Answer
396.9k+ views
Hint: Determine the speed of source along the line of sight of the observer. To do so, you have to determine the angle made by the speed of the locomotive and speed of the locomotive along the line of sight of the observer. Use Doppler’s formula for apparent frequency when both source and observer are moving towards each other.
Formula used:
Apparent frequency, \[{f_{app}} = f\left( {\dfrac{v}{{v - {v_S}}}} \right)\]
Here, f is the original frequency, v is the speed of sound and \[{v_S}\] is the speed of source.
Complete step by step answer:
As we can see this is the case of both observer and source are moving towards each other but the direction is perpendicular to each other. We have to determine the speed of the source along the line of sight of the observer.
Let’s determine the distance AC using Pythagoras theorem as follows,
\[AC = \sqrt {{{\left( {AB} \right)}^2} + {{\left( {BC} \right)}^2}} \]
Substituting 1 mile for AB and 0.60 mile for BC in the above equation, we get,
\[AC = \sqrt {{{\left( 1 \right)}^2} + {{\left( {0.6} \right)}^2}} \]
\[ \Rightarrow AC = 1.166\,{\text{mile}}\]
Let’s determine the angle \[\theta \] as follows,
\[\cos \theta = \dfrac{{AB}}{{AC}}\]
Substituting 1 mile for AB and 1.166 mile for AC in the above equation, we get,
\[\cos \theta = \dfrac{1}{{1.166}}\]
\[ \Rightarrow \cos \theta = 0.857\]
We have the speed of the source along the line of sight is,
\[{v_S} = {v_L}\cos \theta \]
Substituting 80 miles/hr for \[{v_L}\] and 0.857 for \[\cos \theta \] in the above equation, we get,
\[{v_S} = \left( {80\,{\text{mile/hr}}} \right)\left( {\dfrac{{1\,{\text{hr}}}}{{3600\,{\text{s}}}}} \right)\left( {0.857} \right)\]
\[ \Rightarrow {v_S} = \left( {80\,{\text{mile/hr}}} \right)\left( {\dfrac{{1\,{\text{hr}}}}{{3600\,{\text{s}}}}} \right)\left( {0.857} \right)\]
\[ \Rightarrow {v_S} = 0.019\,{\text{miles/s}}\]
Using Doppler’s formula for apparent frequency when both source and observer is moving, we have,
\[{f_{app}} = f\left( {\dfrac{v}{{v - {v_S}}}} \right)\]
Here, f is the original frequency, v is the speed of sound and \[{v_S}\] is the speed of source.
Substituting 400 Hz for f, 0.200 miles/s for v and 0.019 miles/s for \[{v_S}\] in the above equation, we get,
\[{f_{app}} = \left( {400} \right)\left( {\dfrac{{0.200}}{{0.200 - 0.019}}} \right)\]
\[ \Rightarrow {f_{app}} = \left( {400} \right)\left( {1.105} \right)\]
\[ \therefore {f_{app}} = 442\,{\text{Hz}}\]
So, the correct answer is option B.
Note:Make sure that the unit of speed of sound and speed of the observer is the same. In the solution, we have converted the unit from mile/hr to mile/s since the speed of sound is in miles/s. When the source and observer are moving towards each other, the sign of the velocity of source should be negative in the denominator so as the apparent frequency increases.
Formula used:
Apparent frequency, \[{f_{app}} = f\left( {\dfrac{v}{{v - {v_S}}}} \right)\]
Here, f is the original frequency, v is the speed of sound and \[{v_S}\] is the speed of source.
Complete step by step answer:
As we can see this is the case of both observer and source are moving towards each other but the direction is perpendicular to each other. We have to determine the speed of the source along the line of sight of the observer.
![seo images](https://www.vedantu.com/question-sets/6b2a5be0-55ee-4288-8c95-9e90c59347506901628574522783440.png)
Let’s determine the distance AC using Pythagoras theorem as follows,
\[AC = \sqrt {{{\left( {AB} \right)}^2} + {{\left( {BC} \right)}^2}} \]
Substituting 1 mile for AB and 0.60 mile for BC in the above equation, we get,
\[AC = \sqrt {{{\left( 1 \right)}^2} + {{\left( {0.6} \right)}^2}} \]
\[ \Rightarrow AC = 1.166\,{\text{mile}}\]
Let’s determine the angle \[\theta \] as follows,
\[\cos \theta = \dfrac{{AB}}{{AC}}\]
Substituting 1 mile for AB and 1.166 mile for AC in the above equation, we get,
\[\cos \theta = \dfrac{1}{{1.166}}\]
\[ \Rightarrow \cos \theta = 0.857\]
We have the speed of the source along the line of sight is,
\[{v_S} = {v_L}\cos \theta \]
Substituting 80 miles/hr for \[{v_L}\] and 0.857 for \[\cos \theta \] in the above equation, we get,
\[{v_S} = \left( {80\,{\text{mile/hr}}} \right)\left( {\dfrac{{1\,{\text{hr}}}}{{3600\,{\text{s}}}}} \right)\left( {0.857} \right)\]
\[ \Rightarrow {v_S} = \left( {80\,{\text{mile/hr}}} \right)\left( {\dfrac{{1\,{\text{hr}}}}{{3600\,{\text{s}}}}} \right)\left( {0.857} \right)\]
\[ \Rightarrow {v_S} = 0.019\,{\text{miles/s}}\]
Using Doppler’s formula for apparent frequency when both source and observer is moving, we have,
\[{f_{app}} = f\left( {\dfrac{v}{{v - {v_S}}}} \right)\]
Here, f is the original frequency, v is the speed of sound and \[{v_S}\] is the speed of source.
Substituting 400 Hz for f, 0.200 miles/s for v and 0.019 miles/s for \[{v_S}\] in the above equation, we get,
\[{f_{app}} = \left( {400} \right)\left( {\dfrac{{0.200}}{{0.200 - 0.019}}} \right)\]
\[ \Rightarrow {f_{app}} = \left( {400} \right)\left( {1.105} \right)\]
\[ \therefore {f_{app}} = 442\,{\text{Hz}}\]
So, the correct answer is option B.
Note:Make sure that the unit of speed of sound and speed of the observer is the same. In the solution, we have converted the unit from mile/hr to mile/s since the speed of sound is in miles/s. When the source and observer are moving towards each other, the sign of the velocity of source should be negative in the denominator so as the apparent frequency increases.
Recently Updated Pages
How many sigma and pi bonds are present in HCequiv class 11 chemistry CBSE
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
Why Are Noble Gases NonReactive class 11 chemistry CBSE
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
Let X and Y be the sets of all positive divisors of class 11 maths CBSE
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
Let x and y be 2 real numbers which satisfy the equations class 11 maths CBSE
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
Let x 4log 2sqrt 9k 1 + 7 and y dfrac132log 2sqrt5 class 11 maths CBSE
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
Let x22ax+b20 and x22bx+a20 be two equations Then the class 11 maths CBSE
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
Trending doubts
Fill the blanks with the suitable prepositions 1 The class 9 english CBSE
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
At which age domestication of animals started A Neolithic class 11 social science CBSE
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
Which are the Top 10 Largest Countries of the World?
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
Give 10 examples for herbs , shrubs , climbers , creepers
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
Difference between Prokaryotic cell and Eukaryotic class 11 biology CBSE
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
Difference Between Plant Cell and Animal Cell
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
Write a letter to the principal requesting him to grant class 10 english CBSE
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
Change the following sentences into negative and interrogative class 10 english CBSE
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
Fill in the blanks A 1 lakh ten thousand B 1 million class 9 maths CBSE
![arrow-right](/cdn/images/seo-templates/arrow-right.png)