frequency (640 Hz)
- \( v \) = speed of sound (340 m/s)
- \( v_o \) = speed of the observer (0 m/s, since the observer is stationary)
- \( v_s \) = speed of the source (engine)

### Step 1: Convert the speed of the source from km/h to m/s
The speed of the engine is given as 72 km/h. To convert this to m/s, we use the conversion factor:

\[ 1 \text{ km/h} = \frac{1}{3.6} \text{ m/s} \]

Thus,

\[ v_s = 72 \text{ km/h} \times \frac{1 \text{ m/s}}{3.6 \text{ km/h}} = 20 \text{ m/s} \]

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

\[ f' = 640 \times \frac{340 + 0}{340 - 20} \]

### Step 3: Simplify the equation
Now simplify the equation:

\[ f' = 640 \times \frac{340}{320} \]

### Step 4: Calculate the fraction
Next, calculate the fraction:

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

### Step 5: Calculate the apparent frequency
Now, multiply this by the original frequency:

\[ f' = 640 \times 1.0625 = 680 \text{ Hz} \]

### Final Answer
The frequency of the note heard by the person is **680 Hz**.

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