Answer
Verified
100.8k+ views
Hint: If the frequency when the car is approaching is \[{f_1}\] and when the car is leaving is \[{f_2}\]. Get the value of \[{f_1}\] and \[{f_2}\] in terms of \[{f_0}\] . The Doppler effect or Doppler shift can be describing the changes in frequency of sound or light wave produced by a moving source with respect to an observer. Also, we know that the beat frequency is defined as the difference in frequency of two waves. By using this we can get the result.
Formula used:
Frequency when source is approaching is given as,
\[{f_1} = {f_0}\left( {\dfrac{v}{{v - {v_0}}}} \right)\]
Frequency when source is leaving is given as,
\[{f_2} = {f_0}\left( {\dfrac{v}{{v + {v_0}}}} \right)\]
Where \[{f_1}\] and \[{f_2}\] is the frequency required, \[{f_0}\] is the given frequency, \[v\] is the velocity of the observer and \[{v_0}\] is the velocity of sound.
Beat frequency is given as,
\[{f_1} - {f_2}\]
Where \[{f_1} - {f_2}\] represents the change in frequency.
Complete step by step solution:
As we know that the frequency when source is approaching is given as,
\[{f_1} = {f_0}\left( {\dfrac{v}{{v - {v_0}}}} \right) \\ \]
Frequency when source is leaving is given as,
\[{f_2} = {f_0}\left( {\dfrac{v}{{v + {v_0}}}} \right) \\ \]
Now the beat frequency = \[{f_1} - {f_2} \\ \]
\[\text{beat frequency} = {f_0}v\left( {\dfrac{1}{{v - {v_0}}} - \dfrac{1}{{v + {v_0}}}} \right) \\ \]
\[\Rightarrow \text{beat frequency} = {f_0}v\left( {\dfrac{{v + {v_0} - v + {v_0}}}{{{v^2} - v_0^2}}} \right) \\ \]
\[\therefore \text{beat frequency} = \dfrac{{2{f_0}v{v_0}}}{{{v^2} - v_0^2}}\]
Hence option C is the correct answer.
Note:The formula for Doppler Effect is related to the frequency of the sound of an object with its velocity. Doppler Effect is defined as the change in wave frequency during the relative motion between the wave source and its observer. It was given by Christian Johann Doppler. Beats can be determined by subtracting the initial frequency with the frequency observed by the observer.
Formula used:
Frequency when source is approaching is given as,
\[{f_1} = {f_0}\left( {\dfrac{v}{{v - {v_0}}}} \right)\]
Frequency when source is leaving is given as,
\[{f_2} = {f_0}\left( {\dfrac{v}{{v + {v_0}}}} \right)\]
Where \[{f_1}\] and \[{f_2}\] is the frequency required, \[{f_0}\] is the given frequency, \[v\] is the velocity of the observer and \[{v_0}\] is the velocity of sound.
Beat frequency is given as,
\[{f_1} - {f_2}\]
Where \[{f_1} - {f_2}\] represents the change in frequency.
Complete step by step solution:
As we know that the frequency when source is approaching is given as,
\[{f_1} = {f_0}\left( {\dfrac{v}{{v - {v_0}}}} \right) \\ \]
Frequency when source is leaving is given as,
\[{f_2} = {f_0}\left( {\dfrac{v}{{v + {v_0}}}} \right) \\ \]
Now the beat frequency = \[{f_1} - {f_2} \\ \]
\[\text{beat frequency} = {f_0}v\left( {\dfrac{1}{{v - {v_0}}} - \dfrac{1}{{v + {v_0}}}} \right) \\ \]
\[\Rightarrow \text{beat frequency} = {f_0}v\left( {\dfrac{{v + {v_0} - v + {v_0}}}{{{v^2} - v_0^2}}} \right) \\ \]
\[\therefore \text{beat frequency} = \dfrac{{2{f_0}v{v_0}}}{{{v^2} - v_0^2}}\]
Hence option C is the correct answer.
Note:The formula for Doppler Effect is related to the frequency of the sound of an object with its velocity. Doppler Effect is defined as the change in wave frequency during the relative motion between the wave source and its observer. It was given by Christian Johann Doppler. Beats can be determined by subtracting the initial frequency with the frequency observed by the observer.
Recently Updated Pages
Write a composition in approximately 450 500 words class 10 english JEE_Main
Arrange the sentences P Q R between S1 and S5 such class 10 english JEE_Main
Write an article on the need and importance of sports class 10 english JEE_Main
Name the scale on which the destructive energy of an class 11 physics JEE_Main
Choose the exact meaning of the given idiomphrase The class 9 english JEE_Main
Choose the one which best expresses the meaning of class 9 english JEE_Main