A major car is approaching a road crossing with a speed of $108\;km{h^{ - 1}}$. Police standing near the crossing hears the frequency of the car’s horn as $300\;Hz$. What is the real frequency of the horn? (Speed of sound in air = $332\;m{s^{ - 1}}$)
A. $300\;Hz$
B. $332\;Hz$
C. $273\;Hz$
D. $400\;Hz$
Answer
650.1k+ views
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,
$f' = {f_0}\left[ {\dfrac{{{v_{sound}}}}{{{v_{sound}} - {v_{source}}}}} \right]$
Where, $f'$ is the frequency of sound heard, ${f_0}$ is the original frequency of sound, ${v_{sound}}$ is the velocity of sound and ${v_{source}}$ is the velocity of object.
Given data:
The velocity of the car, ${v_{source}} = 108\;km{h^{ - 1}}$
The frequency of sound heard, $f' = 300\;Hz$
The velocity of sound, ${v_{sound}} = 332\;m{s^{ - 1}}$
Complete step by step solution:
The velocity of the car is given by,
${v_{source}} = 108\;km{h^{ - 1}}$
Converting the velocity from $km{h^{ - 1}}$ to $m{s^{ - 1}}$, we get
$
{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
$f' = {f_0}\left[ {\dfrac{{{v_{sound}}}}{{{v_{sound}} - {v_{source}}}}} \right]$
By substituting the given values in the above relation, we get
$
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.
Useful formula:
The doppler effect is given by,
$f' = {f_0}\left[ {\dfrac{{{v_{sound}}}}{{{v_{sound}} - {v_{source}}}}} \right]$
Where, $f'$ is the frequency of sound heard, ${f_0}$ is the original frequency of sound, ${v_{sound}}$ is the velocity of sound and ${v_{source}}$ is the velocity of object.
Given data:
The velocity of the car, ${v_{source}} = 108\;km{h^{ - 1}}$
The frequency of sound heard, $f' = 300\;Hz$
The velocity of sound, ${v_{sound}} = 332\;m{s^{ - 1}}$
Complete step by step solution:
The velocity of the car is given by,
${v_{source}} = 108\;km{h^{ - 1}}$
Converting the velocity from $km{h^{ - 1}}$ to $m{s^{ - 1}}$, we get
$
{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
$f' = {f_0}\left[ {\dfrac{{{v_{sound}}}}{{{v_{sound}} - {v_{source}}}}} \right]$
By substituting the given values in the above relation, we get
$
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.
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