Question

A boy standing on a platform observes the frequency of a train horn as it passes by. The change in the frequency noticed as the train approaches and recedes from him with a velocity of 108 km/h (speed of sound in air = 330 m/s) is:

• A. \( 18.33\% \)
• B. \( 16.67\% \)
• C. \( 21.27\% \)
• D. \( 15.23\% \)

Hint:

Use the Doppler effect formula: \( f' = f \frac{v}{v - v_s} \) for an approaching source and \( f'' = f \frac{v}{v + v_s} \) for a receding source.
- Convert speeds to consistent units (m/s) before applying formulas.

Answer 1

The Correct Option is B

The Doppler effect formula for the apparent frequency when the source is moving towards and away from the observer is: \[ f' = f \frac{v}{v - v_s} \quad \text{(approaching)} \] \[ f'' = f \frac{v}{v + v_s} \quad \text{(receding)} \] where: - \( v = 330 \) m/s (speed of sound), - \( v_s = 108 \) km/h = \( 30 \) m/s (train speed), - \( f \) is the original frequency. 1. Frequency shift when train approaches: \[ f' = f \frac{330}{330 - 30} = f \frac{330}{300} = 1.1f \] 2. Frequency shift when train recedes: \[ f'' = f \frac{330}{330 + 30} = f \frac{330}{360} = 0.9167f \] 3. Total percentage change in frequency: \[ \frac{f' - f''}{f} \times 100 = \left( 1.1 - 0.9167 \right) \times 100 \] \[ = 0.1833 \times 100 = 16.67\% \] Thus, the correct answer is \(\boxed{16.67\%}\).