Answer
64.8k+ views
Hint: To solve this question, we need to use the Doppler Effect formula for the apparent frequency as received by a receiver. We have to substitute the speeds with the proper sign convention for the two cases. On solving the two equations formed, we will get the required speed of the aeroplane.
Formula used: The formula used to solve this question is given by
\[\nu ' = \left( {\dfrac{{v - {v_L}}}{{v - {v_S}}}} \right){\nu _0}\], here $\nu '$ is the apparent frequency as received by a receiver moving with a speed of ${v_L}$ due to a source of sound moving with a speed of ${v_S}$ and emitting a sound of frequency ${\nu _0}$. The velocity of the wave is $v$.
Complete step-by-step solution:
In the given question, the direction of the velocity of the aeroplane is not given to us. Therefore we assume that it is moving away from the radar.
From the Doppler’s Effect formula, we know that the apparent frequency received by an observer is given by
\[\nu ' = \left( {\dfrac{{v - {v_L}}}{{v - {v_S}}}} \right){\nu _0}\]
According to the question, the radio wave has a frequency of $\nu $. Also we know that radio waves are a type of electromagnetic wave which has a speed of $c$. Therefore substituting $v = c$, and ${\nu _0} = \nu $ in the above equation, we get
\[\nu ' = \left( {\dfrac{{c - {v_L}}}{{c - {v_S}}}} \right)\nu \]................(1)
In the first case, the radar is the source and the aeroplane is the receiver. Since the radar is stationary, so we have${v_S} = 0$. Also, since the aeroplane is moving with a speed of ${v_a}$ parallel to the signal, we have \[{v_L} = {v_a}\]. Substituting these in (1), we get the frequency received by the aeroplane as
\[\nu ' = \left( {\dfrac{{c - {v_a}}}{{c - 0}}} \right)\nu \]
$ \Rightarrow \nu ' = \left( {\dfrac{{c - {v_a}}}{c}} \right)\nu $ …………………..(2)
Now, this frequency will be reflected by the aeroplane. So this time the aeroplane becomes the source and the radar becomes the receiver. Now, the source is moving opposite to the signal with a speed of ${v_a}$, so we have \[{v_S} = - {v_a}\]. Also, the receiver, the radar is stationary. So we have ${v_L} = 0$. Substituting these in (1) we get the frequency of the reflected radio wave received by the radar as
$\nu '' = \left( {\dfrac{{c - 0}}{{c + {v_a}}}} \right)\nu '$
$ \Rightarrow \nu '' = \left( {\dfrac{c}{{c + {v_a}}}} \right)\nu '$
Putting (2) in the above, we get
$\nu '' = \left( {\dfrac{c}{{c + {v_a}}}} \right)\left( {\dfrac{{c - {v_a}}}{c}} \right)\nu $
$ \Rightarrow \nu '' = \left( {\dfrac{{c - {v_a}}}{{c + {v_a}}}} \right)\nu $
According to the question, the frequency of the reflected wave is greater than the original frequency, so $\nu '' > \nu $. But from the above equation, we have $\nu '' < \nu $. This means that our assumption of the direction of velocity of the aeroplane is incorrect. So the aeroplane must be moving towards the radar. So we replace \[{v_a}\] by $ - {v_a}$ in the above equation to get
$\nu '' = \left( {\dfrac{{c + {v_a}}}{{c - {v_a}}}} \right)\nu $
Now, we have
$\Delta \nu = \nu '' - \nu $
$ \Rightarrow \Delta \nu = \left( {\dfrac{{c + {v_a}}}{{c - {v_a}}}} \right)\nu - \nu $
Dividing by $\nu $ both the sides, we get
$\dfrac{{\Delta \nu }}{\nu } = \left( {\dfrac{{c + {v_a}}}{{c - {v_a}}}} \right) - 1$
$ \Rightarrow \dfrac{{\Delta \nu }}{\nu } = \dfrac{{2{v_a}}}{{c - {v_a}}}$
According to the question, $\left( {{v_a} < < c} \right)$. So we approximate $c - {v_a} \approx c$ in the above equation to get
$\dfrac{{\Delta \nu }}{\nu } = \dfrac{{2{v_a}}}{c}$
\[ \Rightarrow {v_a} = \dfrac{{c\Delta \nu }}{{2\nu }}\]
Thus, the velocity of the aeroplane is equal to $\dfrac{{c\Delta \nu }}{{2\nu }}$.
Hence, the correct answer is option C.
Note: The direction of velocity is from the source to the receiver. According to the sign convention, all the speeds which are parallel to the direction from the source to the receiver are taken to be positive and those in the opposite direction are taken to be negative. Do not apply the concept of relative velocity while substituting the speeds of the source and the receiver.
Formula used: The formula used to solve this question is given by
\[\nu ' = \left( {\dfrac{{v - {v_L}}}{{v - {v_S}}}} \right){\nu _0}\], here $\nu '$ is the apparent frequency as received by a receiver moving with a speed of ${v_L}$ due to a source of sound moving with a speed of ${v_S}$ and emitting a sound of frequency ${\nu _0}$. The velocity of the wave is $v$.
Complete step-by-step solution:
In the given question, the direction of the velocity of the aeroplane is not given to us. Therefore we assume that it is moving away from the radar.
From the Doppler’s Effect formula, we know that the apparent frequency received by an observer is given by
\[\nu ' = \left( {\dfrac{{v - {v_L}}}{{v - {v_S}}}} \right){\nu _0}\]
According to the question, the radio wave has a frequency of $\nu $. Also we know that radio waves are a type of electromagnetic wave which has a speed of $c$. Therefore substituting $v = c$, and ${\nu _0} = \nu $ in the above equation, we get
\[\nu ' = \left( {\dfrac{{c - {v_L}}}{{c - {v_S}}}} \right)\nu \]................(1)
In the first case, the radar is the source and the aeroplane is the receiver. Since the radar is stationary, so we have${v_S} = 0$. Also, since the aeroplane is moving with a speed of ${v_a}$ parallel to the signal, we have \[{v_L} = {v_a}\]. Substituting these in (1), we get the frequency received by the aeroplane as
\[\nu ' = \left( {\dfrac{{c - {v_a}}}{{c - 0}}} \right)\nu \]
$ \Rightarrow \nu ' = \left( {\dfrac{{c - {v_a}}}{c}} \right)\nu $ …………………..(2)
Now, this frequency will be reflected by the aeroplane. So this time the aeroplane becomes the source and the radar becomes the receiver. Now, the source is moving opposite to the signal with a speed of ${v_a}$, so we have \[{v_S} = - {v_a}\]. Also, the receiver, the radar is stationary. So we have ${v_L} = 0$. Substituting these in (1) we get the frequency of the reflected radio wave received by the radar as
$\nu '' = \left( {\dfrac{{c - 0}}{{c + {v_a}}}} \right)\nu '$
$ \Rightarrow \nu '' = \left( {\dfrac{c}{{c + {v_a}}}} \right)\nu '$
Putting (2) in the above, we get
$\nu '' = \left( {\dfrac{c}{{c + {v_a}}}} \right)\left( {\dfrac{{c - {v_a}}}{c}} \right)\nu $
$ \Rightarrow \nu '' = \left( {\dfrac{{c - {v_a}}}{{c + {v_a}}}} \right)\nu $
According to the question, the frequency of the reflected wave is greater than the original frequency, so $\nu '' > \nu $. But from the above equation, we have $\nu '' < \nu $. This means that our assumption of the direction of velocity of the aeroplane is incorrect. So the aeroplane must be moving towards the radar. So we replace \[{v_a}\] by $ - {v_a}$ in the above equation to get
$\nu '' = \left( {\dfrac{{c + {v_a}}}{{c - {v_a}}}} \right)\nu $
Now, we have
$\Delta \nu = \nu '' - \nu $
$ \Rightarrow \Delta \nu = \left( {\dfrac{{c + {v_a}}}{{c - {v_a}}}} \right)\nu - \nu $
Dividing by $\nu $ both the sides, we get
$\dfrac{{\Delta \nu }}{\nu } = \left( {\dfrac{{c + {v_a}}}{{c - {v_a}}}} \right) - 1$
$ \Rightarrow \dfrac{{\Delta \nu }}{\nu } = \dfrac{{2{v_a}}}{{c - {v_a}}}$
According to the question, $\left( {{v_a} < < c} \right)$. So we approximate $c - {v_a} \approx c$ in the above equation to get
$\dfrac{{\Delta \nu }}{\nu } = \dfrac{{2{v_a}}}{c}$
\[ \Rightarrow {v_a} = \dfrac{{c\Delta \nu }}{{2\nu }}\]
Thus, the velocity of the aeroplane is equal to $\dfrac{{c\Delta \nu }}{{2\nu }}$.
Hence, the correct answer is option C.
Note: The direction of velocity is from the source to the receiver. According to the sign convention, all the speeds which are parallel to the direction from the source to the receiver are taken to be positive and those in the opposite direction are taken to be negative. Do not apply the concept of relative velocity while substituting the speeds of the source and the receiver.
Recently Updated Pages
Write a composition in approximately 450 500 words class 10 english JEE_Main
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
Arrange the sentences P Q R between S1 and S5 such class 10 english JEE_Main
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
What is the common property of the oxides CONO and class 10 chemistry JEE_Main
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
What happens when dilute hydrochloric acid is added class 10 chemistry JEE_Main
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
If four points A63B 35C4 2 and Dx3x are given in such class 10 maths JEE_Main
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
The area of square inscribed in a circle of diameter class 10 maths JEE_Main
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
Other Pages
Excluding stoppages the speed of a bus is 54 kmph and class 11 maths JEE_Main
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
In the ground state an element has 13 electrons in class 11 chemistry JEE_Main
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
Electric field due to uniformly charged sphere class 12 physics JEE_Main
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
A boat takes 2 hours to go 8 km and come back to a class 11 physics JEE_Main
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
According to classical free electron theory A There class 11 physics JEE_Main
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
Differentiate between homogeneous and heterogeneous class 12 chemistry JEE_Main
![arrow-right](/cdn/images/seo-templates/arrow-right.png)