Question

The apparent frequency of the whistle of an engine changes in the ratio 9:8 as the engine passes a stationary observer. If the velocity of the sound is 340 m s\(^{-1}\), then the velocity of the engine is

• A. 40 m s\(^{-1}\)
• B. 20 m s\(^{-1}\)
• C. 340 m s\(^{-1}\)
• D. 180 m s\(^{-1}\)

Hint:

Doppler ratio: \(\dfrac{f_{approach}}{f_{recede}} = \dfrac{v+v_s}{v-v_s}\). Solve for \(v_s\) using cross multiplication.

Answer 1

The Correct Option is B

Step 1: Understanding the Concept:
By Doppler effect: apparent frequency when source approaches: \(f_1 = \dfrac{v}{v-v_s} f\); when receding: \(f_2 = \dfrac{v}{v+v_s} f\).
Step 2: Detailed Explanation:
\[ \frac{f_1}{f_2} = \frac{v + v_s}{v - v_s} = \frac{9}{8} \] \[ 8(v + v_s) = 9(v - v_s) \] \[ 8v + 8v_s = 9v - 9v_s \] \[ 17v_s = v = 340 \] \[ v_s = 20 \text{ m/s} \]
Step 3: Final Answer:
The velocity of the engine is 20 m s\(^{-1}\).