Question

The frequency of the sound of a car horn as received by an observer towards whom the car is moving differs from the frequency of the horn by 2.0%. Assuming that the velocity of sound of air is \[350m{{\operatorname{s}}^{-1}},\] the velocity of car is
A. \[6.0m{{\operatorname{s}}^{-1}}\]
B. \[7.5m{{\operatorname{s}}^{-1}}\]
C. \[7.0m{{\operatorname{s}}^{-1}}\]
D. \[8.5m{{\operatorname{s}}^{-1}}\]

Answer 1

Hint: Question is based on Doppler Effect, first calculate the ratio of frequency and equate the relative velocity.
1. Frequency of car horn \[={{n}_{{}^\circ }}\]
2. Frequency that received by observer \[=n'\]

Complete step by step solution:
Ratio of eq (1) to eq (2) given by the question = \[\dfrac{n'}{{{n}_{{}^\circ }}}\] = 1.02 ….(1)
Use concept of relativity
By Doppler effect= \[\dfrac{n'}{{{n}_{{}^\circ }}}\] = \[\dfrac{V}{V{{V}_{5}}}\] …..(2)
V₅ = velocity of sound
V = velocity of car
Now put the value of eqn (1) to the eq(2)
1.02 = \[\dfrac{V}{V{{V}_{5}}}\]
V = 350 (Given in question)
1.02 = \[\dfrac{350}{350{{V}_{5}}}\]
350 – V₅ = \[\dfrac{350}{1.02}\]
350 – V₅ = 343.13
V₅ = 350 – 343.13
V₅ = 7 m/s (approx.)

Finally velocity of car in air is 7 m/s

Note: (1) Carefully calculate the ratio of frequency, it is \[\dfrac{n'}{{{n}_{{}^\circ }}}\] not \[\dfrac{{{n}_{{}^\circ }}}{n'}\]
(2) Carefully calculate the relative velocity of sound and car.