Question:
A driver in a car, approaching a vertical wall notices that the frequency of his car horn, has changed from 440 Hz to 480 Hz, when it gets reflected from the wall. If the speed of sound in air is 345 m/s, then the speed of the car is :
A driver in a car, approaching a vertical wall notices that the frequency of his car horn, has changed from 440 Hz to 480 Hz, when it gets reflected from the wall. If the speed of sound in air is 345 m/s, then the speed of the car is :
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Apply the Doppler effect twice: wall as stationary observer, then wall as stationary source; the combined ratio is f'/f = (v+u)/(v-u).
Updated On: Nov 5, 2026
- 36 km/hr
- 54 km/hr
- 24 km/hr
- 18 km/hr
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The Correct Option is B
Solution and Explanation
Let the car speed be u and the speed of sound v = 345 m/s.
Frequency received by the wall: \(f_1 = f\,\frac{v}{v - u}\).
Frequency of the reflected sound heard by the driver (moving towards the wall, the wall acting as the source): \(f_2 = f_1\,\frac{v + u}{v} = f\,\frac{v + u}{v - u}\).
\(\frac{480}{440} = \frac{12}{11} = \frac{345 + u}{345 - u}\)
\(11(345 + u) = 12(345 - u) \Rightarrow 23u = 345 \Rightarrow u = 15\ \mathrm{m/s}\).
\(u = 15 \times \frac{18}{5} = 54\ \mathrm{km/h}\).
Answer: (2) 54 km/hr
Frequency received by the wall: \(f_1 = f\,\frac{v}{v - u}\).
Frequency of the reflected sound heard by the driver (moving towards the wall, the wall acting as the source): \(f_2 = f_1\,\frac{v + u}{v} = f\,\frac{v + u}{v - u}\).
\(\frac{480}{440} = \frac{12}{11} = \frac{345 + u}{345 - u}\)
\(11(345 + u) = 12(345 - u) \Rightarrow 23u = 345 \Rightarrow u = 15\ \mathrm{m/s}\).
\(u = 15 \times \frac{18}{5} = 54\ \mathrm{km/h}\).
Answer: (2) 54 km/hr
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