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Question: A major car is approaching a road crossing with a speed of \(108\;km{h^{ - 1}}\). Police standing ne...

A major car is approaching a road crossing with a speed of 108  kmh−1108\;km{h^{ - 1}}. Police standing near the crossing hears the frequency of the car’s horn as 300  Hz300\;Hz. What is the real frequency of the horn? (Speed of sound in air = 332  ms−1332\;m{s^{ - 1}})
A. 300  Hz300\;Hz
B. 332  Hz332\;Hz
C. 273  Hz273\;Hz
D. 400  Hz400\;Hz

Explanation

Solution

Hint: When a car moves with some velocity, it emits a sound wave with some velocity. Hence, police stand in the road side hears the sound of some frequency. There is an effect called Doppler effect, which relates the original frequency, heard frequency, the velocity of the sound and velocity of the object from which the sound emits. By using that effect, the original sound frequency can be calculated.

Useful formula:
The doppler effect is given by,
f′=f0[vsoundvsound−vsource]f' = {f_0}\left[ {\dfrac{{{v_{sound}}}}{{{v_{sound}} - {v_{source}}}}} \right]
Where, f′f' is the frequency of sound heard, f0{f_0} is the original frequency of sound, vsound{v_{sound}} is the velocity of sound and vsource{v_{source}} is the velocity of object.

Given data:
The velocity of the car, vsource=108  kmh−1{v_{source}} = 108\;km{h^{ - 1}}
The frequency of sound heard, f′=300  Hzf' = 300\;Hz
The velocity of sound, vsound=332  ms−1{v_{sound}} = 332\;m{s^{ - 1}}

Complete step by step solution:
The velocity of the car is given by,
vsource=108  kmh−1{v_{source}} = 108\;km{h^{ - 1}}
Converting the velocity from kmh−1km{h^{ - 1}} to ms−1m{s^{ - 1}}, we get
vsource=108  kmh−1 vsource=108×10003600  ms−1 vsource=30  ms−1  {v_{source}} = 108\;km{h^{ - 1}} \\\ {v_{source}} = 108 \times \dfrac{{1000}}{{3600}}\;m{s^{ - 1}} \\\ {v_{source}} = 30\;m{s^{ - 1}} \\\

By using Doppler effect, we get
f′=f0[vsoundvsound−vsource]f' = {f_0}\left[ {\dfrac{{{v_{sound}}}}{{{v_{sound}} - {v_{source}}}}} \right]
By substituting the given values in the above relation, we get
300  Hz=f0[332  ms−1332  ms−1−30  ms−1] 300  Hz=f0[332  ms−1302  ms−1] 300  Hz=f0×1.0993 f0=300  Hz1.0993 f0=272.891  Hz f0≃273  Hz  300\;Hz = {f_0}\left[ {\dfrac{{332\;m{s^{ - 1}}}}{{332\;m{s^{ - 1}} - 30\;m{s^{ - 1}}}}} \right] \\\ 300\;Hz = {f_0}\left[ {\dfrac{{332\;m{s^{ - 1}}}}{{302\;m{s^{ - 1}}}}} \right] \\\ 300\;Hz = {f_0} \times 1.0993 \\\ {f_0} = \dfrac{{300\;Hz}}{{1.0993}} \\\ {f_0} = 272.891\;Hz \\\ {f_0} \simeq 273\;Hz \\\

Hence, the option (C) is correct.

Note: When a moving object emits a sound wave, the frequency of the sound will definitely tend to fall to a level. That frequency will be represented by the Doppler effect. It gives the relation between the original frequency and the heard frequency of the sound.