Question

The frequency of sound heard by a stationary observer is \( f_1 \), when the source of sound is approaching the observer with a speed of 10\% of the speed of sound. If the same source of sound is moving away from the stationary observer with a speed of 20\% of the speed of sound, the frequency of sound heard by the observer is \( f_2 \). Then \( f_1 : f_2 \) is:

• A. \( 1:2 \)
• B. \( 9:11 \)
• C. \( 4:3 \)
• D. \( 1:1 \)

Hint:

The Doppler effect causes a shift in frequency when the source or observer moves. When the source moves towards the observer, the frequency increases, and when it moves away, the frequency decreases.

Answer 1

The Correct Option is C

Step 1: The Doppler effect formula for sound is given by: \[ f' = f \left( \frac{v + v_o}{v - v_s} \right) \] where:
- \( f' \) is the observed frequency,
- \( f \) is the emitted frequency,
- \( v \) is the speed of sound,
- \( v_o \) is the speed of the observer (which is zero here), and
- \( v_s \) is the speed of the source.
Step 2: When the source is approaching the observer, \( v_s = 0.1v \), and the frequency observed is: \[ f_1 = f \left( \frac{v}{v - 0.1v} \right) = f \times \frac{1}{0.9} \] Step 3: When the source is moving away from the observer, \( v_s = 0.2v \), and the frequency observed is: \[ f_2 = f \left( \frac{v}{v + 0.2v} \right) = f \times \frac{1}{1.2} \] Step 4: The ratio \( \frac{f_1}{f_2} \) is: \[ \frac{f_1}{f_2} = \frac{\frac{f}{0.9}}{\frac{f}{1.2}} = \frac{1.2}{0.9} = \frac{4}{3} \]