Question

A car sounding a horn of frequency 1000 Hz passes a stationary observer. The ratio of frequencies of the horn noted by the observer before and after passing of the car is 11:9. The speed of the car is (Speed of sound \( v = 340 \, {ms}^{-1} \)):

• A. \( 34 \, {ms}^{-1} \)
• B. \( 17 \, {ms}^{-1} \)
• C. \( 170 \, {ms}^{-1} \)
• D. \( 340 \, {ms}^{-1} \)

Hint:

Use the Doppler effect formula for sound waves to solve for the speed of the source. Remember that the frequency shift is inversely related to the speed of the source.

Answer 1

The Correct Option is A

We are given that the car passes a stationary observer while sounding a horn with a frequency of 1000 Hz. The ratio of frequencies before and after the car passes the observer is given as 11:9. The formula for the Doppler effect when the source is moving and the observer is stationary is: \[ f' = f \left( \frac{v}{v - v_s} \right) \] where: - \( f' \) is the frequency observed by the observer,
- \( f \) is the frequency of the source (1000 Hz),
- \( v \) is the speed of sound (340 m/s),
- \( v_s \) is the speed of the source (the car’s speed, which we need to find).
Step 1: Set up the equation for frequencies
Before the car passes the observer, the frequency is \( f' = f \left( \frac{v}{v - v_s} \right) \), and after it passes, the frequency is \( f'' = f \left( \frac{v}{v + v_s} \right) \). We are given the ratio of the frequencies before and after passing the observer: \[ \frac{f'}{f''} = \frac{11}{9} \] Substituting the Doppler shift equations for \( f' \) and \( f'' \), we get: \[ \frac{\frac{v}{v - v_s}}{\frac{v}{v + v_s}} = \frac{11}{9} \] Step 2: Solve for the speed of the car
Simplifying the equation: \[ \frac{v + v_s}{v - v_s} = \frac{11}{9} \] Now, cross-multiply to solve for \( v_s \): \[ 9(v + v_s) = 11(v - v_s) \] Expanding both sides: \[ 9v + 9v_s = 11v - 11v_s \] \[ 9v + 9v_s = 11v - 11v_s \] \[ 9v + 9v_s + 11v_s = 11v \] \[ 9v + 20v_s = 11v \] \[ 20v_s = 2v \] \[ v_s = \frac{2v}{20} = \frac{v}{10} \] Substitute \( v = 340 \, {ms}^{-1} \): \[ v_s = \frac{340}{10} = 34 \, {ms}^{-1} \] Thus, the speed of the car is \( 34 \, {ms}^{-1} \).