Question

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?

• A. \( 9:11 \)
• B. \( 1:1 \)
• C. \( 7:9 \)
• D. \( 2:3 \)

Hint:

When two observers move away from a stationary sound source at equal speeds in opposite directions, they experience the same frequency shift. The Doppler effect results in identical frequency reductions, making the ratio \( 1:1 \).

Answer 1

The Correct Option is B

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 \( 1:1 \).