Question

An observer is standing stationary in air moving at a speed of 2m/s. A source of frequency 20Hz is moving in the direction of wind with speed 10m/s towards the observer. What will be the frequency as heard by the observer? Speed of sound in air is 330m/s.

• A. 21.50Hz
• B. 19.4Hz
• C. 21.6Hz
• D. 20Hz

Answer 1

Answer: (B)

19.4Hz

Answer 2

The problem describes a scenario involving the Doppler effect with wind.

Given:

• Source frequency ($f_s$) = 20 Hz
• Speed of sound in still air ($v$) = 330 m/s
• Speed of wind ($v_w$) = 2 m/s
• Speed of source relative to ground ($v_s$) = 10 m/s, moving towards the observer.
• Observer speed relative to ground ($v_o$) = 0 m/s (stationary).

The phrase "A source of frequency 20Hz is moving in the direction of wind with speed 10m/s towards the observer" implies that the wind is blowing from the source towards the observer, thus assisting the sound propagation.

1. Effective speed of sound relative to the ground ($V_{eff}$): Since the wind is in the direction of sound propagation, the effective speed of sound is: $V_{eff} = v + v_w = 330 \text{ m/s} + 2 \text{ m/s} = 332 \text{ m/s}$

2. Doppler effect formula for a stationary observer: The formula for the observed frequency when the observer is stationary and the source is moving is: $f_o = f_s \left( \frac{V_{eff}}{V_{eff} \mp v_s} \right)$ Here, $v_s$ is the speed of the source relative to the ground. The sign in the denominator is '-' if the source is moving towards the observer and '+' if moving away.

3. Calculation based on the given problem statement: Since the source is moving towards the observer, we use the '-' sign in the denominator: $f_o = 20 \text{ Hz} \times \left( \frac{332 \text{ m/s}}{332 \text{ m/s} - 10 \text{ m/s}} \right)$ $f_o = 20 \text{ Hz} \times \left( \frac{332}{322} \right)$ $f_o = 20 \text{ Hz} \times 1.0310559...$ $f_o = 20.6211... \text{ Hz}$

This calculated value (approximately 20.6 Hz) is not among the given options (21.50Hz, 19.4Hz, 21.6Hz, 20Hz). This suggests a possible typo in the question or the options.

4. Re-evaluation assuming a common typo: In many physics problems, a common typo is the direction of motion. If the source were moving away from the observer instead of "towards the observer", the formula would be: $f_o = f_s \left( \frac{V_{eff}}{V_{eff} + v_s} \right)$ $f_o = 20 \text{ Hz} \times \left( \frac{332 \text{ m/s}}{332 \text{ m/s} + 10 \text{ m/s}} \right)$ $f_o = 20 \text{ Hz} \times \left( \frac{332}{342} \right)$ $f_o = 20 \text{ Hz} \times 0.97076...$ $f_o = 19.415... \text{ Hz}$

This value (approximately 19.4 Hz) matches one of the options precisely. Given that the calculated value for the given problem statement (20.62 Hz) is not an option, it is highly probable that there is a typo in the question, and the source was intended to be moving "away from the observer".

The final answer is $\boxed{\text{19.4Hz}}$

Explanation of the solution: The effective speed of sound in the direction of propagation is $V_{eff} = v + v_w = 330 + 2 = 332$ m/s, as the wind is in the direction of sound. If the source moves towards the observer, the Doppler formula is $f_o = f_s \frac{V_{eff}}{V_{eff} - v_s}$. This yields $20 \times \frac{332}{332 - 10} = 20.62$ Hz. This is not among the options. However, if the source was moving away from the observer, the formula would be $f_o = f_s \frac{V_{eff}}{V_{eff} + v_s}$. This yields $20 \times \frac{332}{332 + 10} = 19.415$ Hz, which matches option 19.4Hz. Assuming a typo in the question where the source moves away from the observer, 19.4Hz is the correct answer.

Answer: 19.4Hz Option Number: 2