A source at rest emits sound waves of frequency \( 102 \) Hz. Two observers are moving away from the source of sound in opposite directions each with a speed of \( 10\% \) of the speed of sound. The ratio of the frequencies of sound heard by the observers is?
Show Hint
- \( 9:11 \)
- \( 1:1 \)
- \( 7:9 \)
\( 2:3 \)
The Correct Option is B
Solution and Explanation
Step 1: Applying the Doppler effect formula
The observed frequency when the observer is moving away from the stationary source is given by: \[ f' = f \times \frac{v}{v + v_o} \] where: - \( f = 102 \) Hz (source frequency), - \( v \) is the speed of sound in air, - \( v_o = 0.1 v \) (observer’s speed moving away).
Step 2: Compute the observed frequencies
For observer 1 moving away: \[ f'_1 = 102 \times \frac{v}{v + 0.1v} \] \[ = 102 \times \frac{v}{1.1v} \] \[ = 102 \times \frac{1}{1.1} \] \[ \approx 92.7 \text{ Hz} \] For observer 2 moving in the opposite direction, the same calculation applies: \[ f'_2 = 102 \times \frac{v}{1.1v} = 92.7 \text{ Hz} \]
Step 3: Find the ratio of frequencies
\[ \frac{f'_1}{f'_2} = \frac{92.7}{92.7} = 1:1 \] Thus, the ratio of the frequencies of sound heard by the observers is \( 1:1 \).
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