
A police car horn emits a sound at a frequency $240\,Hz$ when the car is at rest. If the speed of the sound is $330\,m{\sec ^{ - 1}}$, the frequency heard by an observer who is approaching the car at a speed of $11\,m{\sec ^{ - 1}}$, is :
A. $248\,Hz$
B. $244\,Hz$
C. $240\,Hz$
D. $230\,Hz$
Answer
499.8k+ views
Hint:The Doppler effect, also known as the Doppler shift, describes the frequency shifts of any sound or light wave emitted by a moving object in relation to an observer. If an object approaches an observer, the waves it emits become compressed, resulting in a higher frequency.
Complete step by step answer:
The mathematical formula to calculate frequency using Doppler effect is:
${f_0} = \dfrac{{v + {v_0}}}{v}{f_s}$
Where, ${f_0}$= sound frequency of observer, $v$= sound wave speed, ${v_0}$= velocity of observer and ${f_s}$= actual frequency of sound wave.
Now, it is mentioned in the question that
${f_s} = 240\,Hz \\
\Rightarrow v = 330\,m{\sec ^{ - 1}} \\
\Rightarrow {v_0} = 11\,m{\sec ^{ - 1}} \\ $
Substituting these values to the equation we get,
\[{f_0} = 240 \times \left( {\dfrac{{330 + 11}}{{330}}} \right) \\
\Rightarrow {f_0} = 240 \times \dfrac{{341}}{{330}} \\
\Rightarrow {f_0} = 8 \times {\dfrac{{{{341}}}}{{{{11}}}}^{31}} \\
\Rightarrow {f_0} = 8 \times 31 \\
\therefore {f_0} = 248\,Hz \\ \]
Hence, the correct option is A.
Note:The Doppler effect has a variety of practical uses. Meteorologists, for example, use the Doppler effect to detect hurricanes in addition to police radar. The Doppler effect was also used by doctors to detect heart attacks. In astronomy, the Doppler effect is useful because it allows the velocity of light-emitting objects in space, such as stars or galaxies, to be calculated.
Complete step by step answer:
The mathematical formula to calculate frequency using Doppler effect is:
${f_0} = \dfrac{{v + {v_0}}}{v}{f_s}$
Where, ${f_0}$= sound frequency of observer, $v$= sound wave speed, ${v_0}$= velocity of observer and ${f_s}$= actual frequency of sound wave.
Now, it is mentioned in the question that
${f_s} = 240\,Hz \\
\Rightarrow v = 330\,m{\sec ^{ - 1}} \\
\Rightarrow {v_0} = 11\,m{\sec ^{ - 1}} \\ $
Substituting these values to the equation we get,
\[{f_0} = 240 \times \left( {\dfrac{{330 + 11}}{{330}}} \right) \\
\Rightarrow {f_0} = 240 \times \dfrac{{341}}{{330}} \\
\Rightarrow {f_0} = 8 \times {\dfrac{{{{341}}}}{{{{11}}}}^{31}} \\
\Rightarrow {f_0} = 8 \times 31 \\
\therefore {f_0} = 248\,Hz \\ \]
Hence, the correct option is A.
Note:The Doppler effect has a variety of practical uses. Meteorologists, for example, use the Doppler effect to detect hurricanes in addition to police radar. The Doppler effect was also used by doctors to detect heart attacks. In astronomy, the Doppler effect is useful because it allows the velocity of light-emitting objects in space, such as stars or galaxies, to be calculated.
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